Solutions to Axler’s Linear Algebra Done Right

Commutativity of addition in \(\mathbf{C}\) is commutativity in \(\mathbf{R}\), one coordinate at a time. Write \(\alpha = a + bi\) and \(\beta = c + di\) with \(a, b, c, d \in \mathbf{R}\); then by the definition of complex addition in 1.1,

\begin{equation*} \begin{aligned} \alpha + \beta &= (a + c) + (b + d)i \\ &= (c + a) + (d + b)i \\ &= \beta + \alpha, \end{aligned} \end{equation*}

the middle equality because addition in \(\mathbf{R}\) is commutative, and the outer two because complex numbers are equal exactly when their real and imaginary parts agree (1.1).

Associativity of addition in \(\mathbf{C}\) is associativity in \(\mathbf{R}\), one coordinate at a time. Write \(\alpha = a + bi\), \(\beta = c + di\), \(\lambda = e + fi\) with \(a, b, c, d, e, f \in \mathbf{R}\); two applications of the definition of complex addition in 1.1 give

\begin{equation*} \begin{aligned} (\alpha + \beta) + \lambda &= \bigl((a + c) + e\bigr) + \bigl((b + d) + f\bigr)i \\ &= \bigl(a + (c + e)\bigr) + \bigl(b + (d + f)\bigr)i \\ &= \alpha + (\beta + \lambda), \end{aligned} \end{equation*}

the middle equality because addition in \(\mathbf{R}\) is associative, and the outer two because complex numbers are equal exactly when their real and imaginary parts agree (1.1).

Both products expand to the same four real summands in each coordinate. Write \(\alpha = a + bi\), \(\beta = c + di\), \(\lambda = e + fi\) with \(a, b, c, d, e, f \in \mathbf{R}\). Two applications of the definition of complex multiplication in 1.1, expanded by the real distributive and associative laws, give

\begin{equation*} \begin{aligned} (\alpha\beta)\lambda &= \bigl((ac - bd)e - (ad + bc)f\bigr) \\ &\quad + \bigl((ac - bd)f + (ad + bc)e\bigr)i \\ &= (ace - bde - adf - bcf) + (acf - bdf + ade + bce)i \end{aligned} \end{equation*}

and, in the other grouping,

\begin{equation*} \begin{aligned} \alpha(\beta\lambda) &= \bigl(a(ce - df) - b(cf + de)\bigr) \\ &\quad + \bigl(a(cf + de) + b(ce - df)\bigr)i \\ &= (ace - adf - bcf - bde) + (acf + ade + bce - bdf)i. \end{aligned} \end{equation*}

The two real parts are the same four summands in a different order, as are the two imaginary parts, so they agree by commutativity and associativity of addition in \(\mathbf{R}\). Hence \((\alpha\beta)\lambda = \alpha(\beta\lambda)\).

Both sides expand to the same four real summands in each coordinate. Write \(\lambda = e + fi\), \(\alpha = a + bi\), \(\beta = c + di\) with \(a, b, c, d, e, f \in \mathbf{R}\). By the definitions of complex addition and multiplication in 1.1, expanded by the real distributive law,

\begin{equation*} \begin{aligned} \lambda(\alpha + \beta) &= \bigl(e(a + c) - f(b + d)\bigr) \\ &\quad + \bigl(e(b + d) + f(a + c)\bigr)i \\ &= (ea + ec - fb - fd) + (eb + ed + fa + fc)i, \end{aligned} \end{equation*}

while \(\lambda\alpha = (ea - fb) + (eb + fa)i\) and \(\lambda\beta = (ec - fd) + (ed + fc)i\), so

\begin{equation*} \lambda\alpha + \lambda\beta = \bigl((ea - fb) + (ec - fd)\bigr) + \bigl((eb + fa) + (ed + fc)\bigr)i. \end{equation*}

Each coordinate is the same sum of four real numbers in a different order, hence equal by commutativity and associativity of addition in \(\mathbf{R}\). Thus \(\lambda(\alpha + \beta) = \lambda\alpha + \lambda\beta\).

Take \(\beta = (-a) + (-b)i\), where \(\alpha = a + bi\) with \(a, b \in \mathbf{R}\); here \(0\) means \(0 + 0i\).

Existence: by the definition of complex addition in 1.1,

\begin{equation*} \alpha + \beta = \bigl(a + (-a)\bigr) + \bigl(b + (-b)\bigr)i = 0 + 0i = 0. \end{equation*}

Uniqueness: if \(\beta = c + di\) with \(c, d \in \mathbf{R}\) satisfies \(\alpha + \beta = 0\), then \((a + c) + (b + d)i = 0 + 0i\), so \(a + c = 0\) and \(b + d = 0\) since complex numbers are equal exactly when their real and imaginary parts agree (1.1). Uniqueness of additive inverses in \(\mathbf{R}\) then forces \(c = -a\) and \(d = -b\), the \(\beta\) exhibited above.

Take

\begin{equation*} \beta = \frac{a}{a^2 + b^2} + \frac{-b}{a^2 + b^2}\, i, \end{equation*}

where \(\alpha = a + bi\) with \(a, b \in \mathbf{R}\); here \(1\) means \(1 + 0i\). Since \(\alpha \neq 0\), at least one of \(a, b\) is nonzero, so \(a^2 + b^2 > 0\) and the two coefficients are real numbers.

Existence: by the definition of complex multiplication in 1.1,

\begin{equation*} \begin{aligned} \alpha\beta &= \left(a \cdot \frac{a}{a^2 + b^2} - b \cdot \frac{-b}{a^2 + b^2}\right) \\ &\quad + \left(a \cdot \frac{-b}{a^2 + b^2} + b \cdot \frac{a}{a^2 + b^2}\right) i \\ &= \frac{a^2 + b^2}{a^2 + b^2} + \frac{-ab + ab}{a^2 + b^2}\, i \\ &= 1 + 0i = 1. \end{aligned} \end{equation*}

Uniqueness: if \(\beta = c + di\) with \(c, d \in \mathbf{R}\) satisfies \(\alpha\beta = 1\), then comparing real and imaginary parts in \((ac - bd) + (ad + bc)i = 1 + 0i\) gives

\begin{equation*} ac - bd = 1 \qquad\text{and}\qquad ad + bc = 0. \end{equation*}

Taking \(a \cdot (\text{first}) + b \cdot (\text{second})\) yields \((a^2 + b^2)c = a\), and \(-b \cdot (\text{first}) + a \cdot (\text{second})\) yields \((a^2 + b^2)d = -b\). Since \(a^2 + b^2 \neq 0\), this forces

\begin{equation*} c = \frac{a}{a^2 + b^2}, \qquad d = \frac{-b}{a^2 + b^2}, \end{equation*}

the \(\beta\) exhibited above.

Squaring and multiplying once more returns \(1\). Put \(\omega = -\tfrac12 + \tfrac{\sqrt{3}}{2}\, i\), which is the given number in the standard form of 1.1. By the definition of complex multiplication in 1.1, using \(\sqrt{3}\cdot\sqrt{3} = 3\),

\begin{equation*} \begin{aligned} \omega^2 &= \left(\frac{1}{4} - \frac{3}{4}\right) \\ &\quad + \left(-\frac{\sqrt{3}}{4} - \frac{\sqrt{3}}{4}\right) i = -\frac{1}{2} - \frac{\sqrt{3}}{2}\, i, \\ \omega^3 = (\omega^2)\omega &= \left(\frac{1}{4} + \frac{3}{4}\right) \\ &\quad + \left(-\frac{\sqrt{3}}{4} + \frac{\sqrt{3}}{4}\right) i = 1 + 0i = 1, \end{aligned} \end{equation*}

the grouping \(\omega^3 = (\omega^2)\omega\) being legitimate by the associativity of Exercise 1A.3.

Take \(z = \pm\tfrac{\sqrt{2}}{2}(1 + i)\). These are distinct, since their difference \(\sqrt{2}(1 + i)\) is nonzero, and each squares to \(i\):

\begin{equation*} \left( \pm\frac{1 + i}{\sqrt{2}} \right)^{2} = \frac{(1 + i)^2}{2} = \frac{1 + 2i + i^2}{2} = \frac{2i}{2} = i . \end{equation*}

\begin{equation*} x = \tfrac{1}{2}\bigl( (5, 9, -6, 8) - (4, -3, 1, 7) \bigr) = \left( \tfrac{1}{2},\, 6,\, -\tfrac{7}{2},\, \tfrac{1}{2} \right). \end{equation*}

Indeed \(2x = (1, 12, -7, 1)\), and coordinatewise addition (1.13) gives \((4, -3, 1, 7) + (1, 12, -7, 1) = (5, 9, -6, 8)\). (Check!)

The first coordinate forces \(\lambda = 3 + 2i\), and that value fails in the third. Since scalar multiplication acts coordinatewise (1.18) and lists are equal exactly when they agree in each coordinate, such a \(\lambda\) would satisfy \(\lambda(2 - 3i) = 12 - 5i\); as \(2 - 3i \neq 0\) it is invertible (1.3, 1.5), so

\begin{equation*} \lambda = \frac{12 - 5i}{2 - 3i} = \frac{(12 - 5i)(2 + 3i)}{4 + 9} = \frac{39 + 26i}{13} = 3 + 2i . \end{equation*}

But then the third coordinate gives

\begin{equation*} (3 + 2i)(-6 + 7i) = -18 + 9i + 14i^2 = -32 + 9i \neq -32 - 9i , \end{equation*}

a contradiction. Hence no such \(\lambda\) exists.

Associativity in \(\mathbf{F}^n\) is associativity in \(\mathbf{F}\), one coordinate at a time; addition in \(\mathbf{F}\) is associative by 1.3, since \(\mathbf{F}\) is \(\mathbf{R}\) or \(\mathbf{C}\) (1.6). Write \(x = (x_1, \ldots, x_n)\), \(y = (y_1, \ldots, y_n)\), \(z = (z_1, \ldots, z_n)\) with all entries in \(\mathbf{F}\) (1.11). Then

\begin{equation*} \begin{aligned} (x + y) + z &= \bigl( (x_1 + y_1) + z_1, \; \ldots, \; (x_n + y_n) + z_n \bigr) \\ &= \bigl( x_1 + (y_1 + z_1), \; \ldots, \; x_n + (y_n + z_n) \bigr) \\ &= x + (y + z), \end{aligned} \end{equation*}

the outer equalities by the definition 1.13 of addition in \(\mathbf{F}^n\) and the middle one by associativity in \(\mathbf{F}\) applied in each of the \(n\) coordinates.

Associativity in \(\mathbf{F}^n\) is associativity of multiplication in \(\mathbf{F}\), one coordinate at a time; the latter holds by 1.3, since \(\mathbf{F}\) is \(\mathbf{R}\) or \(\mathbf{C}\) (1.6). Write \(x = (x_1, \ldots, x_n)\) with each \(x_k \in \mathbf{F}\) (1.11). Then

\begin{equation*} \begin{aligned} (ab)x &= \bigl( (ab)x_1, \; \ldots, \; (ab)x_n \bigr) \\ &= \bigl( a(bx_1), \; \ldots, \; a(bx_n) \bigr) \\ &= a(bx_1, \ldots, bx_n) = a(bx), \end{aligned} \end{equation*}

the first, third, and fourth equalities by the definition 1.18 of scalar multiplication in \(\mathbf{F}^n\) (the third read from right to left, with scalar \(a\)), and the second by associativity in \(\mathbf{F}\) applied in each of the \(n\) coordinates.

The identity \(1x = x\) is the identity \(1\lambda = \lambda\) in \(\mathbf{F}\), one coordinate at a time; the latter holds by the identities and commutativity statements in 1.3, since \(\mathbf{F}\) is \(\mathbf{R}\) or \(\mathbf{C}\) (1.6). Write \(x = (x_1, \ldots, x_n)\) with each \(x_k \in \mathbf{F}\) (1.11). Then by the definition 1.18 of scalar multiplication,

\begin{equation*} 1x = (1x_1, \ldots, 1x_n) = (x_1, \ldots, x_n) = x , \end{equation*}

the second equality because \(1x_k = x_k\) in \(\mathbf{F}\) for each \(k\).

Distributivity in \(\mathbf{F}^n\) is distributivity in \(\mathbf{F}\), one coordinate at a time; the latter holds by 1.3, since \(\mathbf{F}\) is \(\mathbf{R}\) or \(\mathbf{C}\) (1.6). Write \(x = (x_1, \ldots, x_n)\) and \(y = (y_1, \ldots, y_n)\) with all entries in \(\mathbf{F}\) (1.11). Then

\begin{equation*} \begin{aligned} \lambda(x + y) &= \bigl( \lambda(x_1 + y_1), \; \ldots, \; \lambda(x_n + y_n) \bigr) \\ &= (\lambda x_1 + \lambda y_1, \; \ldots, \; \lambda x_n + \lambda y_n) \\ &= (\lambda x_1, \ldots, \lambda x_n) + (\lambda y_1, \ldots, \lambda y_n) = \lambda x + \lambda y , \end{aligned} \end{equation*}

the first equality by the definitions 1.13 and 1.18 of addition and scalar multiplication in \(\mathbf{F}^n\), the second by distributivity in \(\mathbf{F}\) in each of the \(n\) coordinates, and the last two by 1.13 and 1.18 read from right to left.

Distributivity in \(\mathbf{F}^n\) is distributivity in \(\mathbf{F}\), one coordinate at a time. Write \(x = (x_1, \dots, x_n)\) with each \(x_j \in \mathbf{F}\) (1.11). Then

\begin{equation*} \begin{aligned} (a+b)x &= \bigl((a+b)x_1, \dots, (a+b)x_n\bigr) \\ &= (ax_1 + bx_1, \dots, ax_n + bx_n) \\ &= (ax_1, \dots, ax_n) + (bx_1, \dots, bx_n) = ax + bx, \end{aligned} \end{equation*}

the first and last equalities by the definitions 1.18 and 1.13 of scalar multiplication and addition in \(\mathbf{F}^n\). The second equality is the coordinatewise identity \((a+b)x_j = ax_j + bx_j\): by 1.6 the field \(\mathbf{F}\) is \(\mathbf{R}\) or \(\mathbf{C}\), and 1.3 states distributivity in the form \(\lambda(\alpha + \beta) = \lambda\alpha + \lambda\beta\), so combining it with the commutativity of multiplication in 1.3 gives

\begin{equation*} (a+b)x_j = x_j(a+b) = x_j a + x_j b = a x_j + b x_j . \end{equation*}

Both \(v\) and \(-(-v)\) are additive inverses of \(-v\), and additive inverses are unique (1.27). Indeed \(v + (-v) = 0\) by the definition 1.28 of \(-v\), so \((-v) + v = 0\) by commutativity of addition, which says exactly that \(v\) is an additive inverse of \(-v\); and \(-(-v)\) is one by definition. Hence \(-(-v) = v\).

If \(a \neq 0\) then \(v = 0\), which gives the disjunction. Indeed \(\mathbb{F}\) is \(\mathbb{R}\) or \(\mathbb{C}\) (1.1), so \(a \neq 0\) has an inverse \(\frac{1}{a} \in \mathbb{F}\), and

\begin{equation*} v = 1v = \left( \frac{1}{a} \, a \right) v = \frac{1}{a} (av) = \frac{1}{a} \, 0 = 0 , \end{equation*}

using in turn the multiplicative identity property of 1.20, \(\frac{1}{a} a = 1\) in \(\mathbb{F}\), the associativity property \((bc)v = b(cv)\) of 1.20, the hypothesis \(av = 0\), and 1.31.

The unique solution is \(x = \frac{1}{3}(w - v) = \frac{1}{3}\bigl(w + (-v)\bigr)\), which lies in \(V\) because \(\mathbb{F}\) is \(\mathbb{R}\) or \(\mathbb{C}\), so \(\frac{1}{3} \in \mathbb{F}\).

Existence: by associativity of scalar multiplication and the multiplicative identity property of 1.20, \(3x = (3 \cdot \frac{1}{3})(w - v) = w + (-v)\), so by commutativity and associativity of addition,

\begin{equation*} v + 3x = v + \bigl(w + (-v)\bigr) = w + \bigl(v + (-v)\bigr) = w + 0 = w . \end{equation*}

Uniqueness: if \(v + 3x = w = v + 3x^{\prime}\), then adding \(-v\) on the left and using associativity, \((-v) + v = 0\), and the additive identity property gives \(3x = 3x^{\prime}\); multiplying by \(\frac{1}{3}\) and using associativity of scalar multiplication with the multiplicative identity property gives

\begin{equation*} x = \left( \tfrac{1}{3} \cdot 3 \right) x = \tfrac{1}{3}(3x) = \tfrac{1}{3}(3x^{\prime}) = x^{\prime} . \end{equation*}

The additive identity condition: there is no \(0 \in \emptyset\), since \(\emptyset\) has no elements.

It is the only failure because it is the only requirement of 1.20 asserting the existence of an element of \(V\); the other five (commutativity, the two associativities, additive inverse, multiplicative identity, and the two distributive properties) are all universally quantified over elements of \(V\), hence vacuously true when \(V = \emptyset\). The operations themselves are legitimate: \(\emptyset \times \emptyset = \emptyset\) and \(\mathbb{F} \times \emptyset = \emptyset\), and the empty function serves as each in the sense of 1.19.

Granted the other five conditions of 1.20, the additive inverse condition and the condition \(0v = 0\) each imply the other, so the two definitions have the same objects. (The phrase “the additive identity of \(V\)” is unambiguous in both settings: the proof of 1.26 uses only the additive identity condition and commutativity.)

(i) If \(V\) also satisfies the additive inverse condition, it is a vector space, so \(0v = 0\) for every \(v \in V\) by 1.30.

(ii) If instead \(V\) satisfies \(0v = 0\) for all \(v \in V\), then for \(v \in V\) the vector \((-1)v \in V\) is an additive inverse of \(v\):

\begin{equation*} v + (-1)v = 1v + (-1)v = \bigl(1 + (-1)\bigr) v = 0v = 0 , \end{equation*}

using the multiplicative identity condition, the distributive property \((a + b)v = av + bv\), \(1 + (-1) = 0\) in \(\mathbb{F}\), and the assumed condition.

No: addition on \(W = \mathbb{R} \cup \{\infty, -\infty\}\) is not associative. Take \(u = -\infty\), \(v = \infty\), and \(w = 1\). Then

\begin{equation*} (u + v) + w = \bigl((-\infty) + \infty\bigr) + 1 = 0 + 1 = 1 , \end{equation*}

while

\begin{equation*} u + (v + w) = (-\infty) + (\infty + 1) = (-\infty) + \infty = 0 , \end{equation*}

using \(\infty + 1 = \infty\) from the definition. Since \(1 \neq 0\), the associativity requirement of 1.20 fails, so \(W\) is not a vector space over \(\mathbb{R}\).

The distributive property \((a + b)v = av + bv\) fails too: with \(a = 2\), \(b = -1\), \(v = \infty\),

\begin{equation*} (a + b)v = 1 \cdot \infty = \infty \quad\text{while}\quad av + bv = \infty + (-\infty) = 0 . \end{equation*}

Define the operations pointwise, imitating 1.24: for \(f, g \in V^S\) and \(\lambda \in \mathbb{F}\),

\begin{equation*} (f + g)(x) = f(x) + g(x), \qquad (\lambda f)(x) = \lambda\bigl(f(x)\bigr) \qquad \text{for all } x \in S . \end{equation*}

Both right sides lie in \(V\), so these are an addition and a scalar multiplication on \(V^S\) in the sense of 1.19.

Each requirement of 1.20 then holds because two functions \(S \to V\) are equal exactly when they agree at every \(x \in S\), and at each such \(x\) the requirement is the corresponding property of \(V\). Commutativity is typical:

\begin{equation*} (f + g)(x) = f(x) + g(x) = g(x) + f(x) = (g + f)(x) , \end{equation*}

and associativity of addition, associativity and the multiplicative identity for scalar multiplication, and both distributive properties go the same way. (Check!) For the two existence requirements, take \(\mathbf{0} \in V^S\) with \(\mathbf{0}(x) = 0\) and \(-f \in V^S\) with \((-f)(x) = -\bigl(f(x)\bigr)\) (well defined by 1.27); then \((f + \mathbf{0})(x) = f(x) + 0 = f(x)\) and \(\bigl(f + (-f)\bigr)(x) = 0 = \mathbf{0}(x)\) for every \(x \in S\), so \(f + \mathbf{0} = f\) and \(f + (-f) = \mathbf{0}\).

Hence \(V^S\) is a vector space over \(\mathbb{F}\).

Each requirement of 1.20 for \(V_{\mathbf{C}}\) over \(\mathbf{C}\) reduces to the real vector space properties of \(V\) applied to the two components of \(u + iv\), which is by definition the ordered pair \((u,v)\); so \(u_1 + iv_1 = u_2 + iv_2\) exactly when \(u_1 = u_2\) and \(v_1 = v_2\). Both formulas return elements of \(V_{\mathbf{C}}\), and the representation \(a + bi\) with \(a, b \in \mathbf{R}\) of a complex number is unique, so the operations are an addition and a scalar multiplication in the sense of 1.19. Throughout, \(\alpha u - \beta v\) means \(\alpha u + (-\beta)v\): by 1.28 it means \(\alpha u + \bigl(-(\beta v)\bigr)\), and \(\beta v + (-\beta)v = \bigl(\beta + (-\beta)\bigr)v = 0v = 0\) by distributivity in \(V\) and 1.30, so \(-(\beta v) = (-\beta)v\) by uniqueness of additive inverses (1.27).

Fix \(u, v, u_1, v_1, u_2, v_2 \in V\) and \(\lambda = a + bi\), \(\mu = c + di\) in \(\mathbf{C}\), with \(a, b, c, d \in \mathbf{R}\).

Commutativity and associativity of addition hold componentwise from those properties in \(V\). (Check!) The vector \(\mathbf{0} = 0 + i0\) is an additive identity, since \((u + iv) + (0 + i0) = (u + 0) + i(v + 0) = u + iv\), and \((-u) + i(-v)\) is an additive inverse of \(u + iv\) by the same computation. The multiplicative identity condition holds because \(1 = 1 + 0i\):

\begin{equation*} 1(u + iv) = (1u - 0v) + i(1v + 0u) = u + iv, \end{equation*}

using \(1u = u\) in \(V\) and \(0u = 0v = 0\) (1.30).

Associativity of scalar multiplication: \(\lambda\mu = (ac - bd) + (ad + bc)i\) gives

\begin{equation*} \begin{aligned} (\lambda\mu)(u + iv) &= \bigl((ac-bd)u - (ad+bc)v\bigr) \\ &\quad + i\bigl((ac-bd)v + (ad+bc)u\bigr), \end{aligned} \end{equation*}

while \(\mu(u+iv) = (cu - dv) + i(cv + du)\) gives

\begin{equation*} \begin{aligned} \lambda\bigl(\mu(u+iv)\bigr) &= \bigl(a(cu - dv) - b(cv + du)\bigr) \\ &\quad + i\bigl(a(cv + du) + b(cu - dv)\bigr) \\ &= \bigl((ac)u - (ad)v - (bc)v - (bd)u\bigr) \\ &\quad + i\bigl((ac)v + (ad)u + (bc)u - (bd)v\bigr), \end{aligned} \end{equation*}

by the distributive property \(\alpha(x+y) = \alpha x + \alpha y\) and associativity \(\alpha(\beta u) = (\alpha\beta)u\) in \(V\); the two results agree by \((\alpha + \beta)x = \alpha x + \beta x\) in \(V\) with commutativity and associativity of addition.

First distributive property: with \(x = u_1 + iv_1\) and \(y = u_2 + iv_2\),

\begin{equation*} \begin{aligned} \lambda(x + y) &= \bigl(a(u_1 + u_2) - b(v_1 + v_2)\bigr) \\ &\quad + i\bigl(a(v_1 + v_2) + b(u_1 + u_2)\bigr) \\ &= \bigl((au_1 - bv_1) + (au_2 - bv_2)\bigr) \\ &\quad + i\bigl((av_1 + bu_1) + (av_2 + bu_2)\bigr) = \lambda x + \lambda y, \end{aligned} \end{equation*}

the second equality by \(\alpha(x+y) = \alpha x + \alpha y\) in \(V\) (with the real scalars \(a\) and \(-b\)) and regrouping by commutativity and associativity of addition in \(V\).

Second distributive property: since \(\lambda + \mu = (a + c) + (b + d)i\), with \(x = u + iv\),

\begin{equation*} \begin{aligned} (\lambda + \mu)x &= \bigl((a+c)u - (b+d)v\bigr) + i\bigl((a+c)v + (b+d)u\bigr) \\ &= \bigl((au - bv) + (cu - dv)\bigr) \\ &\quad + i\bigl((av + bu) + (cv + du)\bigr) = \lambda x + \mu x, \end{aligned} \end{equation*}

the second equality by \((\alpha + \beta)x = \alpha x + \beta x\) in \(V\) (with the real scalars \(a, c\) and \(-b, -d\), noting \(-(b+d) = (-b) + (-d)\)) and the same regrouping.

Hence \(V_{\mathbf{C}}\) is a complex vector space.

Subspaces in (a) and (d), not in (b) and (c). Throughout, \(U\) is a subspace exactly when \(0 \in U\) and \(U\) is closed under addition and under scalar multiplication (1.34).

(a) Since \(0 + 2 \cdot 0 + 3 \cdot 0 = 0\), the vector \((0,0,0)\) lies in \(U\); and for \(x, y \in U\) and \(\lambda \in \mathbf{F}\),

\begin{equation*} \begin{aligned} (x_1 + y_1) + 2(x_2 + y_2) + 3(x_3 + y_3) &= (x_1 + 2x_2 + 3x_3) + (y_1 + 2y_2 + 3y_3) \\ &= 0 + 0 = 0, \end{aligned} \end{equation*}

while \(\lambda x_1 + 2\lambda x_2 + 3\lambda x_3 = \lambda(x_1 + 2x_2 + 3x_3) = 0\). So \(x + y\) and \(\lambda x\) lie in \(U\).

(b) The additive identity is missing: \(0 + 2 \cdot 0 + 3 \cdot 0 = 0 \neq 4\).

(c) Closure under addition fails: \((1,1,0)\) and \((0,0,1)\) lie in the set, but their sum \((1,1,1)\) does not, since \(1 \cdot 1 \cdot 1 = 1 \neq 0\).

(d) Since \(0 = 5 \cdot 0\), the vector \((0,0,0)\) lies in \(U\); and if \(x_1 = 5x_3\) and \(y_1 = 5y_3\), then \(x_1 + y_1 = 5(x_3 + y_3)\) and \(\lambda x_1 = \lambda(5x_3) = 5(\lambda x_3)\), which say exactly that \(x + y \in U\) and \(\lambda x \in U\).

All five assertions hold; each is the criterion 1.34 applied to the defining condition, which in every case is preserved by sums and scalar multiples and satisfied by the zero vector.

(a) \(U = \{x \in \mathbf{F}^4 : x_3 = 5x_4 + b\}\) is a subspace if and only if \(b = 0\). For \(b \neq 0\) the zero vector fails the condition, since \(0 = 5 \cdot 0 + b\) forces \(b = 0\). For \(b = 0\) it satisfies it, and \(x_3 = 5x_4\), \(y_3 = 5y_4\) give \(x_3 + y_3 = 5(x_4 + y_4)\) and \(\lambda x_3 = 5(\lambda x_4)\).

(b) \(C[0,1]\) is a subspace of \(\mathbf{R}^{[0,1]}\): the zero function is constant, hence continuous, and sums and scalar multiples of continuous functions are continuous.

(c) The set \(D\) of differentiable functions on \(\mathbf{R}\) is a subspace of \(\mathbf{R}^{\mathbf{R}}\): the zero function is differentiable, and \(f + g\) and \(\lambda f\) are differentiable with \((f+g)^{\prime} = f^{\prime} + g^{\prime}\) and \((\lambda f)^{\prime} = \lambda f^{\prime}\).

(d) \(U = \{f \in \mathbf{R}^{(0,3)} : f \text{ is differentiable and } f^{\prime}(2) = b\}\) is a subspace if and only if \(b = 0\). The zero function has derivative \(0\) at \(2\), so it lies in \(U\) exactly when \(b = 0\); and when \(b = 0\), the derivative rules of (c) give \((f+g)^{\prime}(2) = 0 + 0 = 0\) and \((\lambda f)^{\prime}(2) = \lambda \cdot 0 = 0\).

(e) The sequences in \(\mathbf{C}^{\infty}\) with limit \(0\) form a subspace: \((0, 0, \dots)\) has limit \(0\), and the limit laws give \(\lim_{n \to \infty} (x_n + y_n) = 0\) and \(\lim_{n \to \infty} \lambda x_n = 0\) whenever \(\lim_{n \to \infty} x_n = \lim_{n \to \infty} y_n = 0\).

The set \(U\) in question satisfies the three conditions of 1.34. The zero function is differentiable with \(0^{\prime}(-1) = 0 = 3 \cdot 0(2)\), so \(0 \in U\); and for \(f, g \in U\) and \(\lambda \in \mathbf{R}\), sums and scalar multiples of differentiable functions are differentiable, with \((f+g)^{\prime} = f^{\prime} + g^{\prime}\) and \((\lambda f)^{\prime} = \lambda f^{\prime}\), so

\begin{equation*} (f+g)^{\prime}(-1) = 3f(2) + 3g(2) = 3(f+g)(2), \qquad (\lambda f)^{\prime}(-1) = \lambda \cdot 3f(2) = 3(\lambda f)(2). \end{equation*}

Hence \(f + g \in U\) and \(\lambda f \in U\), and by 1.34 \(U\) is a subspace of \(\mathbf{R}^{(-4,4)}\).

The set \(U\) in question is a subspace exactly when \(b = 0\).

(i) \(b \neq 0\): then \(\int_0^1 0 = 0 \neq b\), so \(U\) omits the additive identity and by 1.34 is not a subspace.

(ii) \(b = 0\): then \(0 \in U\), and for \(f, g \in U\) and \(\lambda \in \mathbf{R}\) the functions \(f+g\) and \(\lambda f\) are continuous, with

\begin{equation*} \int_0^1 (f+g) = \int_0^1 f + \int_0^1 g = 0, \qquad \int_0^1 (\lambda f) = \lambda \int_0^1 f = 0 \end{equation*}

by linearity of the integral. So \(f + g, \lambda f \in U\) and 1.34 makes \(U\) a subspace of \(\mathbf{R}^{[0,1]}\).

No: the scalars of \(\mathbf{C}^2\) are all of \(\mathbf{C}\), and \((1,0) \in \mathbf{R}^2\) with \(i \in \mathbf{C}\) gives

\begin{equation*} i(1,0) = (i, 0) \notin \mathbf{R}^2 . \end{equation*}

Closure under scalar multiplication fails, so by 1.34 \(\mathbf{R}^2\) is not a subspace of the complex vector space \(\mathbf{C}^2\).

(a) Yes, because cubing is injective on \(\mathbf{R}\), so the set is \(\{(a,a,c) : a, c \in \mathbf{R}\}\). Indeed, if \(a, b \in \mathbf{R}\) with \(a^3 = b^3\) then

\begin{equation*} 0 = a^3 - b^3 = (a-b)(a^2 + ab + b^2), \qquad a^2 + ab + b^2 = \Bigl(a + \tfrac{b}{2}\Bigr)^2 + \tfrac{3}{4}b^2, \end{equation*}

and the second factor vanishes only when \(a = b = 0\); otherwise \(a - b = 0\). Either way \(a = b\). The set \(\{(a,a,c)\}\) contains \(0\) and is closed under addition and scalar multiplication, since both operations act coordinatewise and preserve equality of the first two coordinates (Check!), so 1.34 makes it a subspace of \(\mathbf{R}^3\).

(b) No: cubing is not injective on \(\mathbf{C}\). With \(\omega = e^{2\pi i/3}\), so \(\omega^3 = 1\) and \(\omega + \omega^2 = -1\), the vectors \(u = (1,1,0)\) and \(v = (1,\omega,0)\) lie in the set, while their sum \((2, 1+\omega, 0)\) does not, because

\begin{equation*} (1+\omega)^3 = 1 + 3(\omega + \omega^2) + \omega^3 = 2 - 3 = -1 \neq 8 = 2^3 . \end{equation*}

Closure under addition fails, so by 1.34 the set is not a subspace of \(\mathbf{C}^3\).

False. Take \(U = \mathbf{Z}^2 = \{(m,n) \in \mathbf{R}^2 : m, n \in \mathbf{Z}\}\), which is nonempty and closed under addition and negation because \(\mathbf{Z}\) is (Check!). But

\begin{equation*} \tfrac{1}{2}(1,0) = \bigl(\tfrac{1}{2}, 0\bigr) \notin U , \end{equation*}

so closure under scalar multiplication fails and by 1.34 \(U\) is not a subspace of \(\mathbf{R}^2\).

Take the union of the two coordinate axes,

\begin{equation*} U = \{(x_1, x_2) \in \mathbb{R}^2 : x_1 x_2 = 0\}, \end{equation*}

which is nonempty and closed under scalar multiplication because \((\lambda x_1)(\lambda x_2) = \lambda^2 (x_1x_2) = 0\). It is not closed under addition, since \((1,0), (0,1) \in U\) but \((1,0) + (0,1) = (1,1) \notin U\). By 1.34, \(U\) is not a subspace of \(\mathbb{R}^2\).

No: the set \(P\) of periodic functions is not closed under addition. Take

\begin{equation*} f(x) = \cos x, \qquad g(x) = \cos(\sqrt{2}\,x), \qquad h = f + g, \end{equation*}

with periods \(2\pi\) and \(\sqrt{2}\,\pi\), so \(f, g \in P\).

Since \(\cos \leq 1\) we have \(h \leq 2\), and \(h(0) = 2\); moreover \(h(x) = 2\) forces both \(\cos x = 1\) and \(\cos(\sqrt{2}\,x) = 1\), that is \(x = 2\pi m\) and \(\sqrt{2}\,x = 2\pi n\) with \(m, n \in \mathbb{Z}\). Substituting gives \(\sqrt{2}\,m = n\), which forces \(m = 0\) (else \(\sqrt{2} = n/m\) would be rational) and hence \(x = 0\). So

\begin{equation*} \{x \in \mathbb{R} : h(x) = 2\} = \{0\}. \end{equation*}

If \(h\) had period \(p > 0\) then \(h(p) = h(0) = 2\), forcing \(p = 0\), a contradiction. Thus \(f, g \in P\) but \(f + g \notin P\), and by 1.34 \(P\) is not a subspace of \(\mathbb{R}^{\mathbb{R}}\).

Each of the three conditions of 1.34 passes to the intersection because it holds in \(V_1\) and in \(V_2\) separately.

Indeed \(0 \in V_1\) and \(0 \in V_2\), so \(0 \in V_1 \cap V_2\); and if \(u, w \in V_1 \cap V_2\) and \(a \in \mathbb{F}\), then \(u + w\) and \(au\) lie in \(V_1\) (which is closed under both operations) and likewise in \(V_2\), hence in \(V_1 \cap V_2\). By 1.34, \(V_1 \cap V_2\) is a subspace of \(V\).

The argument of Exercise 10 runs verbatim with an arbitrary index set, since each condition of 1.34 need only be checked one member at a time. Let \(\{V_i\}_{i \in I}\) be subspaces of \(V\) with \(I \neq \varnothing\) and put \(W = \bigcap_{i \in I} V_i\).

Every \(V_i\) contains \(0\), so \(0 \in W\). If \(u, w \in W\) and \(a \in \mathbb{F}\), then for each \(i \in I\) we have \(u, w \in V_i\), whence \(u + w \in V_i\) and \(au \in V_i\) by closure in \(V_i\); as \(i\) was arbitrary, \(u + w, au \in W\). By 1.34, \(W\) is a subspace of \(V\).

Let \(U, W\) be subspaces of \(V\).

If \(W \subseteq U\) (or symmetrically \(U \subseteq W\)), then \(U \cup W = U\), a subspace.

Conversely suppose \(U \cup W\) is a subspace and neither inclusion holds, so there are \(u \in U \setminus W\) and \(w \in W \setminus U\). Closure under addition gives \(u + w \in U \cup W\), and both alternatives fail:

(i) \(u + w \in U\) forces \(w = (u+w) - u \in U\), contradicting \(w \notin U\);

(ii) \(u + w \in W\) forces \(u = (u+w) - w \in W\), contradicting \(u \notin W\);

each using that \(U\), respectively \(W\), is closed under subtraction. Hence one of \(U, W\) contains the other.

Let \(V_1, V_2, V_3\) be subspaces of \(V\) and \(W = V_1 \cup V_2 \cup V_3\). If, say, \(V_2 \cup V_3 \subseteq V_1\), then \(W = V_1\) is a subspace.

Conversely suppose \(W\) is a subspace but no \(V_j\) contains the other two.

First, no \(V_j\) lies in the union of the other two: if say \(V_3 \subseteq V_1 \cup V_2\), then \(W = V_1 \cup V_2\) is a subspace, so by Exercise 12 one contains the other, say \(V_2 \subseteq V_1\), whence \(W = V_1\) and \(V_1\) contains \(V_2\) and \(V_3\), contrary to assumption. So we may choose

\begin{equation*} u \in V_1 \setminus (V_2 \cup V_3), \qquad w \in V_2 \setminus (V_1 \cup V_3). \end{equation*}

Next, \(u + \lambda w \in V_3\) for every nonzero \(\lambda \in \mathbb{F}\). Indeed \(u + \lambda w \in W\), and

(i) \(u + \lambda w \in V_1\) would give \(\lambda w = (u + \lambda w) - u \in V_1\), hence \(w = \lambda^{-1}(\lambda w) \in V_1\), contradicting the choice of \(w\);

(ii) \(u + \lambda w \in V_2\) would give \(u = (u + \lambda w) - \lambda w \in V_2\) since \(\lambda w \in V_2\), contradicting the choice of \(u\).

Finally, \(\mathbb{F}\) is \(\mathbb{R}\) or \(\mathbb{C}\), so \(1\) and \(2\) are distinct nonzero scalars; applying the previous paragraph to each and subtracting inside the subspace \(V_3\),

\begin{equation*} w = (u + 2w) - (u + w) \in V_3, \end{equation*}

contradicting \(w \notin V_3\). Hence one of \(V_1, V_2, V_3\) contains the other two.

We have

\begin{equation*} U + W = \{(x, y, 2x) \in \mathbb{F}^3 : x, y \in \mathbb{F}\}, \end{equation*}

that is, \(U + W\) is the set of vectors in \(\mathbb{F}^3\) whose third coordinate is twice the first.

By 1.36 the general element of \(U + W\) is

\begin{equation*} (a, -a, 2a) + (b, b, 2b) = \bigl(a + b,\ b - a,\ 2(a+b)\bigr), \end{equation*}

which has the stated form with \(x = a + b\), \(y = b - a\); this gives the inclusion \(\subseteq\). Conversely, given \(x, y \in \mathbb{F}\), the scalar \(2\) is invertible in \(\mathbb{F}\), so \(a = (x-y)/2\) and \(b = (x+y)/2\) satisfy \(a + b = x\) and \(b - a = y\), and the display exhibits \((x, y, 2x)\) in \(U + W\).

\(U + U = U\). By 1.40 the sum \(U + U\) is the smallest subspace of \(V\) containing \(U\) and \(U\), and that is \(U\) itself, since \(U\) is already a subspace.

Method (2): by 1.36, \(U + U = \{u_1 + u_2 : u_1, u_2 \in U\} \subseteq U\) since \(U\) is closed under addition, while \(u = u + 0 \in U + U\) for each \(u \in U\) since \(0 \in U\).

Yes: \(U + W = W + U\), because addition in \(V\) is commutative (an axiom in 1.20). Indeed, by 1.36 each \(v \in U + W\) has the form

\begin{equation*} v = u + w = w + u \qquad (u \in U,\ w \in W), \end{equation*}

so \(v \in W + U\); hence \(U + W \subseteq W + U\), and interchanging \(U\) and \(W\) gives the reverse inclusion.

Yes: both sides equal

\begin{equation*} S = \{v_1 + v_2 + v_3 : v_1 \in V_1,\ v_2 \in V_2,\ v_3 \in V_3\}, \end{equation*}

the set \(V_1 + V_2 + V_3\) of 1.36, unambiguous because addition in \(V\) is associative (1.20).

Since \(V_1 + V_2\) is a subspace (1.40), applying 1.36 twice writes each \(v \in (V_1+V_2)+V_3\) as \(v = (v_1+v_2)+v_3 = v_1+v_2+v_3 \in S\) with \(v_k \in V_k\); conversely each such \(v_1+v_2+v_3\) equals \((v_1+v_2)+v_3 \in (V_1+V_2)+V_3\). So \((V_1+V_2)+V_3 = S\), and moving the parentheses gives \(V_1 + (V_2+V_3) = S\) by the identical argument.

Yes: \(\{0\}\) is the unique additive identity, and \(\{0\}\) is the only subspace with an additive inverse.

For any subspace \(U\), every element of \(U + \{0\}\) is \(u + 0 = u \in U\), and every \(u \in U\) is \(u + 0 \in U + \{0\}\), so

\begin{equation*} U + \{0\} = \{0\} + U = U \end{equation*}

using 1C.16 for the second equality; thus \(\{0\}\) is a two-sided identity. If \(E\) is another identity, then \(\{0\} + E = \{0\}\) by taking \(U = \{0\}\) in its defining property, while the display with \(U = E\) gives \(\{0\} + E = E\); hence \(E = \{0\}\).

For inverses, suppose \(U + W = \{0\}\). Each \(u \in U\) equals \(u + 0 \in U + W\), so \(U \subseteq \{0\}\) and \(U = \{0\}\); likewise \(W = \{0\}\). Conversely \(\{0\} + \{0\} = \{0\}\), so \(\{0\}\) is its own inverse.

False. In \(V = \mathbf{F}^2\) take

\begin{equation*} V_1 = \{(x, 0) : x \in \mathbf{F}\}, \qquad V_2 = \{(0, y) : y \in \mathbf{F}\}, \qquad U = \mathbf{F}^2, \end{equation*}

each a subspace by 1.34 (Check!). Since \(V_k \subseteq U\), by 1.40 the smallest subspace containing \(V_k\) and \(U\) is \(U\) itself, so

\begin{equation*} V_1 + U = \mathbf{F}^2 = V_2 + U . \end{equation*}

But \((1,0) \in V_1 \setminus V_2\), so \(V_1 \neq V_2\).

Take

\begin{equation*} W = \{(x, 0, y, 0) \in \mathbf{F}^4 : x, y \in \mathbf{F}\}, \end{equation*}

a subspace of \(\mathbf{F}^4\) by 1.34, since it contains \(0\) and both operations act coordinatewise, preserving vanishing of the second and fourth coordinates (Check!).

The sum is all of \(\mathbf{F}^4\): for \((a,b,c,d) \in \mathbf{F}^4\),

\begin{equation*} (a, b, c, d) = (b, b, d, d) + (a - b, 0, c - d, 0) \in U + W . \end{equation*}

It is direct by 1.46, because \(v \in U \cap W\) has the form \(v = (x,x,y,y)\) with second and fourth coordinates \(0\), forcing \(x = y = 0\) and \(v = 0\). Hence \(\mathbf{F}^4 = U \oplus W\).

Take

\begin{equation*} W = \{(0, 0, z_1, z_2, z_3) \in \mathbf{F}^5 : z_1, z_2, z_3 \in \mathbf{F}\}. \end{equation*}

Here \(W\) is a subspace of \(\mathbf{F}^5\) by 1.34: it contains \(0\), and both operations act coordinatewise, preserving vanishing of the first two coordinates (Check!).

The sum is all of \(\mathbf{F}^5\): given \(a = (a_1,\dots,a_5)\), put \(u = (a_1, a_2, a_1+a_2, a_1-a_2, 2a_1) \in U\) (take \(x = a_1\), \(y = a_2\)); then

\begin{equation*} a - u = \bigl(0,\, 0,\, a_3 - a_1 - a_2,\, a_4 - a_1 + a_2,\, a_5 - 2a_1\bigr) \in W, \end{equation*}

so \(a = u + (a-u) \in U + W\). It is direct by 1.46: if \(v \in U \cap W\) then \(v = (x,y,x+y,x-y,2x)\) with first two coordinates \(0\), so \(x = y = 0\) and \(v = 0\). Hence \(\mathbf{F}^5 = U \oplus W\).

Take

\begin{equation*} W_1 = \{(0, 0, z, 0, 0) : z \in \mathbb{F}\}, \qquad W_2 = \{(0, 0, 0, z, 0) : z \in \mathbb{F}\}, \qquad W_3 = \{(0, 0, 0, 0, z) : z \in \mathbb{F}\}. \end{equation*}

Each \(W_k\) is a subspace by 1.34, since both operations act only on the single free coordinate (Check!), and none is \(\{0\}\), each containing the vector with \(1\) in that slot. Writing \(u(x,y) = (x, y, x+y, x-y, 2x)\), the set \(U\) is a subspace too, since

\begin{equation*} u(x,y) + u(x^{\prime},y^{\prime}) = u(x + x^{\prime},\, y + y^{\prime}), \qquad \lambda\, u(x,y) = u(\lambda x, \lambda y), \end{equation*}

and \(0 = u(0,0)\).

The sum is all of \(\mathbb{F}^5\): given \(a = (a_1, \dots, a_5)\), put \(u = u(a_1, a_2) \in U\); then

\begin{equation*} \begin{aligned} a - u &= (0, 0, a_3 - a_1 - a_2, 0, 0) \\ &\quad + (0, 0, 0, a_4 - a_1 + a_2, 0) + (0, 0, 0, 0, a_5 - 2a_1) \end{aligned} \end{equation*}

lies in \(W_1 + W_2 + W_3\), so \(a \in U + W_1 + W_2 + W_3\).

The sum is direct by 1.45: if

\begin{equation*} 0 = u(x,y) + (0, 0, c, 0, 0) + (0, 0, 0, d, 0) + (0, 0, 0, 0, e), \end{equation*}

the first two coordinates give \(x = y = 0\), hence \(u(x,y) = 0\), and then the third, fourth, and fifth give \(c = d = e = 0\). Therefore \(\mathbb{F}^5 = U \oplus W_1 \oplus W_2 \oplus W_3\).

False. In \(V = \mathbb{F}^2\) take

\begin{equation*} U = \{(x, 0) : x \in \mathbb{F}\}, \qquad V_1 = \{(0, y) : y \in \mathbb{F}\}, \qquad V_2 = \{(y, y) : y \in \mathbb{F}\}, \end{equation*}

each a subspace of \(\mathbb{F}^2\) by 1.34, its defining condition being linear in the coordinates (Check!). Both sums exhaust \(\mathbb{F}^2\), since for \((a,b) \in \mathbb{F}^2\)

\begin{equation*} (a, b) = (0, b) + (a, 0) = (b, b) + (a - b, 0), \end{equation*}

and both are direct by 1.46: a vector of \(V_1 \cap U\) has first coordinate \(0\) (from \(V_1\)) and second coordinate \(0\) (from \(U\)), while a vector \((a,b)\) of \(V_2 \cap U\) has \(b = 0\) from \(U\) and \(a = b\) from \(V_2\). So \(\mathbb{F}^2 = V_1 \oplus U = V_2 \oplus U\), yet \((1,1) \in V_2 \setminus V_1\), so \(V_1 \neq V_2\).

Each \(f \in \mathbb{R}^{\mathbb{R}}\) splits as \(f = g + h\) with

\begin{equation*} g(x) = \frac{f(x) + f(-x)}{2} \in V_{\mathrm{e}}, \qquad h(x) = \frac{f(x) - f(-x)}{2} \in V_{\mathrm{o}}, \end{equation*}

since \(g(-x) = g(x)\), \(h(-x) = -h(x)\), and \(g(x) + h(x) = f(x)\) for all \(x\). Hence \(\mathbb{R}^{\mathbb{R}} = V_{\mathrm{e}} + V_{\mathrm{o}}\).

Both \(V_{\mathrm{e}}\) and \(V_{\mathrm{o}}\) are subspaces by 1.34: the zero function is both even and odd, and since the operations are pointwise,

\begin{equation*} (f+g)(-x) = f(-x) + g(-x), \qquad (\lambda f)(-x) = \lambda f(-x), \end{equation*}

so each defining sign condition is inherited by sums and scalar multiples (Check!).

The sum is direct by 1.46: if \(f \in V_{\mathrm{e}} \cap V_{\mathrm{o}}\) then \(f(x) = f(-x) = -f(x)\) for all \(x\), so \(2f(x) = 0\) and \(f = 0\). Therefore \(\mathbb{R}^{\mathbb{R}} = V_{\mathrm{e}} \oplus V_{\mathrm{o}}\).

Take the list

\begin{equation*} (1,0,-1),\quad (0,1,-1),\quad (1,-1,0),\quad (1,1,-2), \end{equation*}

whose entries are pairwise distinct (Check!), and write \(U = \{(x,y,z) \in \mathbf{F}^3 : x+y+z = 0\}\).

Each listed vector lies in \(U\), its coordinates summing to \(0\), and \(U\) is a subspace of \(\mathbf{F}^3\) by 1.34, its defining condition \(x+y+z = 0\) being linear in the coordinates (Check!); so by 2.6, which makes the span the smallest subspace containing the list, the span lies in \(U\). Conversely, if \((x,y,z) \in U\) then \(z = -x-y\), so

\begin{equation*} (x,y,z) = x(1,0,-1) + y(0,1,-1) + 0\,(1,-1,0) + 0\,(1,1,-2) \end{equation*}

lies in the span. Hence the span equals \(U\).

True. Put \(W = \operatorname{span}(v_1 - v_2,\, v_2 - v_3,\, v_3 - v_4,\, v_4)\), a subspace of \(V\) by 2.6. Telescoping from the last vector backwards,

\begin{equation*} \begin{aligned} v_4 &= v_4, \\ v_3 &= (v_3 - v_4) + v_4, \\ v_2 &= (v_2 - v_3) + (v_3 - v_4) + v_4, \\ v_1 &= (v_1 - v_2) + (v_2 - v_3) + (v_3 - v_4) + v_4, \end{aligned} \end{equation*}

so \(v_1, v_2, v_3, v_4 \in W\). By 2.6, \(\operatorname{span}(v_1,v_2,v_3,v_4)\) is the smallest subspace containing them, whence \(V = \operatorname{span}(v_1,v_2,v_3,v_4) \subseteq W \subseteq V\) and \(W = V\).

Each list lies in the span of the other, so by the minimality in 2.6 the two spans contain each other. Writing \(A = \operatorname{span}(v_1,\dots,v_m)\) and \(B = \operatorname{span}(w_1,\dots,w_m)\), both subspaces of \(V\) by 2.6:

\begin{equation*} w_k = 1\,v_1 + \cdots + 1\,v_k + 0\,v_{k+1} + \cdots + 0\,v_m \in A, \end{equation*}

so \(B \subseteq A\); and \(v_1 = w_1 \in B\) while for \(k \geq 2\)

\begin{equation*} v_k = (v_1 + \cdots + v_k) - (v_1 + \cdots + v_{k-1}) = w_k - w_{k-1} \in B, \end{equation*}

so \(A \subseteq B\). Hence \(\operatorname{span}(v_1,\dots,v_m) = \operatorname{span}(w_1,\dots,w_m)\).

(a) The list \(v\) of length one is linearly independent exactly when \(v \neq 0\).

If \(v \neq 0\) and \(a v = 0\) with \(a \neq 0\), then \(a\) is invertible and

\begin{equation*} v = \bigl(a^{-1} a\bigr) v = a^{-1}(a v) = a^{-1}\,0 = 0 \end{equation*}

by 1.31, a contradiction; so \(a = 0\) and the list is linearly independent. If instead \(v = 0\), then \(1 \cdot v = 0\) with \(1 \neq 0\), so the list is linearly dependent (2.17).

(b) The list \(v, w\) is linearly dependent exactly when one of \(v, w\) is a scalar multiple of the other; the assertion is the contrapositive.

Suppose \(v, w\) is linearly dependent, so \(a v + b w = 0\) with \(a, b\) not both \(0\) (2.17).

(i) \(a \neq 0\): multiplying by \(a^{-1}\) gives \(v = -\bigl(a^{-1} b\bigr) w\).

(ii) \(a = 0\): then \(b \neq 0\), so \(w = b^{-1}(b w) = 0 = 0\,v\) by 1.30.

Conversely, if \(v = \lambda w\) then \(1 \cdot v + (-\lambda) w = 0\), and if \(w = \mu v\) then \((-\mu) v + 1 \cdot w = 0\); the coefficient \(1 \neq 0\) makes each combination nontrivial, so the list is linearly dependent (2.17).

Take \(t = 2\), for which

\begin{equation*} (-3)(3,1,4) + 2(2,-3,5) + (5,9,2) = (-9+4+5,\ -3-6+9,\ -12+10+2) = (0,0,0). \end{equation*}

The scalars \(-3, 2, 1\) are not all \(0\), so the list is linearly dependent (2.17), i.e. not linearly independent.

Method (2): solving \(a(3,1,4) + b(2,-3,5) + (5,9,t) = 0\) for the first two coordinates gives \(a = 3b - 9\) and \(3(3b-9) + 2b + 5 = 11b - 22 = 0\), so \(b = 2\) and \(a = -3\); the third coordinate then reads \(-12 + 10 + t = 0\), i.e. \(t = 2\).

If \(c = 8\) the list is linearly dependent, because

\begin{equation*} 2(2,3,1) + 3(1,-1,2) + (-1)(7,3,8) = (4+3-7,\ 6-3-3,\ 2+6-8) = (0,0,0) \end{equation*}

with scalars not all \(0\) (2.17).

Conversely, suppose \(a(2,3,1) + b(1,-1,2) + d(7,3,c) = (0,0,0)\) with \(a, b, d\) not all \(0\) (2.17). Here \((2,3,1), (1,-1,2)\) is linearly independent by Exercise 2A.4(b): in \((1,-1,2) = \lambda(2,3,1)\) the third coordinates force \(\lambda = 2\) while the first force \(2\lambda = 1\), and symmetrically for the other multiple. Hence \(d \neq 0\), since \(d = 0\) would force \(a = b = 0\) as well. Dividing by \(d\) and putting \(\alpha = -a/d\), \(\beta = -b/d\) gives \((7,3,c) = \alpha(2,3,1) + \beta(1,-1,2)\), that is,

\begin{equation*} 2\alpha + \beta = 7, \qquad 3\alpha - \beta = 3, \qquad \alpha + 2\beta = c. \end{equation*}

Adding the first two gives \(\alpha = 2\), then \(\beta = 3\), and so \(c = 2 + 6 = 8\).

(a) Only real scalars are allowed, so suppose \(a, b \in \mathbf{R}\) satisfy

\begin{equation*} a(1+i) + b(1-i) = (a + b) + (a - b)i = 0. \end{equation*}

A complex number vanishes exactly when its real and imaginary parts do, and \(a \pm b\) are real, so \(a + b = 0\) and \(a - b = 0\); adding and subtracting gives \(a = b = 0\). The list is linearly independent (2.15).

(b) Now \(-i\) is an allowed scalar, and

\begin{equation*} (-i)(1+i) + (-1)(1-i) = (1 - i) - (1 - i) = 0 \end{equation*}

is a nontrivial vanishing combination, so the list is linearly dependent (2.17).

Suppose \(a_1, a_2, a_3, a_4 \in \mathbb{F}\) satisfy

\begin{equation*} \begin{aligned} 0 &= a_1(v_1 - v_2) + a_2(v_2 - v_3) + a_3(v_3 - v_4) + a_4 v_4 \\ &= a_1 v_1 + (a_2 - a_1) v_2 + (a_3 - a_2) v_3 + (a_4 - a_3) v_4 . \end{aligned} \end{equation*}

Linear independence of \(v_1, v_2, v_3, v_4\) forces every coefficient to vanish (2.15):

\begin{equation*} a_1 = 0, \qquad a_2 - a_1 = 0, \qquad a_3 - a_2 = 0, \qquad a_4 - a_3 = 0 , \end{equation*}

and reading these in order gives \(a_1 = a_2 = a_3 = a_4 = 0\). Hence the list is linearly independent.

True. Suppose \(a_1, \ldots, a_m \in \mathbb{F}\) satisfy

\begin{equation*} \begin{aligned} 0 &= a_1(5v_1 - 4v_2) + a_2 v_2 + a_3 v_3 + \cdots + a_m v_m \\ &= (5a_1) v_1 + (a_2 - 4a_1) v_2 + a_3 v_3 + \cdots + a_m v_m . \end{aligned} \end{equation*}

Linear independence of \(v_1, \ldots, v_m\) makes every coefficient \(0\) (2.15):

\begin{equation*} 5a_1 = 0, \qquad a_2 - 4a_1 = 0, \qquad a_3 = \cdots = a_m = 0 . \end{equation*}

Since \(\mathbb{F}\) is \(\mathbb{R}\) or \(\mathbb{C}\), we have \(5 \neq 0\), so \(a_1 = 0\) and then \(a_2 = 4a_1 = 0\). Hence all the scalars are \(0\) and the list is linearly independent.

True. Suppose \(a_1, \ldots, a_m \in \mathbb{F}\) satisfy

\begin{equation*} 0 = a_1(\lambda v_1) + \cdots + a_m(\lambda v_m) = (a_1 \lambda) v_1 + \cdots + (a_m \lambda) v_m , \end{equation*}

the second equality by commutativity and associativity of scalar multiplication. Linear independence of \(v_1, \ldots, v_m\) gives \(a_k \lambda = 0\) for every \(k\) (2.15), and \(\lambda \neq 0\) is invertible in \(\mathbb{F}\), so \(a_k = \lambda^{-1}(a_k\lambda) = 0\) for every \(k\). Hence \(\lambda v_1, \ldots, \lambda v_m\) is linearly independent.

False. In \(V = \mathbb{R}^2\) with \(m = 2\), take

\begin{equation*} v_1 = (1,0), \quad v_2 = (0,1), \qquad w_1 = (-1,0), \quad w_2 = (0,-1). \end{equation*}

Each of \(v_1, v_2\) and \(w_1, w_2\) is linearly independent, since \(a_1(1,0) + a_2(0,1) = (a_1, a_2)\) and \(b_1(-1,0) + b_2(0,-1) = (-b_1,-b_2)\) vanish only for zero scalars. But \(v_1 + w_1 = v_2 + w_2 = (0,0)\), and a list containing \(0\) is linearly dependent (2.18).

By 2.17 there are \(a_1, \ldots, a_m \in \mathbb{F}\), not all \(0\), with

\begin{equation*} 0 = \sum_{k=1}^{m} a_k(v_k + w) = a_1 v_1 + \cdots + a_m v_m + s w , \qquad s = a_1 + \cdots + a_m . \end{equation*}

Here \(s \neq 0\): otherwise \(a_1 v_1 + \cdots + a_m v_m = 0\) with the \(a_k\) not all \(0\), contradicting linear independence of \(v_1, \ldots, v_m\) (2.15). Dividing by \(s\),

\begin{equation*} w = -(a_1/s) v_1 - \cdots - (a_m/s) v_m \in \operatorname{span}(v_1, \ldots, v_m). \end{equation*}

If \(w \in \operatorname{span}(v_1, \ldots, v_m)\), say \(w = c_1 v_1 + \cdots + c_m v_m\), then

\begin{equation*} c_1 v_1 + \cdots + c_m v_m + (-1) w = 0 \end{equation*}

has last scalar \(-1 \neq 0\), so \(v_1, \ldots, v_m, w\) is linearly dependent (2.17). That is the contrapositive of one implication.

Conversely, suppose \(w \notin \operatorname{span}(v_1, \ldots, v_m)\) and \(a_1 v_1 + \cdots + a_m v_m + b w = 0\). Were \(b \neq 0\), inverting \(b\) would give

\begin{equation*} w = -(a_1/b) v_1 - \cdots - (a_m/b) v_m \in \operatorname{span}(v_1, \ldots, v_m), \end{equation*}

contrary to hypothesis. So \(b = 0\), leaving \(a_1 v_1 + \cdots + a_m v_m = 0\), whence \(a_1 = \cdots = a_m = 0\) by linear independence of \(v_1, \ldots, v_m\) (2.15). All the scalars vanish, so \(v_1, \ldots, v_m, w\) is linearly independent.

Each list is recovered from the other: \(v_1 = w_1\) and \(v_k = w_k - w_{k-1}\) for \(2 \leq k \leq m\), by cancellation in the partial sums.

(i) Suppose \(v_1, \ldots, v_m\) is linearly independent and \(a_1 w_1 + \cdots + a_m w_m = 0\). Since \(v_j\) occurs in \(w_k\) exactly when \(k \geq j\),

\begin{equation*} 0 = \sum_{k=1}^{m} a_k w_k = \sum_{j=1}^{m} \Bigl( \sum_{k=j}^{m} a_k \Bigr) v_j , \end{equation*}

so \(\sum_{k=j}^{m} a_k = 0\) for each \(j\) (2.15); subtracting the equation for \(j+1\) from that for \(j\) gives \(a_j = 0\) for \(j < m\), and \(j = m\) gives \(a_m = 0\). Hence \(w_1, \ldots, w_m\) is linearly independent.

(ii) Suppose \(w_1, \ldots, w_m\) is linearly independent and \(b_1 v_1 + \cdots + b_m v_m = 0\). Substituting \(v_1 = w_1\) and \(v_k = w_k - w_{k-1}\),

\begin{equation*} 0 = \sum_{k=1}^{m} b_k w_k - \sum_{k=1}^{m-1} b_{k+1} w_k = \sum_{k=1}^{m-1} (b_k - b_{k+1}) w_k + b_m w_m , \end{equation*}

so \(b_1 = b_2 = \cdots = b_m\) and \(b_m = 0\) (2.15); hence all \(b_k = 0\) and \(v_1, \ldots, v_m\) is linearly independent.

Every linearly independent list in \(\mathcal{P}_4(\mathbf{F})\) has length at most \(5\), so no list of six can be linearly independent.

Indeed \(1, z, z^2, z^3, z^4\) spans \(\mathcal{P}_4(\mathbf{F})\), since each \(p \in \mathcal{P}_4(\mathbf{F})\) has \(\deg p \le 4\) and so is written as \(p(z) = a_0 + a_1 z + a_2 z^2 + a_3 z^3 + a_4 z^4\) by 2.11 (padding with zero coefficients; \(p = 0\) takes all \(a_k = 0\)). Thus \(\mathcal{P}_4(\mathbf{F})\) is finite-dimensional (2.9) with a spanning list of length \(5\), and 2.22 bounds every linearly independent list by that length.

The list \(1, z, z^2, z^3, z^4\) is linearly independent in \(\mathcal{P}_4(\mathbf{F})\) and has length \(5\), so by 2.22 no spanning list of length \(4\) can exist.

Linear independence is 2.16(b): if \(a_0 + a_1 z + a_2 z^2 + a_3 z^3 + a_4 z^4 = 0\) for every \(z \in \mathbf{F}\), then all \(a_k = 0\). (That condition involves only the listed vectors and \(\mathbf{F}\), so it reads the same in the subspace \(\mathcal{P}_4(\mathbf{F})\) as in \(\mathcal{P}(\mathbf{F})\).) A spanning list \(q_1, q_2, q_3, q_4\) would make \(\mathcal{P}_4(\mathbf{F})\) finite-dimensional (2.9) with \(5 \le 4\) by 2.22, which is false.

(\(\Leftarrow\)) Given such a sequence, a spanning list \(w_1, \dots, w_n\) of \(V\) (2.9) would force \(n + 1 \le n\) by 2.22 applied to the linearly independent list \(v_1, \dots, v_{n+1}\). So \(V\) is not finite-dimensional, i.e. infinite-dimensional (2.13).

(\(\Rightarrow\)) Suppose \(V\) is infinite-dimensional and choose \(v_m\) recursively.

(i) \(V \ne \{0\}\), since otherwise the empty list spans \(V\); pick \(v_1 \ne 0\), linearly independent by 2.16(c).

(ii) Given \(v_1, \dots, v_m\) linearly independent, no list spans \(V\), so \(\operatorname{span}(v_1, \dots, v_m) \ne V\) and we may choose

\begin{equation*} v_{m+1} \in V \setminus \operatorname{span}(v_1, \dots, v_m). \end{equation*}

By Exercise 2A.13 the list \(v_1, \dots, v_{m+1}\) is linearly independent.

The recursion never halts, and every initial segment is linearly independent.

Take \(e_k \in \mathbf{F}^\infty\) to be the sequence with \(1\) in coordinate \(k\) and \(0\) elsewhere. For every \(m\),

\begin{equation*} a_1 e_1 + \cdots + a_m e_m = (a_1, \dots, a_m, 0, 0, \dots), \end{equation*}

which is the zero sequence only if \(a_1 = \cdots = a_m = 0\); so \(e_1, \dots, e_m\) is linearly independent. By Exercise 2A.17, \(\mathbf{F}^\infty\) is infinite-dimensional.

Take \(f_k \in C[0,1]\) to be \(f_k(x) = x^k\) for \(k = 0, 1, 2, \dots\), each continuous on \([0,1]\).

For every \(m\) the list \(f_0, \dots, f_m\) is linearly independent: if \(a_0 f_0 + \cdots + a_m f_m = 0\) in \(C[0,1]\), then the polynomial \(p(z) = a_0 + a_1 z + \cdots + a_m z^m\) vanishes at every point of \([0,1]\), hence has infinitely many zeros in \(\mathbf{R}\), so \(p = 0\) by 4.8 (a nonzero \(p\) of degree \(d\) has at most \(d\) zeros for \(d \ge 1\), and none for \(d = 0\)). Uniqueness of polynomial coefficients then gives \(a_0 = \cdots = a_m = 0\).

Applying Exercise 2A.17 to the sequence \(f_0, f_1, f_2, \dots\) shows \(C[0,1]\) is infinite-dimensional.

The \(m + 1\) polynomials all lie in

\begin{equation*} U = \{p \in \mathcal{P}_m(\mathbf{F}) : p(2) = 0\}, \end{equation*}

which is spanned by a list of length \(m\); so 2.22 forbids their linear independence.

\(U\) is a subspace of \(\mathcal{P}_m(\mathbf{F})\) by 1.34, since \(0 \in U\) and \((p + \lambda q)(2) = p(2) + \lambda q(2) = 0\) for \(p, q \in U\). For \(j \in \{1, \dots, m\}\) put \(q_j(z) = z^j - 2^j\); each has degree \(j \le m\) and \(q_j(2) = 0\), so \(q_j \in U\) and \(\operatorname{span}(q_1, \dots, q_m) \subseteq U\). Conversely, writing \(p \in U\) as \(p(z) = a_0 + a_1 z + \cdots + a_m z^m\) (2.11, 2.12),

\begin{equation*} p(z) = p(z) - p(2) = a_1 q_1(z) + a_2 q_2(z) + \cdots + a_m q_m(z), \end{equation*}

so \(U = \operatorname{span}(q_1, \dots, q_m)\) and \(U\) is finite-dimensional (2.9). (For \(m = 0\) both sides are \(\{0\}\).)

Linear independence depends only on the vectors and \(\mathbf{F}\), so were \(p_0, \dots, p_m\) linearly independent in \(\mathcal{P}_m(\mathbf{F})\) it would be a linearly independent list of length \(m + 1\) in \(U\), giving \(m + 1 \le m\) by 2.22.

Only the zero vector space \(\{0\}\), whose unique basis is the empty list.

The empty list is linearly independent (2.15) with span \(\{0\}\) (2.4), hence a basis of \(\{0\}\) by 2.26; every nonempty list in \(\{0\}\) reads \(0, \ldots, 0\) and is linearly dependent (2.18). So \(\{0\}\) has exactly one basis.

Conversely, let \(v_1, \ldots, v_n\) be the unique basis of \(V\) and suppose \(n \ge 1\). Then \(v_1 \ne 0\) by linear independence, and \(2v_1, v_2, \ldots, v_n\) is again a basis: it spans, since \(v_1 = \tfrac12(2v_1)\) puts \(V = \operatorname{span}(v_1, \ldots, v_n)\) inside its span (\(\tfrac12 \in \mathbf{F}\) as \(\mathbf{F}\) is \(\mathbf{R}\) or \(\mathbf{C}\)); and \(a_1(2v_1) + a_2v_2 + \cdots + a_nv_n = 0\) gives \(2a_1 = a_2 = \cdots = a_n = 0\), so all \(a_k = 0\). But \(2v_1 \ne v_1\) since \(2v_1 - v_1 = v_1 \ne 0\), so this is a second basis, a contradiction. Hence \(n = 0\) and \(V = \operatorname{span}() = \{0\}\).

Throughout, a list is a basis exactly when it is linearly independent and spans (2.26).

(a) \(a_1 e_1 + \cdots + a_n e_n = (a_1, \ldots, a_n)\), so every \((x_1, \ldots, x_n) \in \mathbf{F}^n\) equals \(x_1 e_1 + \cdots + x_n e_n\) and the combination vanishes only for \(a_1 = \cdots = a_n = 0\). Hence \(e_1, \ldots, e_n\) is a basis of \(\mathbf{F}^n\).

(b) \(a(1,2) + b(3,5) = (x,y)\) reads \(a + 3b = x\), \(2a + 5b = y\), whose unique solution is \(b = 2x - y\), \(a = -5x + 3y\); and

\begin{equation*} (-5x+3y)(1,2) + (2x-y)(3,5) = (x, y). \quad \text{(Check!)} \end{equation*}

So representations exist and are unique, making \((1,2),(3,5)\) a basis of \(\mathbf{F}^2\) by 2.28, of length two like the standard basis in (a).

(c) \(a(1,2,-4) + b(7,-5,6) = (0,0,0)\) gives \(a + 7b = 0\) and \(2a - 5b = 0\), hence \(-19b = 0\) and \(a = b = 0\): linearly independent. It does not span, since \(a(1,2,-4) + b(7,-5,6) = (0,0,1)\) forces \(a = b = 0\) from the first two coordinates and then \(0 \neq 1\) in the third. So it is not a basis of \(\mathbf{F}^3\).

(d) It spans \(\mathbf{F}^2\), since the sublist \((1,2),(3,5)\) already does by (b) and enlarging a list only enlarges its span. It is linearly dependent: the formulas of (b) at \((x,y) = (4,13)\) give

\begin{equation*} 19(1,2) - 5(3,5) - 1\cdot(4,13) = (0,0), \end{equation*}

a nontrivial vanishing combination (2.17). Hence not a basis.

(e) \(a(1,1,0) + b(0,0,1) = (a,a,b)\), which lies in \(U = \{(x,x,y)\}\), and conversely \((x,x,y) = x(1,1,0) + y(0,0,1)\); so \(U = \operatorname{span}\big((1,1,0),(0,0,1)\big)\) is a subspace (2.6) that the list spans. It vanishes only for \(a = b = 0\), so the list is a basis of \(U\).

(f) \(a(1,-1,0) + b(1,0,-1) = (a+b,-a,-b)\), whose coordinates sum to \(0\), so the span lies in \(U = \{x + y + z = 0\}\); conversely \((x,y,z) \in U\) has \(x = -y - z\) and

\begin{equation*} (-y)(1,-1,0) + (-z)(1,0,-1) = (-y-z,\; y,\; z) = (x,y,z). \end{equation*}

Hence \(U\) is that span, a subspace by 2.6; and \((a+b,-a,-b) = 0\) forces \(a = b = 0\). So the list is a basis of \(U\).

(g) By definition every \(p \in \mathcal{P}_m(\mathbf{F})\) is \(p(z) = a_0 + a_1 z + \cdots + a_m z^m\), i.e. \(1, z, \ldots, z^m\) spans \(\mathcal{P}_m(\mathbf{F})\); it is linearly independent by 2.16(b). Hence it is a basis.

Finally, \((7,5), (-4,9)\) is a basis of \(\mathbf{F}^2\): the system \(7a - 4b = x\), \(5a + 9b = y\) has determinant \(83 \neq 0\) in \(\mathbf{F}\), so it has the unique solution

\begin{equation*} a = \frac{9x + 4y}{83}, \qquad b = \frac{7y - 5x}{83} \quad \text{(Check!)}, \end{equation*}

and 2.28 applies. That \((1,2),(3,5)\) is a basis was shown in (b).

(a) A basis of \(U\) is

\begin{equation*} (3,1,0,0,0),\quad (0,0,7,1,0),\quad (0,0,0,0,1), \end{equation*}

each vector lying in \(U\) (Check!). It spans: the constraints \(x_1 = 3x_2\), \(x_3 = 7x_4\) leave \(x_2 = t\), \(x_4 = s\), \(x_5 = r\) free, and

\begin{equation*} (3t, t, 7s, s, r) = t(3,1,0,0,0) + s(0,0,7,1,0) + r(0,0,0,0,1). \end{equation*}

It is linearly independent, since that combination equals \((3a,a,7b,b,c)\), whose second, fourth, and fifth coordinates give \(a = b = c = 0\). So it is a basis by 2.26.

(b) Adjoin \((1,0,0,0,0)\) and \((0,0,1,0,0)\), giving the list

\begin{equation*} (3,1,0,0,0),\ (0,0,7,1,0),\ (0,0,0,0,1),\ (1,0,0,0,0),\ (0,0,1,0,0). \end{equation*}

The combination \(a(3,1,0,0,0) + b(0,0,7,1,0) + c(0,0,0,0,1) + d(1,0,0,0,0) + e(0,0,1,0,0)\) equals \((3a + d,\, a,\, 7b + e,\, b,\, c)\). It spans \(\mathbf{R}^5\): take \(a = x_2\), \(b = x_4\), \(c = x_5\), \(d = x_1 - 3x_2\), \(e = x_3 - 7x_4\). It is linearly independent: if that vector is \(0\), coordinates two, four, five give \(a = b = c = 0\), and then coordinates one and three give \(d = e = 0\). So this length-five list is a basis of \(\mathbf{R}^5\) extending (a).

(c) Take

\begin{equation*} \begin{aligned} W &= \operatorname{span}\big((1,0,0,0,0), (0,0,1,0,0)\big) \\ &= \{(x_1, 0, x_3, 0, 0) : x_1, x_3 \in \mathbf{R}\}, \end{aligned} \end{equation*}

a subspace by 2.6. By (b) every vector of \(\mathbf{R}^5\) splits as a combination of the first three listed vectors, which lies in \(U\), plus one of the last two, which lies in \(W\); so \(U + W = \mathbf{R}^5\). And \((x_1,0,x_3,0,0) \in U\) forces \(x_1 = 3 \cdot 0 = 0\) and \(x_3 = 7 \cdot 0 = 0\), so \(U \cap W = \{0\}\). Hence \(\mathbf{R}^5 = U \oplus W\) by 1.46.

(a) A basis of \(U\) is

\begin{equation*} (1,6,0,0,0),\quad (0,0,-2,1,0),\quad (0,0,-3,0,1), \end{equation*}

each vector lying in \(U\) (Check!). It spans: the constraints \(z_2 = 6z_1\), \(z_3 = -2z_4 - 3z_5\) leave \(z_1 = t\), \(z_4 = s\), \(z_5 = r\) free, and

\begin{equation*} (t, 6t, -2s - 3r, s, r) = t(1,6,0,0,0) + s(0,0,-2,1,0) + r(0,0,-3,0,1). \end{equation*}

It is linearly independent, since that combination equals \((a, 6a, -2b - 3c, b, c)\), whose first, fourth, and fifth coordinates give \(a = b = c = 0\). So it is a basis by 2.26.

(b) Adjoin \((0,1,0,0,0)\) and \((0,0,1,0,0)\), giving

\begin{equation*} (1,6,0,0,0),\ (0,0,-2,1,0),\ (0,0,-3,0,1),\ (0,1,0,0,0),\ (0,0,1,0,0). \end{equation*}

The combination \(a(1,6,0,0,0) + b(0,0,-2,1,0) + c(0,0,-3,0,1) + d(0,1,0,0,0) + e(0,0,1,0,0)\) equals \((a,\; 6a + d,\; -2b - 3c + e,\; b,\; c)\). It spans \(\mathbf{C}^5\): take \(a = z_1\), \(b = z_4\), \(c = z_5\), \(d = z_2 - 6z_1\), \(e = z_3 + 2z_4 + 3z_5\). It is linearly independent: if that vector is \(0\), coordinates one, four, five give \(a = b = c = 0\), and then coordinates two and three give \(d = e = 0\). So it is a basis of \(\mathbf{C}^5\) extending (a).

(c) Take

\begin{equation*} \begin{aligned} W &= \operatorname{span}\big((0,1,0,0,0), (0,0,1,0,0)\big) \\ &= \{(0, z_2, z_3, 0, 0) : z_2, z_3 \in \mathbf{C}\}, \end{aligned} \end{equation*}

a subspace by 2.6. By (b) every vector of \(\mathbf{C}^5\) splits as a combination of the first three listed vectors, which lies in \(U\), plus one of the last two, which lies in \(W\); so \(U + W = \mathbf{C}^5\). And \((0,z_2,z_3,0,0) \in U\) forces \(z_2 = 6 \cdot 0 = 0\) and \(z_3 = 0\), so \(U \cap W = \{0\}\). Hence \(\mathbf{C}^5 = U \oplus W\) by 1.46.

Reduce the concatenated bases of \(U\) and \(W\) to a basis of \(V\) (2.30).

In detail: \(U\) and \(W\) are finite-dimensional (2.25), so they have bases \(u_1, \ldots, u_m\) and \(w_1, \ldots, w_n\) (2.31). Each \(v \in V\) is \(v = u + w\) with \(u \in U\), \(w \in W\) since \(V = U + W\), and expanding \(u\) and \(w\) in those bases gives

\begin{equation*} v = a_1u_1 + \cdots + a_mu_m + b_1w_1 + \cdots + b_nw_n, \end{equation*}

so \(u_1, \ldots, u_m, w_1, \ldots, w_n\) spans \(V\). By 2.30 some sublist of it is a basis of \(V\), and every vector of that sublist lies in \(U \cup W\).

False: take

\begin{equation*} p_0(z) = 1, \qquad p_1(z) = z, \qquad p_2(z) = z^2 + z^3, \qquad p_3(z) = z^3 , \end{equation*}

of degrees \(0, 1, 3, 3\), none equal to \(2\).

This list spans, since

\begin{equation*} a + bz + cz^2 + dz^3 = a\,p_0 + b\,p_1 + c\,p_2 + (d - c)\,p_3 , \end{equation*}

and \(1, z, z^2, z^3\) spans \(\mathcal{P}_3(\mathbf{F})\) (Example 2.27(g)). It is linearly independent: \(a\,p_0 + b\,p_1 + c\,p_2 + d\,p_3 = 0\) reads \(a + bz + cz^2 + (c+d)z^3 = 0\) for all \(z\), so \(a = b = c = 0\) and \(c + d = 0\) by 2.16(b), whence \(d = 0\). So it is a basis by 2.26.

By 2.26 it suffices that the list be linearly independent and span \(V\).

Linear independence: regrouping \(a(v_1 + v_2) + b(v_2 + v_3) + c(v_3 + v_4) + d v_4 = 0\) gives

\begin{equation*} a v_1 + (a + b) v_2 + (b + c) v_3 + (c + d) v_4 = 0 , \end{equation*}

so \(a = a + b = b + c = c + d = 0\) by linear independence of the basis \(v_1, v_2, v_3, v_4\); reading in order, \(a = b = c = d = 0\).

Spanning: writing \(S\) for the span of the new list, which is a subspace (2.6),

\begin{equation*} v_4 \in S, \quad v_3 = (v_3 + v_4) - v_4 \in S, \quad v_2 = (v_2 + v_3) - v_3 \in S, \end{equation*}

and then \(v_1 = (v_1 + v_2) - v_2 \in S\). Hence \(V = \operatorname{span}(v_1, v_2, v_3, v_4) \subseteq S\), so \(S = V\).

False: in \(V = \mathbf{F}^4\) take \(v_1, v_2, v_3, v_4\) to be the standard basis \(e_1, e_2, e_3, e_4\) and

\begin{equation*} \begin{aligned} U &= \operatorname{span}(e_1,\, e_2,\, e_3 + e_4) \\ &= \{(x_1, x_2, x_3, x_4) \in \mathbf{F}^4 : x_3 = x_4\}, \end{aligned} \end{equation*}

the two descriptions agreeing because \(e_1, e_2, e_3 + e_4\) all satisfy \(x_3 = x_4\), a subspace condition, while \(x_3 = x_4\) gives \((x_1,x_2,x_3,x_4) = x_1 e_1 + x_2 e_2 + x_3(e_3+e_4)\).

The hypotheses hold: \(e_1, e_2 \in U\) (third and fourth coordinates both \(0\)), while \(e_3, e_4 \notin U\) (their third and fourth coordinates differ). But \(e_3 + e_4 \in U\) has third coordinate \(1\), whereas every vector in \(\operatorname{span}(e_1, e_2)\) has third coordinate \(0\); so \(v_1, v_2\) does not span \(U\) and is not a basis of \(U\) (2.26).

With \(w_0 = 0\) one has \(v_k = w_k - w_{k-1}\) for every \(k\), so each list lies in the span of the other and \[\operatorname{span}(v_1, \ldots, v_m) = \operatorname{span}(w_1, \ldots, w_m);\] in particular one list spans \(V\) exactly when the other does. Expanding each side in the other list (with \(b_{m+1} = 0\)),

\begin{equation*} \begin{aligned} \sum_{k=1}^{m} a_k w_k &= \sum_{j=1}^{m} \Bigl( \sum_{k=j}^{m} a_k \Bigr) v_j, \\ \sum_{k=1}^{m} b_k v_k &= \sum_{k=1}^{m} (b_k - b_{k+1}) w_k . \end{aligned} \end{equation*}

(i) If \(v_1, \ldots, v_m\) is linearly independent and \(\sum_k a_k w_k = 0\), the first identity forces \(\sum_{k=j}^{m} a_k = 0\) for every \(j\); subtracting the equation for \(j+1\) from that for \(j\) gives \(a_j = 0\).

(ii) If \(w_1, \ldots, w_m\) is linearly independent and \(\sum_k b_k v_k = 0\), the second identity forces \(b_k = b_{k+1}\) for every \(k\), and \(b_{m+1} = 0\) then gives \(b_1 = \cdots = b_m = 0\).

So the lists are linearly independent together, and by 2.26 one is a basis of \(V\) exactly when the other is.

By 2.26 it suffices to check spanning and linear independence of the concatenated list.

Spanning. Each \(v \in V\) is \(v = u + w\) with \(u \in U\), \(w \in W\), since \(V = U + W\); expanding \(u\) and \(w\) in their bases gives

\begin{equation*} v = a_1 u_1 + \cdots + a_m u_m + b_1 w_1 + \cdots + b_n w_n . \end{equation*}

Independence. If that sum is \(0\), then \(u + w = 0\) with \(u = \sum_j a_j u_j \in U\) and \(w = \sum_k b_k w_k \in W\), so \(u = w = 0\) by 1.45 (the sum \(U + W\) is direct); independence of each of the two bases makes every \(a_j\) and every \(b_k\) zero.

Everything follows from the identity \((\ast)\): for \(a_k, b_k \in \mathbf{R}\), with each \(v_k\) read as \(v_k + i0\),

\begin{equation*} \sum_{k=1}^{n} (a_k + b_k i)(v_k + i0) = \Bigl( \sum_{k=1}^{n} a_k v_k \Bigr) + i \Bigl( \sum_{k=1}^{n} b_k v_k \Bigr), \end{equation*}

since \((a + bi)(v + i0) = (av) + i(bv)\) and addition in \(V_{\mathbf{C}}\) is componentwise (Exercise 8 in Section 1B). As every complex scalar is \(a + bi\) with \(a, b \in \mathbf{R}\), \((\ast)\) covers all complex linear combinations of the list.

Spanning. Given \(u + iv \in V_{\mathbf{C}}\), expand \(u = \sum_k a_k v_k\) and \(v = \sum_k b_k v_k\) over \(\mathbf{R}\); then \((\ast)\) gives \(\sum_k (a_k + b_k i) v_k = u + iv\).

Independence. If \(\sum_k (a_k + b_k i) v_k = 0 + i0\), then \((\ast)\) and componentwise equality in \(V_{\mathbf{C}} = V \times V\) give \(\sum_k a_k v_k = \sum_k b_k v_k = 0\) in \(V\); independence over \(\mathbf{R}\) makes every real \(a_k, b_k\) zero.

Hence \(v_1, \ldots, v_n\) is a basis of \(V_{\mathbf{C}}\) over \(\mathbf{C}\) by 2.26.

A subspace of \(\mathbb{R}^2\) has dimension \(0\), \(1\), or \(2\), and those three cases are exactly the three listed sets.

A line through the origin is precisely the span of one nonzero vector: if \(L = \{w + \lambda v : \lambda \in \mathbb{R}\}\) with \(v \neq 0\) and \(0 \in L\), then \(w = -\lambda_0 v\) for some \(\lambda_0\), so

\begin{equation*} L = \{(\lambda - \lambda_0)v : \lambda \in \mathbb{R}\} = \operatorname{span}(v); \end{equation*}

conversely \(\operatorname{span}(v) = \{0 + \lambda v : \lambda \in \mathbb{R}\}\) is such a line when \(v \neq 0\). Each listed set is therefore a subspace, spans being subspaces (2.6). Conversely a subspace \(U\) is finite-dimensional (2.25) with \(\dim U \leq \dim \mathbb{R}^2 = 2\) by 2.37 and 2.36, and:

(i) \(\dim U = 0\): the empty list is a basis, so \(U = \{0\}\) by 2.4.

(ii) \(\dim U = 1\): a basis of \(U\) is a single \(v \neq 0\) (a list containing \(0\) is dependent), so \(U = \operatorname{span}(v)\) is a line through the origin.

(iii) \(\dim U = 2 = \dim \mathbb{R}^2\): then \(U = \mathbb{R}^2\) by 2.39.

A subspace of \(\mathbb{R}^3\) has dimension \(0\), \(1\), \(2\), or \(3\), and those four cases are exactly the four listed sets.

Lines through the origin are the \(1\)-dimensional subspaces. If \(L = \{w + \lambda v : \lambda \in \mathbb{R}\}\) with \(v \neq 0\) and \(0 \in L\), then \(w = -\lambda_0 v\) for some \(\lambda_0\), so \(L = \operatorname{span}(v)\), of dimension \(1\) since \(v\) alone is linearly independent. Conversely a \(1\)-dimensional subspace is \(\operatorname{span}(v) = \{0 + \lambda v : \lambda \in \mathbb{R}\}\) for a basis vector \(v \neq 0\).

Planes through the origin are the \(2\)-dimensional subspaces. If \(P = \{w + \lambda_1 v_1 + \lambda_2 v_2 : \lambda_i \in \mathbb{R}\}\) with \(v_1, v_2\) linearly independent and \(0 \in P\), then \(w = -(\mu_1 v_1 + \mu_2 v_2)\) for some \(\mu_1, \mu_2\), so

\begin{equation*} \begin{aligned} P &= \{(\lambda_1 - \mu_1)v_1 + (\lambda_2 - \mu_2)v_2 : \lambda_i \in \mathbb{R}\}\\ &= \operatorname{span}(v_1, v_2), \end{aligned} \end{equation*}

a subspace (2.6) with basis \(v_1, v_2\), so \(\dim P = 2\). Conversely a \(2\)-dimensional subspace is the span of its (linearly independent) basis \(v_1, v_2\), hence such a plane.

Each listed set is thus a subspace; conversely a subspace \(U\) of \(\mathbb{R}^3\) is finite-dimensional (2.25) with \(\dim U \leq \dim \mathbb{R}^3 = 3\) by 2.37 and 2.36, and:

(i) \(\dim U = 0\): the empty list is a basis, so \(U = \{0\}\).

(ii) \(\dim U = 1\): \(U\) is a line through the origin.

(iii) \(\dim U = 2\): \(U\) is a plane through the origin.

(iv) \(\dim U = 3 = \dim \mathbb{R}^3\): then \(U = \mathbb{R}^3\) by 2.39.

(a) A basis of \(U\) is \((x-6),\ (x-6)^2,\ (x-6)^3,\ (x-6)^4\). Writing \(\mathcal{P}_4 = \mathcal{P}_4(\mathbb{F})\), of dimension \(5\) by 2.36, the shifted powers

\begin{equation*} 1,\ (x-6),\ (x-6)^2,\ (x-6)^3,\ (x-6)^4 \end{equation*}

are linearly independent (in a vanishing combination the coefficient of \(x^4\) gives \(c_4 = 0\), then \(x^3\) gives \(c_3 = 0\), and so downward), so having length \(5 = \dim \mathcal{P}_4\) they are a basis of \(\mathcal{P}_4\) by 2.38. Expanding \(p\) in them,

\begin{equation*} p = c_0 + c_1(x-6) + c_2(x-6)^2 + c_3(x-6)^3 + c_4(x-6)^4 , \end{equation*}

gives \(p(6) = c_0\), so \(p \in U\) if and only if \(c_0 = 0\). The four listed polynomials thus span \(U\) and are independent as a sublist; \(\dim U = 4\).

(b) Adjoin the constant polynomial \(1\), recovering the basis of \(\mathcal{P}_4\) above.

(c) Take \(W = \operatorname{span}(1)\). For \(p \in \mathcal{P}_4\), let \(c\) be the constant polynomial \(p(6)\); then \(p = (p - c) + c\) with \(p - c \in U\) and \(c \in W\), giving \(\mathcal{P}_4 = U + W\). If \(q \in U \cap W\), then \(q\) is a constant with \(q(6) = 0\), so \(q = 0\); by 1.46 the sum is direct, and \(\mathcal{P}_4(\mathbb{F}) = U \oplus W\).

(a) A basis of \(U\) is \(1,\ (x-6),\ (x-6)^3,\ (x-6)^4\). Writing \(\mathcal{P}_4 = \mathcal{P}_4(\mathbb{R})\), of dimension \(5\) by 2.36, the shifted powers

\begin{equation*} 1,\ (x-6),\ (x-6)^2,\ (x-6)^3,\ (x-6)^4 \end{equation*}

are a basis of \(\mathcal{P}_4\) by 2C.3 and 2.38. Expanding \(p = \sum_k c_k (x-6)^k\) and differentiating twice,

\begin{equation*} p^{\prime\prime} = 2c_2 + 6c_3(x-6) + 12c_4(x-6)^2 , \end{equation*}

so \(p^{\prime\prime}(6) = 2c_2\) and \(p \in U\) if and only if \(c_2 = 0\). The four listed polynomials thus span \(U\) and are independent as a sublist; \(\dim U = 4\).

(b) Adjoin \((x-6)^2\), recovering the basis of \(\mathcal{P}_4\) above.

(c) Take \(W = \operatorname{span}\bigl((x-6)^2\bigr)\). For \(p \in \mathcal{P}_4\) set \(\lambda = \tfrac{1}{2}p^{\prime\prime}(6)\); since \(\bigl(\lambda(x-6)^2\bigr)^{\prime\prime} = 2\lambda = p^{\prime\prime}(6)\),

\begin{equation*} p = \bigl(p - \lambda (x-6)^2\bigr) + \lambda (x-6)^2 \end{equation*}

has first summand in \(U\) and second in \(W\), so \(\mathcal{P}_4 = U + W\). If \(q \in U \cap W\), then \(q = c(x-6)^2\) with \(q^{\prime\prime}(6) = 2c = 0\), so \(q = 0\); by 1.46 the sum is direct, and \(\mathcal{P}_4(\mathbb{R}) = U \oplus W\).

(a) A basis of \(U\) is \(1,\ (x-2)(x-5),\ x(x-2)(x-5),\ x^2(x-2)(x-5)\). Writing \(\mathcal{P}_4 = \mathcal{P}_4(\mathbb{F})\), of dimension \(5\) by 2.36, the list

\begin{equation*} 1,\quad x-2,\quad (x-2)(x-5),\quad x(x-2)(x-5),\quad x^2(x-2)(x-5) \end{equation*}

is monic of degrees \(0,1,2,3,4\), hence linearly independent (in a vanishing combination the coefficient of \(x^4\) gives \(c_4 = 0\), then \(x^3\) gives \(c_3 = 0\), and so downward), so having length \(5 = \dim\mathcal{P}_4\) it is a basis of \(\mathcal{P}_4\) by 2.38. Expand \(p \in \mathcal{P}_4\) in it with coefficients \(c_0, \ldots, c_4\); every term after the second vanishes at \(2\) and at \(5\), so

\begin{equation*} p(2) = c_0, \qquad p(5) = c_0 + 3c_1 , \end{equation*}

so \(p \in U\) if and only if \(3c_1 = 0\), i.e. (as \(3\) is invertible in \(\mathbb{F}\)) if and only if \(c_1 = 0\). Hence \(U\) is the span of the four listed polynomials, each of which does lie in \(U\), and they are linearly independent as a sublist; \(\dim U = 4\).

(b) Adjoin \(x-2\), recovering the basis of \(\mathcal{P}_4\) above.

(c) Take \(W = \operatorname{span}(x-2)\). Given \(p \in \mathcal{P}_4\), set \(\lambda = \tfrac{1}{3}\bigl(p(5) - p(2)\bigr)\) and \(q = p - \lambda(x-2)\); then \(q(2) = p(2)\) and

\begin{equation*} q(5) = p(5) - 3\lambda = p(2) = q(2), \end{equation*}

so \(p = q + \lambda(x-2)\) lies in \(U + W\). If \(r \in U \cap W\), then \(r = c(x-2)\) and \(r(2) = r(5)\) gives \(0 = 3c\), so \(r = 0\); by 1.46 the sum is direct: \(\mathcal{P}_4(\mathbb{F}) = U \oplus W\).

(a) A basis of \(U\) is \(1,\ (x-2)(x-5)(x-6),\ x(x-2)(x-5)(x-6)\). Writing \(\mathcal{P}_4 = \mathcal{P}_4(\mathbb{F})\), of dimension \(5\) by 2.36, the list

\begin{equation*} 1,\quad x-2,\quad (x-2)(x-5),\quad (x-2)(x-5)(x-6),\quad x(x-2)(x-5)(x-6) \end{equation*}

is monic of degrees \(0,1,2,3,4\), hence linearly independent (the coefficient of \(x^4\) gives \(c_4 = 0\), then \(x^3\) gives \(c_3 = 0\), and so downward), so having length \(5 = \dim\mathcal{P}_4\) it is a basis of \(\mathcal{P}_4\) by 2.38. Expand \(p \in \mathcal{P}_4\) in it with coefficients \(c_0, \dots, c_4\); the last two terms vanish at each of \(2, 5, 6\), so

\begin{equation*} p(2) = c_0, \qquad p(5) = c_0 + 3c_1, \qquad p(6) = c_0 + 4c_1 + 4c_2 , \end{equation*}

and \(p \in U\) if and only if \(3c_1 = 0\) and \(4c_1 + 4c_2 = 0\), i.e. (as \(3\) and \(4\) are invertible in \(\mathbb{F}\)) if and only if \(c_1 = c_2 = 0\). Hence \(U\) is the span of the three listed polynomials, all of which lie in \(U\) (the constant takes one value everywhere, the other two vanish at \(2, 5, 6\)), and they are independent as a sublist; \(\dim U = 3\).

(b) Adjoin \(x-2\) and \((x-2)(x-5)\), recovering after reordering the basis of \(\mathcal{P}_4\) above.

(c) Take \(W = \operatorname{span}\bigl(x-2,\ (x-2)(x-5)\bigr)\). The basis in (b) spans \(\mathcal{P}_4\) and each of its vectors lies in \(U\) or in \(W\), so \(\mathcal{P}_4 = U + W\). If \(q \in U \cap W\), then by (a) and the definition of \(W\),

\begin{equation*} a_0 + a_3 (x-2)(x-5)(x-6) + a_4\, x(x-2)(x-5)(x-6) = b_1(x-2) + b_2(x-2)(x-5) \end{equation*}

for some scalars; subtracting writes \(0\) in the basis from (b), so all coefficients vanish by 2.28 and \(q = 0\). By 1.46 the sum is direct: \(\mathcal{P}_4(\mathbb{F}) = U \oplus W\).

(a) A basis of \(U\) is \(x,\ x^2 - \tfrac{1}{3},\ x^3,\ x^4 - \tfrac{1}{5}\). Writing \(\mathcal{P}_4 = \mathcal{P}_4(\mathbb{R})\), of dimension \(5\) by 2.36, the list

\begin{equation*} 1,\quad x,\quad x^2 - \tfrac{1}{3},\quad x^3,\quad x^4 - \tfrac{1}{5} \end{equation*}

is monic of degrees \(0,1,2,3,4\), hence linearly independent (the coefficient of \(x^4\) forces \(c_4 = 0\), then \(x^3\) forces \(c_3 = 0\), and so downward), so having length \(5 = \dim\mathcal{P}_4\) it is a basis of \(\mathcal{P}_4\) by 2.38. Since \(\int_{-1}^{1} x^{k}\,dx\) equals \(2/(k+1)\) for \(k\) even and \(0\) for \(k\) odd,

\begin{equation*} \int_{-1}^{1} 1 = 2, \qquad \int_{-1}^{1} x = \int_{-1}^{1} x^3 = 0, \end{equation*}

while \(\int_{-1}^{1}(x^2 - \tfrac13) = \tfrac23 - \tfrac23 = 0\) and \(\int_{-1}^{1}(x^4 - \tfrac15) = \tfrac25 - \tfrac25 = 0\). So expanding \(p\) in the basis with coefficients \(c_0, \dots, c_4\) gives \(\int_{-1}^{1} p = 2c_0\), and \(p \in U\) if and only if \(c_0 = 0\). Hence \(U\) is the span of the four listed polynomials, each of which lies in \(U\), and they are independent as a sublist; \(\dim U = 4\).

(b) Adjoin the constant polynomial \(1\), recovering the basis of \(\mathcal{P}_4\) above.

(c) Take \(W = \operatorname{span}(1)\). Given \(p \in \mathcal{P}_4\), set \(c = \tfrac{1}{2}\int_{-1}^{1} p\); then \(p = (p - c) + c\) with \(c \in W\) and

\begin{equation*} \int_{-1}^{1}(p - c) = \int_{-1}^{1} p - 2c = 0 , \end{equation*}

so \(p - c \in U\) and \(\mathcal{P}_4 = U + W\). If \(q \in U \cap W\), then \(q\) is a constant \(a\) with \(0 = \int_{-1}^{1} a = 2a\), so \(q = 0\); by 1.46 the sum is direct: \(\mathcal{P}_4(\mathbb{R}) = U \oplus W\).

The list \(v_1 - v_m, \dots, v_{m-1} - v_m\) is a linearly independent list of length \(m-1\) lying in \(U = \operatorname{span}(v_1 + w, \dots, v_m + w)\), which gives the bound. (For \(m = 1\) the claim is only \(\dim U \ge 0\), so assume \(m \ge 2\).)

Each \(v_k - v_m = (v_k + w) - (v_m + w)\) lies in \(U\), and if \(\sum_{k=1}^{m-1} a_k (v_k - v_m) = 0\), regrouping gives \[a_1 v_1 + \dots + a_{m-1} v_{m-1} - (a_1 + \dots + a_{m-1}) v_m = 0,\] so linear independence of \(v_1, \dots, v_m\) forces \(a_1 = \dots = a_{m-1} = 0\).

Since \(U\) is spanned by a finite list it is finite-dimensional, and a basis of \(U\) is a spanning list of \(U\) of length \(\dim U\); by 2.22 the independent list above is no longer, i.e. \[m - 1 \le \dim U = \dim \operatorname{span}(v_1 + w, \dots, v_m + w).\]

Each \(p_k\) has degree \(k \le m\), so \(p_0, \dots, p_m\) is a list of \(m+1 = \dim \mathcal{P}_m(\mathbb{F})\) vectors in \(\mathcal{P}_m(\mathbb{F})\) (2.36); by 2.38 only linear independence needs checking.

Suppose \(a_0 p_0 + \dots + a_m p_m = 0\) with not every \(a_k\) zero, and let \(k\) be the largest index with \(a_k \ne 0\), so that

\begin{equation*} a_0 p_0 + a_1 p_1 + \dots + a_k p_k = 0 . \end{equation*}

Writing \(p_j(z) = c_{j,0} + \dots + c_{j,j} z^{\,j}\) with \(c_{j,j} \ne 0\), the terms with \(j < k\) have degree \(j < k\) and so contribute nothing to the coefficient of \(z^k\) on the left, which is therefore \(a_k c_{k,k} \ne 0\). But the coefficients of a polynomial are uniquely determined (as noted after 4.8: a polynomial with a nonzero coefficient has only finitely many zeros, while \(\mathbb{F}\) is infinite), so the left side being the zero function forces that coefficient to be \(0\), a contradiction.

Hence every \(a_k = 0\), and \(p_0, p_1, \dots, p_m\) is a basis of \(\mathcal{P}_m(\mathbb{F})\).

Each \(p_k\) has degree \(k + (m-k) = m\), so \(p_0, \dots, p_m\) is a list of \(m+1 = \dim \mathcal{P}_m(\mathbb{F})\) vectors in \(\mathcal{P}_m(\mathbb{F})\) (2.36); by 2.38 only linear independence needs checking.

Suppose \(a_0 p_0 + \dots + a_m p_m = 0\) with not every \(a_k\) zero, and let \(j\) be the smallest index with \(a_j \ne 0\), so that \(\sum_{k=j}^{m} a_k x^k (1-x)^{m-k} = 0\). By the binomial theorem, \[x^k (1 - x)^{m-k} = \sum_{i=0}^{m-k} \binom{m-k}{i} (-1)^i x^{k+i},\] so every monomial of \(p_k\) has degree at least \(k\); hence the terms with \(k > j\) contribute nothing to the coefficient of \(x^{\,j}\) on the left, which is therefore \(a_j \ne 0\). Since the coefficients of a polynomial are uniquely determined (as noted after 4.8, and used in 2C.9), the left side being the zero function forces \(a_j = 0\), a contradiction.

Hence every \(a_k = 0\), and \(p_0, \dots, p_m\) is a basis of \(\mathcal{P}_m(\mathbb{F})\).

Take the first two vectors \(u_1, u_2\) of a basis of \(U \cap W\), a list of length at least \(2\). Indeed \(U \cap W\) and \(U + W\) are subspaces of \(\mathbb{C}^6\) (Exercise 10 in Section 1C, and 1.40), hence finite-dimensional by 2.25 with \(\dim(U + W) \le \dim \mathbb{C}^6 = 6\) by 2.37 and 2.36, so 2.43 gives \[\dim(U \cap W) = \dim U + \dim W - \dim(U + W) \ge 4 + 4 - 6 = 2 .\] A sublist of a linearly independent list is linearly independent, so \(u_1, u_2\) is linearly independent. Were \(u_2 = \lambda u_1\), then \(\lambda u_1 + (-1)u_2 = 0\) would be a vanishing combination with a nonzero coefficient, and symmetrically for \(u_1 = \lambda u_2\). Hence neither of the vectors \(u_1, u_2 \in U \cap W\) is a scalar multiple of the other.

The intersection \(U \cap W\) is trivial, so 1.46 makes the sum direct. Indeed \(U \cap W\) is a subspace of \(\mathbb{R}^8\) (Exercise 10 in Section 1C), hence finite-dimensional by 2.25, and since \(\dim(U + W) = \dim \mathbb{R}^8 = 8\) by hypothesis and 2.36, applying 2.43 to \(U\) and \(W\) gives \[8 = 3 + 5 - \dim(U \cap W), \quad\text{so}\quad \dim(U \cap W) = 0 .\] A space of dimension \(0\) has the empty list as a basis, whose span is \(\{0\}\); thus \(U \cap W = \{0\}\) and \(\mathbb{R}^8 = U + W = U \oplus W\).

The intersection has dimension at least \(1\). Both \(U + W\) and \(U \cap W\) are subspaces of \(\mathbb{R}^9\) (1.40 and Exercise 10 in Section 1C), hence finite-dimensional by 2.25 with \(\dim(U + W) \le \dim \mathbb{R}^9 = 9\) by 2.37 and 2.36, so 2.43 gives \[\dim(U \cap W) = \dim U + \dim W - \dim(U + W) \ge 5 + 5 - 9 = 1 .\] Since \(\dim\{0\} = 0\) (the empty list being a basis), \(U \cap W \ne \{0\}\).

Two applications of 2.43 force \(\dim(V_1 \cap V_2 \cap V_3) \ge 1\). All the subspaces below are subspaces of \(V\) (Exercise 10 in Section 1C for intersections, 1.40 for sums), hence finite-dimensional by 2.25 with dimension at most \(\dim V = 10\) by 2.37. So

\begin{equation*} \begin{aligned} \dim(V_1 \cap V_2) &= \dim V_1 + \dim V_2 - \dim(V_1 + V_2) \\ &\ge 7 + 7 - 10 = 4 , \end{aligned} \end{equation*}

and, since \((V_1 \cap V_2) \cap V_3 = V_1 \cap V_2 \cap V_3\),

\begin{equation*} \begin{aligned} \dim(V_1 \cap V_2 \cap V_3) &= \dim(V_1 \cap V_2) + \dim V_3 - \dim\bigl((V_1 \cap V_2) + V_3\bigr) \\ &\ge 4 + 7 - 10 = 1 . \end{aligned} \end{equation*}

Since \(\dim\{0\} = 0\) (the empty list is a basis of \(\{0\}\)), \(V_1 \cap V_2 \cap V_3 \ne \{0\}\).

For any subspaces \(A, B\) of \(V\) (all finite-dimensional by 2.25),

\begin{equation*} \dim(A \cap B) = \dim A + \dim B - \dim(A + B) \geq \dim A + \dim B - \dim V, \end{equation*}

by 2.43 together with \(\dim(A + B) \leq \dim V\), which holds by 2.37 since \(A + B\) is a subspace of \(V\) (1.40). Applying this with \(A = V_1, B = V_2\), then with \(A = V_1 \cap V_2, B = V_3\),

\begin{equation*} \begin{aligned} \dim(V_1 \cap V_2 \cap V_3) &\geq \dim(V_1 \cap V_2) + \dim V_3 - \dim V\\ &\geq \dim V_1 + \dim V_2 + \dim V_3 - 2 \dim V, \end{aligned} \end{equation*}

which is positive by hypothesis. Since \(\dim\{0\} = 0\) (the empty list is its basis), \(V_1 \cap V_2 \cap V_3 \neq \{0\}\).

Extend a basis \(u_1, \dots, u_m\) of \(U\) to a basis \(u_1, \dots, u_m, v_1, \dots, v_{n-m}\) of \(V\) (2.32; the extension has length \(\dim V = n\) by 2.34) and take, for \(k \in \{1, \dots, n-m\}\),

\begin{equation*} \begin{aligned} W_k = \operatorname{span}(u_1, \dots, u_m,\; &v_1, \dots, v_{k-1},\\ &v_{k+1}, \dots, v_{n-m}), \end{aligned} \end{equation*}

the span of that basis with \(v_k\) deleted. There are indeed \(n - m \geq 1\) of these, since \(m \leq n\) by 2.37 while \(m = n\) would give \(U = V\) by 2.39.

Each such spanning list has length \(n-1\) and is linearly independent, being a sublist of a basis; so it is a basis of \(W_k\) and \(\dim W_k = n - 1\).

Since \(u_1, \dots, u_m\) occurs in every spanning list, \(U \subseteq W_1 \cap \dots \cap W_{n-m}\). Conversely, let \(v\) lie in that intersection and write \(v = \sum_j a_j u_j + \sum_k b_k v_k\) in the basis of \(V\). Fixing \(k\), membership in \(W_k\) also expresses \(v\) in that same basis with coefficient \(0\) on \(v_k\), so \(b_k = 0\) by the uniqueness in 2.28; as \(k\) was arbitrary, \(v = \sum_j a_j u_j \in U\). Hence \[W_1 \cap \dots \cap W_{n-m} = U .\]

Concatenating bases does it. Put \(n_k = \dim V_k\), choose a basis \(v_{k,1}, \dots, v_{k,n_k}\) of each \(V_k\) (2.31), and let \(L\) be the concatenated list

\begin{equation*} \begin{aligned} &v_{1,1}, \dots, v_{1,n_1}, \; v_{2,1}, \dots, v_{2,n_2},\\ &\dots, \; v_{m,1}, \dots, v_{m,n_m}, \end{aligned} \end{equation*}

of length \(n_1 + \dots + n_m\).

\(L\) spans \(V_1 + \dots + V_m\): every element of the sum is \(u_1 + \dots + u_m\) with \(u_k \in V_k\) (1.36), and expanding each \(u_k\) in its basis exhibits it as a linear combination of vectors of \(L\); conversely each vector of \(L\) lies in some \(V_k\), hence in the subspace \(V_1 + \dots + V_m\) (1.40).

So \(V_1 + \dots + V_m\) has a finite spanning list and is therefore finite-dimensional; any basis of it is a linearly independent list in it, so by 2.22 that basis is no longer than \(L\): \[\dim(V_1 + \dots + V_m) \leq n_1 + \dots + n_m = \dim V_1 + \dots + \dim V_m.\]

Take \(V_k = \operatorname{span}(v_k)\), where \(v_1, \dots, v_n\) is a basis of \(V\) (2.31), of length \(\dim V = n\).

Each \(V_k\) is one-dimensional. No \(v_k\) is \(0\), since a list containing \(0\) is dependent; and a one-vector list \(v_k \neq 0\) is linearly independent, as \(a v_k = 0\) with \(a \neq 0\) gives \(v_k = a^{-1}(a v_k) = 0\). So \(v_k\) is a basis of \(V_k\).

The sum is \(V\). Each \(v \in V\) is \(v = a_1 v_1 + \dots + a_n v_n\) with \(a_k v_k \in V_k\), so \(V \subseteq V_1 + \dots + V_n\), and the reverse inclusion holds since each \(V_k\) is a subspace of \(V\).

The sum is direct. If \(w_1 + \dots + w_n = 0\) with \(w_k = a_k v_k \in V_k\), then

\begin{equation*} a_1 v_1 + \dots + a_n v_n = 0, \end{equation*}

so every \(a_k = 0\) by linear independence, whence every \(w_k = 0\); apply 1.45.

Hence \(V = V_1 \oplus \dots \oplus V_n\) with each \(V_k\) one-dimensional.

The formula is false, the three lines

\begin{equation*} \begin{aligned} V_1 &= \operatorname{span}\bigl((1,0)\bigr), \\ V_2 &= \operatorname{span}\bigl((0,1)\bigr), \\ V_3 &= \operatorname{span}\bigl((1,1)\bigr) \end{aligned} \end{equation*}

in \(\mathbf{R}^2\) being a counterexample.

Why one might guess it. The table preceding these exercises matches a finite set \(S\) with a space \(V\), \(\#S\) with \(\dim V\), union with sum (each the smallest object containing the two, by 1.40), and intersection with intersection. Under that dictionary two-set inclusion-exclusion, \(\#(S_1 \cup S_2) = \#S_1 + \#S_2 - \#(S_1 \cap S_2)\), becomes the true theorem 2.43; translating the three-set counting formula term by term gives the displayed guess.

The guess is false. Each \(V_i\) above is the span of one nonzero vector, so \(\dim V_i = 1\). Their sum contains \((1,0)\) and \((0,1)\), hence equals \(\mathbf{R}^2\), and the left side is \(2\). Every pairwise intersection is \(\{0\}\): a vector of \(V_1 \cap V_3\) is both \((a,0)\) and \((b,b)\), forcing \(b = 0\), and comparing the vanishing coordinate likewise kills \(V_1 \cap V_2\) (where \((a,0) = (0,b)\)) and \(V_2 \cap V_3\) (where \((0,a) = (b,b)\)). So \(V_1 \cap V_2 \cap V_3 = \{0\}\) as well, all four intersection terms have dimension \(0\), and the right side is

\begin{equation*} 1 + 1 + 1 - 0 - 0 - 0 + 0 = 3 \neq 2 . \end{equation*}

Average the three ways of grouping the sum. Every space below is a subspace of the ambient finite-dimensional space (1.40 for sums, 1.34 for intersections) and so is finite-dimensional by 2.25, letting 2.43 be applied freely. Applying it first to \(V_1 + V_2\) and \(V_3\), then to \(V_1\) and \(V_2\),

\begin{equation*} \begin{aligned} \dim(V_1 + V_2 + V_3) &= \dim(V_1 + V_2) + \dim V_3 - \dim\bigl((V_1 + V_2) \cap V_3\bigr)\\ &= \dim V_1 + \dim V_2 + \dim V_3 - \dim(V_1 \cap V_2)\\ &\quad - \dim\bigl((V_1 + V_2) \cap V_3\bigr) . \end{aligned} \end{equation*}

Call this identity (1). Since addition of subspaces is commutative and associative, the same computation applied to the groupings \((V_1 + V_3) + V_2\) and \((V_2 + V_3) + V_1\) gives identities (2) and (3), namely (1) with the index swaps \(2 \leftrightarrow 3\) and \(1 \leftrightarrow 3\); their subtracted terms are \(\dim(V_1 \cap V_3)\), \(\dim((V_1 + V_3) \cap V_2)\) and \(\dim(V_2 \cap V_3)\), \(\dim((V_2 + V_3) \cap V_1)\).

All three identities have left side \(\dim(V_1 + V_2 + V_3)\), so adding them and dividing by \(3\) leaves that side unchanged while averaging the right sides: the three pairwise-intersection terms become the first displayed fraction of the statement and the three sum-intersection terms the second. That is the claimed formula.

If \(b = c = 0\), then \(T(x,y,z) = (2x - 4y + 3z,\; 6x)\) satisfies the two conditions in 3.1: for \(u = (x_1,y_1,z_1)\), \(v = (x_2,y_2,z_2)\) and \(\lambda \in \mathbb{R}\), regrouping coordinatewise gives

\begin{equation*} \begin{aligned} T(u + v) &= \bigl(2(x_1 + x_2) - 4(y_1 + y_2) + 3(z_1 + z_2),\; 6(x_1 + x_2)\bigr)\\ &= (2x_1 - 4y_1 + 3z_1,\; 6x_1) + (2x_2 - 4y_2 + 3z_2,\; 6x_2) = Tu + Tv,\\ T(\lambda v) &= (2\lambda x - 4\lambda y + 3\lambda z,\; 6\lambda x) = \lambda(Tv). \end{aligned} \end{equation*}

Conversely, suppose \(T\) is linear. Then \(T(0,0,0) = (0,0)\) by 3.10, while the defining formula gives \(T(0,0,0) = (b, 0)\); comparing first coordinates, \(b = 0\). For \(c\), homogeneity with \(\lambda = 2\) and \(v = (1,1,1)\) compares the second coordinates of

\begin{equation*} T(2,2,2) = (2 + b,\; 12 + 8c), \qquad 2\,T(1,1,1) = (2 + 2b,\; 12 + 2c), \end{equation*}

since \(c \cdot 2 \cdot 2 \cdot 2 = 8c\); thus \(12 + 8c = 12 + 2c\) and \(c = 0\).

If \(b = c = 0\), then

\begin{equation*} Tp = \left(3p(4) + 5p^{\prime}(6),\; \int_{-1}^{2} x^3 p(x)\,dx\right) \end{equation*}

satisfies the two conditions in 3.1: evaluation at a point, differentiation and the integral are each additive and homogeneous, so every coordinate of \(T(p+q)\) is the sum of the corresponding coordinates of \(Tp\) and \(Tq\), and every coordinate of \(T(\lambda p)\) is \(\lambda\) times that of \(Tp\), addition and scalar multiplication in \(\mathbb{R}^2\) being coordinatewise.

Conversely, suppose \(T\) is linear. Take \(p\) the constant polynomial \(1\) (so \(p^{\prime} = 0\)); the first coordinates of \(Tp\) and of \(T(2p)\) are \(3 + b\) and \(3 \cdot 2 + b \cdot 2 \cdot 2 = 6 + 4b\), so homogeneity with \(\lambda = 2\) gives

\begin{equation*} 6 + 4b = 2(3 + b) = 6 + 2b , \end{equation*}

whence \(b = 0\). Now take \(q\) the constant polynomial \(\pi/2\), and note \(\int_{-1}^{2} x^3\,dx = \tfrac{16}{4} - \tfrac{1}{4} = \tfrac{15}{4}\). Since \(\sin(\pi/2) = 1\) and \(\sin \pi = 0\), the second coordinates of \(Tq\) and of \(T(2q)\) are \(\tfrac{15\pi}{8} + c\) and \(\tfrac{15\pi}{4}\), so homogeneity with \(\lambda = 2\) gives

\begin{equation*} \frac{15\pi}{4} = 2\left(\frac{15\pi}{8} + c\right) = \frac{15\pi}{4} + 2c , \end{equation*}

whence \(c = 0\).

Take \(A_{j,k}\) to be the \(j^{\text{th}}\) coordinate of \(Te_k\), where \(e_1, \ldots, e_n\) is the standard basis of \(\mathbb{F}^n\) (2.27); that is,

\begin{equation*} Te_k = (A_{1,k},\, A_{2,k},\, \ldots,\, A_{m,k}) \qquad (k = 1, \ldots, n). \end{equation*}

Since \((x_1, \ldots, x_n) = x_1 e_1 + \cdots + x_n e_n\), linearity of \(T\) and the coordinatewise vector space operations on \(\mathbb{F}^m\) give

\begin{equation*} \begin{aligned} T(x_1, \ldots, x_n) &= x_1 (Te_1) + \cdots + x_n (Te_n) \\ &= x_1 (A_{1,1}, \ldots, A_{m,1}) + \cdots + x_n (A_{1,n}, \ldots, A_{m,n}) \\ &= (A_{1,1}x_1 + \cdots + A_{1,n}x_n,\; \ldots,\; A_{m,1}x_1 + \cdots + A_{m,n}x_n), \end{aligned} \end{equation*}

the last step because the \(j^{\text{th}}\) coordinate of the middle expression is \(A_{j,1}x_1 + \cdots + A_{j,n}x_n\).

Suppose \(a_1, \ldots, a_m \in \mathbb{F}\) satisfy \(a_1 v_1 + \cdots + a_m v_m = 0\). Applying \(T\) and using linearity, with \(T0 = 0\) by 3.10,

\begin{equation*} a_1 (Tv_1) + \cdots + a_m (Tv_m) = T(a_1 v_1 + \cdots + a_m v_m) = T0 = 0. \end{equation*}

Linear independence of \(Tv_1, \ldots, Tv_m\) now forces \(a_1 = \cdots = a_m = 0\). Hence \(v_1, \ldots, v_m\) is linearly independent.

Every condition in 1.20 follows by evaluating both sides at an arbitrary \(v \in V\) and quoting the corresponding property of \(W\), the operations being \((S+T)(v) = Sv + Tv\) and \((\lambda T)(v) = \lambda(Tv)\) (3.5).

Closure, for \(u, v \in V\) and \(\mu \in \mathbb{F}\):

\begin{equation*} \begin{aligned} (S+T)(u+v) &= (Su + Sv) + (Tu + Tv) = (S+T)(u) + (S+T)(v), \\ (S+T)(\mu v) &= \mu(Sv) + \mu(Tv) = \mu\bigl((S+T)(v)\bigr), \\ (\lambda T)(u+v) &= \lambda(Tu) + \lambda(Tv) = (\lambda T)(u) + (\lambda T)(v), \\ (\lambda T)(\mu v) &= (\lambda\mu)(Tv) = (\mu\lambda)(Tv) = \mu((\lambda T)v), \end{aligned} \end{equation*}

using additivity and homogeneity of \(S, T\), distributivity and associativity of scalar multiplication in \(W\), and commutativity in \(\mathbb{F}\); so \(S+T, \lambda T \in \mathcal{L}(V,W)\). The remaining axioms, evaluated at \(v\):

\begin{equation*} \begin{aligned} (S+T)(v) &= Sv + Tv = Tv + Sv = (T+S)(v), \\ \bigl((R+S)+T\bigr)(v) &= (Rv+Sv)+Tv = Rv+(Sv+Tv) = \bigl(R+(S+T)\bigr)(v), \\ ((\lambda\mu)T)v &= \lambda\bigl(\mu(Tv)\bigr) = \bigl(\lambda(\mu T)\bigr)v, \\ (T + 0)(v) &= Tv + 0 = Tv, \\ \bigl(T + (-1)T\bigr)(v) &= Tv + (-1)(Tv) = 0, \\ (1T)(v) &= 1(Tv) = Tv, \\ \bigl(\lambda(S+T)\bigr)(v) &= \lambda(Sv)+\lambda(Tv) = (\lambda S+\lambda T)(v), \\ \bigl((\lambda + \mu)T\bigr)(v) &= \lambda(Tv) + \mu(Tv) = (\lambda T + \mu T)(v), \end{aligned} \end{equation*}

where the additive identity is the zero linear map of Example 3.3, the additive inverse \(-T = (-1)T\) works by 1.32 applied in \(W\), and each right-hand step is the matching axiom of \(W\) or of \(\mathbb{F}\). Hence \(\mathcal{L}(V,W)\) is a vector space over \(\mathbb{F}\).

Each identity follows by evaluating both sides at a point of the common domain, using \((ST)(u) = S(Tu)\) from 3.7.

Associativity: with \(T_3 \in \mathcal{L}(V_1, V_2)\), \(T_2 \in \mathcal{L}(V_2, V_3)\), \(T_1 \in \mathcal{L}(V_3, V_4)\), both sides are maps \(V_1 \to V_4\), and for \(u \in V_1\),

\begin{equation*} \bigl((T_1T_2)T_3\bigr)(u) = T_1\bigl(T_2(T_3 u)\bigr) = \bigl(T_1(T_2T_3)\bigr)(u). \end{equation*}

Identity: for \(T \in \mathcal{L}(V,W)\), with \(I\) the identity operator on \(V\) in \(TI\) and on \(W\) in \(IT\),

\begin{equation*} (TI)(v) = T(Iv) = Tv = I(Tv) = (IT)(v) \qquad (v \in V). \end{equation*}

Distributivity: for \(T, T_1, T_2 \in \mathcal{L}(U,V)\), \(S, S_1, S_2 \in \mathcal{L}(V,W)\), and \(u \in U\),

\begin{equation*} \begin{aligned} \bigl((S_1 + S_2)T\bigr)(u) &= S_1(Tu) + S_2(Tu) = (S_1T + S_2T)(u), \\ \bigl(S(T_1+T_2)\bigr)(u) &= S(T_1u+T_2u) = S(T_1u)+S(T_2u) = (ST_1+ST_2)(u), \end{aligned} \end{equation*}

the second line using additivity of \(S\).

Take \(\lambda \in \mathbb{F}\) with \(Tu = \lambda u\), where \(u\) is a basis of the one-dimensional space \(V\); such a \(\lambda\) exists because \(Tu \in V = \operatorname{span}(u)\). Every \(v \in V\) has the form \(v = au\) with \(a \in \mathbb{F}\), so homogeneity of \(T\) gives

\begin{equation*} Tv = T(au) = a(Tu) = a(\lambda u) = \lambda(au) = \lambda v. \end{equation*}

Define \(\varphi \colon \mathbb{R}^2 \to \mathbb{R}\) by

\begin{equation*} \varphi(x, y) = \begin{cases} \dfrac{x^3}{x^2 + y^2} & \text{if } (x,y) \neq (0,0),\\ 0 & \text{if } (x,y) = (0,0). \end{cases} \end{equation*}

Homogeneity: if \(a = 0\) or \(v = (0,0)\) then \(av = (0,0)\), so both \(\varphi(av)\) and \(a\varphi(v)\) equal \(0\); otherwise \(av = (ax, ay) \neq (0,0)\) and

\begin{equation*} \varphi(av) = \frac{a^3x^3}{a^2(x^2+y^2)} = a\cdot\frac{x^3}{x^2+y^2} = a\varphi(v). \end{equation*}

Additivity fails at \(u = (1,0)\), \(v = (0,1)\):

\begin{equation*} \varphi(u+v) = \varphi(1,1) = \tfrac12 \neq 1 + 0 = \varphi(u) + \varphi(v), \end{equation*}

so \(\varphi\) is not linear by 3.1.

Define \(\varphi \colon \mathbb{C} \to \mathbb{C}\) to be complex conjugation:

\begin{equation*} \varphi(z) = \overline{z}, \end{equation*}

so that \(\varphi(a + bi) = a - bi\) for all \(a, b \in \mathbb{R}\).

Additivity: for \(w = a + bi\) and \(z = c + di\) with \(a, b, c, d \in \mathbb{R}\),

\begin{equation*} \varphi(w + z) = (a + c) - (b + d)i = (a - bi) + (c - di) = \varphi(w) + \varphi(z). \end{equation*}

Homogeneity fails because the scalars here include \(i\): taking \(\lambda = i\) and \(z = 1\),

\begin{equation*} \varphi(\lambda z) = \overline{i} = -i \neq i = i\,\overline{1} = \lambda\varphi(z), \end{equation*}

so \(\varphi\) is not linear by 3.1.

False: take \(q(x) = x^2\), so that \((Tp)(x) = (p(x))^2\), and let \(p\) be the constant polynomial \(1\). Then \(T(2p)\) is the constant polynomial \(4\) while \(2(Tp)\) is the constant polynomial \(2\):

\begin{equation*} \begin{aligned} \bigl(T(2p)\bigr)(x) &= \bigl((2p)(x)\bigr)^2 = 4, \\ \bigl(2(Tp)\bigr)(x) &= 2\bigl(p(x)\bigr)^2 = 2. \end{aligned} \end{equation*}

So homogeneity fails, and \(T\) is not linear by 3.1.

(\(\Rightarrow\)) If \(T = \lambda I\), then for all \(S \in \mathcal{L}(V)\) and \(v \in V\), homogeneity of \(S\) gives

\begin{equation*} (ST)v = S(\lambda v) = \lambda (Sv) = (TS)v . \end{equation*}

(\(\Leftarrow\)) Suppose \(ST = TS\) for every \(S \in \mathcal{L}(V)\); we may assume \(V \neq \{0\}\), since otherwise \(T = 0 = 0 \cdot I\). Fix \(v \neq 0\). If \(Tv \notin \operatorname{span}(v)\), then \(v, Tv\) is linearly independent (in a relation \(av + b(Tv) = 0\), \(b \neq 0\) would put \(Tv\) in \(\operatorname{span}(v)\), and then \(av = 0\) with \(v \neq 0\) forces \(a = 0\)), so by 2.32 it extends to a basis \(v, Tv, u_3, \dots, u_n\) of the finite-dimensional space \(V\), and 3.4 supplies \(S \in \mathcal{L}(V)\) with \(Sv = v\) and \(S(Tv) = Su_3 = \dots = Su_n = 0\), whence

\begin{equation*} (ST)v = S(Tv) = 0 \neq Tv = T(Sv) = (TS)v, \end{equation*}

since \(Tv \neq 0\) as a member of a linearly independent list. This contradiction gives \(Tv = a_v v\) for a scalar \(a_v \in \mathbf{F}\), unique because \(v \neq 0\). For nonzero \(v, w\) the scalar \(a_v\) agrees with \(a_w\):

(i) \(v, w\) linearly dependent, say \(w = cv\) with \(c \neq 0\):

\begin{equation*} a_w w = Tw = c\,Tv = c\,a_v v = a_v w, \qquad \text{so } a_w = a_v . \end{equation*}

(ii) \(v, w\) linearly independent, so \(v + w \neq 0\):

\begin{equation*} a_{v+w} v + a_{v+w} w = T(v+w) = Tv + Tw = a_v v + a_w w, \end{equation*}

and independence of \(v, w\) forces \(a_v = a_{v+w} = a_w\).

Writing \(\lambda\) for this common value, \(Tv = \lambda v\) for every nonzero \(v \in V\), while \(T0 = 0 = \lambda \cdot 0\) by 3.10. Hence \(T = \lambda I\).

Additivity fails at the pair \(u, w\), where \(u \in U\) satisfies \(Su \neq 0\) (available since \(S \neq 0\)) and \(w \in V\) satisfies \(w \notin U\) (available since \(U \neq V\)). Indeed \(u + w \notin U\), since \(u + w \in U\) would give \(w = (u+w) - u \in U\) by closure of the subspace \(U\) under addition and additive inverses. Hence \(T(u+w) = 0\), while \(Tu = Su\) and \(Tw = 0\), so

\begin{equation*} T(u + w) = 0 \neq Su = Tu + Tw . \end{equation*}

Therefore \(T\) is not a linear map by 3.1.

Take \(T \in \mathcal{L}(V,W)\) to be the map furnished by the linear map lemma 3.4 on an extended basis: \(U\) is finite-dimensional by 2.25 and so has a basis \(u_1, \dots, u_m\) (2.31, the empty list if \(U = \{0\}\)), which is linearly independent in \(V\) and hence extends by 2.32 to a basis \(u_1, \dots, u_m, v_1, \dots, v_n\) of \(V\); define

\begin{equation*} Tu_k = Su_k \quad (k = 1, \dots, m), \qquad Tv_j = 0 \quad (j = 1, \dots, n) . \end{equation*}

Every \(u \in U\) is \(u = c_1 u_1 + \dots + c_m u_m\) with \(c_k \in \mathbf{F}\), so linearity of \(T\) and then of \(S\) give

\begin{equation*} Tu = c_1 (Tu_1) + \dots + c_m (Tu_m) = c_1 (Su_1) + \dots + c_m (Su_m) = Su , \end{equation*}

which for \(m = 0\) reads \(T0 = 0 = S0\) (3.10). Hence \(T\) extends \(S\).

We exhibit a linearly independent list of arbitrary length \(m\) in the vector space \(\mathcal{L}(V,W)\) (3.6); by 2.22 no finite list can then span it, so it is infinite-dimensional (2.13).

First, \(W\) contains a linearly independent list \(w_1, \dots, w_m\): starting from the empty list and given a linearly independent \(w_1, \dots, w_k\), we have \(\operatorname{span}(w_1, \dots, w_k) \neq W\) (else this finite list would span \(W\), contradicting infinite-dimensionality), and any \(w_{k+1}\) chosen outside that span keeps the list independent, since in a relation \(a_1 w_1 + \dots + a_{k+1} w_{k+1} = 0\) we must have \(a_{k+1} = 0\) (otherwise \(w_{k+1} \in \operatorname{span}(w_1, \dots, w_k)\)) and then \(a_1 = \dots = a_k = 0\).

Next, let \(v_1, \dots, v_p\) be a basis of \(V\) (2.31), with \(p \geq 1\) since \(\dim V > 0\), and use the linear map lemma 3.4 to define \(T_1, \dots, T_m \in \mathcal{L}(V,W)\) by

\begin{equation*} T_j v_1 = w_j \qquad \text{and} \qquad T_j v_k = 0 \ \text{ for } k = 2, \dots, p . \end{equation*}

If \(a_1 T_1 + \dots + a_m T_m = 0\), then evaluating at \(v_1\) using the operations 3.5 gives

\begin{equation*} a_1 w_1 + \dots + a_m w_m = 0 , \end{equation*}

so \(a_1 = \dots = a_m = 0\) and \(T_1, \dots, T_m\) is linearly independent, as required.

Take \(w_j = w\) and \(w_k = 0\) for \(k \neq j\), where \(w \in W\) is nonzero (available since \(W \neq \{0\}\)) and \(j\) is an index with \(a_j \neq 0\) in a dependence relation with scalars \(a_1, \dots, a_m \in \mathbf{F}\) not all zero,

\begin{equation*} a_1 v_1 + \dots + a_m v_m = 0 . \end{equation*}

If some \(T \in \mathcal{L}(V,W)\) satisfied \(Tv_k = w_k\) for every \(k\), then applying \(T\) to this relation and using \(T0 = 0\) (3.10) together with \(w_k = 0\) for \(k \neq j\) would give

\begin{equation*} 0 = T(a_1 v_1 + \dots + a_m v_m) = a_1 w_1 + \dots + a_m w_m = a_j w , \end{equation*}

forcing \(w = 0\) since \(a_j \neq 0\), a contradiction. So no such \(T\) exists.

Fix a basis \(v_1, \dots, v_n\) of \(V\), where \(n = \dim V \geq 2\) (so that \(v_2\) is available), and let the linear map lemma 3.4 supply \(S, T \in \mathcal{L}(V)\) with

\begin{equation*} \begin{aligned} S v_1 &= v_1, & S v_k &= 0 \quad \text{for } k = 2, \dots, n, \\ T v_1 &= v_2, & T v_k &= 0 \quad \text{for } k = 2, \dots, n . \end{aligned} \end{equation*}

Evaluating both compositions at \(v_1\),

\begin{equation*} (ST)v_1 = S v_2 = 0 \neq v_2 = T v_1 = (TS) v_1 , \end{equation*}

where \(v_2 \neq 0\) as a member of a linearly independent list. Hence \(ST \neq TS\).

It suffices to put \(I\) inside any two-sided ideal \(\mathcal{E} \neq \{0\}\): then every \(T \in \mathcal{L}(V)\) equals \(TI\) by the identity property in 3.8 and lies in \(\mathcal{E}\) by the ideal property, so \(\mathcal{E} = \mathcal{L}(V)\). (That \(\{0\}\) and \(\mathcal{L}(V)\) are two-sided ideals is immediate from \(T0 = 0T = 0\).)

Fix \(E \in \mathcal{E}\) with \(E \neq 0\) and \(u \in V\) with \(Eu \neq 0\). Then \(V \neq \{0\}\), so \(V\) has a basis \(v_1, \dots, v_n\) with \(n = \dim V \geq 1\) finite (2.31), and the linearly independent one-term list \(Eu\) extends by 2.32 to a basis \(u_1, \dots, u_n\) of \(V\) with \(u_1 = Eu\) (of length \(n\), every basis of \(V\) having length \(\dim V\)). For each \(j\), the linear map lemma 3.4 defines \(A_j, B_j \in \mathcal{L}(V)\) by

\begin{equation*} \begin{aligned} A_j v_j &= u, & A_j v_i &= 0 \ \text{ for } i \neq j, \\ B_j u_1 &= v_j, & B_j u_i &= 0 \ \text{ for } i = 2, \dots, n . \end{aligned} \end{equation*}

Two applications of the ideal property give \(B_j E A_j \in \mathcal{E}\) (unambiguous by associativity, 3.8), and on the basis, using \(E0 = 0\) (3.10),

\begin{equation*} \begin{aligned} (B_j E A_j) v_i &= 0 \qquad (i \neq j), \\ (B_j E A_j) v_j &= B_j (Eu) = B_j u_1 = v_j . \end{aligned} \end{equation*}

Since \(\mathcal{E}\) is a subspace and \(V\) is finite-dimensional, the finite sum \(R := B_1 E A_1 + \dots + B_n E A_n\) lies in \(\mathcal{E}\), and the displayed values give \(R v_i = v_i\) for each \(i\). So \(R\) and \(I\) agree on a basis, whence \(R = I\) by the uniqueness half of 3.4, and \(I \in \mathcal{E}\).

Take the linear map (Check!) \(T \in \mathcal{L}(\mathbf{F}^5, \mathbf{F}^2)\) given by

\begin{equation*} T(z_1, z_2, z_3, z_4, z_5) = (z_1, z_2). \end{equation*}

Then \(Tz = 0\) exactly when \(z_1 = z_2 = 0\), so \(\operatorname{null} T = \{(0,0,z_3,z_4,z_5)\}\) has basis \((0,0,1,0,0), (0,0,0,1,0), (0,0,0,0,1)\) and \(\dim \operatorname{null} T = 3\); and \(T(w_1, w_2, 0, 0, 0) = (w_1, w_2)\) gives \(\operatorname{range} T = \mathbf{F}^2\), so \(\dim \operatorname{range} T = 2\).

For every \(v \in V\), the vector \(S(Tv)\) lies in \(\operatorname{range} S \subseteq \operatorname{null} T\), so \(T(S(Tv)) = 0\) and hence

\begin{equation*} (ST)^2 v = S\Bigl(T\bigl(S(Tv)\bigr)\Bigr) = S(0) = 0 \end{equation*}

by 3.10. Thus \((ST)^2 = 0\).

(a) Surjectivity of \(T\). Indeed

\begin{equation*} \begin{aligned} \operatorname{range} T &= \{z_1 v_1 + \cdots + z_m v_m : z_k \in \mathbf{F}\} \\ &= \operatorname{span}(v_1, \dots, v_m), \end{aligned} \end{equation*}

so \(\operatorname{range} T = V\), that is \(T\) surjective (3.19), says exactly that \(v_1, \dots, v_m\) spans \(V\).

(b) Injectivity of \(T\). Indeed, writing \(z = (z_1, \dots, z_m)\),

\begin{equation*} \operatorname{null} T = \{z \in \mathbf{F}^m : z_1 v_1 + \cdots + z_m v_m = 0\}, \end{equation*}

so \(\operatorname{null} T = \{0\}\), that is \(T\) injective (3.15), says exactly that the only scalars with \(z_1 v_1 + \cdots + z_m v_m = 0\) are \(z_1 = \cdots = z_m = 0\), i.e. that the list is linearly independent.

The set \(E = \{T : \dim \operatorname{null} T > 2\}\) is not closed under addition, hence is not a subspace. Take the linear maps (Check!) \(S, T \in \mathcal{L}(\mathbf{R}^5, \mathbf{R}^4)\) given by

\begin{equation*} S(x_1, \dots, x_5) = (x_1, x_2, 0, 0), \qquad T(x_1, \dots, x_5) = (0, 0, x_3, x_4). \end{equation*}

Then \(Sx = 0\) iff \(x_1 = x_2 = 0\) and \(Tx = 0\) iff \(x_3 = x_4 = 0\), so

\begin{equation*} \begin{aligned} \operatorname{null} S &= \{(0,0,x_3,x_4,x_5)\} = \operatorname{span}(e_3,e_4,e_5), \\ \operatorname{null} T &= \{(x_1,x_2,0,0,x_5)\} = \operatorname{span}(e_1,e_2,e_5), \end{aligned} \end{equation*}

each of dimension \(3 > 2\), the spanning lists being sublists of the standard basis \(e_1, \dots, e_5\) of \(\mathbf{R}^5\) (2.27); so \(S, T \in E\). But \((S+T)(x_1, \dots, x_5) = (x_1, x_2, x_3, x_4)\) vanishes only when \(x_1 = x_2 = x_3 = x_4 = 0\), so \(\operatorname{null}(S+T) = \operatorname{span}(e_5)\) has dimension \(1\) and \(S + T \notin E\).

Take the linear map (Check!) \(T \in \mathcal{L}(\mathbf{R}^4)\) given by

\begin{equation*} T(x_1, x_2, x_3, x_4) = (0, 0, x_1, x_2). \end{equation*}

Then \(Tx = 0\) exactly when \(x_1 = x_2 = 0\), while every output has its first two coordinates \(0\) and \(T(a, b, 0, 0) = (0, 0, a, b)\) for all \(a, b \in \mathbf{R}\); hence

\begin{equation*} \operatorname{null} T = \{(0,0,x_3,x_4)\} = \{(0,0,a,b)\} = \operatorname{range} T . \end{equation*}

If \(T \in \mathcal{L}(\mathbf{R}^5)\) satisfied \(\operatorname{range} T = \operatorname{null} T\), then these equal subspaces would share a common dimension \(k\), a nonnegative integer, and the fundamental theorem of linear maps 3.21 (applicable since \(\mathbf{R}^5\) is finite-dimensional) would give

\begin{equation*} 5 = \dim \mathbf{R}^5 = \dim \operatorname{null} T + \dim \operatorname{range} T = 2k \end{equation*}

which is impossible. Hence no such \(T\) exists.

The set \(E\) of non-injective maps is not closed under addition, hence is not a subspace. Fix bases \(v_1, \dots, v_n\) of \(V\) and \(w_1, \dots, w_m\) of \(W\), where \(2 \leq n = \dim V \leq m = \dim W\), so that \(w_1, \dots, w_n\) is a linearly independent sublist; the linear map lemma 3.4 supplies \(S, T \in \mathcal{L}(V,W)\) with

\begin{equation*} \begin{aligned} S v_1 &= 0, & S v_j &= w_j \quad (j = 2, \dots, n),\\ T v_1 &= w_1, & T v_j &= 0 \quad (j = 2, \dots, n). \end{aligned} \end{equation*}

Here \(S v_1 = 0\) and \(T v_2 = 0\) (the vector \(v_2\) exists because \(n \geq 2\)), with \(v_1, v_2 \neq 0\) as members of a basis, so \(\operatorname{null} S\) and \(\operatorname{null} T\) are nonzero and \(S, T \in E\) by 3.15. But \((S+T)v_j = w_j\) for every \(j\), so any \(v = a_1 v_1 + \cdots + a_n v_n \in \operatorname{null}(S+T)\) satisfies

\begin{equation*} 0 = (S+T)v = a_1 w_1 + \cdots + a_n w_n , \end{equation*}

forcing \(a_1 = \cdots = a_n = 0\) and \(v = 0\). Hence \(S + T\) is injective by 3.15, and \(S + T \notin E\).

The set \(E\) of non-surjective maps is not closed under addition, hence is not a subspace. Fix bases \(v_1, \dots, v_n\) of \(V\) and \(w_1, \dots, w_m\) of \(W\), where \(n = \dim V \geq m = \dim W \geq 2\), and let the linear map lemma 3.4 supply \(S, T \in \mathcal{L}(V,W)\) with

\begin{equation*} \begin{aligned} S v_k &= w_k \ \ (1 \leq k \leq m-1), & S v_k &= 0 \ \ (m \leq k \leq n), \\ T v_m &= w_m, & T v_k &= 0 \ \ (k \neq m), \end{aligned} \end{equation*}

which is legitimate because \(m - 1 \geq 1\) and \(m \leq n\). Expanding \(v = a_1 v_1 + \dots + a_n v_n\) and using linearity gives

\begin{equation*} \begin{aligned} \operatorname{range} S &= \operatorname{span}(w_1, \dots, w_{m-1}), \\ \operatorname{range} T &= \operatorname{span}(w_m), \end{aligned} \end{equation*}

of dimensions \(m - 1\) and \(1\), both smaller than \(m = \dim W\) (the spanning lists are linearly independent, being sublists of a basis), so \(\operatorname{range} S, \operatorname{range} T \neq W\) and \(S, T \in E\). But \((S+T)v_k = w_k\) for \(k = 1, \dots, m\), so the subspace \(\operatorname{range}(S+T)\) of \(W\) (3.18) contains the basis \(w_1, \dots, w_m\) and hence equals \(W\); thus \(S + T\) is surjective and \(S + T \notin E\).

Suppose \(a_1, \dots, a_n \in \mathbf{F}\) satisfy \(a_1 T v_1 + \dots + a_n T v_n = 0\). By linearity of \(T\) this says \(a_1 v_1 + \dots + a_n v_n \in \operatorname{null} T\), and \(\operatorname{null} T = \{0\}\) by 3.15 since \(T\) is injective, so

\begin{equation*} a_1 v_1 + \dots + a_n v_n = 0. \end{equation*}

Linear independence of \(v_1, \dots, v_n\) now forces \(a_1 = \dots = a_n = 0\). Hence \(T v_1, \dots, T v_n\) is linearly independent in \(W\).

Each \(T v_k\) lies in \(\operatorname{range} T\), which is a subspace of \(W\) by 3.18 and therefore contains every linear combination of them; hence \(\operatorname{span}(T v_1, \dots, T v_n) \subseteq \operatorname{range} T\). Conversely, every \(w \in \operatorname{range} T\) is \(w = Tv\) with \(v = a_1 v_1 + \dots + a_n v_n\) for some \(a_k \in \mathbf{F}\), the list \(v_1, \dots, v_n\) spanning \(V\), so linearity gives

\begin{equation*} w = T(a_1 v_1 + \dots + a_n v_n) = a_1 T v_1 + \dots + a_n T v_n \end{equation*}

in \(\operatorname{span}(T v_1, \dots, T v_n)\). The two inclusions give \(\operatorname{span}(T v_1, \dots, T v_n) = \operatorname{range} T\).

Take \(U = \operatorname{span}(w_1, \dots, w_n)\), where \(u_1, \dots, u_m\) is a basis of \(\operatorname{null} T\) (finite-dimensional by 2.25, being a subspace of the finite-dimensional \(V\)), extended by 2.32 to a basis

\begin{equation*} u_1, \dots, u_m, w_1, \dots, w_n \end{equation*}

of \(V\).

If \(v \in U \cap \operatorname{null} T\), writing \(v\) in each of the two bases and subtracting gives

\begin{equation*} a_1 u_1 + \dots + a_m u_m - b_1 w_1 - \dots - b_n w_n = 0, \end{equation*}

so every coefficient vanishes by independence of the extended list; hence \(U \cap \operatorname{null} T = \{0\}\).

For the range, \(\{Tu : u \in U\} \subseteq \operatorname{range} T\) is immediate, and given \(x = Tv\) with \(v = a_1 u_1 + \dots + a_m u_m + b_1 w_1 + \dots + b_n w_n\), the vector \(u = b_1 w_1 + \dots + b_n w_n \in U\) satisfies

\begin{equation*} Tu = Tv - (a_1 T u_1 + \dots + a_m T u_m) = Tv = x \end{equation*}

since each \(u_j \in \operatorname{null} T\). Hence \(\operatorname{range} T = \{Tu : u \in U\}\).

Here \(\dim \operatorname{null} T = 2\): a vector lies in \(\operatorname{null} T\) exactly when it has the form

\begin{equation*} (5x_2, x_2, 7x_4, x_4) = x_2 (5, 1, 0, 0) + x_4 (0, 0, 7, 1), \end{equation*}

and \((5,1,0,0), (0,0,7,1)\) is linearly independent (read off the second and fourth coordinates), hence a basis of \(\operatorname{null} T\). Since \(\dim \mathbf{F}^4 = 4\), the fundamental theorem of linear maps (3.21) gives

\begin{equation*} \dim \operatorname{range} T = 4 - 2 = 2 = \dim \mathbf{F}^2 . \end{equation*}

As \(\operatorname{range} T\) is a subspace of \(\mathbf{F}^2\) (by 3.18) of full dimension, 2.39 gives \(\operatorname{range} T = \mathbf{F}^2\); that is, \(T\) is surjective.

The fundamental theorem of linear maps (3.21), applied to \(T\) on the finite-dimensional space \(\mathbf{R}^8\), gives

\begin{equation*} \dim \operatorname{range} T = 8 - \dim \operatorname{null} T = 8 - \dim U = 5 . \end{equation*}

Since \(\operatorname{range} T\) is a subspace of \(\mathbf{R}^5\) (by 3.18) with \(\dim \operatorname{range} T = 5 = \dim \mathbf{R}^5\), 2.39 gives \(\operatorname{range} T = \mathbf{R}^5\). Hence \(T\) is surjective.

Such a map would need a three-dimensional range inside \(\mathbf{F}^2\). Write \(N\) for the displayed set; a vector lies in \(N\) exactly when it has the form

\begin{equation*} (3x_2, x_2, x_3, x_3, x_3) = x_2 (3, 1, 0, 0, 0) + x_3 (0, 0, 1, 1, 1), \end{equation*}

and \((3,1,0,0,0), (0,0,1,1,1)\) is linearly independent (read off the second and third coordinates), so it is a basis of \(N\) and \(\dim N = 2\).

If \(T \in \mathcal{L}(\mathbf{F}^5, \mathbf{F}^2)\) had \(\operatorname{null} T = N\), then 3.21 applied to \(T\) on the finite-dimensional \(\mathbf{F}^5\) would give

\begin{equation*} \dim \operatorname{range} T = 5 - 2 = 3 , \end{equation*}

contradicting \(\dim \operatorname{range} T \le \dim \mathbf{F}^2 = 2\) (by 2.37, since \(\operatorname{range} T\) is a subspace of \(\mathbf{F}^2\) by 3.18).

The list \(u_1, \dots, u_m, v_1, \dots, v_n\) spans \(V\), where \(T \in \mathcal{L}(V)\) is the given map, \(u_1, \dots, u_m\) is a basis of \(\operatorname{null} T\) and \(w_1, \dots, w_n\) a basis of \(\operatorname{range} T\) (both exist by 2.31, these spaces being finite-dimensional by hypothesis), and \(v_k \in V\) is chosen with \(T v_k = w_k\).

Indeed, given \(v \in V\), write \(Tv = b_1 w_1 + \dots + b_n w_n\); then linearity gives

\begin{equation*} T\bigl(v - (b_1 v_1 + \dots + b_n v_n)\bigr) = Tv - (b_1 w_1 + \dots + b_n w_n) = 0, \end{equation*}

so \(v - (b_1 v_1 + \dots + b_n v_n) = a_1 u_1 + \dots + a_m u_m\) for some scalars \(a_j\), whence

\begin{equation*} v = a_1 u_1 + \dots + a_m u_m + b_1 v_1 + \dots + b_n v_n . \end{equation*}

Thus \(V\) is spanned by a finite list, so \(V\) is finite-dimensional.

(i) If \(T \in \mathcal{L}(V, W)\) is injective, then \(\operatorname{null} T = \{0\}\) (by 3.15), so 3.21 and 2.37 give

\begin{equation*} \dim V = \dim \operatorname{range} T \le \dim W , \end{equation*}

the inequality because \(\operatorname{range} T\) is a subspace of the finite-dimensional \(W\) (by 3.18).

(ii) Conversely, let \(n = \dim V \le m = \dim W\), let \(v_1, \dots, v_n\) be a basis of \(V\) and \(w_1, \dots, w_m\) a basis of \(W\), and use the linear map lemma (3.4) to get \(T \in \mathcal{L}(V, W)\) with \(T v_k = w_k\) for \(k = 1, \dots, n\) (legitimate since \(n \le m\)). If \(v = c_1 v_1 + \dots + c_n v_n \in \operatorname{null} T\), then

\begin{equation*} 0 = Tv = c_1 w_1 + \dots + c_n w_n , \end{equation*}

so all \(c_k = 0\) by linear independence of the sublist \(w_1, \dots, w_n\); hence \(v = 0\) and \(T\) is injective by 3.15.

(i) If \(T \in \mathcal{L}(V, W)\) is surjective, then \(\operatorname{range} T = W\), so 3.21 gives

\begin{equation*} \dim V = \dim \operatorname{null} T + \dim W \ge \dim W . \end{equation*}

(ii) Conversely, let \(n = \dim V \ge m = \dim W\), let \(v_1, \dots, v_n\) be a basis of \(V\) and \(w_1, \dots, w_m\) a basis of \(W\), and use the linear map lemma (3.4) to get \(T \in \mathcal{L}(V, W)\) with

\begin{equation*} T v_k = \begin{cases} w_k & \text{if } 1 \le k \le m, \\ 0 & \text{if } m < k \le n \end{cases} \end{equation*}

(legitimate since \(m \le n\)). Then \(\operatorname{range} T\) is a subspace of \(W\) (by 3.18) containing \(w_1, \dots, w_m\), hence containing \(\operatorname{span}(w_1, \dots, w_m) = W\) by 2.6. So \(T\) is surjective.

(i) If \(T \in \mathcal{L}(V, W)\) has \(\operatorname{null} T = U\), then 3.21 and 2.37 (applied to the subspace \(\operatorname{range} T\) of the finite-dimensional \(W\), by 3.18) give

\begin{equation*} \dim V = \dim U + \dim \operatorname{range} T \le \dim U + \dim W , \end{equation*}

which rearranges to \(\dim U \ge \dim V - \dim W\).

(ii) Conversely, suppose \(\dim U \ge \dim V - \dim W\). Extend a basis \(u_1, \dots, u_j\) of \(U\) (finite-dimensional by 2.25) to a basis \(u_1, \dots, u_j, v_1, \dots, v_n\) of \(V\) (by 2.32); then \(n = \dim V - \dim U \le \dim W\), so a basis \(w_1, \dots, w_m\) of \(W\) has the linearly independent sublist \(w_1, \dots, w_n\). By 3.4 there is \(T \in \mathcal{L}(V, W)\) with

\begin{equation*} T u_i = 0 \quad (i = 1, \dots, j), \qquad T v_k = w_k \quad (k = 1, \dots, n) . \end{equation*}

Then \(U = \operatorname{span}(u_1, \dots, u_j) \subseteq \operatorname{null} T\), the latter being a subspace by 3.13; and if \(v = a_1 u_1 + \dots + a_j u_j + b_1 v_1 + \dots + b_n v_n\) lies in \(\operatorname{null} T\), then

\begin{equation*} 0 = Tv = b_1 w_1 + \dots + b_n w_n , \end{equation*}

forcing every \(b_k = 0\), so \(v \in U\). Hence \(\operatorname{null} T = U\).

(i) If \(ST = I\) and \(Tu = Tv\), then applying \(S\) gives

\begin{equation*} u = (ST)u = S(Tu) = S(Tv) = (ST)v = v , \end{equation*}

so \(T\) is injective.

(ii) Conversely, suppose \(T\) is injective. Extend a basis \(w_1, \dots, w_n\) of \(\operatorname{range} T\) (a subspace of the finite-dimensional \(W\), by 3.18 and 2.25) to a basis \(w_1, \dots, w_m\) of \(W\) (by 2.32), choose \(v_k \in V\) with \(T v_k = w_k\) for \(k \le n\), and use the linear map lemma (3.4) to get \(S \in \mathcal{L}(W, V)\) with

\begin{equation*} S w_k = \begin{cases} v_k & \text{if } 1 \le k \le n, \\ 0 & \text{if } n < k \le m. \end{cases} \end{equation*}

Given \(v \in V\), write \(Tv = c_1 w_1 + \dots + c_n w_n\). Then

\begin{equation*} (ST)v = c_1 v_1 + \dots + c_n v_n, \qquad T(c_1 v_1 + \dots + c_n v_n) = Tv , \end{equation*}

so injectivity of \(T\) gives \(c_1 v_1 + \dots + c_n v_n = v\) and hence \((ST)v = v\). Thus \(ST = I\).

(i) If \(TS = I\), then every \(w \in W\) satisfies

\begin{equation*} w = (TS)w = T(Sw) \in \operatorname{range} T , \end{equation*}

so \(\operatorname{range} T = W\); that is, \(T\) is surjective.

(ii) Conversely, suppose \(T\) is surjective. Let \(w_1, \dots, w_m\) be a basis of \(W\) (by 2.31, \(W\) being finite-dimensional), choose \(v_k \in V\) with \(T v_k = w_k\) (possible since \(\operatorname{range} T = W\)), and use the linear map lemma (3.4) to get \(S \in \mathcal{L}(W, V)\) with \(S w_k = v_k\) for \(k = 1, \dots, m\). Then

\begin{equation*} (TS) w_k = T(S w_k) = T v_k = w_k \end{equation*}

for every \(k\), so \(TS\) and \(I\) are linear maps on \(W\) agreeing on a basis of \(W\); by the uniqueness assertion of 3.4 they are equal, so \(TS = I\).

Write \(E = \{v \in V : Tv \in U\}\). Then \(E\) is a subspace of \(V\) by 1.34: \(T0 = 0 \in U\) (by 3.10, and since \(U\) is a subspace of \(W\)), while for \(v_1, v_2, v \in E\) and \(\lambda \in \mathbf{F}\),

\begin{equation*} T(v_1 + v_2) = Tv_1 + Tv_2 \in U, \qquad T(\lambda v) = \lambda\, Tv \in U , \end{equation*}

\(U\) being closed under addition and scalar multiplication.

Now apply the fundamental theorem of linear maps (3.21) to the restriction \(T|_E \in \mathcal{L}(E, W)\), whose domain is finite-dimensional by 2.25:

\begin{equation*} \dim E = \dim \operatorname{null} (T|_E) + \dim \operatorname{range} (T|_E) . \end{equation*}

Here \(\operatorname{null}(T|_E) = \operatorname{null} T\), since \(Tv = 0\) forces \(Tv = 0 \in U\) and hence \(v \in E\); and \(\operatorname{range}(T|_E) = U \cap \operatorname{range} T\), since \(Tv \in U\) for every \(v \in E\), while any \(w \in U \cap \operatorname{range} T\) equals \(Tv\) for some \(v\), and that \(v\) lies in \(E\) because \(Tv = w \in U\). Substituting,

\begin{equation*} \dim E = \dim \operatorname{null} T + \dim (U \cap \operatorname{range} T) , \end{equation*}

which is the asserted formula.

Apply the fundamental theorem of linear maps (3.21) to the restriction

\begin{equation*} R = T|_{\operatorname{null} ST} \in \mathcal{L}(\operatorname{null} ST,\ V), \end{equation*}

whose domain is finite-dimensional (a subspace of the finite-dimensional \(U\), by 3.13 and 2.25):

\begin{equation*} \dim\operatorname{null} ST = \dim\operatorname{null} R + \dim\operatorname{range} R. \end{equation*}

Now \(\operatorname{null} R \subseteq \operatorname{null} T\), since \(Ru = Tu\); and \(\operatorname{range} R \subseteq \operatorname{null} S\), since \(v = Tu\) with \(u \in \operatorname{null} ST\) gives

\begin{equation*} Sv = S(Tu) = (ST)u = 0. \end{equation*}

Each containment sits inside a finite-dimensional space (\(\operatorname{null} T\) inside \(U\), and \(\operatorname{null} S\) inside \(V\), by 3.13 and 2.25), so 2.37 gives \(\dim \operatorname{null} R \le \dim \operatorname{null} T\) and \(\dim \operatorname{range} R \le \dim \operatorname{null} S\). Substituting,

\begin{equation*} \dim\operatorname{null} ST \le \dim\operatorname{null} S + \dim\operatorname{null} T. \end{equation*}

(i) \(\operatorname{range} ST \subseteq \operatorname{range} S\), since \((ST)u = S(Tu)\) with \(Tu \in V\). Both are subspaces of \(W\) (by 3.18) and \(\operatorname{range} S\) is finite-dimensional (by 3.21 applied to \(S\) on the finite-dimensional \(V\)), so 2.37 gives \(\dim \operatorname{range} ST \le \dim \operatorname{range} S\).

(ii) Restrict \(S\) to the subspace \(\operatorname{range} T\) of \(V\) (a subspace by 3.18), obtaining \(R = S|_{\operatorname{range} T}\), whose range is

\begin{equation*} \operatorname{range} R = \{S(Tu) : u \in U\} = \operatorname{range} ST , \end{equation*}

since \(v\) ranges over \(\operatorname{range} T\) exactly when \(v = Tu\) for some \(u \in U\). As \(\operatorname{range} T\) is finite-dimensional (by 3.21 applied to \(T\) on the finite-dimensional \(U\)), 3.21 applied to \(R\) gives

\begin{equation*} \begin{aligned} \dim \operatorname{range} T &= \dim \operatorname{null} R + \dim \operatorname{range} ST \\ &\ge \dim \operatorname{range} ST . \end{aligned} \end{equation*}

Being at most each of the two dimensions, \(\dim \operatorname{range} ST\) is at most their minimum.

(a) The hypothesis \(ST = 0\) says \(\operatorname{range} T \subseteq \operatorname{null} S\), since \(S(Tv) = (ST)v = 0\) for every \(v \in V\). Both are subspaces of the five-dimensional \(V\) (by 3.18 and 3.13), so 2.37 and the fundamental theorem of linear maps (3.21) applied to \(S\) give

\begin{equation*} \begin{aligned} \dim \operatorname{range} T &\le \dim \operatorname{null} S \\ &= 5 - \dim \operatorname{range} S , \end{aligned} \end{equation*}

that is, \(\dim \operatorname{range} S + \dim \operatorname{range} T \le 5\). Hence the two ranges cannot both have dimension at least \(3\), so Exercise 3B.23 applied to \(TS\) (legitimate, as \(V\) is finite-dimensional) gives

\begin{equation*} \begin{aligned} \dim \operatorname{range} TS &\le \min\{\dim \operatorname{range} T, \dim \operatorname{range} S\} \\ &\le 2 . \end{aligned} \end{equation*}

(b) Take \(S, T \in \mathcal{L}(\mathbf{F}^5)\) given by

\begin{equation*} \begin{aligned} S(x_1, \dots, x_5) &= (x_1, x_2, 0, 0, 0), \\ T(x_1, \dots, x_5) &= (0, 0, x_1, x_2, 0), \end{aligned} \end{equation*}

each linear because every output coordinate is a linear function of the input. Then

\begin{equation*} (ST)(x_1, \dots, x_5) = S(0, 0, x_1, x_2, 0) = 0 , \end{equation*}

so \(ST = 0\), while \((TS)(x_1, \dots, x_5) = T(x_1, x_2, 0, 0, 0) = (0, 0, x_1, x_2, 0)\), so

\begin{equation*} \operatorname{range} TS = \operatorname{span}\big((0,0,1,0,0),\ (0,0,0,1,0)\big) \end{equation*}

has dimension \(2\), those two vectors being linearly independent.

(i) If \(T = ES\) and \(v \in \operatorname{null} S\), then

\begin{equation*} Tv = (ES)v = E(Sv) = E0 = 0 \end{equation*}

by 3.10, so \(\operatorname{null} S \subseteq \operatorname{null} T\).

(ii) Conversely, suppose \(\operatorname{null} S \subseteq \operatorname{null} T\). Let \(w_1, \dots, w_m\) be a basis of \(\operatorname{range} S\) (a subspace of \(W\) by 3.18, hence finite-dimensional by 2.25; the empty list if \(\operatorname{range} S = \{0\}\)), choose \(v_k \in V\) with \(S v_k = w_k\), extend by 2.32 to a basis \(w_1, \dots, w_n\) of \(W\), and use the linear map lemma (3.4) to get \(E \in \mathcal{L}(W)\) with

\begin{equation*} E w_k = T v_k \ \ (k \le m), \qquad E w_k = 0 \ \ (m < k \le n). \end{equation*}

Given \(v \in V\), write \(Sv = a_1 w_1 + \dots + a_m w_m\) and put \(u = a_1 v_1 + \dots + a_m v_m\). Then \(Su = a_1 w_1 + \dots + a_m w_m = Sv\), so \(v - u \in \operatorname{null} S \subseteq \operatorname{null} T\) and hence \(Tv = Tu\). Therefore

\begin{equation*} (ES)v = a_1 E w_1 + \dots + a_m E w_m = a_1 T v_1 + \dots + a_m T v_m = Tu = Tv , \end{equation*}

so \(ES = T\).

(i) If \(S = TE\), then every \(w = Sv\) in \(\operatorname{range} S\) satisfies

\begin{equation*} w = (TE)v = T(Ev) \in \operatorname{range} T , \end{equation*}

so \(\operatorname{range} S \subseteq \operatorname{range} T\).

(ii) Conversely, suppose \(\operatorname{range} S \subseteq \operatorname{range} T\). Let \(v_1, \dots, v_n\) be a basis of \(V\) (which exists by 2.31, \(V\) being finite-dimensional), choose \(u_k \in V\) with \(T u_k = S v_k\) (possible because \(S v_k \in \operatorname{range} S \subseteq \operatorname{range} T\)), and use the linear map lemma (3.4) to get \(E \in \mathcal{L}(V)\) with \(E v_k = u_k\) for \(k = 1, \dots, n\). Then

\begin{equation*} (TE) v_k = T(E v_k) = T u_k = S v_k \end{equation*}

for every \(k\), so \(TE\) and \(S\) are linear maps from \(V\) to \(W\) agreeing on a basis of \(V\); by the uniqueness assertion of 3.4 they are equal, so \(S = TE\).

Every \(v \in V\) splits as \(v = (v - Pv) + Pv\), with \(Pv \in \operatorname{range} P\) and

\begin{equation*} P(v - Pv) = Pv - P(Pv) = Pv - P^2 v = 0 , \end{equation*}

so \(v - Pv \in \operatorname{null} P\); hence \(V = \operatorname{null} P + \operatorname{range} P\), both summands being subspaces of \(V\) (by 3.13 and 3.18).

If \(v \in \operatorname{null} P \cap \operatorname{range} P\), say \(v = Pu\), then

\begin{equation*} v = Pu = P^2 u = P(Pu) = Pv = 0 , \end{equation*}

so \(\operatorname{null} P \cap \operatorname{range} P = \{0\}\). By 1.46 the sum is therefore direct:

\begin{equation*} V = \operatorname{null} P \oplus \operatorname{range} P . \end{equation*}

It suffices to show that \(D_m = D|_{\mathcal{P}_m(\mathbf{R})}\) maps onto \(\mathcal{P}_{m-1}(\mathbf{R})\) for every \(m \ge 1\), since every \(q \in \mathcal{P}(\mathbf{R})\) lies in \(\mathcal{P}_{m-1}(\mathbf{R})\) for some \(m \ge 1\) (take \(m = \deg q + 1\), or \(m = 1\) if \(q = 0\)). Throughout, \(\dim \mathcal{P}_m(\mathbf{R}) = m+1\) (by 2.36), \(\deg 0 = -\infty\), and nonconstant means \(\deg p \ge 1\).

First, \(D1\) is constant: both \(x\) and \(x+1\) are nonconstant of degree \(1\), so \(\deg Dx = \deg D(x+1) = 0\), and linearity gives

\begin{equation*} D1 = D(x+1) - Dx , \end{equation*}

a difference of two constants. Hence \(D\) maps \(\mathcal{P}_m(\mathbf{R})\) into \(\mathcal{P}_{m-1}(\mathbf{R})\) for each \(m \ge 1\): a nonconstant \(p \in \mathcal{P}_m(\mathbf{R})\) has \(\deg Dp = \deg p - 1 \le m-1\), while a constant \(p\) has \(Dp\) constant, so \(Dp \in \mathcal{P}_0(\mathbf{R}) \subseteq \mathcal{P}_{m-1}(\mathbf{R})\).

Also \(\operatorname{null} D \subseteq \mathcal{P}_0(\mathbf{R})\), since a nonconstant \(p\) has \(\deg Dp = \deg p - 1 \ge 0\) and hence \(Dp \ne 0\). So \(D_m \in \mathcal{L}\big(\mathcal{P}_m(\mathbf{R}),\ \mathcal{P}_{m-1}(\mathbf{R})\big)\) has \(\dim \operatorname{null} D_m \le \dim \mathcal{P}_0(\mathbf{R}) = 1\) (by 2.37), and 3.21 applied to \(D_m\) on the finite-dimensional \(\mathcal{P}_m(\mathbf{R})\) gives

\begin{equation*} \dim\operatorname{range} D_m \ge (m+1) - 1 = m = \dim\mathcal{P}_{m-1}(\mathbf{R}). \end{equation*}

Since \(\operatorname{range} D_m\) is a subspace of \(\mathcal{P}_{m-1}(\mathbf{R})\) (by 3.18), 2.37 forces equality here, so 2.39 gives \(\operatorname{range} D_m = \mathcal{P}_{m-1}(\mathbf{R})\), and \(D\) is surjective.

Fix \(m \ge 0\) with \(p \in \mathcal{P}_m(\mathbb{R})\) and apply the fundamental theorem of linear maps to \(T \in \mathcal{L}\big(\mathcal{P}_{m+1}(\mathbb{R}),\ \mathcal{P}_m(\mathbb{R})\big)\) given by

\begin{equation*} Tq = 5q^{\prime\prime} + 3q^{\prime} , \end{equation*}

which indeed lands in \(\mathcal{P}_m(\mathbb{R})\) (differentiation lowers degree by at least one) and is linear (differentiation is linear).

Here \(\operatorname{null} T = \mathcal{P}_0(\mathbb{R})\), of dimension \(1\): every constant polynomial lies in it, and conversely \(5q^{\prime\prime} + 3q^{\prime} = 0\) makes \(r = q^{\prime}\) satisfy \(r = -\tfrac53 r^{\prime}\), which is impossible for \(r \ne 0\) since then \(\deg r^{\prime} < \deg r\); so \(q^{\prime} = 0\). Hence 3.21 gives

\begin{equation*} \dim \operatorname{range} T = (m+2) - 1 = m+1 = \dim \mathcal{P}_m(\mathbb{R}) , \end{equation*}

using \(\dim \mathcal{P}_k(\mathbb{R}) = k+1\) (by 2.36). As \(\operatorname{range} T\) is a subspace of \(\mathcal{P}_m(\mathbb{R})\) (by 3.18) of full dimension, 2.39 gives \(\operatorname{range} T = \mathcal{P}_m(\mathbb{R})\), which contains \(p\); so some \(q \in \mathcal{P}_{m+1}(\mathbb{R})\) has \(5q^{\prime\prime} + 3q^{\prime} = p\).

Write \(U = \{au : a \in \mathbb{F}\} = \operatorname{span}(u)\), a subspace of \(V\) by 2.6, and note that the hypothesis \(u \notin \operatorname{null}\varphi\) says \(\varphi(u) \neq 0\).

Given \(v \in V\), put \(a = \varphi(v)/\varphi(u) \in \mathbb{F}\) (legitimate since \(\varphi(u) \ne 0\)); linearity of \(\varphi\) gives

\begin{equation*} \varphi(v - au) = \varphi(v) - a\,\varphi(u) = 0 , \end{equation*}

so \(v = (v - au) + au \in \operatorname{null}\varphi + U\), whence \(V = \operatorname{null}\varphi + U\).

If \(w \in (\operatorname{null}\varphi) \cap U\), say \(w = au\), then

\begin{equation*} 0 = \varphi(w) = a\,\varphi(u) , \end{equation*}

so \(a = 0\) and \(w = 0\); thus \((\operatorname{null}\varphi) \cap U = \{0\}\). By 1.46, applied to the subspaces \(\operatorname{null}\varphi\) (a subspace by 3.13) and \(U\), the sum is direct:

\begin{equation*} V = \operatorname{null} \varphi \oplus \{au : a \in \mathbb{F}\} . \end{equation*}

(i) If \(T \in \mathcal{L}(V, W)\) has \(\operatorname{null} T = X\) and \(\operatorname{range} T = Y\), then the fundamental theorem of linear maps (3.21), applicable since \(V\) is finite-dimensional, gives

\begin{equation*} \dim V = \dim \operatorname{null} T + \dim \operatorname{range} T = \dim X + \dim Y . \end{equation*}

(ii) Conversely, suppose \(\dim X + \dim Y = \dim V\); write \(n = \dim V\) and \(m = \dim X\) (finite by 2.25), so \(\dim Y = n - m\). Extend a basis \(x_1, \dots, x_m\) of \(X\) to a basis

\begin{equation*} x_1, \dots, x_m, v_1, \dots, v_{n-m} \end{equation*}

of \(V\) (by 2.32, with exactly \(n-m\) vectors added since every basis of \(V\) has length \(n\), by 2.34), take a basis \(y_1, \dots, y_{n-m}\) of \(Y \subseteq W\), and use the linear map lemma (3.4) to get \(T \in \mathcal{L}(V, W)\) with

\begin{equation*} T x_k = 0 \quad (k \le m), \qquad T v_j = y_j \quad (j \le n-m) . \end{equation*}

Then \(Tv = \sum_j b_j y_j\) for \(v = \sum_k a_k x_k + \sum_j b_j v_j\), and each \(y_j = T v_j\), so

\begin{equation*} \operatorname{range} T = \operatorname{span}(y_1, \dots, y_{n-m}) = Y . \end{equation*}

Also \(X = \operatorname{span}(x_1, \dots, x_m) \subseteq \operatorname{null} T\), the latter being a subspace by 3.13, while 3.21 gives

\begin{equation*} \dim \operatorname{null} T = n - (n - m) = m = \dim X ; \end{equation*}

hence \(X = \operatorname{null} T\) by 2.39.

The set \(\operatorname{null}\varphi\) is a two-sided ideal of \(\mathcal{L}(V)\): it is a subspace of \(\mathcal{L}(V)\) (by 3.13), and for \(E \in \operatorname{null}\varphi\) and \(T \in \mathcal{L}(V)\) the multiplicativity of \(\varphi\) gives

\begin{equation*} \varphi(TE) = \varphi(T)\varphi(E) = 0 = \varphi(E)\varphi(T) = \varphi(ET) , \end{equation*}

so \(TE, ET \in \operatorname{null}\varphi\). Since \(V\) is finite-dimensional, Exercise 3A.17 says the only two-sided ideals of \(\mathcal{L}(V)\) are \(\{0\}\) and \(\mathcal{L}(V)\).

The alternative \(\operatorname{null}\varphi = \{0\}\) is impossible: it would make \(\varphi\) injective (by 3.15), whereas \(\dim V > 1\) and 3.72 (with \(W = V\)) give

\begin{equation*} \dim \mathcal{L}(V) = (\dim V)^2 \ge 4 > 1 = \dim \mathbb{F} , \end{equation*}

so no linear map from \(\mathcal{L}(V)\) to \(\mathbb{F}\) is injective (by 3.22).

Hence \(\operatorname{null}\varphi = \mathcal{L}(V)\), that is, \(\varphi = 0\).

Throughout, each element of \(V_{\mathbb{C}}\) is written \(u + iv\) with \(u, v \in V\) uniquely determined, and by Exercise 1B.8 the operations are

\begin{equation*} (u_1 + iv_1) + (u_2 + iv_2) = (u_1 + u_2) + i(v_1 + v_2), \end{equation*}

\begin{equation*} (a + bi)(u + iv) = (au - bv) + i(av + bu) \qquad (a, b \in \mathbb{R}), \end{equation*}

and likewise in \(W_{\mathbb{C}}\); in particular \(w_1 + iw_2 = 0\) exactly when \(w_1 = w_2 = 0\).

(a) Additivity and complex homogeneity:

\begin{equation*} \begin{aligned} T_{\mathbb{C}}\big((u_1 + iv_1) + (u_2 + iv_2)\big) &= T(u_1 + u_2) + i\,T(v_1 + v_2) \\ &= (Tu_1 + iTv_1) + (Tu_2 + iTv_2) \\ &= T_{\mathbb{C}}(u_1 + iv_1) + T_{\mathbb{C}}(u_2 + iv_2), \end{aligned} \end{equation*}

\begin{equation*} \begin{aligned} T_{\mathbb{C}}\big((a+bi)(u+iv)\big) &= T(au - bv) + i\,T(av + bu) \\ &= (a\,Tu - b\,Tv) + i(a\,Tv + b\,Tu) \\ &= (a+bi)\,T_{\mathbb{C}}(u + iv), \end{aligned} \end{equation*}

each middle equality by \(\mathbb{R}\)-linearity of \(T\) and each last one by the definition of the operations in \(W_{\mathbb{C}}\). Hence \(T_{\mathbb{C}} \in \mathcal{L}(V_{\mathbb{C}}, W_{\mathbb{C}})\).

(b) Here \(\operatorname{null} T_{\mathbb{C}} = (\operatorname{null} T)_{\mathbb{C}}\), since \(Tu + iTv = 0\) exactly when \(Tu = 0\) and \(Tv = 0\). So if \(\operatorname{null} T = \{0\}\), then \(u + iv \in \operatorname{null} T_{\mathbb{C}}\) forces \(u = v = 0\); conversely, if \(\operatorname{null} T_{\mathbb{C}} = \{0\}\), then \(u \in \operatorname{null} T\) gives \(T_{\mathbb{C}}(u + i0) = Tu + i\,T0 = 0\), so \(u + i0 = 0\) and \(u = 0\). Now 3.15 converts both statements into injectivity.

(c) If \(\operatorname{range} T = W\), then given \(w_1 + iw_2 \in W_{\mathbb{C}}\) choose \(u, v \in V\) with \(Tu = w_1\) and \(Tv = w_2\), so that

\begin{equation*} T_{\mathbb{C}}(u + iv) = Tu + iTv = w_1 + iw_2 , \end{equation*}

giving \(\operatorname{range} T_{\mathbb{C}} = W_{\mathbb{C}}\). Conversely, if \(\operatorname{range} T_{\mathbb{C}} = W_{\mathbb{C}}\), then each \(w \in W\) satisfies \(w + i0 = T_{\mathbb{C}}(u + iv) = Tu + iTv\) for some \(u, v \in V\); comparing coordinates gives \(Tu = w\), so \(\operatorname{range} T = W\).

Fix bases \(v_1, \dots, v_n\) of \(V\) and \(w_1, \dots, w_m\) of \(W\), write \(A = \mathcal{M}(T)\), so that \(T v_k = \sum_{j=1}^m A_{j,k} w_j\) for each \(k\), and let \(K\) be the set of indices \(k\) whose column of \(A\) has at least one nonzero entry. If \(k \notin K\) then \(T v_k = 0\), so

\begin{equation*} \begin{aligned} \operatorname{range} T &= \operatorname{span}(T v_1, \dots, T v_n) \\ &= \operatorname{span}(T v_k : k \in K), \end{aligned} \end{equation*}

the first equality because \(v_1, \dots, v_n\) spans \(V\) and \(T\) is linear. Reducing this spanning list of length \(|K|\) to a basis (2.30) gives \(\dim \operatorname{range} T \le |K|\).

Finally, picking one nonzero entry from each column indexed by \(K\) exhibits \(|K|\) distinct nonzero entries of \(A\) (distinct columns give distinct entries), so

\begin{equation*} \begin{aligned} \dim \operatorname{range} T &\le |K| \\ &\le \#\{\text{nonzero entries of } \mathcal{M}(T)\} . \end{aligned} \end{equation*}

Write \(n = \dim V \ge 1\) and \(m = \dim W \ge 1\).

(i) If every entry of \(\mathcal{M}(T)\) equals \(1\) with respect to bases \(v_1, \dots, v_n\) of \(V\) and \(w_1, \dots, w_m\) of \(W\), then \(T v_k = w_1 + \dots + w_m = w\) for every \(k\), and \(w \ne 0\) since \(w_1, \dots, w_m\) is linearly independent. As \(v_1, \dots, v_n\) spans \(V\),

\begin{equation*} \begin{aligned} \operatorname{range} T &= \operatorname{span}(T v_1, \dots, T v_n) \\ &= \operatorname{span}(w) , \end{aligned} \end{equation*}

which has dimension \(1\).

(ii) Conversely, suppose \(\dim \operatorname{range} T = 1\), say \(\operatorname{range} T = \operatorname{span}(w)\) with \(w \ne 0\). Extend to a basis \(u_1, \dots, u_m\) of \(W\) with \(u_1 = w\) (by 2.32) and set

\begin{equation*} w_1 = u_1 - u_2 - \dots - u_m, \qquad w_j = u_j \ \ (2 \le j \le m). \end{equation*}

Then \(\operatorname{span}(w_1, \dots, w_m)\) contains \(u_2, \dots, u_m\) and \(w_1 + u_2 + \dots + u_m = u_1\), so it equals \(W\); having length \(m = \dim W\), this list is a basis of \(W\) (by 2.42), and \(w_1 + \dots + w_m = u_1 = w\).

For \(V\), choose \(v_1\) with \(T v_1 = w\) and a basis \(z_1, \dots, z_{n-1}\) of \(\operatorname{null} T\), which has dimension \(n - 1\) by 3.21. The list \(v_1, z_1, \dots, z_{n-1}\) is linearly independent (applying \(T\) to a vanishing combination gives \(a w = 0\), so \(a = 0\), and then the remaining coefficients vanish), hence a basis of \(V\) by 2.38. Set

\begin{equation*} v_k = v_1 + z_{k-1} \qquad (2 \le k \le n) . \end{equation*}

Then \(\operatorname{span}(v_1, \dots, v_n)\) contains \(v_1\) and each \(v_k - v_1 = z_{k-1}\), so it equals \(V\), making \(v_1, \dots, v_n\) a basis of \(V\) (by 2.42). Finally

\begin{equation*} T v_k = T v_1 + T z_{k-1} = w = w_1 + \dots + w_m \end{equation*}

for \(2 \le k \le n\), and \(T v_1 = w\) as well; so all entries of \(\mathcal{M}(T)\) with respect to these bases equal \(1\).

Both identities follow by reading off coefficients in column \(k\), which are unique because \(w_1, \dots, w_m\) is a basis (2.28). Write \(A = \mathcal{M}(S)\), \(C = \mathcal{M}(T)\), and fix \(k \in \{1, \dots, n\}\).

(a) Since \(S + T \in \mathcal{L}(V,W)\) (3.5, 3.6),

\begin{equation*} \begin{aligned} (S + T) v_k &= S v_k + T v_k = \sum_{j=1}^{m} A_{j,k}\, w_j + \sum_{j=1}^{m} C_{j,k}\, w_j \\ &= \sum_{j=1}^{m} (A_{j,k} + C_{j,k})\, w_j , \end{aligned} \end{equation*}

so \(\mathcal{M}(S+T)_{j,k} = A_{j,k} + C_{j,k} = (A + C)_{j,k}\) by 3.34. Hence \(\mathcal{M}(S + T) = \mathcal{M}(S) + \mathcal{M}(T)\).

(b) Since \(\lambda T \in \mathcal{L}(V,W)\) (3.5, 3.6),

\begin{equation*} (\lambda T) v_k = \lambda (T v_k) = \lambda \sum_{j=1}^{m} C_{j,k}\, w_j = \sum_{j=1}^{m} (\lambda C_{j,k})\, w_j , \end{equation*}

so \(\mathcal{M}(\lambda T)_{j,k} = \lambda C_{j,k} = (\lambda C)_{j,k}\) by 3.36. Hence \(\mathcal{M}(\lambda T) = \lambda \mathcal{M}(T)\).

Take the basis of \(\mathcal{P}_3(\mathbf{R})\) to be

\begin{equation*} v_1 = x, \qquad v_2 = \tfrac{x^2}{2}, \qquad v_3 = \tfrac{x^3}{3}, \qquad v_4 = 1, \end{equation*}

and the basis of \(\mathcal{P}_2(\mathbf{R})\) to be the standard one,

\begin{equation*} w_1 = 1, \qquad w_2 = x, \qquad w_3 = x^2 . \end{equation*}

The list \(v_1, v_2, v_3, v_4\) spans \(\mathcal{P}_3(\mathbf{R})\), since its span contains the scalar multiples \(1, x, x^2, x^3\); having length \(4 = \dim \mathcal{P}_3(\mathbf{R})\), it is a basis by 2.42. Differentiating,

\begin{equation*} \begin{aligned} D v_1 &= (x)^{\prime} = 1 &&= 1 w_1 + 0 w_2 + 0 w_3, \\ D v_2 &= \left(\tfrac{x^2}{2}\right)^{\prime} = x &&= 0 w_1 + 1 w_2 + 0 w_3, \\ D v_3 &= \left(\tfrac{x^3}{3}\right)^{\prime} = x^2 &&= 0 w_1 + 0 w_2 + 1 w_3, \\ D v_4 &= (1)^{\prime} = 0 &&= 0 w_1 + 0 w_2 + 0 w_3 . \end{aligned} \end{equation*}

Reading the coefficients off as columns,

\begin{equation*} \mathcal{M}(D) = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \end{pmatrix}. \end{equation*}

Take the basis of \(V\) to be \(v_1, \dots, v_r, u_1, \dots, u_{n-r}\), where \(n = \dim V\), \(r = \dim \operatorname{range} T\), the list \(u_1, \dots, u_{n-r}\) is a basis of \(\operatorname{null} T\) (its length is \(n - r\) by 3.21), and \(v_1, \dots, v_r\) are vectors extending it to a basis of \(V\) (2.32), reordered to put the \(v\)’s first. The proof of 3.21 shows \(Tv_1, \dots, Tv_r\) is a basis of \(\operatorname{range} T\); set

\begin{equation*} w_k = T v_k \quad (1 \le k \le r), \end{equation*}

a linearly independent list in \(W\), and extend it by 2.32 to a basis \(w_1, \dots, w_m\) of \(W\), where \(m = \dim W\).

With respect to these bases, column \(k\) of \(A = \mathcal{M}(T)\) is read off from

\begin{equation*} T v_k = w_k \quad (1 \le k \le r), \qquad T u_i = 0 \quad (1 \le i \le n - r), \end{equation*}

so \(A_{k,k} = 1\) for \(1 \le k \le r\) and every other entry of \(A\) is \(0\), as required.

Choose the basis of \(W\) to have \(T v_1\) as its first vector when \(T v_1 \ne 0\). The first column of \(A = \mathcal{M}(T)\) holds the coefficients in \(T v_1 = A_{1,1} w_1 + \cdots + A_{n,1} w_n\), unique by 2.28, so with \(n = \dim W\):

(i) \(T v_1 = 0\). Let \(w_1, \dots, w_n\) be any basis of \(W\) (one exists by 2.31). Then \(T v_1 = 0 \cdot w_1 + \cdots + 0 \cdot w_n\), so the first column is entirely \(0\) – permitted, since the \(1\) is only possibly there.

(ii) \(T v_1 \ne 0\). Then the one-vector list \(w_1 = T v_1\) is linearly independent, so it extends to a basis \(w_1, \dots, w_n\) of \(W\) (2.32). Now

\begin{equation*} T v_1 = w_1 = 1 \cdot w_1 + 0 \cdot w_2 + \cdots + 0 \cdot w_n , \end{equation*}

so \(A_{1,1} = 1\) and \(A_{j,1} = 0\) for \(j \ge 2\).

Take \(v_1\) with \(\varphi(v_1) = 1\) and \(v_2, \dots, v_m\) a basis of \(\operatorname{null}\varphi\), where \(\varphi = \pi \circ T\) and \(\pi \colon W \to \mathbf{F}\) sends \(a_1 w_1 + \cdots + a_n w_n\) to \(a_1\). Here \(\pi\) is linear because the coefficients of a vector in the basis \(w_1, \dots, w_n\) are unique and depend additively and homogeneously on the vector (2.28), so \(\varphi \in \mathcal{L}(V, \mathbf{F})\) by 3.7. Writing \(m = \dim V\) and \(A = \mathcal{M}(T)\) with respect to \(v_1, \dots, v_m\) and \(w_1, \dots, w_n\), applying \(\pi\) to \(T v_k = \sum_{j=1}^{n} A_{j,k} w_j\) gives

\begin{equation*} \varphi(v_k) = A_{1,k} , \end{equation*}

so the first row of \(\mathcal{M}(T)\) is the list \(\varphi(v_1), \dots, \varphi(v_m)\).

(i) \(\varphi = 0\). Any basis of \(V\) (2.31) makes the entire first row \(0\), which is permitted.

(ii) \(\varphi \ne 0\). Then \(\operatorname{range}\varphi\) is a nonzero subspace of the \(1\)-dimensional space \(\mathbf{F}\), hence equals \(\mathbf{F}\) (2.37, 2.39), so \(\dim \operatorname{null}\varphi = m - 1\) by 3.21 and \(\varphi(v_1) = 1\) is attainable. The list \(v_1, \dots, v_m\) above is linearly independent: applying \(\varphi\) to \(c_1 v_1 + \cdots + c_m v_m = 0\) gives \(c_1 = 0\), and then independence of \(v_2, \dots, v_m\) gives \(c_2 = \cdots = c_m = 0\). Having length \(m = \dim V\), it is a basis of \(V\) (2.38), and

\begin{equation*} A_{1,1} = \varphi(v_1) = 1, \qquad A_{1,k} = \varphi(v_k) = 0 \ \text{ for } k \ge 2 . \end{equation*}

Both sides are \(1\)-by-\(p\) matrices (by 3.41 and 3.44), so it suffices to compare entries in column \(k\). Fix \(j\) and \(k\); since \((A_{j,\cdot})_{1,r} = A_{j,r}\), two applications of 3.41 give

\begin{equation*} \bigl(A_{j,\cdot}B\bigr)_{1,k} = \sum_{r=1}^{n} \bigl(A_{j,\cdot}\bigr)_{1,r} B_{r,k} = \sum_{r=1}^{n} A_{j,r} B_{r,k} = (AB)_{j,k} = \bigl((AB)_{j,\cdot}\bigr)_{1,k}. \end{equation*}

Hence \((AB)_{j,\cdot} = A_{j,\cdot}\,B\).

Both sides are \(1\)-by-\(p\) matrices (by 3.41, 3.44, 3.36, 3.34), so it suffices to compare entries in column \(k\). Fix \(k\); since the entry in row \(1\), column \(k\), of \(B_{r,\cdot}\) is \(B_{r,k}\), the definitions of matrix multiplication (3.41), scalar multiplication (3.36), and matrix addition (3.34) give

\begin{equation*} (aB)_{1,k} = \sum_{r=1}^{n} a_r B_{r,k} = \bigl(a_1 B_{1,\cdot} + \cdots + a_n B_{n,\cdot}\bigr)_{1,k}. \end{equation*}

Hence \(aB = a_1 B_{1,\cdot} + \cdots + a_n B_{n,\cdot}\).

Take

\begin{equation*} A = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}, \qquad B = \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}. \end{equation*}

By 3.41 (Check!),

\begin{equation*} AB = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}, \qquad BA = \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}, \end{equation*}

which differ in row \(1\), column \(1\). Hence \(AB \ne BA\).

The sizes force themselves: \(A(B+C)\) making sense means \(B\) and \(C\) have a common size \(n\)-by-\(p\) (3.34) and \(A\) is \(m\)-by-\(n\), so \(AB\) and \(AC\) are both defined and both \(m\)-by-\(p\) (3.41), hence addable, with \(AB + AC\) of the same size \(m\)-by-\(p\) as \(A(B+C)\). Likewise \((D+E)F\) making sense means \(D\) and \(E\) are both \(m\)-by-\(n\) and \(F\) is \(n\)-by-\(p\), so \(DF\), \(EF\), and \(DF + EF\) are all defined and \(m\)-by-\(p\).

It therefore suffices to compare entries in row \(j\), column \(k\). Fix such \(j\) and \(k\). By 3.41, then 3.34, then distributivity in \(\mathbf{F}\) and splitting of a finite sum,

\begin{equation*} \begin{aligned} \bigl(A(B+C)\bigr)_{j,k} &= \sum_{r=1}^{n} A_{j,r}\,(B+C)_{r,k} \\ &= \sum_{r=1}^{n} A_{j,r}\,\bigl(B_{r,k} + C_{r,k}\bigr) \\ &= \sum_{r=1}^{n} \bigl(A_{j,r}B_{r,k} + A_{j,r}C_{r,k}\bigr) \\ &= \sum_{r=1}^{n} A_{j,r}B_{r,k} + \sum_{r=1}^{n} A_{j,r}C_{r,k} \\ &= (AB)_{j,k} + (AC)_{j,k} \\ &= (AB + AC)_{j,k}, \end{aligned} \end{equation*}

where the last two equalities use 3.41 and 3.34 again. Hence \(A(B+C) = AB + AC\). The same three steps on the other side give

\begin{equation*} \begin{aligned} \bigl((D+E)F\bigr)_{j,k} &= \sum_{r=1}^{n} \bigl(D_{j,r} + E_{j,r}\bigr) F_{r,k} \\ &= \sum_{r=1}^{n} D_{j,r}F_{r,k} + \sum_{r=1}^{n} E_{j,r}F_{r,k} = (DF + EF)_{j,k}, \end{aligned} \end{equation*}

so \((D+E)F = DF + EF\).

Associativity of matrix multiplication is associativity of composition, transported through 3.43. First the sizes: \((AB)C\) making sense forces \(A\) to be \(m\)-by-\(n\), \(B\) to be \(n\)-by-\(p\), and \(C\) to be \(p\)-by-\(q\), so by 3.41 the products \(BC\) (\(n\)-by-\(q\)) and then \(A(BC)\) (\(m\)-by-\(q\)) are defined, of the same size as \((AB)C\).

Throw the matrices out: for an \(m\)-by-\(n\) matrix \(A\) define \(T_A \in \mathcal{L}(\mathbf{F}^n, \mathbf{F}^m)\) by

\begin{equation*} T_A(x_1,\dots,x_n) = \Bigl(\sum_{r=1}^{n} A_{1,r}x_r,\ \dots,\ \sum_{r=1}^{n} A_{m,r}x_r\Bigr), \end{equation*}

which is linear because each coordinate is a fixed linear combination of \(x_1, \dots, x_n\) (Check!). Sending the standard basis vector \(e_k\) to \(T_A e_k = \sum_{j=1}^{m} A_{j,k} f_j\) exhibits column \(k\) of \(A\) as column \(k\) of \(\mathcal{M}(T_A)\), so \(\mathcal{M}(T_A) = A\) with respect to standard bases (3.31).

Form \(T_A, T_B, T_C\) this way. Two applications of 3.43, legitimate because standard bases are used throughout, give

\begin{equation*} \mathcal{M}\bigl((T_A T_B)T_C\bigr) = (AB)C, \qquad \mathcal{M}\bigl(T_A (T_B T_C)\bigr) = A(BC). \end{equation*}

Composition of functions is associative, so \((T_A T_B)T_C = T_A(T_B T_C)\); equal maps have equal matrices, whence \((AB)C = A(BC)\).

Method (2): fix \(j \in \{1,\dots,m\}\) and \(k \in \{1,\dots,q\}\) and compute entries. By 3.41 twice, then interchanging two finite sums,

\begin{equation*} \begin{aligned} \bigl((AB)C\bigr)_{j,k} &= \sum_{s=1}^{p} (AB)_{j,s} C_{s,k} = \sum_{s=1}^{p} \Bigl(\sum_{r=1}^{n} A_{j,r}B_{r,s}\Bigr) C_{s,k} = \sum_{s=1}^{p}\sum_{r=1}^{n} A_{j,r}B_{r,s}C_{s,k} \\ &= \sum_{r=1}^{n}\sum_{s=1}^{p} A_{j,r}B_{r,s}C_{s,k} = \sum_{r=1}^{n} A_{j,r}\Bigl(\sum_{s=1}^{p} B_{r,s}C_{s,k}\Bigr) = \sum_{r=1}^{n} A_{j,r}(BC)_{r,k} \\ &= \bigl(A(BC)\bigr)_{j,k}. \end{aligned} \end{equation*}

Since \(j\) and \(k\) were arbitrary, \((AB)C = A(BC)\).

Group \(A^3 = A^2 A\) (legitimate: matrix multiplication is associative by Exercise 3C.12, and every product here is \(n\)-by-\(n\) by 3.41) and apply 3.41 twice. Fix \(j\) and \(k\); then

\begin{equation*} \begin{aligned} (A^3)_{j,k} = (A^2 A)_{j,k} &= \sum_{r=1}^{n} (A^2)_{j,r}\, A_{r,k} = \sum_{r=1}^{n} \Bigl(\sum_{p=1}^{n} A_{j,p}A_{p,r}\Bigr) A_{r,k} \\ &= \sum_{r=1}^{n}\sum_{p=1}^{n} A_{j,p}A_{p,r}A_{r,k} = \sum_{p=1}^{n}\sum_{r=1}^{n} A_{j,p}A_{p,r}A_{r,k}, \end{aligned} \end{equation*}

the third equality by distributivity in \(\mathbf{F}\) and the last by interchanging two finite sums.

Write \(\Phi(A) = A^{\mathrm{t}}\), an \(n\)-by-\(m\) matrix by 3.54, and check the two conditions of 3.1 entrywise: both sides of each identity are \(n\)-by-\(m\) matrices, so it suffices to fix \(j \in \{1,\dots,m\}\), \(k \in \{1,\dots,n\}\) and compare entries in row \(k\), column \(j\).

Additivity. For \(A, B \in \mathbf{F}^{m,n}\), using 3.54 for the outer equalities and 3.34 for the inner ones,

\begin{equation*} \bigl((A+B)^{\mathrm{t}}\bigr)_{k,j} = (A+B)_{j,k} = A_{j,k} + B_{j,k} = \bigl(A^{\mathrm{t}} + B^{\mathrm{t}}\bigr)_{k,j}, \end{equation*}

so \(\Phi(A+B) = \Phi(A) + \Phi(B)\).

Homogeneity. For \(\lambda \in \mathbf{F}\), the same two citations with 3.36 in place of 3.34 give

\begin{equation*} \bigl((\lambda A)^{\mathrm{t}}\bigr)_{k,j} = (\lambda A)_{j,k} = \lambda A_{j,k} = \bigl(\lambda A^{\mathrm{t}}\bigr)_{k,j}, \end{equation*}

so \(\Phi(\lambda A) = \lambda\,\Phi(A)\). Hence \(\Phi \in \mathcal{L}(\mathbf{F}^{m,n}, \mathbf{F}^{n,m})\).

Both sides are \(p\)-by-\(m\) matrices – \(AC\) is \(m\)-by-\(p\) by 3.41 and transposes by 3.54, while \(C^{t}\) is \(p\)-by-\(n\) and \(A^{t}\) is \(n\)-by-\(m\), so \(C^{t}A^{t}\) is defined – so it suffices to compare entries in row \(k\), column \(j\). Fix \(k \in \{1,\dots,p\}\) and \(j \in \{1,\dots,m\}\); using 3.54 to strip transposes and 3.41 for each product,

\begin{equation*} \begin{aligned} \bigl((AC)^{t}\bigr)_{k,j} = (AC)_{j,k} &= \sum_{r=1}^{n} A_{j,r} C_{r,k} = \sum_{r=1}^{n} (C^{t})_{k,r} (A^{t})_{r,j} \\ &= (C^{t} A^{t})_{k,j}, \end{aligned} \end{equation*}

the middle equality because multiplication in \(\mathbf{F}\) is commutative. Hence \((AC)^{t} = C^{t}A^{t}\).

Both conditions say that the columns of \(A\) all lie on one line: write \(U = \operatorname{span}(A_{\cdot,1}, \dots, A_{\cdot,n}) \subseteq \mathbf{F}^{m,1}\), so that \(\dim U\) is the rank of \(A\) (by 3.52 and 3.58).

(i) Rank \(1\) implies the factorization. Then \(\dim U = 1\), so \(U = \operatorname{span}( c)\) for some nonzero \(c \in \mathbf{F}^{m,1}\), say with entries \(c_1, \dots, c_m\). Each column \(A_{\cdot,k}\) lies in \(U\), so \(A_{\cdot,k} = d_k c\) for some \(d_k \in \mathbf{F}\); comparing row \(j\) (using 3.44) gives

\begin{equation*} A_{j,k} = d_k c_j = c_j d_k \qquad (1 \le j \le m,\ 1 \le k \le n). \end{equation*}

(ii) The factorization implies rank \(1\). Let \(c \in \mathbf{F}^{m,1}\) have entries \(c_1, \dots, c_m\). Since row \(j\) of \(A_{\cdot,k}\) is \(A_{j,k} = c_j d_k\), which is row \(j\) of \(d_k c\), every column satisfies \(A_{\cdot,k} = d_k c\), so

\begin{equation*} U \subseteq \operatorname{span}( c), \qquad \dim U \le \dim \operatorname{span}( c) \le 1 \end{equation*}

by 2.37 (a one-element list spans a space of dimension at most \(1\), by 2.30). Since \(A \ne 0\), some column \(A_{\cdot,k}\) is a nonzero element of \(U\), so \(\dim U \ge 1\). Hence \(\dim U = 1\).

Everything rests on the dictionary \(x \mapsto u = x_1 u_1 + \dots + x_n u_n\) from \(\mathbf{F}^{n,1}\) onto \(V\), under which \(Tu = 0\) exactly when \(Ax = 0\). Here \(A = \mathcal{M}(T)\) is \(n\)-by-\(n\) (both lists are bases, so \(\dim V = n\)) and \(Tu_k = \sum_{j} A_{j,k} v_j\) by 3.31, whence linearity of \(T\) gives

\begin{equation*} Tu = \sum_{k=1}^{n} x_k \, T u_k = \sum_{j=1}^{n} (Ax)_{j,1} \, v_j, \qquad Ax = x_1 A_{\cdot,1} + \dots + x_n A_{\cdot,n}, \end{equation*}

the second identity by 3.50. Since \(v_1, \dots, v_n\) is linearly independent, \(Tu = 0\) iff \(Ax = 0\); since \(u_1, \dots, u_n\) is a basis, \(u = 0\) iff \(x = 0\), and every \(u \in V\) arises from some \(x\).

(a) \(\Longleftrightarrow\) (b). By 3.15, \(T\) is injective iff \(\operatorname{null} T = \{0\}\), which by the dictionary holds iff the only \(x\) with \(x_1 A_{\cdot,1} + \dots + x_n A_{\cdot,n} = Ax = 0\) is \(x = 0\) – that is, iff the columns of \(A\) are linearly independent.

(b) \(\Longleftrightarrow\) (c). The columns form a list of length \(n\) in \(\mathbf{F}^{n,1}\), and \(\dim \mathbf{F}^{n,1} = n\) (3.40); such a list is a basis as soon as it is linearly independent (2.38) or spanning (2.42), so each of (b), (c) implies the other.

(c) \(\Longleftrightarrow\) (d). Writing \(U\) for the span of the columns, (c) says \(U = \mathbf{F}^{n,1}\), which by 2.39 and 3.40 holds iff \(\dim U = n\), i.e. iff the column rank of \(A\) is \(n\) (3.52, 3.58). The same argument in \(\mathbf{F}^{1,n}\) shows (d) holds iff the row rank of \(A\) is \(n\). The two ranks are equal by 3.57.

(d) \(\Longleftrightarrow\) (e). As in (b) \(\Longleftrightarrow\) (c), applied to the \(n\) rows in \(\mathbf{F}^{1,n}\).

Take \(T \in \mathcal{L}(V,W)\) itself as the inverse of \(T^{-1} \in \mathcal{L}(W,V)\). Invertibility of \(T\) (3.59) gives

\begin{equation*} T T^{-1} = I_W \qquad \text{and} \qquad T^{-1} T = I_V, \end{equation*}

which are precisely the two identities 3.59 demands of an inverse of \(T^{-1}\). Hence \(T^{-1}\) is invertible, and \((T^{-1})^{-1} = T\) by uniqueness of the inverse (3.60).

Take \(T^{-1} S^{-1} \in \mathcal{L}(W,U)\) as the inverse; it and \(ST \in \mathcal{L}(U,W)\) are linear because compositions of linear maps are linear (3.8). Associativity of composition gives

\begin{equation*} (T^{-1} S^{-1})(ST) = T^{-1}(S^{-1} S)T = T^{-1} I_V T = T^{-1} T = I_U, \end{equation*}

\begin{equation*} (ST)(T^{-1} S^{-1}) = S(T T^{-1})S^{-1} = S I_V S^{-1} = S S^{-1} = I_W. \end{equation*}

Hence \(ST\) is invertible (3.59), and \((ST)^{-1} = T^{-1} S^{-1}\) by uniqueness of the inverse (3.60).

We run the cycle (a) \(\Rightarrow\) (b) \(\Rightarrow\) (c) \(\Rightarrow\) (a), writing \(n = \dim V\).

(a) \(\Rightarrow\) (b). Let \(v_1, \dots, v_n\) be a basis of \(V\). If \(a_1 Tv_1 + \dots + a_n Tv_n = 0\), then \(T(a_1 v_1 + \dots + a_n v_n) = 0\), and \(T\) is injective (3.63) with \(\operatorname{null} T = \{0\}\) (3.15), so \(a_1 v_1 + \dots + a_n v_n = 0\) and all \(a_k = 0\); thus \(Tv_1, \dots, Tv_n\) is linearly independent. It spans \(V\) because for \(v \in V\), writing \(T^{-1} v = a_1 v_1 + \dots + a_n v_n\) and applying \(T\) gives

\begin{equation*} v = T(T^{-1} v) = a_1 Tv_1 + \dots + a_n Tv_n . \end{equation*}

Hence \(Tv_1, \dots, Tv_n\) is a basis of \(V\).

(b) \(\Rightarrow\) (c). Apply (b) to a basis of \(V\), which exists by 2.31.

(c) \(\Rightarrow\) (a). By Exercise 10 in Section 3B, \(\operatorname{range} T = \operatorname{span}(Tv_1, \dots, Tv_n) = V\), the last equality because \(Tv_1, \dots, Tv_n\) is a basis of \(V\). So \(T\) is surjective, hence injective (3.65, as \(V\) is finite-dimensional), hence invertible (3.63).

The set \(\mathcal{N}\) of noninvertible operators is not closed under addition, hence is not a subspace (1.34). Fix a basis \(v_1, \dots, v_n\) of \(V\) (2.31), where \(n = \dim V \ge 2\), and let \(S, T \in \mathcal{L}(V)\) be given by the linear map lemma (3.4) via

\begin{equation*} S v_1 = v_1, \quad S v_k = 0 \ (k \ge 2); \qquad T v_1 = 0, \quad T v_k = v_k \ (k \ge 2). \end{equation*}

Since \(n \ge 2\), the basis vectors \(v_1, v_2\) exist and are nonzero, while \(S v_2 = 0\) and \(T v_1 = 0\); so neither \(S\) nor \(T\) is injective (3.15), hence neither is invertible (3.63), and \(S, T \in \mathcal{N}\). But \(S + T\) agrees with \(I\) on the basis,

\begin{equation*} (S + T)v_1 = v_1 + 0 = v_1, \qquad (S+T)v_k = 0 + v_k = v_k \ \ (k \ge 2), \end{equation*}

so \(S + T = I\), which is invertible. Hence \(S + T \notin \mathcal{N}\).

(i) Necessity. If such an invertible \(T\) exists and \(u \in U\) has \(Su = 0\), then \(Tu = Su = 0\), and \(T\) is injective (3.63) with \(\operatorname{null} T = \{0\}\) (3.15), so \(u = 0\). Thus \(\operatorname{null} S = \{0\}\) and \(S\) is injective (3.15).

(ii) Sufficiency. Suppose \(S\) is injective. Let \(u_1, \dots, u_m\) be a basis of the (finite-dimensional, by 2.25) subspace \(U\) and extend it to a basis \(u_1, \dots, u_m, v_1, \dots, v_p\) of \(V\) (2.32). The list \(Su_1, \dots, Su_m\) is linearly independent: \(a_1 Su_1 + \dots + a_m Su_m = 0\) gives \(a_1 u_1 + \dots + a_m u_m \in \operatorname{null} S = \{0\}\), whence all \(a_k = 0\). Extend it to a basis

\begin{equation*} Su_1, \dots, Su_m, w_1, \dots, w_q \end{equation*}

of \(V\) (2.32); since every basis of \(V\) has length \(\dim V\) (2.34), \(q = p\). So the linear map lemma (3.4) supplies \(T \in \mathcal{L}(V)\) with

\begin{equation*} T u_k = S u_k \ \ (k = 1, \dots, m), \qquad T v_j = w_j \ \ (j = 1, \dots, p). \end{equation*}

This \(T\) carries a basis of \(V\) to a basis of \(V\), so it is invertible by Exercise 3D.3 ((c) \(\Rightarrow\) (a)); and for \(u = a_1 u_1 + \dots + a_m u_m \in U\), linearity gives

\begin{equation*} Tu = a_1 Su_1 + \dots + a_m Su_m = S(a_1 u_1 + \dots + a_m u_m) = Su . \end{equation*}

(i) Necessity. If \(S = ET\) with \(E\) invertible, then \(Tv = 0\) gives \(Sv = E0 = 0\), while \(Sv = 0\) gives \(E(Tv) = 0\) and hence \(Tv = 0\), since \(E\) is injective (3.63) with \(\operatorname{null} E = \{0\}\) (3.15). So \(\operatorname{null} S = \operatorname{null} T\).

(ii) Sufficiency. Suppose \(\operatorname{null} S = \operatorname{null} T\), and define \(\varphi \colon \operatorname{range} T \to \operatorname{range} S\) by \(\varphi(Tv) = Sv\). This is unambiguous precisely because of the hypothesis: \(Tv_1 = Tv_2\) forces \(v_1 - v_2 \in \operatorname{null} T = \operatorname{null} S\), hence \(Sv_1 = Sv_2\). It is linear, since \(\varphi(\lambda Tv_1 + Tv_2) = S(\lambda v_1 + v_2) = \lambda\varphi(Tv_1) + \varphi(Tv_2)\); surjective onto \(\operatorname{range} S\) by construction; and injective, because \(\varphi(Tv) = Sv = 0\) puts \(v \in \operatorname{null} S = \operatorname{null} T\), so \(Tv = 0\) (3.15). Thus \(\varphi\) is an isomorphism (3.63, 3.69).

Let \(u_1, \dots, u_m\) be a basis of \(\operatorname{range} T\) (finite-dimensional as a subspace of \(W\), by 2.25) and extend it to a basis \(u_1, \dots, u_n\) of \(W\) (2.32), where \(n = \dim W\). The list \(\varphi u_1, \dots, \varphi u_m\) is again linearly independent (apply the injective \(\varphi\) to a vanishing combination), so extend it to a basis

\begin{equation*} \varphi u_1, \dots, \varphi u_m, y_{m+1}, \dots, y_n \end{equation*}

of \(W\) (2.32; the count \(n - m\) because every basis of \(W\) has length \(n\), by 2.34). The linear map lemma (3.4) now gives \(E \in \mathcal{L}(W)\) with

\begin{equation*} E u_k = \varphi u_k \ \ (k \le m), \qquad E u_k = y_k \ \ (k > m), \end{equation*}

which carries a basis of \(W\) to a basis of \(W\) and so is invertible by Exercise 3D.3 ((c) \(\Rightarrow\) (a)). Finally, for \(v \in V\) write \(Tv = c_1 u_1 + \dots + c_m u_m\); then linearity of \(E\) and of \(\varphi\) give

\begin{equation*} E(Tv) = c_1 \varphi u_1 + \dots + c_m \varphi u_m = \varphi(Tv) = Sv , \end{equation*}

so \(S = ET\).

(i) Necessity. If \(S = TE\) with \(E\) invertible, then \(E\) is surjective (3.63), so \(\operatorname{range} E = V\) and

\begin{equation*} \operatorname{range} S = \{ T(Ev) : v \in V \} = \{ Tu : u \in V \} = \operatorname{range} T . \end{equation*}

(ii) Sufficiency. Suppose \(\operatorname{range} S = \operatorname{range} T\); we build \(E\) from adapted bases. The fundamental theorem of linear maps (3.21), applicable since \(V\) is finite-dimensional, gives

\begin{equation*} \dim \operatorname{null} S = \dim V - \dim \operatorname{range} S = \dim \operatorname{null} T ; \end{equation*}

call this common value \(m\) and set \(r = \dim V - m\). Take a basis \(u_1, \dots, u_m\) of \(\operatorname{null} S\) extended to a basis \(u_1, \dots, u_m, v_1, \dots, v_r\) of \(V\) (2.32, 2.34), and a basis \(p_1, \dots, p_m\) of \(\operatorname{null} T\) (2.25, 2.31).

Then \(Sv_1, \dots, Sv_r\) is a basis of \(\operatorname{range} S\): it spans, since \(Su_k = 0\) makes \(Sv = a_1 Sv_1 + \dots + a_r Sv_r\) for \(v = \sum_k b_k u_k + \sum_j a_j v_j\); and if \(\sum_j a_j Sv_j = 0\) then \(\sum_j a_j v_j \in \operatorname{null} S = \operatorname{span}(u_1, \dots, u_m)\), so

\begin{equation*} b_1 u_1 + \dots + b_m u_m - a_1 v_1 - \dots - a_r v_r = 0 \end{equation*}

for some \(b_k\), forcing every \(a_j = 0\) by independence of the basis of \(V\).

Since \(Sv_j \in \operatorname{range} S = \operatorname{range} T\), choose \(q_j \in V\) with \(T q_j = S v_j\). The list \(p_1, \dots, p_m, q_1, \dots, q_r\) has length \(m + r = \dim V\), so by 2.38 it is a basis of \(V\) once independent: applying \(T\) to a vanishing combination kills the \(p_k\) and leaves \(\sum_j a_j Sv_j = 0\), so all \(a_j = 0\) by the previous paragraph, and then all \(b_k = 0\). The linear map lemma (3.4) therefore gives \(E \in \mathcal{L}(V)\) with

\begin{equation*} E u_k = p_k \ \ (k = 1, \dots, m), \qquad E v_j = q_j \ \ (j = 1, \dots, r), \end{equation*}

invertible by Exercise 3D.3 ((c) \(\Rightarrow\) (a)) since it carries a basis of \(V\) to a basis of \(V\). Finally \(TE\) and \(S\) agree on the basis \(u_1, \dots, u_m, v_1, \dots, v_r\), namely

\begin{equation*} (TE)u_k = T p_k = 0 = S u_k, \qquad (TE)v_j = T q_j = S v_j , \end{equation*}

so \(S = TE\).

(i) Necessity. If \(S = E_2 T E_1\) with both \(E_i\) invertible, then for \(v \in V\), using injectivity of \(E_2\) (3.63) for the middle step,

\begin{equation*} Sv = 0 \iff T E_1 v = 0 \iff E_1 v \in \operatorname{null} T, \end{equation*}

so \(\operatorname{null} S = \{E_1^{-1} u : u \in \operatorname{null} T\}\). Thus \(E_1^{-1}\) restricted to \(\operatorname{null} T\) is linear, injective and onto \(\operatorname{null} S\), hence an isomorphism (3.63); both null spaces are finite-dimensional (2.25), so \(\dim \operatorname{null} S = \dim \operatorname{null} T\) by 3.70.

(ii) Sufficiency. Let \(n\) be the common null-space dimension and \(k = \dim V - n\), so that

\begin{equation*} \dim \operatorname{range} S = k = \dim \operatorname{range} T \end{equation*}

by the fundamental theorem of linear maps (3.21). Extend a basis \(u_1, \dots, u_n\) of \(\operatorname{null} S\) to a basis \(u_1, \dots, u_n, x_1, \dots, x_k\) of \(V\), and a basis \(p_1, \dots, p_n\) of \(\operatorname{null} T\) to a basis \(p_1, \dots, p_n, y_1, \dots, y_k\) of \(V\) (2.32). Since \(Su_j = 0\), the list \(Sx_1, \dots, Sx_k\) spans \(\operatorname{range} S\) and has length \(k = \dim \operatorname{range} S\), so it is a basis of \(\operatorname{range} S\) (2.42); likewise \(Ty_1, \dots, Ty_k\) is a basis of \(\operatorname{range} T\). Extend both to bases of \(W\) (2.32),

\begin{equation*} S x_1, \dots, S x_k, w_1, \dots, w_r \qquad\text{and}\qquad T y_1, \dots, T y_k, z_1, \dots, z_r , \end{equation*}

with the same \(r = \dim W - k\) in each. The linear map lemma (3.4) now gives \(E_1 \in \mathcal{L}(V)\) and \(E_2 \in \mathcal{L}(W)\) with

\begin{equation*} E_1 u_j = p_j, \quad E_1 x_i = y_i; \qquad E_2 (T y_i) = S x_i, \quad E_2 z_j = w_j . \end{equation*}

Each carries a basis to a basis, hence is invertible by Exercise 3D.3 ((c) \(\Rightarrow\) (a)). Finally \(S\) and \(E_2 T E_1\) agree on the basis \(u_1, \dots, u_n, x_1, \dots, x_k\) of \(V\):

\begin{equation*} E_2 T E_1 u_j = E_2 T p_j = 0 = S u_j, \qquad E_2 T E_1 x_i = E_2 T y_i = S x_i , \end{equation*}

so \(S = E_2 T E_1\).

Take any subspace \(U\) of \(V\) with \(V = \operatorname{null} T \oplus U\), which exists by 2.33 since \(V\) is finite-dimensional and \(\operatorname{null} T\) is a subspace of \(V\). The restriction \(T|_U \in \mathcal{L}(U,W)\) inherits additivity and homogeneity from \(T\), and:

(i) Injective. If \(u \in U\) and \(Tu = 0\), then \(u \in \operatorname{null} T \cap U = \{0\}\), the intersection being trivial because the sum is direct (1.46); so \(\operatorname{null}(T|_U) = \{0\}\) and 3.15 applies.

(ii) Surjective. Given \(w \in W\), surjectivity of \(T\) supplies \(v \in V\) with \(Tv = w\); writing \(v = x + u\) with \(x \in \operatorname{null} T\) and \(u \in U\),

\begin{equation*} w = Tv = Tx + Tu = 0 + Tu = T|_U(u). \end{equation*}

Hence \(T|_U\) is invertible (3.63), i.e. an isomorphism of \(U\) onto \(W\).

(a) The zero map lies in \(\mathcal{E}\) (its null space is \(V\)), and for \(S, T \in \mathcal{E}\), \(\lambda \in \mathbf{F}\), \(u \in U\),

\begin{equation*} (S + T)u = 0 + 0 = 0, \qquad (\lambda T)u = \lambda 0 = 0, \end{equation*}

so \(S + T\) and \(\lambda T\) lie in \(\mathcal{E}\). Hence \(\mathcal{E}\) is a subspace of \(\mathcal{L}(V,W)\) by 1.34.

(b) The formula is

\begin{equation*} \dim \mathcal{E} = (\dim V - \dim U)(\dim W). \end{equation*}

Following the hint, let \(\Phi(T) = T|_U\) map \(\mathcal{L}(V,W)\) to \(\mathcal{L}(U,W)\); it is linear because restriction preserves the pointwise operations, \((S+T)|_U = S|_U + T|_U\) and \((\lambda T)|_U = \lambda (T|_U)\).

Its null space is \(\mathcal{E}\), since \(T|_U = 0\) says exactly \(U \subseteq \operatorname{null} T\). It is surjective: given \(R \in \mathcal{L}(U,W)\), extend a basis \(u_1, \dots, u_m\) of \(U\) (finite-dimensional by 2.25) to a basis \(u_1, \dots, u_m, v_1, \dots, v_n\) of \(V\) (2.32) and take the \(T \in \mathcal{L}(V,W)\) given by the linear map lemma (3.4) with

\begin{equation*} Tu_j = R u_j \quad (j = 1, \dots, m), \qquad T v_k = 0 \quad (k = 1, \dots, n); \end{equation*}

then \(T|_U\) and \(R\) agree on a basis of \(U\), so \(\Phi(T) = R\). Since \(\mathcal{L}(V,W)\) is finite-dimensional (3.72), the fundamental theorem of linear maps (3.21) applied to \(\Phi\) gives

\begin{equation*} \dim \mathcal{E} = \dim \mathcal{L}(V,W) - \dim \mathcal{L}(U,W) = (\dim V - \dim U)(\dim W), \end{equation*}

the last step evaluating both terms by 3.72.

(\(\Longleftarrow\)) If \(S\) and \(T\) are invertible then \(T^{-1}S^{-1}\) is a two-sided inverse of \(ST\) (Check!), so \(ST\) is invertible.

(\(\Longrightarrow\)) Suppose \(ST\) is invertible; then \(ST\) is injective and surjective (by 3.63).

(i) \(S\) is surjective: every \((ST)v = S(Tv)\) lies in \(\operatorname{range} S\), so

\begin{equation*} V = \operatorname{range}(ST) \subseteq \operatorname{range} S \subseteq V. \end{equation*}

Hence \(S\) is invertible, since a surjective operator on a finite-dimensional \(V\) is invertible (by 3.65).

(ii) \(T\) is injective: if \(Tv = 0\) then \((ST)v = S0 = 0\), so \(v = 0\) by injectivity of \(ST\) and 3.15. Hence \(T\) is invertible (by 3.65 again, \(V\) finite-dimensional).

Both \(S\) and \(U\) are invertible, and then \(T = S^{-1}U^{-1}\) forces \(T^{-1} = US\).

(i) \(U\) is invertible: if \(Uv = 0\) then \(v = (STU)v = S(T0) = 0\), so \(U\) is injective (by 3.15) and hence invertible, \(V\) being finite-dimensional (by 3.65).

(ii) \(S\) is invertible: every \((STU)v = S\bigl((TU)v\bigr)\) lies in \(\operatorname{range} S\), so

\begin{equation*} V = \operatorname{range} I = \operatorname{range}(STU) \subseteq \operatorname{range} S \subseteq V, \end{equation*}

making \(S\) surjective and hence invertible (by 3.65).

Multiplying \(STU = I\) by \(S^{-1}\) on the left and \(U^{-1}\) on the right gives \(T = S^{-1}U^{-1}\), a product of invertible maps, hence invertible with

\begin{equation*} T^{-1} = (S^{-1}U^{-1})^{-1} = (U^{-1})^{-1}(S^{-1})^{-1} = US, \end{equation*}

by Exercises 1 and 2 in this section.

On \(V = \mathbf{F}^\infty\) take \(S = T = B\) and \(U = F^2\), where \(B\) and \(F\) are the backward and forward shifts

\begin{equation*} B(x_1, x_2, x_3, \dots) = (x_2, x_3, x_4, \dots), \quad F(x_1, x_2, \dots) = (0, x_1, x_2, \dots), \end{equation*}

both linear by Example 3.3. Then \(STU = B^2F^2 = I\), since

\begin{equation*} B^2 F^2 (x_1, x_2, \dots) = B^2 (0, 0, x_1, x_2, \dots) = (x_1, x_2, \dots). \end{equation*}

But \(T = B\) is not invertible: \(B(1, 0, 0, \dots) = 0\), so \(B\) is not injective (by 3.15) and hence not invertible (by 3.63).

True: \(S\) is even invertible. Because \(V\) is finite-dimensional and the operator \(RST\) is surjective, \(RST\) is invertible (by 3.65). Writing \(RST = R(ST)\), Exercise 11 gives that \(R\) and \(ST\) are invertible; applying Exercise 11 to the invertible \(ST\) gives that \(S\) (and \(T\)) is invertible. In particular \(S\) is injective (by 3.63).

\(T\) is injective, and injectivity converts the spanning of \(Tv_1, \dots, Tv_m\) into that of \(v_1, \dots, v_m\). Indeed \(V\) is finite-dimensional, being spanned by the finite list \(Tv_1, \dots, Tv_m\), and each \(Tv_k\) lies in the subspace \(\operatorname{range} T\) (by 3.18), so

\begin{equation*} V = \operatorname{span}(Tv_1, \ldots, Tv_m) \subseteq \operatorname{range} T \subseteq V ; \end{equation*}

thus \(T\) is surjective and hence injective (by 3.65, \(V\) finite-dimensional).

Now let \(v \in V\). Because \(Tv_1, \ldots, Tv_m\) spans \(V\), there are \(a_1, \ldots, a_m \in \mathbf{F}\) with

\begin{equation*} Tv = a_1 Tv_1 + \cdots + a_m Tv_m = T(a_1 v_1 + \cdots + a_m v_m), \end{equation*}

so injectivity gives \(v = a_1 v_1 + \cdots + a_m v_m\). Hence \(v_1, \ldots, v_m\) spans \(V\).

Take \(A\) to be the \(m\)-by-\(n\) matrix whose \(k^{\text{th}}\) column is \(Te_k\), where \(e_1, \ldots, e_n\) is the standard basis of \(\mathbf{F}^{n,1}\) (so \(e_k\) has \(k^{\text{th}}\) entry \(1\) and other entries \(0\)); that is, \(A_{j,k} = (Te_k)_j\). For \(x \in \mathbf{F}^{n,1}\) with entries \(x_1, \ldots, x_n\) we have \(x = x_1e_1 + \cdots + x_ne_n\), so linearity of \(T\) gives

\begin{equation*} \begin{aligned} Tx &= x_1 Te_1 + \cdots + x_n Te_n \\ &= x_1 A_{\cdot,1} + \cdots + x_n A_{\cdot,n} = Ax, \end{aligned} \end{equation*}

the last equality being 3.50 (\(Ax\) is that linear combination of the columns of \(A\)).

Write \(n = \dim V\) and \(U = \operatorname{null} S\), a finite-dimensional subspace of \(V\) (by 2.25); by 3.72, \(\dim \mathcal{L}(V) = n^2\).

(a) The null space of \(\mathcal{A}\) is the set of \(T\) with range inside \(U\):

\begin{equation*} \operatorname{null} \mathcal{A} = \{T \in \mathcal{L}(V) : \operatorname{range} T \subseteq U\}, \end{equation*}

since \(ST = 0\) says \(Tv \in \operatorname{null} S\) for every \(v \in V\). Enlarging the target space from \(U\) to \(V\) gives a map \(\Phi : \mathcal{L}(V, U) \to \mathcal{L}(V)\), \(\bigl(\Phi( R)\bigr)v = Rv\), which is linear and injective (if \(\Phi( R) = 0\) then \(Rv = 0\) for all \(v\)), with range exactly \(\operatorname{null}\mathcal{A}\): each \(\Phi( R)\) has range in \(U\), and conversely any such \(T\) is \(\Phi( R)\) for \(R \in \mathcal{L}(V,U)\) defined by \(Rv = Tv\). So \(\Phi\) is an isomorphism onto \(\operatorname{null}\mathcal{A}\), and 3.72 (both \(V\) and \(U\) finite-dimensional) gives

\begin{equation*} \dim \operatorname{null} \mathcal{A} = \dim \mathcal{L}(V, U) = (\dim V)(\dim \operatorname{null} S). \end{equation*}

(b) Apply 3.21 to \(\mathcal{A}\) on the finite-dimensional space \(\mathcal{L}(V)\), then to \(S\) on \(V\):

\begin{equation*} \begin{aligned} \dim \operatorname{range} \mathcal{A} &= n^2 - n\,(\dim \operatorname{null} S) = n\bigl(n - \dim \operatorname{null} S\bigr) \\ &= (\dim V)(\dim \operatorname{range} S). \end{aligned} \end{equation*}

Evaluation at \(1\) is an isomorphism: define \(\Phi : \mathcal{L}(\mathbf{F}, V) \to V\) by \(\Phi(T) = T1\). Linearity of \(\Phi\) is immediate from the definitions of addition and scalar multiplication in \(\mathcal{L}(\mathbf{F},V)\) (Check!).

\(\Phi\) is injective: if \(T1 = 0\) then homogeneity of \(T\) gives

\begin{equation*} T\lambda = T(\lambda \cdot 1) = \lambda \, T1 = 0 \qquad \text{for all } \lambda \in \mathbf{F}, \end{equation*}

so \(T = 0\) and \(\operatorname{null}\Phi = \{0\}\) (by 3.15).

\(\Phi\) is surjective: given \(v \in V\), the map \(T_v \lambda = \lambda v\) is linear (Check!) and \(\Phi(T_v) = 1 \cdot v = v\).

Hence \(\Phi\) is invertible (by 3.63), so \(\mathcal{L}(\mathbf{F}, V)\) and \(V\) are isomorphic (3.69).

(\(\Longleftarrow\)) If \(T = \lambda I\) then \(Tv_k = \lambda v_k\) for every basis \(v_1, \ldots, v_n\), so \(\mathcal{M}(T)\) is the diagonal matrix \(\lambda I\) no matter which basis is used.

(\(\Longrightarrow\)) Suppose \(\mathcal{M}(T, (w_1, \ldots, w_n)) = A\) for every basis \(w_1, \ldots, w_n\), and let \(n = \dim V \geq 1\) (if \(n = 0\) then \(T = 0 = 0I\)). Fix a basis \(v_1, \ldots, v_n\).

(i) \(A\) is diagonal. Fix \(k\) and set \(u_k = 2v_k\) and \(u_i = v_i\) for \(i \neq k\); this is again a basis (Check!; here \(2 \neq 0\) in \(\mathbf{F} = \mathbf{R}\) or \(\mathbf{C}\)). Column \(k\) of \(A\) read in each of the two bases gives

\begin{equation*} Tv_k = \sum_{j} A_{j,k}v_j \quad\text{and}\quad 2\,Tv_k = Tu_k = 2A_{k,k}v_k + \sum_{j \neq k} A_{j,k}v_j . \end{equation*}

Twice the first minus the second yields \(\sum_{j \neq k} A_{j,k}v_j = 0\), so \(A_{j,k} = 0\) for \(j \neq k\) by linear independence. Write \(\lambda_k = A_{k,k}\); then \(Tw_k = \lambda_k w_k\) for every basis \(w_1, \ldots, w_n\) and every \(k\).

(ii) The \(\lambda_k\) agree. For \(k \geq 2\), swapping \(v_1\) and \(v_k\) gives a basis with first vector \(v_k\), so \(Tv_k = \lambda_1 v_k\); comparing with \(Tv_k = \lambda_k v_k\) and \(v_k \neq 0\) gives \(\lambda_k = \lambda_1 =: \lambda\).

Thus \(T\) and \(\lambda I\) agree on the basis \(v_1, \ldots, v_n\), hence \(T = \lambda I\) (by 3.4).

Fix \(m\) with \(q \in \mathcal{P}_m(\mathbf{R})\) (take \(m = \deg q\), or \(m = 0\) if \(q = 0\)) and define \(T : \mathcal{P}_m(\mathbf{R}) \to \mathcal{P}_m(\mathbf{R})\) by

\begin{equation*} (Tp)(x) = (x^2 + x) p^{\prime\prime}(x) + 2x \, p^{\prime}(x) + p(3). \end{equation*}

This does land in \(\mathcal{P}_m(\mathbf{R})\), since \(\deg p \leq m\) gives \(\deg\bigl((x^2+x)p^{\prime\prime}\bigr) \leq m\) and \(\deg(2x p^{\prime}) \leq m\) while \(p(3)\) is constant; and \(T\) is linear because differentiation, evaluation at \(3\), and multiplication by a fixed polynomial are all linear (Check!).

\(T\) is injective: let \(p \neq 0\) have degree \(d\) and leading coefficient \(a_d \neq 0\).

(i) \(d = 0\): then \(p^{\prime} = p^{\prime\prime} = 0\), so \(Tp = p(3) = a_0 \neq 0\).

(ii) \(d \geq 1\): as \(p(3)\) contributes only to the constant term, the coefficient of \(x^d\) in \(Tp\) is

\begin{equation*} d(d-1) a_d + 2d \, a_d = a_d \, d(d+1) \neq 0, \end{equation*}

so \(Tp \neq 0\). (For \(d = 1\) the first term is \(0\), matching \(p^{\prime\prime} = 0\); for \(d \geq 2\) it is the \(x^2 \cdot d(d-1)a_d x^{d-2}\) term.)

Hence \(\operatorname{null} T = \{0\}\), so \(T\) is injective (by 3.15) and therefore surjective, being an operator on the finite-dimensional \(\mathcal{P}_m(\mathbf{R})\) (by 3.65). Thus some \(p \in \mathcal{P}_m(\mathbf{R}) \subseteq \mathcal{P}(\mathbf{R})\) satisfies \(Tp = q\), which is the asserted identity.

Both statements are about the operator \(T \in \mathcal{L}(\mathbf{F}^n)\) defined by

\begin{equation*} T(x_1, \ldots, x_n) = \Big( \sum_{k=1}^{n} A_{1,k} x_k , \ \ldots , \ \sum_{k=1}^{n} A_{n,k} x_k \Big), \end{equation*}

which is linear because each coordinate is (Check!).

(i) Since \(x_1, \ldots, x_n\) solves the homogeneous system exactly when \((x_1, \ldots, x_n) \in \operatorname{null} T\), statement (a) says \(\operatorname{null} T = \{0\}\), that is, \(T\) is injective (by 3.15).

(ii) Since the system with right side \(c_1, \ldots, c_n\) is solvable exactly when \((c_1, \ldots, c_n) \in \operatorname{range} T\), statement (b) says \(\operatorname{range} T = \mathbf{F}^n\), that is, \(T\) is surjective (3.19).

Because the equations and the variables are equal in number, \(T\) is an operator on the finite-dimensional space \(\mathbf{F}^n\), so 3.65 applies and gives that \(T\) is injective if and only if \(T\) is surjective. Hence (a) holds if and only if (b) holds.

Every matrix below is taken with respect to \(v_1, \dots, v_n\) in both roles, so \(\mathcal{M}(I) = I\) (each \(Iv_k = v_k\), by 3.79) and \(\mathcal{M}(SR) = \mathcal{M}(S)\mathcal{M}( R)\) for \(S, R \in \mathcal{L}(V)\) (by 3.81, with this one basis in all three roles). Write \(A = \mathcal{M}(T)\); note \(\dim V = n\).

(\(\Longrightarrow\)) If \(T\) is invertible then \(T^{-1} \in \mathcal{L}(V)\) and

\begin{equation*} \begin{aligned} A\,\mathcal{M}(T^{-1}) &= \mathcal{M}(T T^{-1}) = \mathcal{M}(I) = I, \\ \mathcal{M}(T^{-1})\,A &= \mathcal{M}(T^{-1} T) = \mathcal{M}(I) = I, \end{aligned} \end{equation*}

so \(A\) is invertible with \(A^{-1} = \mathcal{M}(T^{-1})\) (3.80).

(\(\Longleftarrow\)) If \(A\) is invertible, then because \(\mathcal{M} \colon \mathcal{L}(V) \to \mathbf{F}^{n,n}\) is an isomorphism (by 3.71, with \(v_1, \dots, v_n\) as basis of domain and target) there is \(S \in \mathcal{L}(V)\) with \(\mathcal{M}(S) = A^{-1}\), whence

\begin{equation*} \mathcal{M}(ST) = \mathcal{M}(S)\,\mathcal{M}(T) = A^{-1}A = I = \mathcal{M}(I). \end{equation*}

Injectivity of \(\mathcal{M}\) gives \(ST = I\), and then \(TS = I\) by 3.68 (\(V\) finite-dimensional). Hence \(T\) is invertible.

The two matrices have the same columns, because \(Tv_k = u_k\). Write \(A = \mathcal{M}(T, (v_1, \dots, v_n))\) and \(C = \mathcal{M}(I, (u_1, \dots, u_n), (v_1, \dots, v_n))\); by the definition 3.31 of the matrix of a linear map, their entries are the scalars with

\begin{equation*} \begin{aligned} T v_k &= A_{1,k} v_1 + \cdots + A_{n,k} v_n, \\ u_k = I u_k &= C_{1,k} v_1 + \cdots + C_{n,k} v_n, \end{aligned} \end{equation*}

for each \(k = 1, \dots, n\). Fix \(k\); since \(Tv_k = u_k\) and a vector has only one representation as a linear combination of the basis \(v_1, \dots, v_n\) (by 2.28), we get \(A_{j,k} = C_{j,k}\) for every \(j\). Hence \(A = C\).

Transfer to operators on \(\mathbf{F}^n\), where 3.68 applies. Let \(n\) be the common size of \(A\) and \(B\), and take every matrix with respect to the standard basis \(e_1, \dots, e_n\) of \(\mathbf{F}^n\) in both roles, so that \(\mathcal{M}(I) = I\) (each \(Ie_k = e_k\), by 3.79) and \(\mathcal{M}(RQ) = \mathcal{M}( R)\mathcal{M}(Q)\) for operators \(R, Q\) (by 3.81, with this one basis in all three roles). Since \(\mathcal{M} \colon \mathcal{L}(\mathbf{F}^n) \to \mathbf{F}^{n,n}\) is an isomorphism (by 3.71, with \(e_1, \dots, e_n\) as basis of domain and target), choose \(S, T \in \mathcal{L}(\mathbf{F}^n)\) with \(\mathcal{M}(S) = A\) and \(\mathcal{M}(T) = B\). Then

\begin{equation*} \mathcal{M}(ST) = \mathcal{M}(S)\,\mathcal{M}(T) = AB = I = \mathcal{M}(I), \end{equation*}

so \(ST = I\) by injectivity of \(\mathcal{M}\). As \(\mathbf{F}^n\) is finite-dimensional, 3.68 gives \(TS = I\), and hence

\begin{equation*} BA = \mathcal{M}(T)\,\mathcal{M}(S) = \mathcal{M}(TS) = \mathcal{M}(I) = I. \end{equation*}

Write \(G = \{(v, Tv) : v \in V\}\) for the graph. Because \(T\) is a function, for \(v \in V\) and \(w \in W\)

\begin{equation*} (v, w) \in G \iff w = Tv \end{equation*}

(an element \((u, Tu)\) of \(G\) equals \((v,w)\) only if \(u = v\), forcing \(w = Tv\)); both directions below use this. By 1.34, \(G\) is a subspace exactly when \((0,0) \in G\) and \(G\) is closed under addition and scalar multiplication.

(\(\Longrightarrow\)) If \(T\) is linear then \(T0 = 0\) (by 3.10), so \((0,0) = (0, T0) \in G\), and additivity and homogeneity of \(T\) give

\begin{equation*} \begin{aligned} (u, Tu) + (v, Tv) &= (u + v, Tu + Tv) = \bigl(u+v, T(u+v)\bigr) \in G, \\ \lambda (v, Tv) &= (\lambda v, \lambda Tv) = \bigl(\lambda v, T(\lambda v)\bigr) \in G. \end{aligned} \end{equation*}

Hence \(G\) is a subspace of \(V \times W\).

(\(\Longleftarrow\)) If \(G\) is a subspace then for \(u, v \in V\) and \(\lambda \in \mathbf{F}\), closure under the two operations gives

\begin{equation*} (u + v,\; Tu + Tv) \in G \qquad\text{and}\qquad (\lambda v,\; \lambda Tv) \in G, \end{equation*}

which by the equivalence above say \(Tu + Tv = T(u+v)\) and \(\lambda Tv = T(\lambda v)\). So \(T\) is linear (3.1).

Apply 3.21 to the \(k\)-th coordinate projection

\begin{equation*} P_k : V_1 \times \cdots \times V_m \to V_k, \qquad P_k(v_1, \ldots, v_m) = v_k , \end{equation*}

which is linear because the operations on the product are slotwise (3.87) (Check!) and surjective because \(P_k(0, \ldots, 0, v, 0, \ldots, 0) = v\) for each \(v \in V_k\) (with \(v\) in slot \(k\)). Its domain is finite-dimensional by hypothesis, so 3.21 gives that \(\operatorname{range} P_k = V_k\) is finite-dimensional. As \(k\) was arbitrary, each \(V_k\) is finite-dimensional.

Restriction to the slots is an isomorphism. Write \(\mathcal{V} = V_1 \times \cdots \times V_m\) and let \(\iota_k : V_k \to \mathcal{V}\) be the \(k\)-th inclusion \(\iota_k(v) = (0, \ldots, 0, v, 0, \ldots, 0)\), which is linear because the operations on \(\mathcal{V}\) are slotwise (3.87) and satisfies

\begin{equation*} (v_1, \ldots, v_m) = \iota_1(v_1) + \cdots + \iota_m(v_m). \end{equation*}

Define \(\Phi(T) = (T \circ \iota_1, \ldots, T \circ \iota_m)\) for \(T \in \mathcal{L}(\mathcal{V}, W)\); each \(T \circ \iota_k\) lies in \(\mathcal{L}(V_k, W)\) by 3.7, so \(\Phi\) maps \(\mathcal{L}(\mathcal{V}, W)\) into \(\mathcal{L}(V_1, W) \times \cdots \times \mathcal{L}(V_m, W)\), and \(\Phi\) is linear because the operations on both sides are computed pointwise and slotwise (3.5, 3.87) (Check!).

\(\Phi\) is injective: if \(T \circ \iota_k = 0\) for every \(k\), then the display above and linearity of \(T\) give

\begin{equation*} T(v_1, \ldots, v_m) = (T \circ \iota_1)(v_1) + \cdots + (T \circ \iota_m)(v_m) = 0, \end{equation*}

so \(T = 0\) and \(\operatorname{null}\Phi = \{0\}\) (by 3.15).

\(\Phi\) is surjective: given \((T_1, \ldots, T_m)\), the map \(T(v_1, \ldots, v_m) = T_1v_1 + \cdots + T_mv_m\) is linear by additivity and homogeneity of each \(T_k\) together with slotwise operations (Check!), and

\begin{equation*} (T \circ \iota_k)(v) = T_1 0 + \cdots + T_k v + \cdots + T_m 0 = T_k v \end{equation*}

since \(T_j0 = 0\) (3.10); thus \(\Phi(T) = (T_1, \ldots, T_m)\).

Hence \(\Phi\) is invertible (by 3.63), so the two spaces are isomorphic (3.69).

Recording the component maps is an isomorphism. Write \(\mathcal{W} = W_1 \times \cdots \times W_m\) and let \(P_k : \mathcal{W} \to W_k\) be the \(k\)-th coordinate projection, linear as in Exercise 2 of this section, so that

\begin{equation*} w = \big(P_1 w, \ldots, P_m w\big) \qquad \text{for all } w \in \mathcal{W}. \end{equation*}

Define \(\Psi(T) = (P_1 \circ T, \ldots, P_m \circ T)\) for \(T \in \mathcal{L}(V, \mathcal{W})\); each \(P_k \circ T\) lies in \(\mathcal{L}(V, W_k)\) by 3.7, so \(\Psi\) maps \(\mathcal{L}(V, \mathcal{W})\) into \(\mathcal{L}(V, W_1) \times \cdots \times \mathcal{L}(V, W_m)\), and \(\Psi\) is linear because the operations on both sides are pointwise and slotwise (3.5, 3.87) and each \(P_k\) is linear (Check!).

\(\Psi\) is injective: if \(P_k(Tv) = 0\) for every \(k\) and every \(v \in V\), then the display above applied to \(w = Tv\) gives

\begin{equation*} Tv = \big(P_1(Tv), \ldots, P_m(Tv)\big) = (0, \ldots, 0), \end{equation*}

so \(T = 0\) and \(\operatorname{null}\Psi = \{0\}\) (by 3.15).

\(\Psi\) is surjective: given \((T_1, \ldots, T_m)\), the map \(Tv = (T_1v, \ldots, T_mv)\) is linear by additivity and homogeneity of each \(T_k\) together with slotwise operations in \(\mathcal{W}\) (Check!), and \((P_k \circ T)v = T_kv\) for all \(k\) and \(v\), so \(\Psi(T) = (T_1, \ldots, T_m)\).

Hence \(\Psi\) is invertible (by 3.63), so the two spaces are isomorphic (3.69).

Evaluation on the standard basis is an isomorphism: with \(e_1, \ldots, e_m\) the standard basis of \(\mathbf{F}^m\), define

\begin{equation*} \Phi : \mathcal{L}(\mathbf{F}^m, V) \to V^m, \qquad \Phi(T) = (Te_1, \ldots, Te_m), \end{equation*}

linear because the operations on \(\mathcal{L}(\mathbf{F}^m, V)\) are pointwise (3.5) and those on \(V^m\) slotwise (3.87) (Check!).

\(\Phi\) is injective: if \(Te_k = 0\) for each \(k\), then for every \((x_1, \ldots, x_m) \in \mathbf{F}^m\),

\begin{equation*} T(x_1, \ldots, x_m) = x_1 Te_1 + \cdots + x_m Te_m = 0, \end{equation*}

so \(T = 0\) and \(\operatorname{null}\Phi = \{0\}\) (by 3.15).

\(\Phi\) is surjective: given \((v_1, \ldots, v_m) \in V^m\), the linear map lemma 3.4 applied to the basis \(e_1, \ldots, e_m\) provides \(T \in \mathcal{L}(\mathbf{F}^m, V)\) with \(Te_k = v_k\) for each \(k\), so \(\Phi(T) = (v_1, \ldots, v_m)\).

Hence \(\Phi\) is invertible (by 3.63), so \(\mathcal{L}(\mathbf{F}^m, V)\) and \(V^m\) are isomorphic (3.69).

We prove \(U \subseteq W\); symmetry then finishes. Since \(0 \in U\), we have \(v \in v + U = x + W\), so \(v = x + w_0\) for some \(w_0 \in W\). Given \(u \in U\), also \(v + u \in v + U = x + W\), say \(v + u = x + w\) with \(w \in W\), whence

\begin{equation*} u = (x + w) - v = (x + w) - (x + w_0) = w - w_0 \in W, \end{equation*}

as \(W\) is closed under additive inverses and addition. Thus \(U \subseteq W\). Interchanging the pairs \((v, U)\) and \((x, W)\), which the symmetric hypothesis \(v + U = x + W\) permits, gives \(W \subseteq U\). Hence \(U = W\).

The translates of \(U\) are exactly the level sets of \(\varphi(x,y,z) = 2x + 3y + 5z\). This \(\varphi : \mathbf{R}^3 \to \mathbf{R}\) is linear (Check!), so \(U = \operatorname{null}\varphi\) is a subspace of \(\mathbf{R}^3\) (by 3.13); write \(A_c = \{p \in \mathbf{R}^3 : \varphi(p) = c\}\), which is the set displayed in the exercise. Since \(p \in v + U\) means \(p - v \in U = \operatorname{null}\varphi\), linearity of \(\varphi\) gives, for \(v, p \in \mathbf{R}^3\),

\begin{equation*} p \in v + U \iff \varphi(p - v) = 0 \iff \varphi(p) = \varphi(v), \end{equation*}

that is,

\begin{equation*} v + U = A_{\varphi(v)} \qquad \text{for every } v \in \mathbf{R}^3 . \end{equation*}

(\(\Longrightarrow\)) If \(A\) is a translate of \(U\), say \(A = v + U\) (3.97), then \(A = A_c\) with \(c = \varphi(v)\).

(\(\Longleftarrow\)) If \(A = A_c\), then \(v = \left(\tfrac{c}{2}, 0, 0\right)\) has \(\varphi(v) = c\), so \(v + U = A_c = A\) and \(A\) is a translate of \(U\).

(a) If \(A = \{x \in V : Tx = c\}\) is nonempty, fix \(x_0 \in A\); then \(A = x_0 + \operatorname{null} T\), since linearity of \(T\) and \(Tx_0 = c\) give, for \(x \in V\),

\begin{equation*} \begin{aligned} x \in A &\iff T(x - x_0) = Tx - c = 0 \iff x - x_0 \in \operatorname{null} T \\ &\iff x \in x_0 + \operatorname{null} T . \end{aligned} \end{equation*}

As \(\operatorname{null} T\) is a subspace of \(V\) (by 3.13), \(A\) is then a translate of it (3.97).

(b) The system 3.27 says \(Tx = c\), where \(c = (c_1, \ldots, c_m)\) and \(T \colon \mathbf{F}^n \to \mathbf{F}^m\) is the linear map of 3.25,

\begin{equation*} T(x_1, \ldots, x_n) = \Big( \sum_{k=1}^{n} A_{1,k} x_k, \; \ldots, \; \sum_{k=1}^{n} A_{m,k} x_k \Big) . \end{equation*}

So by (a) its solution set is either empty or a translate of \(\operatorname{null} T\), a subspace of \(\mathbf{F}^n\) (by 3.13).

(\(\Longrightarrow\)) If \(A = x + U\) with \(U\) a subspace, write \(v = x + u_1\) and \(w = x + u_2\) with \(u_1, u_2 \in U\); then for \(\lambda \in \mathbf{F}\),

\begin{equation*} \begin{aligned} \lambda v + (1 - \lambda) w &= \lambda (x + u_1) + (1 - \lambda)(x + u_2) \\ &= x + \big(\lambda u_1 + (1 - \lambda) u_2\big) \in x + U = A, \end{aligned} \end{equation*}

since \(U\) is closed under scalar multiplication and addition.

(\(\Longleftarrow\)) Assume \((\ast)\): \(\lambda v + (1 - \lambda)w \in A\) for all \(v, w \in A\) and all \(\lambda \in \mathbf{F}\). Fix \(x \in A\) (possible as \(A \neq \varnothing\)) and set \(U = A - x = \{a - x : a \in A\}\), so that \(A = x + U\); we check the three conditions of 1.34 for \(U\).

(i) Additive identity: \(0 = x - x \in U\), since \(x \in A\).

(ii) Scalar multiples: for \(u = a - x \in U\) and \(\lambda \in \mathbf{F}\), condition \((\ast)\) with \(v = a\), \(w = x\) gives

\begin{equation*} x + \lambda u = \lambda a + (1 - \lambda) x \in A, \qquad\text{so}\qquad \lambda u \in A - x = U . \end{equation*}

(iii) Sums: for \(u_1 = a - x\) and \(u_2 = b - x\) in \(U\), condition \((\ast)\) with \(\lambda = \tfrac{1}{2}\) (a scalar, as \(\mathbf{F}\) is \(\mathbf{R}\) or \(\mathbf{C}\)) gives \(\tfrac{1}{2}a + \tfrac{1}{2}b \in A\), hence

\begin{equation*} \tfrac{1}{2} u_1 + \tfrac{1}{2} u_2 = \big(\tfrac{1}{2} a + \tfrac{1}{2} b\big) - x \in U, \end{equation*}

and applying (ii) with the scalar \(2\) yields \(u_1 + u_2 \in U\).

Thus \(U\) is a subspace and \(A = x + U\) is a translate of it.

If \(A_1 \cap A_2 \neq \varnothing\), fix \(x \in A_1 \cap A_2\); then \(A_1 \cap A_2 = x + (U_1 \cap U_2)\), a translate of the subspace \(U_1 \cap U_2\) of \(V\) (Exercise 1C.10). Indeed \(x - v \in U_1\) and \(x - w \in U_2\), so 3.101 gives \(A_1 = x + U_1\) and \(A_2 = x + U_2\), whence for \(y \in V\),

\begin{equation*} \begin{aligned} y \in A_1 \cap A_2 &\iff y - x \in U_1 \ \text{ and } \ y - x \in U_2 \\ &\iff y - x \in U_1 \cap U_2 \iff y \in x + (U_1 \cap U_2). \end{aligned} \end{equation*}

Otherwise \(A_1 \cap A_2\) is the empty set.

Write \(\operatorname{supp} x = \{k \in \mathbf{Z}^+ : x_k \neq 0\}\), so that \(U\) consists of the \(x \in \mathbf{F}^\infty\) with \(\operatorname{supp} x\) finite.

(a) The three conditions of 1.34 follow from

\begin{equation*} \operatorname{supp} 0 = \varnothing, \qquad \operatorname{supp}(\lambda x) \subseteq \operatorname{supp} x, \end{equation*}

\begin{equation*} \operatorname{supp}(x + y) \subseteq \operatorname{supp} x \cup \operatorname{supp} y, \end{equation*}

together with the finiteness of a union of two finite sets. Hence \(U\) is a subspace of \(\mathbf{F}^\infty\).

(b) For each \(j \in \mathbf{Z}^+\) let

\begin{equation*} S_j = \{ 2^{\,j-1}(2k - 1) : k \in \mathbf{Z}^+ \}, \end{equation*}

and let \(v_j \in \mathbf{F}^\infty\) have \(k\)-th coordinate \(1\) for \(k \in S_j\) and \(0\) otherwise. Since every \(m \in \mathbf{Z}^+\) is \(2^a q\) with \(q\) odd in exactly one way, \(m\) lies in \(S_{a+1}\) alone; so the \(S_j\) are pairwise disjoint, and each is infinite.

Fix \(n \in \mathbf{Z}^+\); then \(v_1 + U, \ldots, v_n + U\) is linearly independent in \(\mathbf{F}^\infty/U\). For if \(a_1(v_1 + U) + \cdots + a_n(v_n + U) = 0 + U\), then by the definition 3.102 of the quotient operations and by 3.101,

\begin{equation*} x = a_1 v_1 + \cdots + a_n v_n \in U, \end{equation*}

while disjointness of \(S_1, \ldots, S_n\) makes exactly one of \((v_1)_k, \ldots, (v_n)_k\) equal to \(1\) for \(k \in S_j\), so \(x_k = a_j\) there. Were some \(a_j \neq 0\), the infinite set \(S_j\) would lie in \(\operatorname{supp} x\), contradicting \(x \in U\); hence \(a_1 = \cdots = a_n = 0\).

So \(\mathbf{F}^\infty/U\) has linearly independent lists of every length, which a finite-dimensional space cannot (by 2.22). Hence \(\mathbf{F}^\infty/U\) is infinite-dimensional.

\(A = v_1 + U\), where \(U = \operatorname{span}(v_2 - v_1, \ldots, v_m - v_1)\) is a subspace of \(V\) by 2.6; this single identity gives (a), and gives (c) because \(U\) is spanned by a list of length \(m - 1\), which contains a basis of \(U\) by 2.30, so \(\dim U \leq m - 1 < m\).

(a) and (c). If \(a = \lambda_1 v_1 + \cdots + \lambda_m v_m\) with \(\sum_{j} \lambda_j = 1\), then

\begin{equation*} \begin{aligned} a - v_1 &= \sum_{j=1}^{m} \lambda_j v_j - \Big( \sum_{j=1}^{m} \lambda_j \Big) v_1 \\ &= \sum_{j=1}^{m} \lambda_j (v_j - v_1) = \sum_{j=2}^{m} \lambda_j (v_j - v_1) \in U , \end{aligned} \end{equation*}

the last step because the \(j = 1\) term vanishes; hence \(A \subseteq v_1 + U\). Conversely, if \(u = c_2 (v_2 - v_1) + \cdots + c_m (v_m - v_1)\), then

\begin{equation*} v_1 + u = \Big( 1 - \sum_{j=2}^{m} c_j \Big) v_1 + c_2 v_2 + \cdots + c_m v_m , \end{equation*}

whose coefficients sum to \(1\), so \(v_1 + u \in A\). Thus \(A = v_1 + U\).

(b) Write \(B = x + W\) with \(x \in V\) and \(W\) a subspace of \(V\); since each \(v_j \in B\), we have \(v_j - x \in W\). For \(a = \lambda_1 v_1 + \cdots + \lambda_m v_m \in A\) with \(\sum_j \lambda_j = 1\),

\begin{equation*} a - x = \sum_{j=1}^{m} \lambda_j v_j - \Big( \sum_{j=1}^{m} \lambda_j \Big) x = \sum_{j=1}^{m} \lambda_j (v_j - x) \in W , \end{equation*}

so \(a \in x + W = B\). Hence \(A \subseteq B\).

The map \(S(u, x) = u + \Psi x\) is an isomorphism from \(U \times (V/U)\) onto \(V\), where \(\Psi\) is defined as follows. Let \(v_1 + U, \ldots, v_n + U\) be a basis of \(V/U\) (its vectors have this form by 3.99), and let \(\Psi \in \mathcal{L}(V/U, V)\) be the linear map with \(\Psi(v_j + U) = v_j\) for each \(j\), which exists by the linear map lemma 3.4 applied to that basis.

Since the operations on \(U \times (V/U)\) are coordinatewise (3.87) and \(\Psi\) is linear, \(S\) is linear (Check!).

\(S\) is surjective: given \(v \in V\), write \(v + U = a_1(v_1 + U) + \cdots + a_n(v_n + U)\), which equals \((a_1 v_1 + \cdots + a_n v_n) + U\) by 3.102, so \(u := v - (a_1 v_1 + \cdots + a_n v_n) \in U\) by 3.101; with \(x = v + U\), linearity of \(\Psi\) gives \(\Psi x = a_1 v_1 + \cdots + a_n v_n\) and hence

\begin{equation*} S(u, x) = u + \Psi x = v . \end{equation*}

\(S\) is injective: if \(u + \Psi x = 0\), write \(x = a_1(v_1 + U) + \cdots + a_n(v_n + U)\), so \(\Psi x = a_1 v_1 + \cdots + a_n v_n = -u \in U\); then 3.101 and 3.102 give

\begin{equation*} x = (a_1 v_1 + \cdots + a_n v_n) + U = 0 + U , \end{equation*}

whence \(\Psi x = 0\) and \(u = 0\). So \(\operatorname{null} S = \{0\}\) and \(S\) is injective by 3.15.

Being linear, injective, and surjective, \(S\) is invertible by 3.63, hence an isomorphism by 3.69.

The list spans \(V/U\) and is linearly independent, hence is a basis.

Spanning. Every element of \(V/U\) is \(v + U\) for some \(v \in V\) (3.99); as \(V = U + W\), write \(v = u + w\) with \(u \in U\) and \(w = a_1 w_1 + \cdots + a_m w_m \in W\). Then \(v - (a_1 w_1 + \cdots + a_m w_m) = u \in U\), so 3.101 and 3.102 give

\begin{equation*} v + U = (a_1 w_1 + \cdots + a_m w_m) + U = a_1 (w_1 + U) + \cdots + a_m (w_m + U). \end{equation*}

Linear independence. Suppose \(a_1 (w_1 + U) + \cdots + a_m (w_m + U) = 0 + U\). By 3.102 the left side is \((a_1 w_1 + \cdots + a_m w_m) + U\), so 3.101 gives

\begin{equation*} a_1 w_1 + \cdots + a_m w_m \in U \cap W = \{0\} , \end{equation*}

the last equality by 1.46 since \(V = U \oplus W\) (the vector lies in \(W\) because \(W\) is a subspace containing each \(w_k\)). Linear independence of the basis \(w_1, \ldots, w_m\) of \(W\) now forces \(a_1 = \cdots = a_m = 0\).

The list spans \(V\) and is linearly independent, hence is a basis of \(V\) (2.26). Let \(\pi \colon V \to V/U\) be the quotient map \(\pi(v) = v + U\) (3.104); by 3.101, \(\operatorname{null} \pi = U\).

Spanning. Given \(v \in V\), write \(v + U = a_1(v_1 + U) + \dots + a_m(v_m + U)\), which by 3.102 equals \((a_1 v_1 + \dots + a_m v_m) + U\), so 3.101 gives \(v - (a_1 v_1 + \dots + a_m v_m) \in U\); expanding that vector in the basis \(u_1, \dots, u_n\) of \(U\) yields

\begin{equation*} v = a_1 v_1 + \dots + a_m v_m + b_1 u_1 + \dots + b_n u_n . \end{equation*}

Linear independence. Suppose \(a_1 v_1 + \dots + a_m v_m + b_1 u_1 + \dots + b_n u_n = 0\). Applying \(\pi\) and using \(u_j \in \operatorname{null} \pi\) gives

\begin{equation*} a_1(v_1 + U) + \dots + a_m(v_m + U) = 0 + U , \end{equation*}

so \(a_1 = \dots = a_m = 0\) by linear independence of the basis \(v_1 + U, \dots, v_m + U\). The displayed equation then reduces to \(b_1 u_1 + \dots + b_n u_n = 0\), and linear independence of \(u_1, \dots, u_n\) gives \(b_1 = \dots = b_n = 0\).

Pick \(v \in V\) with \(\varphi(v) \neq 0\), which exists because \(\varphi \neq 0\); then the one-vector list \(v + \operatorname{null}\varphi\) is a basis of \(V/(\operatorname{null}\varphi)\), so that space has dimension \(1\). (Note \(V\) is not assumed finite-dimensional, so 3.105 is unavailable.)

It spans: for every \(x \in V\) we have \(\varphi\bigl(x - \frac{\varphi(x)}{\varphi(v)}v\bigr) = \varphi(x) - \varphi(x) = 0\), so by 3.101,

\begin{equation*} x + \operatorname{null} \varphi = \frac{\varphi(x)}{\varphi(v)} \, \bigl( v + \operatorname{null} \varphi \bigr). \end{equation*}

It is linearly independent: \(\lambda(v + \operatorname{null}\varphi) = 0 + \operatorname{null}\varphi\) gives \(\lambda v \in \operatorname{null}\varphi\), so \(\lambda \varphi(v) = 0\) and hence \(\lambda = 0\).

Method (2): since \(\varphi\bigl(\frac{\lambda}{\varphi(v)}v\bigr) = \lambda\) for every \(\lambda \in \mathbf{F}\), we have \(\operatorname{range}\varphi = \mathbf{F}\), so 3.107(d) makes \(\widetilde{\varphi}\) of 3.106 an isomorphism from \(V/(\operatorname{null}\varphi)\) onto \(\mathbf{F}\). Its inverse \(\widetilde{\varphi}^{\,-1}\) is an isomorphism too (3.69, 3.63), and an isomorphism carries a basis to a basis; transporting the basis \(1\) of \(\mathbf{F}\) (the direction that matters, since a basis of the quotient is what we lack) gives the one-vector basis \(\widetilde{\varphi}^{\,-1}(1)\).

Take \(\varphi = \alpha \circ \pi\), where \(\pi \colon V \to V/U\) is the quotient map \(\pi(v) = v + U\) (3.104) and \(\alpha \in (V/U)^{\prime}\) is the linear functional determined by \(\alpha(v + U) = 1\) for a one-vector basis \(v + U\) of \(V/U\) (such a basis exists because \(\dim V/U = 1\), and \(\alpha\) exists by the linear map lemma 3.4). Then \(\varphi \in \mathcal{L}(V, \mathbf{F})\), since a composition of linear maps is linear (3.7).

By 3.101, \(\operatorname{null} \pi = U\). Also \(\alpha\) is injective: \(\alpha\bigl(\lambda(v+U)\bigr) = \lambda\), so \(\operatorname{null}\alpha = \{0\}\) and 3.15 applies. Hence

\begin{equation*} \operatorname{null} \varphi = \{ x \in V : \alpha(\pi x) = 0 \} = \{ x \in V : \pi x = 0 \} = \operatorname{null} \pi = U . \end{equation*}

Throughout, \(\pi \colon V \to V/U\) is the quotient map \(\pi(v) = v + U\) (3.104), which is linear with \(\operatorname{null} \pi = U\) by 3.101.

(a) The restriction \(\pi|_W \in \mathcal{L}(W, V/U)\) is surjective: given \(v + U \in V/U\), write \(v = u + w\) with \(u \in U\) and \(w \in W\) (possible as \(V = U + W\)), so \(v - w \in U\) and hence \(v + U = w + U = \pi|_W(w)\) by 3.101. Since \(W\) is finite-dimensional, the fundamental theorem of linear maps 3.21 gives

\begin{equation*} \dim W = \dim (U \cap W) + \dim V/U \geq \dim V/U , \end{equation*}

using \(\operatorname{null} \pi|_W = W \cap \operatorname{null}\pi = U \cap W\).

(b) Take \(W = \operatorname{span}(v_1, \dots, v_n)\), where \(v_1 + U, \dots, v_n + U\) is a basis of \(V/U\) and \(n = \dim V/U\) (such a basis exists by 2.31, and its vectors have this form because \(\pi\) is surjective).

The list \(v_1, \dots, v_n\) is linearly independent, since applying \(\pi\) to \(a_1 v_1 + \dots + a_n v_n = 0\) gives \(a_1(v_1 + U) + \dots + a_n(v_n + U) = 0 + U\) and hence \(a_1 = \dots = a_n = 0\); so \(\dim W = n = \dim V/U\).

\(V = U + W\): for \(v \in V\), expanding \(v + U\) in the basis and using 3.102 gives \(v + U = (a_1 v_1 + \dots + a_n v_n) + U\), so by 3.101

\begin{equation*} v = \bigl( v - (a_1 v_1 + \dots + a_n v_n) \bigr) + (a_1 v_1 + \dots + a_n v_n) \in U + W . \end{equation*}

\(U \cap W = \{0\}\): if \(w = a_1 v_1 + \dots + a_n v_n\) lies in \(U\), then \(\pi(w) = a_1(v_1 + U) + \dots + a_n(v_n + U) = 0 + U\), so every \(a_j = 0\) and \(w = 0\).

By 1.46 the sum is direct, so \(V = U \oplus W\) with \(\dim W = \dim V/U\).

If such an \(S\) exists, then each \(u \in U\) has \(\pi(u) = 0 + U\) by 3.101, so

\begin{equation*} Tu = S(\pi(u)) = S(0) = 0 \end{equation*}

by 3.10; hence \(U \subseteq \operatorname{null} T\).

Conversely, suppose \(U \subseteq \operatorname{null} T\) and set \(S(v + U) = Tv\). This is well defined (the point of the hypothesis): if \(v + U = v^{\prime} + U\), then \(v - v^{\prime} \in U \subseteq \operatorname{null} T\) by 3.101, so \(Tv = Tv^{\prime}\). It is linear because addition and scalar multiplication on \(V/U\) are given by 3.102 and \(T\) is linear:

\begin{equation*} S\bigl( (v + U) + \lambda(v^{\prime} + U) \bigr) = T(v + \lambda v^{\prime}) = S(v + U) + \lambda S(v^{\prime} + U). \end{equation*}

Finally \((S \circ \pi)(v) = S(v + U) = Tv\) for every \(v \in V\), so \(T = S \circ \pi\).

A linear functional \(\varphi \in V^{\prime}\) that is not the zero map is surjective. Indeed, choose \(u \in V\) with \(a := \varphi(u) \neq 0\); since \(a\) is an invertible element of the field \(\mathbf{F}\), homogeneity of \(\varphi\) gives, for every \(\lambda \in \mathbf{F}\),

\begin{equation*} \varphi\!\left(\frac{\lambda}{a}\,u\right) = \frac{\lambda}{a}\,\varphi(u) = \lambda , \end{equation*}

so \(\operatorname{range}\varphi = \mathbf{F}\).

Method (2): \(\operatorname{range}\varphi\) is a subspace of \(\mathbf{F}\) by 3.18, and \(\dim \mathbf{F} = 1\) (the list \(1\) is a basis), so \(\dim \operatorname{range}\varphi \leq 1\) by 2.37. If that dimension is \(0\) then \(\varphi\) is the zero map; if it is \(1\) then \(\operatorname{range}\varphi = \mathbf{F}\) by 2.39.

Take

\begin{equation*} \varphi_1(f) = f(0), \qquad \varphi_2(f) = f(1), \qquad \varphi_3(f) = f(0) + f(1). \end{equation*}

Each evaluation \(f \mapsto f(t)\) is a linear functional on \(\mathbf{R}^{[0,1]}\), because the operations there are pointwise: \((f + \lambda g)(t) = f(t) + \lambda g(t)\). This gives \(\varphi_1\) (take \(t = 0\)) and \(\varphi_2\) (take \(t = 1\)), and \(\varphi_3 = \varphi_1 + \varphi_2\) is linear as a sum of linear maps (3.6).

They are pairwise distinct: let \(f\) equal \(1\) at \(0\) and \(0\) on \((0,1]\), and let \(g\) equal \(1\) at \(1\) and \(0\) on \([0,1)\). Then

\begin{equation*} \bigl(\varphi_1(f), \varphi_2(f), \varphi_3(f)\bigr) = (1, 0, 1), \qquad \bigl(\varphi_1(g), \varphi_2(g), \varphi_3(g)\bigr) = (0, 1, 1), \end{equation*}

so \(f\) separates \(\varphi_1\) from \(\varphi_2\) and \(\varphi_2\) from \(\varphi_3\), while \(g\) separates \(\varphi_1\) from \(\varphi_3\).

Take \(\varphi\) to be the first functional in the dual basis of a basis of \(V\) whose first vector is \(v\). In detail: the one-vector list \(v\) is linearly independent (if \(av = 0\) with \(a \neq 0\), then \(v = a^{-1}(av) = 0\), contrary to \(v \neq 0\)), so since \(V\) is finite-dimensional it extends by 2.32 to a basis \(v_1, \dots, v_n\) of \(V\) with \(v_1 = v\). By the linear map lemma 3.4 there is \(\varphi \in \mathcal{L}(V, \mathbf{F}) = V^{\prime}\) with

\begin{equation*} \varphi(v_1) = 1 \qquad \text{and} \qquad \varphi(v_j) = 0 \quad \text{for } j = 2, \dots, n , \end{equation*}

and then \(\varphi(v) = \varphi(v_1) = 1\).

Extend a basis \(u_1, \dots, u_m\) of \(U\) to a basis \(u_1, \dots, u_m, v_{m+1}, \dots, v_n\) of \(V\) (by 2.32), and let \(\varphi \in V^{\prime}\) be the linear functional supplied by the linear map lemma 3.4 with

\begin{equation*} \varphi(u_j) = 0 \ (1 \le j \le m), \quad \varphi(v_{m+1}) = 1, \quad \varphi(v_k) = 0 \ (m+2 \le k \le n). \end{equation*}

The vector \(v_{m+1}\) really is present: \(U\) is finite-dimensional by 2.25 with \(\dim U \le \dim V\) by 2.37, and \(\dim U = \dim V\) would give \(U = V\) by 2.39, so \(m = \dim U < \dim V = n\).

Then \(\varphi\) vanishes on \(U\), since every \(u \in U\) is \(a_1 u_1 + \cdots + a_m u_m\) and linearity gives \(\varphi(u) = 0\); and \(\varphi \neq 0\) because \(\varphi(v_{m+1}) = 1\).

Each \(\varphi_j = \psi_j \circ S\) is a composition of linear maps and hence a linear functional on \(V\) by 3.7. Here \(S \in \mathcal{L}(V, \operatorname{range} T)\) is \(T\) with its target narrowed to \(\operatorname{range} T\), and \(\psi_1, \dots, \psi_m \in (\operatorname{range} T)^{\prime}\) is the dual basis of \(w_1, \dots, w_m\); by 3.114 the scalar \(\psi_j(w)\) is the \(j\)-th coefficient of \(w\) in that basis, which for \(w = Tv\) is exactly \(\varphi_j(v)\) by the defining equation.

Method (2): compare coefficients. Additivity of \(T\) gives

\begin{equation*} \begin{aligned} \varphi_1(u+v)w_1 + \cdots + \varphi_m(u+v)w_m &= T(u+v) = Tu + Tv \\ &= \bigl(\varphi_1(u) + \varphi_1(v)\bigr)w_1

• \cdots + \bigl(\varphi_m(u) + \varphi_m(v)\bigr)w_m , \end{aligned} \end{equation*}

and the uniqueness of coefficients relative to the basis \(w_1, \dots, w_m\) (2.28) forces \(\varphi_j(u+v) = \varphi_j(u) + \varphi_j(v)\). Homogeneity is the same computation started from \(T(\lambda v) = \lambda\, Tv\).

If \(\beta = c\varphi\) and \(\varphi(v) = 0\), then \(\beta(v) = c\,\varphi(v) = 0\), so \(\operatorname{null}\varphi \subseteq \operatorname{null}\beta\).

Conversely, suppose \(\operatorname{null}\varphi \subseteq \operatorname{null}\beta\).

(i) \(\varphi = 0\). Then \(V = \operatorname{null}\varphi \subseteq \operatorname{null}\beta\), so \(\beta = 0 = 0 \cdot \varphi\).

(ii) \(\varphi \neq 0\). By Exercise 3F.1 a nonzero linear functional is surjective, so choose \(u \in V\) with \(\varphi(u) = 1\) and put \(c = \beta(u)\). For \(v \in V\), the vector \(w = v - \varphi(v)u\) satisfies \(\varphi(w) = \varphi(v) - \varphi(v) = 0\), so \(w \in \operatorname{null}\varphi \subseteq \operatorname{null}\beta\); applying \(\beta\) to \(v = w + \varphi(v)u\) then gives

\begin{equation*} \beta(v) = \beta(w) + \varphi(v)\,\beta(u) = c\,\varphi(v) = (c\varphi)(v) . \end{equation*}

Hence \(\beta = c\varphi\).

Write \(V = V_1 \times \cdots \times V_m\) and let \(\iota_j \colon V_j \to V\) be the inclusion \(\iota_j(v) = (0, \dots, 0, v, 0, \dots, 0)\) with \(v\) in slot \(j\). Then

\begin{equation*} \Gamma \colon V^{\prime} \to V_1{}^{\prime} \times \cdots \times V_m{}^{\prime}, \qquad \Gamma(\varphi) = (\varphi \circ \iota_1, \dots, \varphi \circ \iota_m), \end{equation*}

is an isomorphism. Each \(\iota_j\) is linear because the operations on \(V\) are coordinatewise (3.87, 3.89), so each \(\varphi \circ \iota_j\) lies in \(V_j{}^{\prime}\) by 3.7; and \(\Gamma\) is linear because addition and scalar multiplication in \(V^{\prime}\) are pointwise (3.5) and in the product are coordinatewise (Check!).

Its two-sided inverse is

\begin{equation*} \Lambda(\varphi_1, \dots, \varphi_m) \colon (v_1, \dots, v_m) \mapsto \varphi_1(v_1) + \cdots + \varphi_m(v_m), \end{equation*}

which lies in \(V^{\prime}\) since the operations on \(V\) are coordinatewise and each \(\varphi_j\) is linear (Check!). Indeed, for \(v \in V_j\), using \(\varphi_k(0) = 0\) (3.10),

\begin{equation*} \bigl(\Lambda(\varphi_1, \dots, \varphi_m) \circ \iota_j\bigr)(v) = \varphi_j(v) , \end{equation*}

so \(\Gamma \circ \Lambda\) is the identity; and since \((v_1, \dots, v_m) = \iota_1(v_1) + \cdots + \iota_m(v_m)\), linearity of \(\varphi\) gives

\begin{equation*} \bigl(\Lambda(\Gamma\varphi)\bigr)(v_1, \dots, v_m) = \varphi(\iota_1 v_1) + \cdots + \varphi(\iota_m v_m) = \varphi(v_1, \dots, v_m) , \end{equation*}

so \(\Lambda \circ \Gamma\) is the identity. Hence \(\Gamma\) is invertible (3.59), that is, an isomorphism (3.69).

Both composites are identity maps, so by 3.59 each of \(\Gamma, \Lambda\) is an inverse of the other (and inverses are unique by 3.60). Both are linear: \(\Gamma\) because each \(\varphi_j\) is linear and the operations on \(\mathbf{F}^n\) are coordinatewise, \(\Lambda\) because \(\sum_k (a_k + \lambda b_k)v_k = \sum_k a_k v_k + \lambda \sum_k b_k v_k\).

For \(v \in V\), the expansion 3.114 of \(v\) in the basis \(v_1, \dots, v_n\) gives

\begin{equation*} (\Lambda \circ \Gamma)(v) = \varphi_1(v) v_1 + \cdots + \varphi_n(v) v_n = v . \end{equation*}

For \((a_1, \dots, a_n) \in \mathbf{F}^n\), putting \(v = a_1 v_1 + \cdots + a_n v_n\) and using \(\varphi_j(v_k) = 1\) for \(k = j\) and \(0\) otherwise (3.112), linearity gives \(\varphi_j(v) = \sum_k a_k \varphi_j(v_k) = a_j\), so

\begin{equation*} (\Gamma \circ \Lambda)(a_1, \dots, a_n) = (\varphi_1(v), \dots, \varphi_n(v)) = (a_1, \dots, a_n). \end{equation*}

Each \(\varphi_k\) is a linear functional on \(\mathcal{P}_m(\mathbf{R})\) satisfying \(\varphi_k(x^j) = 1\) if \(j = k\) and \(\varphi_k(x^j) = 0\) if \(j \neq k\), which is exactly the defining condition 3.112 for the dual basis of \(1, x, \dots, x^m\).

Linearity: \(k\)-fold differentiation, evaluation at \(0\), and division by the fixed nonzero scalar \(k!\) are each linear, so

\begin{equation*} \varphi_k(p + \lambda q) = \frac{p^{(k)}(0) + \lambda\, q^{(k)}(0)}{k!} = \varphi_k(p) + \lambda \varphi_k(q). \end{equation*}

Duality: repeated differentiation gives \((x^j)^{(k)} = j(j-1)\cdots(j-k+1)\, x^{\,j-k}\) for \(k \leq j\), and \((x^j)^{(k)} = 0\) for \(k > j\).

(i) \(k > j\): the derivative is \(0\), so \(\varphi_k(x^j) = 0\).

(ii) \(k < j\): the derivative has the factor \(x^{\,j-k}\) with \(j - k \geq 1\), which vanishes at \(0\), so \(\varphi_k(x^j) = 0\).

(iii) \(k = j\): the derivative is the constant \(k(k-1)\cdots 1 = k!\), so \(\varphi_k(x^k) = k!/k! = 1\).

(a) Each \((x-5)^j\) has degree \(j \leq m\), so the list lies in \(\mathcal{P}_m(\mathbf{R})\), and it has length \(m+1 = \dim \mathcal{P}_m(\mathbf{R})\) (2.36); by 2.38 it is therefore enough to check linear independence. Suppose \(a_0 + a_1(x-5) + \cdots + a_m(x-5)^m = 0\) with not all \(a_j\) zero, and let \(k\) be the largest index with \(a_k \neq 0\). The terms with \(j < k\) have degree at most \(k-1\), while \((x-5)^k\) has leading coefficient \(1\), so the coefficient of \(x^k\) in the sum is \(a_k \neq 0\) and the sum is not the zero polynomial, a contradiction.

(b) The dual basis is \(\varphi_0, \dots, \varphi_m\) with

\begin{equation*} \varphi_k(p) = \frac{p^{(k)}(5)}{k!} , \end{equation*}

that is, Exercise 3F.9 with derivatives evaluated at \(5\) instead of \(0\).

Each \(\varphi_k\) is linear, since differentiating \(k\) times, evaluating at \(5\), and dividing by the fixed nonzero scalar \(k!\) are linear operations. For the duality condition 3.112, repeated differentiation gives \(\big((x-5)^j\big)^{(k)} = j(j-1)\cdots(j-k+1)(x-5)^{\,j-k}\) for \(k \leq j\), and \(0\) for \(k > j\).

(i) \(k > j\): the derivative is \(0\), so \(\varphi_k\big((x-5)^j\big) = 0\).

(ii) \(k < j\): the factor \((x-5)^{\,j-k}\) with \(j - k \geq 1\) vanishes at \(5\), so \(\varphi_k\big((x-5)^j\big) = 0\).

(iii) \(k = j\): the derivative is the constant \(k!\), so \(\varphi_k\big((x-5)^k\big) = k!/k! = 1\).

Both sides are linear functionals on \(V\) agreeing on the basis \(v_1, \dots, v_n\), hence are equal by the uniqueness half of the linear map lemma 3.4. Indeed, \(\eta := \psi(v_1)\varphi_1 + \cdots + \psi(v_n)\varphi_n\) lies in \(V^{\prime}\) because \(V^{\prime}\) is a vector space (3.110), and the dual-basis relations \(\varphi_k(v_j) = 1\) for \(k = j\) and \(0\) otherwise (3.112) give

\begin{equation*} \eta(v_j) = \sum_{k=1}^{n} \psi(v_k)\varphi_k(v_j) = \psi(v_j) \end{equation*}

for each \(j\).

Both identities come from evaluating at an arbitrary \(\varphi \in W^{\prime}\) and then at an arbitrary \(v \in V\), using the definition 3.118 of the dual map, \(R^{\prime}(\varphi) = \varphi \circ R\).

(a)

\begin{equation*} \begin{aligned} \big((S+T)^{\prime}(\varphi)\big)(v) &= \varphi\big((S+T)v\big) = \varphi(Sv) + \varphi(Tv) \\ &= \big(S^{\prime}(\varphi)\big)(v) + \big(T^{\prime}(\varphi)\big)(v) = \big((S^{\prime} + T^{\prime})(\varphi)\big)(v), \end{aligned} \end{equation*}

using 3.5 for \((S+T)v = Sv + Tv\), the additivity of \(\varphi\), and pointwise addition in \(V^{\prime}\) and in \(\mathcal{L}(W^{\prime}, V^{\prime})\).

(b)

\begin{equation*} \begin{aligned} \big((\lambda T)^{\prime}(\varphi)\big)(v) &= \varphi(\lambda\, Tv) = \lambda\, \varphi(Tv) \\ &= \lambda \big(T^{\prime}(\varphi)\big)(v) = \big((\lambda T^{\prime})(\varphi)\big)(v), \end{aligned} \end{equation*}

using 3.5 for \((\lambda T)v = \lambda\, Tv\), the homogeneity of \(\varphi\), and pointwise scalar multiplication in \(V^{\prime}\) and in \(\mathcal{L}(W^{\prime}, V^{\prime})\).

For every \(\varphi \in V^{\prime}\), the definition 3.118 of the dual map gives \(I^{\prime}(\varphi) = \varphi \circ I = \varphi\), since

\begin{equation*} (\varphi \circ I)(v) = \varphi(Iv) = \varphi(v) \qquad \text{for every } v \in V . \end{equation*}

Thus \(I^{\prime} \in \mathcal{L}(V^{\prime}, V^{\prime})\) fixes every element of \(V^{\prime}\), so it is the identity operator on \(V^{\prime}\).

(a) \(T^{\prime}(\varphi_1)\) and \(T^{\prime}(\varphi_2)\) are the linear functionals on \(\mathbf{R}^3\) given by

\begin{equation*} \big(T^{\prime}(\varphi_1)\big)(x, y, z) = 4x + 5y + 6z, \qquad \big(T^{\prime}(\varphi_2)\big)(x, y, z) = 7x + 8y + 9z , \end{equation*}

that is, the two rows of the formula for \(T\). Indeed \(T^{\prime}(\varphi_j) = \varphi_j \circ T\) by 3.118, and by 3.113 the functionals \(\varphi_1, \varphi_2\) select the first and second coordinates of \(\mathbf{R}^2\).

(b) By 3.113 again, \(\psi_1, \psi_2, \psi_3\) select the coordinates of \(\mathbf{R}^3\), so the functionals \(4\psi_1 + 5\psi_2 + 6\psi_3\) and \(7\psi_1 + 8\psi_2 + 9\psi_3\) take \((x,y,z)\) to \(4x + 5y + 6z\) and to \(7x + 8y + 9z\). Comparing with (a),

\begin{equation*} T^{\prime}(\varphi_1) = 4\psi_1 + 5\psi_2 + 6\psi_3 \qquad \text{and} \qquad T^{\prime}(\varphi_2) = 7\psi_1 + 8\psi_2 + 9\psi_3 . \end{equation*}

(a) \(T^{\prime}(\varphi)\) is the linear functional on \(\mathcal{P}(\mathbb{R})\) taking \(p\) to \(8p(4) + 16\,p^{\prime}(4) + p^{\prime\prime\prime}(4)\). Indeed \((T^{\prime}(\varphi))(p) = \varphi(Tp) = (Tp)^{\prime}(4)\) by 3.118, and the product rule gives

\begin{equation*} (Tp)^{\prime}(x) = 2x\,p(x) + x^2 p^{\prime}(x) + p^{\prime\prime\prime}(x) , \end{equation*}

which at \(x = 4\) is the stated value.

(b) \((T^{\prime}(\varphi))(x^3) = 19/6\). For \(p(x) = x^3\) we have \(p^{\prime\prime}(x) = 6x\), so \((Tp)(x) = x^5 + 6x\) and

\begin{equation*} (T^{\prime}(\varphi))(x^3) = \int_0^1 (x^5 + 6x)\,dx = \left[\frac{x^6}{6} + 3x^2\right]_0^1 = \frac{1}{6} + 3 = \frac{19}{6}. \end{equation*}

If \(T = 0\), then \(\bigl(T^{\prime}(\varphi)\bigr)(v) = \varphi(Tv) = \varphi(0) = 0\) for all \(\varphi \in W^{\prime}\) and \(v \in V\), so \(T^{\prime} = 0\).

Conversely, suppose \(T \neq 0\) and choose \(v \in V\) with \(Tv \neq 0\). The one-vector list \(Tv\) is linearly independent, so since \(W\) is finite-dimensional it extends by 2.32 to a basis \(w_1 = Tv, w_2, \ldots, w_m\) of \(W\); let \(\psi_1, \ldots, \psi_m \in W^{\prime}\) be its dual basis (3.112). Then

\begin{equation*} \bigl(T^{\prime}(\psi_1)\bigr)(v) = \psi_1(Tv) = \psi_1(w_1) = 1 \neq 0 , \end{equation*}

so \(T^{\prime} \neq 0\). Contrapositively, \(T^{\prime} = 0\) implies \(T = 0\).

Method (2) for the second direction: \(T^{\prime} = 0\) gives \(\operatorname{null} T^{\prime} = W^{\prime}\), and \(\operatorname{null} T^{\prime} = (\operatorname{range} T)^0\) by 3.128(a); since \(\operatorname{range} T\) is a subspace of the finite-dimensional space \(W\), 3.127(b) then forces \(\operatorname{range} T = \{0\}\), i.e. \(T = 0\).

The equivalence is 3.63 applied at both ends of the chain

\begin{equation*} \begin{aligned} T \text{ is invertible} \ &\iff\ T \text{ is injective and } T \text{ is surjective}\\ &\iff\ T^{\prime} \text{ is surjective and } T^{\prime} \text{ is injective}\\ &\iff\ T^{\prime} \text{ is invertible}, \end{aligned} \end{equation*}

the middle equivalence being 3.131 (\(T\) injective \(\iff\) \(T^{\prime}\) surjective) and 3.129 (\(T\) surjective \(\iff\) \(T^{\prime}\) injective), whose hypothesis that \(V\) and \(W\) are finite-dimensional holds here.

The map in question is \(\Lambda T = T^{\prime}\), which lands in \(\mathcal{L}(W^{\prime}, V^{\prime})\) by 3.118; we show it is linear, injective, and of equal-dimensional domain and codomain.

Linearity is 3.120(a) and 3.120(b): for \(S, T \in \mathcal{L}(V, W)\) and \(\lambda \in \mathbb{F}\),

\begin{equation*} \begin{aligned} \Lambda(S + T) &= S^{\prime} + T^{\prime} = \Lambda S + \Lambda T, \\ \Lambda(\lambda T) &= \lambda T^{\prime} = \lambda\,\Lambda T . \end{aligned} \end{equation*}

If \(\Lambda T = 0\), then \(T^{\prime} = 0\), so \(T = 0\) by Exercise 16 of this section (which applies since \(W\) is finite-dimensional); hence \(\operatorname{null}\Lambda = \{0\}\) and \(\Lambda\) is injective by 3.15. Finally 3.111 gives \(\dim V^{\prime} = \dim V\) and \(\dim W^{\prime} = \dim W\), so 3.72 twice yields

\begin{equation*} \dim \mathcal{L}(W^{\prime}, V^{\prime}) = (\dim W)(\dim V) = \dim \mathcal{L}(V, W). \end{equation*}

An injective linear map between finite-dimensional spaces of equal dimension is invertible (3.65), so \(\Lambda\) is an isomorphism of \(\mathcal{L}(V, W)\) onto \(\mathcal{L}(W^{\prime}, V^{\prime})\).

The two conditions on \(\varphi \in V^{\prime}\) are verbatim translations of each other. For fixed \(\varphi \in V^{\prime}\),

\begin{equation*} \begin{aligned} \varphi \in U^0 \ &\iff\ \varphi(u) = 0 \text{ for every } u \in U\\ &\iff\ u \in \operatorname{null}\varphi \text{ for every } u \in U\\ &\iff\ U \subseteq \operatorname{null} \varphi , \end{aligned} \end{equation*}

the first step by the definition 3.121 of the annihilator, the second by the definition 3.11 of the null space (each \(u \in U\) lies in \(V\), as \(U \subseteq V\)), and the third by the definition of inclusion. Since both sets are subsets of \(V^{\prime}\), they are equal.

Write \(X = \{v \in V : \varphi(v) = 0 \text{ for every } \varphi \in U^0\}\).

(i) \(U \subseteq X\): if \(u \in U\) and \(\varphi \in U^0\), then \(\varphi(u) = 0\) by 3.121.

(ii) \(X \subseteq U\), by contraposition. Suppose \(v \in V\) and \(v \notin U\). The subspace \(U\) is finite-dimensional (2.25); let \(u_1, \ldots, u_m\) be a basis of it (the empty list if \(U = \{0\}\)). Then \(u_1, \ldots, u_m, v\) is linearly independent, since a relation \(a_1 u_1 + \cdots + a_m u_m + bv = 0\) with \(b \neq 0\) would put \(v \in U\), while \(b = 0\) forces \(a_1 = \cdots = a_m = 0\). Extend it to a basis

\begin{equation*} u_1, \ldots, u_m,\ v,\ w_1, \ldots, w_k \end{equation*}

of \(V\) (2.32) and let \(\varphi\) be the member of the dual basis (3.112) dual to \(v\), so that

\begin{equation*} \varphi(u_1) = \cdots = \varphi(u_m) = 0, \qquad \varphi(v) = 1 . \end{equation*}

Vanishing on a basis of \(U\), the functional \(\varphi\) vanishes on \(U\), so \(\varphi \in U^0\); as \(\varphi(v) = 1 \neq 0\), we get \(v \notin X\).

Hence \(U = X\).

(a) Annihilation reverses inclusions. If \(U \subseteq W\) and \(\varphi \in W^0\), then \(\varphi\) vanishes on \(W\) and hence on the subset \(U\), so \(\varphi \in U^0\); thus \(W^0 \subseteq U^0\).

Conversely, suppose \(W^0 \subseteq U^0\) and \(u \in U\). Every \(\varphi \in W^0\) then lies in \(U^0\), so \(\varphi(u) = 0\). Since \(V\) is finite-dimensional and \(W\) is a subspace of \(V\), Exercise 20 of this section gives

\begin{equation*} W = \{v \in V : \varphi(v) = 0 \text{ for every } \varphi \in W^0\}, \end{equation*}

so \(u \in W\). Hence \(U \subseteq W\).

(b) If \(U = W\), then trivially \(U^0 = W^0\). Conversely, \(W^0 = U^0\) gives both \(W^0 \subseteq U^0\) and \(U^0 \subseteq W^0\), so (a) applied each way gives \(U \subseteq W\) and \(W \subseteq U\), whence \(U = W\).

(a) Each side consists of the \(\varphi \in V^{\prime}\) vanishing on both \(U\) and \(W\). Indeed \(U \subseteq U + W\) and \(W \subseteq U + W\) (as \(0\) lies in each), so \(\varphi \in (U + W)^0\) forces \(\varphi \in U^0 \cap W^0\); conversely if \(\varphi \in U^0 \cap W^0\), then for \(u \in U\) and \(w \in W\),

\begin{equation*} \varphi(u + w) = \varphi(u) + \varphi(w) = 0 , \end{equation*}

so \(\varphi \in (U + W)^0\). (Finite-dimensionality is not used here.)

(b) One inclusion is immediate: \(\varphi \in U^0\) vanishes on \(U\) and hence on \(U \cap W\), so \(U^0 \subseteq (U \cap W)^0\), and likewise \(W^0 \subseteq (U \cap W)^0\); since \((U \cap W)^0\) is a subspace of \(V^{\prime}\) (3.124), it is closed under addition, giving

\begin{equation*} U^0 + W^0 \subseteq (U \cap W)^0 . \end{equation*}

For the dimensions, put \(n = \dim V\); then \(\dim V^{\prime} = n\) by 3.111, so 2.43 applies to the subspaces \(U^0, W^0\) of \(V^{\prime}\) and all dimensions below are finite. By 2.43, then (a), then 3.125,

\begin{equation*} \begin{aligned} \dim(U^0 + W^0) &= \dim U^0 + \dim W^0 - \dim(U^0 \cap W^0) \\ &= \dim U^0 + \dim W^0 - \dim\bigl((U + W)^0\bigr) \\ &= (n - \dim U) + (n - \dim W) - \bigl(n - \dim(U + W)\bigr) \\ &= n - \dim U - \dim W + \dim(U + W). \end{aligned} \end{equation*}

Now 2.43 applied to \(U\) and \(W\) inside \(V\) gives \(\dim(U + W) = \dim U + \dim W - \dim(U \cap W)\), and substituting turns the last line into

\begin{equation*} \dim(U^0 + W^0) = n - \dim(U \cap W) = \dim\bigl((U \cap W)^0\bigr), \end{equation*}

the last equality being 3.125 for the subspace \(U \cap W\). A subspace of matching finite dimension is the whole space (2.39), so \((U \cap W)^0 = U^0 + W^0\).

Write \(W = (\operatorname{null} \varphi_1) \cap \cdots \cap (\operatorname{null} \varphi_m)\), a subspace of \(V\); all three sets equal \(W^0\).

(b) \(=\) (c): Exercise 19 of this section says \(W^0 = \{\varphi \in V^{\prime} : W \subseteq \operatorname{null}\varphi\}\), which is (c).

(a) \(=\) (b): two ingredients.

(i) \((\operatorname{null}\psi)^0 = \operatorname{span}(\psi)\) for every \(\psi \in V^{\prime}\). If \(\psi = 0\), both sides are \(\{0\}\) (here \(V^0 = \{0\}\) by 3.127(a)). If \(\psi \neq 0\), then \(\operatorname{range}\psi\) is a nonzero subspace of the one-dimensional space \(\mathbf{F}\), so \(\dim\operatorname{range}\psi = 1\) and 3.21 then 3.125 give

\begin{equation*} \begin{aligned} \dim \operatorname{null}\psi &= \dim V - 1, \\ \dim (\operatorname{null}\psi)^0 &= 1 . \end{aligned} \end{equation*}

Since \(\psi\) vanishes on \(\operatorname{null}\psi\), the one-dimensional space \(\operatorname{span}(\psi)\) sits inside the one-dimensional space \((\operatorname{null}\psi)^0\), and 2.39 gives equality.

(ii) \((U_1 \cap \cdots \cap U_m)^0 = U_1^{\,0} + \cdots + U_m^{\,0}\) for subspaces \(U_1, \ldots, U_m\) of \(V\), by induction on \(m\): trivial for \(m = 1\), and for \(m > 1\) Exercise 22(b) applied to the subspaces \(U_1 \cap \cdots \cap U_{m-1}\) and \(U_m\) gives

\begin{equation*} (U_1 \cap \cdots \cap U_m)^0 = (U_1 \cap \cdots \cap U_{m-1})^0 + U_m^{\,0}, \end{equation*}

to which the induction hypothesis applies.

Taking \(U_k = \operatorname{null}\varphi_k\) in (ii) and then (i) termwise,

\begin{equation*} \begin{aligned} W^0 &= \operatorname{span}(\varphi_1) + \cdots + \operatorname{span}(\varphi_m) \\ &= \operatorname{span}(\varphi_1, \ldots, \varphi_m), \end{aligned} \end{equation*}

the last step because both sides consist of all \(a_1\varphi_1 + \cdots + a_m\varphi_m\) with \(a_k \in \mathbf{F}\).

Both parts rest on \(\operatorname{null}\Gamma = U^0\), where \(U = \operatorname{span}(v_1, \ldots, v_m)\). Indeed \(\Gamma(\varphi) = 0\) says \(\varphi(v_k) = 0\) for each \(k\), which forces \(\varphi(a_1v_1 + \cdots + a_mv_m) = 0\) by linearity, so \(\varphi \in U^0\); conversely \(\varphi \in U^0\) kills each \(v_k \in U\).

(a) By 3.15, the displayed equality, and 3.127(a),

\begin{equation*} \begin{aligned} \Gamma \text{ is injective} \ &\iff\ \operatorname{null} \Gamma = \{0\} \\ &\iff\ U^0 = \{0\} \ \iff\ U = V , \end{aligned} \end{equation*}

and \(U = V\) says exactly that \(v_1, \ldots, v_m\) spans \(V\).

(b) By 3.21 applied to \(\Gamma \in \mathcal{L}(V^{\prime}, \mathbf{F}^m)\), then 3.111, then \(\operatorname{null}\Gamma = U^0\) and 3.125,

\begin{equation*} \begin{aligned} \dim \operatorname{range} \Gamma &= \dim V^{\prime} - \dim U^0 \\ &= \dim V - \dim U^0 \\ &= \dim V - (\dim V - \dim U) \\ &= \dim U . \end{aligned} \end{equation*}

Since \(\operatorname{range}\Gamma\) is a subspace of the \(m\)-dimensional space \(\mathbf{F}^m\), 2.39 says \(\Gamma\) is surjective exactly when \(\dim\operatorname{range}\Gamma = m\), that is, when \(\dim U = m\). Finally \(v_1, \ldots, v_m\) is a length-\(m\) spanning list of \(U\), so it is linearly independent exactly when \(\dim U = m\): independence makes it a basis of \(U\), and \(\dim U = m\) makes it a basis by 2.42.

Both parts rest on \(\operatorname{null}\Gamma = W\), where \(W = (\operatorname{null}\varphi_1) \cap \cdots \cap (\operatorname{null}\varphi_m)\): indeed \(\Gamma(v) = 0\) says \(\varphi_k(v) = 0\) for every \(k\), that is, \(v \in W\). Exercise 23 of this section supplies

\begin{equation*} W^0 = \operatorname{span}(\varphi_1, \ldots, \varphi_m). \end{equation*}

(a) By 3.15, then \(\operatorname{null}\Gamma = W\), then 3.127(b) for the subspace \(W\) of \(V\),

\begin{equation*} \begin{aligned} \Gamma \text{ is injective} \ &\iff\ \operatorname{null} \Gamma = \{0\} \\ &\iff\ W = \{0\} \ \iff\ W^0 = V^{\prime} , \end{aligned} \end{equation*}

and by the display \(W^0 = V^{\prime}\) says exactly that \(\varphi_1, \ldots, \varphi_m\) spans \(V^{\prime}\).

(b) By 3.21 applied to \(\Gamma\), then \(\operatorname{null}\Gamma = W\) and 3.125,

\begin{equation*} \begin{aligned} \dim \operatorname{range} \Gamma &= \dim V - \dim W = \dim W^0 \\ &= \dim \operatorname{span}(\varphi_1, \ldots, \varphi_m), \end{aligned} \end{equation*}

the last equality again by Exercise 23. Since \(\operatorname{range}\Gamma\) is a subspace of the \(m\)-dimensional space \(\mathbf{F}^m\), 2.39 says \(\Gamma\) is surjective exactly when this dimension equals \(m\). Finally \(\varphi_1, \ldots, \varphi_m\) is a length-\(m\) spanning list of its own span, so its span has dimension \(m\) exactly when the list is linearly independent (independence makes it a basis of the span; conversely 2.42 makes it a basis).

Write \(W = \{v \in V : \varphi(v) = 0 \text{ for every } \varphi \in \Omega\}\); we must show \(W^0 = \Omega\).

(i) \(\Omega = \{0\}\). Then \(W = V\), so \(W^0 = V^0 = \{0\} = \Omega\) by 3.127(a).

(ii) \(\Omega \neq \{0\}\). Since \(\dim V^{\prime} = \dim V < \infty\) (3.111), the subspace \(\Omega\) is finite-dimensional (2.25); let \(\varphi_1, \ldots, \varphi_m\) be a basis of it, \(m \geq 1\). Then

\begin{equation*} W = (\operatorname{null} \varphi_1) \cap \cdots \cap (\operatorname{null} \varphi_m): \end{equation*}

each \(\varphi_k\) lies in \(\Omega\), giving \(\subseteq\), and if \(\varphi_k(v) = 0\) for all \(k\), then every \(\varphi = a_1\varphi_1 + \cdots + a_m\varphi_m \in \Omega\) satisfies \(\varphi(v) = 0\), giving \(\supseteq\). Now Exercise 23 of this section gives

\begin{equation*} W^0 = \operatorname{span}(\varphi_1, \ldots, \varphi_m) = \Omega . \end{equation*}

The set in question is \(\operatorname{null}\varphi\), and both it and \(\operatorname{range} T\) turn out to be \(5\)-dimensional subspaces of \(V = \mathcal{P}_5(\mathbf{R})\), one inside the other. Here \(\dim V = 6\), with basis \(1, x, \ldots, x^5\).

Since \(\varphi(1) = 1\), we have \(\varphi \neq 0\), so \(\operatorname{range}\varphi\) is a nonzero subspace of \(\mathbf{R}\) and hence one-dimensional; also \(\dim\operatorname{span}(\varphi) = 1\). By 3.21,

\begin{equation*} \dim \operatorname{null} \varphi = 6 - 1 = 5 . \end{equation*}

By 3.128(a), \((\operatorname{range} T)^0 = \operatorname{null} T^{\prime} = \operatorname{span}(\varphi)\). In particular \(\varphi\) annihilates \(\operatorname{range} T\) (3.121), so \(\operatorname{range} T \subseteq \operatorname{null}\varphi\), and 3.125 applied to the subspace \(\operatorname{range} T\) of \(V\) gives

\begin{equation*} \dim \operatorname{range} T = 6 - \dim \operatorname{span}(\varphi) = 5 . \end{equation*}

By 2.39 inside \(\operatorname{null}\varphi\), therefore \(\operatorname{range} T = \operatorname{null}\varphi = \{p \in \mathcal{P}_5(\mathbf{R}) : p(8) = 0\}\).

Put \(W = (\operatorname{null}\varphi_1) \cap \cdots \cap (\operatorname{null}\varphi_m)\). Exercise 23 of this section gives \(W^0 = \operatorname{span}(\varphi_1, \ldots, \varphi_m)\), which has dimension \(m\) since the linearly independent list \(\varphi_1, \ldots, \varphi_m\) is a basis of its span. Hence 3.125 gives

\begin{equation*} \dim W = \dim V - \dim W^0 = (\dim V) - m . \end{equation*}

Method (2): define \(\Gamma \in \mathcal{L}(V, \mathbf{F}^m)\) by \(\Gamma(v) = (\varphi_1(v), \ldots, \varphi_m(v))\) as in Exercise 25 of this section, so that \(\operatorname{null}\Gamma = W\) (a vector is killed by \(\Gamma\) exactly when each \(\varphi_k\) kills it). Linear independence of \(\varphi_1, \ldots, \varphi_m\) makes \(\Gamma\) surjective by Exercise 25(b), so \(\dim \operatorname{range}\Gamma = m\) and 3.21 gives

\begin{equation*} \dim V = \dim W + m . \end{equation*}

Both parts are containment plus a dimension count, using throughout that a nonzero \(\alpha \in X^{\prime}\) has \(\operatorname{range}\alpha = \mathbf{F}\) and hence, by 3.21,

\begin{equation*} \dim \operatorname{null} \alpha = \dim X - 1 . \end{equation*}

(a) By 3.128(a) the hypothesis reads \((\operatorname{range} T)^0 = \operatorname{span}(\varphi)\).

(i) \(\varphi = 0\): then \((\operatorname{range} T)^0 = \{0\}\), so \(\operatorname{range} T = W\) by 3.127(a), while \(\operatorname{null}\varphi = W\) too.

(ii) \(\varphi \neq 0\): then \(\varphi \in (\operatorname{range} T)^0\) gives \(\operatorname{range} T \subseteq \operatorname{null}\varphi\) (3.121), and \(\dim\operatorname{span}(\varphi) = 1\), so 3.125 for the subspace \(\operatorname{range} T\) of \(W\) gives

\begin{equation*} \dim \operatorname{range} T = \dim W - 1 = \dim \operatorname{null} \varphi , \end{equation*}

whence \(\operatorname{range} T = \operatorname{null}\varphi\) by 2.39.

(b) By 3.130(b) the hypothesis reads \((\operatorname{null} T)^0 = \operatorname{span}(\psi)\).

(i) \(\psi = 0\): then \((\operatorname{null} T)^0 = \{0\}\), so \(\operatorname{null} T = V = \operatorname{null}\psi\) by 3.127(a).

(ii) \(\psi \neq 0\): then \(\psi \in (\operatorname{null} T)^0\) gives \(\operatorname{null} T \subseteq \operatorname{null}\psi\), and \(\dim\operatorname{span}(\psi) = 1\), so 3.125 for the subspace \(\operatorname{null} T\) of \(V\) gives

\begin{equation*} \dim \operatorname{null} T = \dim V - 1 = \dim \operatorname{null} \psi , \end{equation*}

whence \(\operatorname{null} T = \operatorname{null}\psi\) by 2.39.

Take \(v_j = \Gamma^{-1}(e_j)\), where \(e_1, \ldots, e_n\) is the standard basis of \(\mathbf{F}^n\) and \(\Gamma \in \mathcal{L}(V, \mathbf{F}^n)\) is defined by

\begin{equation*} \Gamma(v) = \bigl( \varphi_1(v), \ldots, \varphi_n(v) \bigr) ; \end{equation*}

here \(\dim V = \dim V^{\prime} = n\) by 3.111.

\(\Gamma\) is invertible. If \(v \in \operatorname{null}\Gamma\), then \(\varphi_k(v) = 0\) for each \(k\), so every \(\varphi = a_1\varphi_1 + \cdots + a_n\varphi_n \in V^{\prime}\) satisfies \(\varphi(v) = 0\) (the list spans \(V^{\prime}\)); were \(v \neq 0\), extending \(v\) to a basis of \(V\) (2.32) and applying the linear map lemma (3.4) would produce \(\varphi \in V^{\prime}\) with \(\varphi(v) = 1\), a contradiction. So \(\operatorname{null}\Gamma = \{0\}\), and \(\Gamma\) is injective by 3.15, hence invertible by 3.65 and 3.63 since \(\dim V = n = \dim \mathbf{F}^n\).

Then \(v_1, \ldots, v_n\) is linearly independent: applying \(\Gamma\) to \(a_1v_1 + \cdots + a_nv_n = 0\) gives \(a_1e_1 + \cdots + a_ne_n = 0\), so all \(a_j = 0\). Being independent of length \(n = \dim V\), it is a basis of \(V\) (2.38). Finally \(\Gamma(v_j) = e_j\) says \(\varphi_k(v_j)\) is \(1\) for \(k = j\) and \(0\) otherwise, which by 3.112 is exactly the statement that \(\varphi_1, \ldots, \varphi_n\) is the dual basis of \(v_1, \ldots, v_n\).

Everything follows from \(i^{\prime}(\varphi) = \varphi \circ i = \varphi|_U\), the restriction of \(\varphi\) to \(U\), by the definition 3.118 of the dual map.

(a) For \(\varphi \in V^{\prime}\), \(\varphi \in \operatorname{null} i^{\prime}\) says \(\varphi|_U = 0\), that is, \(\varphi(u) = 0\) for every \(u \in U\), which is \(\varphi \in U^0\) by 3.121. (No finite-dimensionality needed.)

(b) One inclusion holds since \(i^{\prime} \in \mathcal{L}(V^{\prime}, U^{\prime})\). Conversely, let \(\varphi \in U^{\prime}\). The subspace \(U\) is finite-dimensional (2.25); take a basis \(u_1, \ldots, u_m\) of \(U\) and extend to a basis \(u_1, \ldots, u_n\) of \(V\) (2.32). The linear map lemma (3.4) gives \(\psi \in V^{\prime}\) with

\begin{equation*} \psi(u_j) = \begin{cases} \varphi(u_j) & \text{if } 1 \le j \le m, \\ 0 & \text{if } m < j \le n. \end{cases} \end{equation*}

Then \(\psi|_U\) and \(\varphi\) agree on a basis of \(U\), hence on \(U\), so \(i^{\prime}(\psi) = \varphi\).

(c) In the notation of 3.106, \(\widetilde{i^{\prime}}\) maps \(V^{\prime}/(\operatorname{null} i^{\prime}) = V^{\prime}/U^0\) to \(U^{\prime}\) by \(\varphi + U^0 \mapsto i^{\prime}(\varphi)\), the domain being identified via (a). It is injective by 3.107(b), and surjective because \(\operatorname{range}\widetilde{i^{\prime}} = \operatorname{range} i^{\prime} = U^{\prime}\) by 3.107(c) and (b). Hence it is invertible (3.63), so an isomorphism from \(V^{\prime}/U^0\) onto \(U^{\prime}\).

(a) Each \(\Lambda v\) is linear on \(V^{\prime}\), because for \(\varphi, \psi \in V^{\prime}\) and \(\lambda \in \mathbf{F}\) the definitions of addition and scalar multiplication in \(V^{\prime} = \mathcal{L}(V, \mathbf{F})\) give

\begin{equation*} \begin{aligned} (\Lambda v)(\varphi + \psi) &= (\varphi + \psi)(v) = \varphi(v) + \psi(v), \\ (\Lambda v)(\lambda\varphi) &= (\lambda\varphi)(v) = \lambda\,\varphi(v) ; \end{aligned} \end{equation*}

these are \((\Lambda v)(\varphi) + (\Lambda v)(\psi)\) and \(\lambda (\Lambda v)(\varphi)\), so \(\Lambda v \in V^{\prime\prime}\). And \(\Lambda\) itself is linear, since for all \(u, v \in V\), \(\lambda \in \mathbf{F}\) and every \(\varphi \in V^{\prime}\),

\begin{equation*} \begin{aligned} \bigl( \Lambda(u + v) \bigr)(\varphi) &= \varphi(u) + \varphi(v), \\ \bigl( \Lambda(\lambda v) \bigr)(\varphi) &= \lambda\,\varphi(v), \end{aligned} \end{equation*}

which are the values at \(\varphi\) of \(\Lambda u + \Lambda v\) and of \(\lambda \Lambda v\). Thus \(\Lambda \in \mathcal{L}(V, V^{\prime\prime})\).

(b) Both \(T^{\prime\prime} \circ \Lambda\) and \(\Lambda \circ T\) map \(V\) to \(V^{\prime\prime}\), so fix \(v \in V\) and \(\varphi \in V^{\prime}\) and compare values at \(\varphi\). Using the definition 3.118 of the dual map twice,

\begin{equation*} \begin{aligned} \bigl( (T^{\prime\prime} \circ \Lambda)(v) \bigr)(\varphi) &= \bigl( T^{\prime\prime}(\Lambda v) \bigr)(\varphi) \\ &= \bigl( (\Lambda v) \circ T^{\prime} \bigr)(\varphi) \\ &= (\Lambda v)\bigl( T^{\prime}(\varphi) \bigr) \\ &= (\Lambda v)(\varphi \circ T) \\ &= (\varphi \circ T)(v) \\ &= \varphi(Tv) \\ &= \bigl( \Lambda(Tv) \bigr)(\varphi) \\ &= \bigl( (\Lambda \circ T)(v) \bigr)(\varphi) , \end{aligned} \end{equation*}

where the second equality uses \(T^{\prime\prime}(\alpha) = \alpha \circ T^{\prime}\), the fourth uses \(T^{\prime}(\varphi) = \varphi \circ T\), and the fifth and seventh use the definition of \(\Lambda\). As \(\varphi\) and \(v\) were arbitrary, \(T^{\prime\prime} \circ \Lambda = \Lambda \circ T\).

(c) Applying 3.111 twice gives \(\dim V^{\prime\prime} = \dim V^{\prime} = \dim V < \infty\). If \(v \in \operatorname{null}\Lambda\), then \(\varphi(v) = 0\) for every \(\varphi \in V^{\prime}\); were \(v \neq 0\), extending \(v\) to a basis of \(V\) (2.32) and applying the linear map lemma (3.4) would give \(\varphi \in V^{\prime}\) with \(\varphi(v) = 1\), a contradiction. So \(\operatorname{null}\Lambda = \{0\}\) and \(\Lambda\) is injective by 3.15, hence surjective by 3.65 and invertible by 3.63. Thus \(\Lambda\) is an isomorphism from \(V\) onto \(V^{\prime\prime}\).

All three parts run off the formula \(\bigl(\pi^{\prime}(\varphi)\bigr)(v) = \varphi(v + U)\), from \(\pi(v) = v + U\) and the definition 3.118 of the dual map. No finite-dimensionality is used.

(a) If \(\varphi \in \operatorname{null}\pi^{\prime}\), then \(\varphi(v + U) = 0\) for every \(v \in V\); since \(\pi\) is surjective, this says \(\varphi\) vanishes on \(V/U\), so \(\varphi = 0\). Hence \(\operatorname{null}\pi^{\prime} = \{0\}\) and 3.15 gives injectivity.

(b) For \(\varphi \in (V/U)^{\prime}\) and \(u \in U\) we have \(u + U = 0 + U\) by 3.101, so

\begin{equation*} \bigl( \pi^{\prime}(\varphi) \bigr)(u) = \varphi(0 + U) = 0 \end{equation*}

(\(0 + U\) is the additive identity of \(V/U\) and \(\varphi\) is linear, 3.10); thus \(\pi^{\prime}(\varphi) \in U^0\) by 3.121.

Conversely, let \(\psi \in U^0\) and define \(\varphi \colon V/U \to \mathbf{F}\) by \(\varphi(v + U) = \psi(v)\). This is well defined, since \(v + U = w + U\) gives \(v - w \in U\) (3.101) and hence \(\psi(v) - \psi(w) = \psi(v - w) = 0\); and it is linear, by the definitions of the operations on \(V/U\) together with linearity of \(\psi\) (Check!). Then \(\bigl(\pi^{\prime}(\varphi)\bigr)(v) = \psi(v)\) for all \(v\), so \(\psi \in \operatorname{range}\pi^{\prime}\). Hence \(\operatorname{range}\pi^{\prime} = U^0\).

(c) By 3.124, \(U^0\) is a subspace of \(V^{\prime}\), and by (b) we may regard \(\pi^{\prime}\) as a map into \(U^0\); it is injective by (a) and surjective onto \(U^0\) by (b), hence invertible (3.63). So \(\pi^{\prime}\) is an isomorphism from \((V/U)^{\prime}\) onto \(U^0\).

Write \(z = a + bi\) and \(w = c + di\) with \(a, b, c, d \in \mathbb{R}\), so that by 4.2

\begin{equation*} \begin{aligned} \bar{z} &= a - bi, \quad &|z| &= \sqrt{a^2 + b^2}, \\ \bar{w} &= c - di, \quad &|w| &= \sqrt{c^2 + d^2}. \end{aligned} \end{equation*}

(a) \(z + \bar{z} = (a + bi) + (a - bi) = 2a = 2\operatorname{Re} z\).

(b) \(z - \bar{z} = (a + bi) - (a - bi) = 2bi = 2(\operatorname{Im} z)i\).

(c) Using \(i^2 = -1\),

\begin{equation*} z\bar{z} = (a + bi)(a - bi) = a^2 + b^2 = |z|^2 . \end{equation*}

(d) Since \(w + z = (c + a) + (d + b)i\) with real parts and imaginary parts as shown,

\begin{equation*} \overline{w + z} = (c + a) - (d + b)i = (c - di) + (a - bi) = \bar{w} + \bar{z}. \end{equation*}

And \(wz = (ca - db) + (cb + da)i\) with \(ca - db, cb + da \in \mathbb{R}\), so

\begin{equation*} \overline{wz} = (ca - db) - (cb + da)i = (c - di)(a - bi) = \bar{w}\,\bar{z} . \end{equation*}

(e) \(\bar{z} = a + (-b)i\) with \(a, -b \in \mathbb{R}\), so \(\bar{\bar{z}} = a - (-b)i = a + bi = z\).

(f) From \(b^2 \ge 0\) and \(a^2 \ge 0\), with the square root increasing on \([0, \infty)\),

\begin{equation*} \begin{aligned} |\operatorname{Re} z| &= \sqrt{a^2} \le \sqrt{a^2 + b^2} = |z|, \\ |\operatorname{Im} z| &= \sqrt{b^2} \le \sqrt{a^2 + b^2} = |z| . \end{aligned} \end{equation*}

(g) By (e), \(\operatorname{Re}\bar z = a\) and \(\operatorname{Im}\bar z = -b\), so \(|\bar{z}| = \sqrt{a^2 + b^2} = |z|\).

(h) By (c) three times, (d) once, and commutativity,

\begin{equation*} \begin{aligned} |wz|^2 &= (wz)\overline{(wz)} = (w\bar{w})(z\bar{z}) \\ &= |w|^2 |z|^2 = \bigl(|w|\,|z|\bigr)^2 , \end{aligned} \end{equation*}

and both \(|wz|\) and \(|w|\,|z|\) are nonnegative, so \(|wz| = |w|\,|z|\).

The triangle inequality (4.4) applied to \(w = (w - z) + z\) gives

\begin{equation*} |w| \le |w - z| + |z| , \qquad\text{so}\qquad |w| - |z| \le |w - z| . \end{equation*}

Interchanging \(w\) and \(z\) gives \(|z| - |w| \le |z - w| = |w - z|\), the last equality because multiplicativity of absolute value (4.4, Exercise 4.1(h)) gives \(|-u| = |-1|\,|u| = |u|\), as \(|-1| = 1\). Since \(\bigl|\,|w| - |z|\,\bigr|\) is one of the two left sides, it is at most \(|w - z|\).

Fix \(v \in V\) and write \(\varphi(v) = \alpha + \beta i\) with \(\alpha, \beta \in \mathbb{R}\), so \(\sigma(v) = \alpha\). Complex homogeneity of \(\varphi\) at the scalar \(i\) gives

\begin{equation*} \varphi(iv) = i\varphi(v) = i(\alpha + \beta i) = -\beta + \alpha i , \end{equation*}

whose real part is \(-\beta\) since \(-\beta, \alpha \in \mathbb{R}\); thus \(\sigma(iv) = -\beta\) and

\begin{equation*} \sigma(v) - i\sigma(iv) = \alpha - i(-\beta) = \alpha + \beta i = \varphi(v) . \end{equation*}

No: take \(p(z) = z^m\) and \(q(z) = 1 - z^m\). Both lie in the set \(U\) in question, since each has \(z^m\)-coefficient \(\pm 1 \ne 0\) and no higher power, so \(\deg p = \deg q = m\). But

\begin{equation*} (p + q)(z) = z^m + (1 - z^m) = 1 , \end{equation*}

the nonzero constant polynomial, of degree \(0 \ne m\) since \(m\) is positive. So \(p + q \notin U\) and \(U\) is not closed under addition, hence not a subspace of \(\mathcal{P}(\mathbb{F})\).

No: take \(p(z) = z^2 + z\) and \(q(z) = -z^2\). Both have degree \(2\), so both lie in the set \(U\) in question, but

\begin{equation*} (p + q)(z) = (z^2 + z) + (-z^2) = z \end{equation*}

is nonzero of odd degree \(1\). So \(p + q \notin U\), and as in Exercise 4.4 closure under addition fails, hence \(U\) is not a subspace of \(\mathcal{P}(\mathbb{F})\).

Take

\begin{equation*} p(z) = (z - \lambda_1)^{\,n - m + 1}(z - \lambda_2)\cdots(z - \lambda_m) , \end{equation*}

which reads \(p(z) = (z - \lambda_1)^{\,n}\) when \(m = 1\); the exponent \(n - m + 1\) is a positive integer because \(m \le n\).

Degree: the \(n\) factors are each monic of degree \(1\), and a product of monic polynomials is monic with degree the sum of the degrees (multiplying out produces the top power with coefficient \(1\) and nothing higher), so \(\deg p = (n - m + 1) + (m - 1) = n\); in particular \(p \ne 0\).

Zeros: \(p(\lambda_k) = 0\) for each \(k\), since the factor \(\lambda_k - \lambda_k\) occurs. Conversely, if \(p(\lambda) = 0\), then because \(\mathbb{F}\) has no zero divisors some factor \(\lambda - \lambda_k\) vanishes, so \(\lambda \in \{\lambda_1, \ldots, \lambda_m\}\). Thus the zero set of \(p\) is exactly \(\{\lambda_1, \ldots, \lambda_m\}\).

The evaluation map \(T \colon \mathcal{P}_m(\mathbb{F}) \to \mathbb{F}^{m+1}\),

\begin{equation*} Tp = \bigl(p(z_1), \ldots, p(z_{m+1})\bigr) , \end{equation*}

is invertible; existence and uniqueness are then its surjectivity and injectivity at \((w_1, \ldots, w_{m+1})\). It is linear because addition and scalar multiplication of functions are pointwise.

\(T\) is injective. If \(Tp = 0\), then \(p\) vanishes at the \(m + 1\) distinct points \(z_1, \ldots, z_{m+1}\). Were \(p \ne 0\), then either \(\deg p = 0\), so \(p\) is a nonzero constant with no zeros at all, or \(\deg p = k\) with \(1 \le k \le m\), and 4.8 allows \(p\) at most \(k \le m\) zeros; both contradict having \(m + 1\) zeros. So \(\operatorname{null} T = \{0\}\) and 3.15 applies.

Since \(1, z, \ldots, z^m\) is a basis of \(\mathcal{P}_m(\mathbb{F})\) (2.36),

\begin{equation*} \dim \mathcal{P}_m(\mathbb{F}) = m + 1 = \dim \mathbb{F}^{m+1}, \end{equation*}

so the injective map \(T\) between finite-dimensional spaces of equal dimension is invertible by 3.65. Hence there is exactly one \(p \in \mathcal{P}_m(\mathbb{F})\) with \(Tp = (w_1, \ldots, w_{m+1})\), that is, with \(p(z_k) = w_k\) for each \(k\).

The equivalence is read off the factorization \(p(z) = c(z - \lambda_1)\cdots(z - \lambda_m)\) with \(c \ne 0\) supplied by 4.13, whose zero set is \(\{\lambda_1, \ldots, \lambda_m\}\); so \(p\) has \(m\) distinct zeros exactly when \(\lambda_1, \ldots, \lambda_m\) are distinct. Throughout, the product rule \((fg)^{\prime} = f^{\prime}g + fg^{\prime}\) holds on \(\mathcal{P}(\mathbb{C})\): both sides are bilinear in \((f, g)\), so it suffices to check monomials, where it reads \((j+k)z^{j+k-1} = jz^{j-1}z^k + z^j kz^{k-1}\).

If \(m = 0\), then \(p\) is a nonzero constant with no zeros at all, so \(p\) has \(m = 0\) distinct zeros and shares no zero with \(p^{\prime}\); both sides hold. Assume \(m \ge 1\).

(i) \(\lambda_1, \ldots, \lambda_m\) distinct. The product rule applied repeatedly gives

\begin{equation*} p^{\prime}(z) = c \sum_{k=1}^{m} \; \prod_{j \neq k} (z - \lambda_j) , \end{equation*}

and at \(z = \lambda_k\) every summand but the \(k\)th contains the factor \(\lambda_k - \lambda_k = 0\), leaving

\begin{equation*} p^{\prime}(\lambda_k) = c \prod_{j \neq k} (\lambda_k - \lambda_j) \neq 0 \end{equation*}

since \(c \ne 0\) and the \(\lambda\)’s are distinct. So no zero of \(p\) is a zero of \(p^{\prime}\).

(ii) Two of the \(\lambda\)’s coincide, say \(\lambda_1 = \lambda_2 =: \lambda\). With \(g(z) = c(z - \lambda_3)\cdots(z - \lambda_m)\) we get \(p(z) = (z - \lambda)^2 g(z)\), so the product rule gives

\begin{equation*} p^{\prime}(z) = 2(z - \lambda)g(z) + (z - \lambda)^2 g^{\prime}(z) , \end{equation*}

whence \(p^{\prime}(\lambda) = 0 = p(\lambda)\) and \(\lambda\) is a common zero. Contrapositively, no common zero forces \(m\) distinct zeros.

The zero is \(\lambda_1\), the first linear factor in the real factorization. Suppose \(p \in \mathcal{P}(\mathbb{R})\) has odd degree \(n\), so \(n \ge 1\) and \(p\) is nonconstant; by 4.16 there are reals \(c, \lambda_1, \ldots, \lambda_m, b_1, \ldots, b_M, c_1, \ldots, c_M\) with \(b_k^2 < 4c_k\) for each \(k\) and

\begin{equation*} \begin{aligned} p(x) = c(x - \lambda_1) &\cdots (x - \lambda_m) \\ &\times (x^2 + b_1 x + c_1) \cdots (x^2 + b_M x + c_M) . \end{aligned} \end{equation*}

Here \(c \ne 0\), since otherwise the right side would be the zero polynomial. Comparing degrees,

\begin{equation*} n = m + 2M , \end{equation*}

so \(m\) is odd because \(n\) is; in particular \(m \ge 1\), the factor \(x - \lambda_1\) is present, and \(p(\lambda_1) = 0\).

\(Tp\) is the quotient \(q\) in the factorization \(p(x) - p(3) = (x-3)q(x)\).

(i) If \(p\) is constant then \((Tp)(x) = 0\) for \(x \neq 3\) and \((Tp)(3) = p^{\prime}(3) = 0\), so \(Tp = 0 \in \mathcal{P}(\mathbb{R})\).

(ii) If \(\deg p = m \geq 1\), then \(x \mapsto p(x) - p(3)\) has degree \(m\) and vanishes at \(3\), so 4.6 supplies \(q \in \mathcal{P}(\mathbb{R})\) with \(\deg q = m - 1\) and

\begin{equation*} p(x) - p(3) = (x - 3)q(x) \qquad \text{for all } x \in \mathbb{R}. \end{equation*}

Dividing gives \((Tp)(x) = q(x)\) for \(x \neq 3\); differentiating gives \(p^{\prime}(x) = q(x) + (x-3)q^{\prime}(x)\), so \(x = 3\) yields \((Tp)(3) = p^{\prime}(3) = q(3)\). Hence \(Tp = q \in \mathcal{P}(\mathbb{R})\).

For linearity, fix \(p_1, p_2 \in \mathcal{P}(\mathbb{R})\) and \(\alpha \in \mathbb{R}\) and compare values at each \(x\). For \(x \neq 3\),

\begin{equation*} \begin{aligned} \bigl(T(\alpha p_1 + p_2)\bigr)(x) &= \frac{\alpha p_1(x) + p_2(x) - \alpha p_1(3) - p_2(3)}{x - 3} \\ &= \alpha (Tp_1)(x) + (Tp_2)(x), \end{aligned} \end{equation*}

while at \(x = 3\), since differentiation is linear,

\begin{equation*} \bigl(T(\alpha p_1 + p_2)\bigr)(3) = \alpha p_1^{\prime}(3) + p_2^{\prime}(3) = \alpha (Tp_1)(3) + (Tp_2)(3). \end{equation*}

Thus \(T(\alpha p_1 + p_2) = \alpha \, Tp_1 + Tp_2\), so \(T\) is linear.

Writing \(p(z) = \sum_{k=0}^{m} a_k z^k\) and \(\bar{p}(z) = \sum_{k=0}^{m} \overline{a_k} z^k\), we have \(q = p \, \bar{p}\), a product of polynomials and hence a polynomial.

Indeed, conjugation is additive and multiplicative and \(\overline{\bar{z}} = z\) (4.4), so

\begin{equation*} \overline{p(\bar z)} = \sum_{k=0}^{m} \overline{a_k} \bigl(\overline{\bar z}\bigr)^{k} = \sum_{k=0}^{m} \overline{a_k} z^k = \bar{p}(z). \end{equation*}

Expanding the product and collecting powers of \(z\),

\begin{equation*} q(z) = \sum_{n=0}^{2m} c_n z^n, \qquad c_n = \sum_{\substack{j + k = n \\ 0 \leq j,\, k \leq m}} a_j \overline{a_k}, \end{equation*}

and by the uniqueness of coefficients (a consequence of 4.8) these are the coefficients of \(q\). The swap \((j,k) \mapsto (k,j)\) carries the index set to itself, so conjugating and relabeling gives

\begin{equation*} \overline{c_n} = \sum_{j+k=n} \overline{a_j} a_k = \sum_{j+k=n} a_j \overline{a_k} = c_n , \end{equation*}

whence \(c_n \in \mathbb{R}\) (4.4) for every \(n\). Thus \(q\) is a polynomial with real coefficients.

The polynomial of imaginary parts vanishes identically. Write \(p(z) = \sum_{k=0}^{m} a_k z^k\) with \(a_k \in \mathbb{C}\) and set

\begin{equation*} u(x) = \sum_{k=0}^{m} (\operatorname{Im} a_k) x^k \in \mathcal{P}_m(\mathbb{R}). \end{equation*}

For \(x \in \mathbb{R}\) each \(x^k\) is real, and \(\operatorname{Im}\) is additive with \(\operatorname{Im}(a t) = t \operatorname{Im} a\) for \(t \in \mathbb{R}\), so

\begin{equation*} \operatorname{Im} p(x) = \sum_{k=0}^{m} \operatorname{Im}\bigl(a_k x^k\bigr) = u(x). \end{equation*}

By hypothesis \(u(x_0) = u(x_1) = \cdots = u(x_m) = 0\), so \(u\) has \(m+1\) distinct zeros. Were \(u \neq 0\), its degree \(d\) would satisfy \(0 \leq d \leq m\), and \(u\) would have at most \(d \leq m\) zeros (none if \(d = 0\); by 4.8 if \(d \geq 1\)) – a contradiction. Hence \(u = 0\), so by uniqueness of coefficients \(\operatorname{Im} a_k = 0\) for every \(k\); all coefficients of \(p\) are real.

Method (2): put \(w_k = p(x_k) \in \mathbb{R}\). Exercise 4.7 over \(\mathbb{R}\) gives \(\tilde{p} \in \mathcal{P}_m(\mathbb{R})\) with \(\tilde{p}(x_k) = w_k\) for each \(k\); viewing \(\tilde{p} \in \mathcal{P}_m(\mathbb{C})\), the uniqueness assertion of Exercise 4.7 over \(\mathbb{C}\) forces \(p = \tilde{p}\).

With \(m = \deg p\), the cosets

\begin{equation*} 1 + U, \; z + U, \; z^2 + U, \; \ldots, \; z^{m-1} + U \end{equation*}

form a basis of \(\mathcal{P}(\mathbb{F})/U\), whence \(\dim \mathcal{P}(\mathbb{F})/U = m = \deg p\); this answers (b) and (a). (For \(m = 0\) the list is empty and \(U = \mathcal{P}(\mathbb{F})\).) Here \(U\) is indeed a subspace, since \(0 = p \cdot 0\), \(pq_1 + pq_2 = p(q_1+q_2)\) and \(\lambda(pq) = p(\lambda q)\).

Spanning: given \(r \in \mathcal{P}(\mathbb{F})\), the division algorithm 4.9 (legitimate as \(p \neq 0\)) yields \(q, s \in \mathcal{P}(\mathbb{F})\) with \(r = pq + s\) and \(\deg s < m\), so \(s = a_0 + a_1 z + \cdots + a_{m-1} z^{m-1}\); as \(pq \in U\) and the quotient map is linear (3.104),

\begin{equation*} r + U = s + U = a_0(1+U) + a_1(z+U) + \cdots + a_{m-1}\bigl(z^{m-1}+U\bigr). \end{equation*}

Independence: if that combination equals \(0 + U\), then \(s = a_0 + \cdots + a_{m-1}z^{m-1}\) lies in \(U\) by 3.101, say \(s = pq\). Were \(q \neq 0\) we would get \(\deg s = m + \deg q \geq m\), contradicting \(\deg s \leq m-1\); so \(q = 0\), hence \(s = 0\), and uniqueness of coefficients (4.8) forces \(a_0 = \cdots = a_{m-1} = 0\).

Here \(m, n \geq 1\), and \(T\) does land in \(\mathcal{P}_{m+n-1}(\mathbb{C})\) because \(\deg(rp) \leq (n-1)+m\) and \(\deg(sq) \leq (m-1)+n\). Throughout we use \(\deg(uv) = \deg u + \deg v\), which makes \(\mathcal{P}(\mathbb{C})\) free of zero divisors and so permits cancellation of a nonzero factor.

(a) Suppose \(T(r,s) = 0\), that is, \(rp = -sq\). Since \(q\) is nonconstant, 4.13 gives \(c \neq 0\) and \(\lambda_1, \ldots, \lambda_n \in \mathbb{C}\) (repetitions allowed) with

\begin{equation*} q(z) = c(z - \lambda_1) \cdots (z - \lambda_n), \end{equation*}

and each \(\lambda_k\), being a zero of \(q\), satisfies \(p(\lambda_k) \neq 0\). Assume \(r \neq 0\); we show by induction on \(j\) that for \(j = 0, 1, \ldots, n\) there is \(r_j \neq 0\) with

\begin{equation*} r(z) = (z - \lambda_1)\cdots(z - \lambda_j)\, r_j(z) . \end{equation*}

Take \(r_0 = r\). Given \(r_j\) with \(j < n\), substituting this factorization of \(r\) into \(rp = -sq\) and cancelling the nonzero factor \((z-\lambda_1)\cdots(z-\lambda_j)\) leaves

\begin{equation*} r_j(z) p(z) = -c\, s(z)\, (z - \lambda_{j+1}) \cdots (z - \lambda_n); \end{equation*}

evaluating at \(z = \lambda_{j+1}\) kills the right side, and \(p(\lambda_{j+1}) \neq 0\) then forces \(r_j(\lambda_{j+1}) = 0\), so 4.6 yields \(r_{j+1}\) of degree \(\deg r_j - 1\) (hence nonzero) with \(r_j(z) = (z-\lambda_{j+1}) r_{j+1}(z)\). At \(j = n\) this gives \(\deg r = n + \deg r_n \geq n\), contradicting \(\deg r \leq n-1\). So \(r = 0\), whence \(sq = 0\) with \(q \neq 0\) gives \(s = 0\). Thus \(\operatorname{null} T = \{(0,0)\}\) and \(T\) is injective (3.15).

(b) Since \(\dim \mathcal{P}_k(\mathbb{C}) = k+1\) and dimension adds over products (3.92),

\begin{equation*} \dim \bigl(\mathcal{P}_{n-1}(\mathbb{C}) \times \mathcal{P}_{m-1}(\mathbb{C})\bigr) = n + m = \dim \mathcal{P}_{m+n-1}(\mathbb{C}), \end{equation*}

so the injective \(T\) between spaces of equal finite dimension is surjective by 3.65.

(c) The constant polynomial \(1\) lies in \(\mathcal{P}_{m+n-1}(\mathbb{C})\), so by (b) there is a pair \((r,s) \in \mathcal{P}_{n-1}(\mathbb{C}) \times \mathcal{P}_{m-1}(\mathbb{C})\) with \(rp + sq = T(r,s) = 1\).

In both parts take \(u \in U\); invariance (5.2) asks that \(Tu \in U\).

(a) If \(U \subseteq \operatorname{null} T\), then \(Tu = 0 \in U\), since the subspace \(U\) contains \(0\).

(b) If \(\operatorname{range} T \subseteq U\), then \(Tu \in \operatorname{range} T \subseteq U\).

\(T\) sends a decomposition of \(v\) to a decomposition of \(Tv\). Let \(v \in V_1 + \dots + V_m\) (a subspace, by 1.40) and write \(v = v_1 + \dots + v_m\) with \(v_k \in V_k\). By additivity of \(T\),

\begin{equation*} Tv = Tv_1 + \dots + Tv_m , \end{equation*}

and \(Tv_k \in V_k\) for each \(k\) since \(V_k\) is invariant. Hence \(Tv \in V_1 + \dots + V_m\).

Membership in an intersection is checked one index at a time. Let \(\{U_a\}_{a \in A}\) be such a collection and \(U = \bigcap_{a \in A} U_a\). Then \(U\) is a subspace by 1.34: it contains \(0\), and \(u, w \in U\), \(\lambda \in \mathbf{F}\) give \(u + w, \lambda u \in U_a\) for every \(a\), hence in \(U\). Now for \(u \in U\) and any \(a \in A\) we have \(u \in U_a\), so invariance of \(U_a\) gives \(Tu \in U_a\); as \(a\) was arbitrary,

\begin{equation*} Tu \in \bigcap_{a \in A} U_a = U . \end{equation*}

True. Suppose \(U \neq \{0\}\) and pick \(u \in U\) with \(u \neq 0\); the one-vector list \(u\) is then linearly independent, so 2.32 extends it to a basis \(u, e_2, \dots, e_n\) of \(V\). Given any \(v \in V\), the linear map lemma 3.4 supplies \(T \in \mathcal{L}(V)\) with

\begin{equation*} Tu = v, \qquad Te_2 = \dots = Te_n = 0 . \end{equation*}

By hypothesis \(U\) is invariant under this \(T\), and \(u \in U\), so \(v = Tu \in U\). Hence \(V \subseteq U\), that is, \(U = V\).

\(T\) has no eigenvalues. Indeed, \(T(x,y) = \lambda(x,y)\) says

\begin{equation*} -3y = \lambda x, \qquad x = \lambda y , \end{equation*}

and substituting the second into the first gives \((\lambda^2 + 3) y = 0\). If \(y = 0\) then \(x = \lambda y = 0\), so an eigenvector (necessarily nonzero, 5.5) has \(y \neq 0\) and hence \(\lambda^2 = -3\), impossible for \(\lambda \in \mathbf{R}\).

The eigenvalues are \(1\) and \(-1\), with eigenvectors the nonzero vectors \((w,w)\) and \((w,-w)\) respectively.

The equation \(T(w,z) = \lambda(w,z)\) reads

\begin{equation*} z = \lambda w, \qquad w = \lambda z , \end{equation*}

whence \((\lambda^2 - 1) w = 0\). If \(w = 0\) then \(z = \lambda w = 0\), excluded for an eigenvector, so \(\lambda^2 = 1\), leaving \(\lambda = \pm 1\).

(i) \(\lambda = 1\): both equations say \(z = w\), and \(T(w,w) = (w,w)\), so

\begin{equation*} \operatorname{null}(T - I) = \operatorname{span}\bigl( (1,1) \bigr) . \end{equation*}

(ii) \(\lambda = -1\): both say \(z = -w\), and \(T(w,-w) = -(w,-w)\), so

\begin{equation*} \operatorname{null}(T + I) = \operatorname{span}\bigl( (1,-1) \bigr) . \end{equation*}

Each null space contains a nonzero vector, so both \(\pm 1\) really are eigenvalues.

The eigenvalues are \(0\) and \(5\), with eigenvectors the nonzero vectors \((z_1,0,0)\) and \((0,0,z_3)\) respectively.

The equation \(T(z_1,z_2,z_3) = \lambda(z_1,z_2,z_3)\) reads

\begin{equation*} 2z_2 = \lambda z_1, \qquad 0 = \lambda z_2, \qquad 5z_3 = \lambda z_3 . \end{equation*}

(i) \(\lambda = 0\): these say \(z_2 = z_3 = 0\), so

\begin{equation*} \operatorname{null} T = \operatorname{span}\bigl( (1,0,0) \bigr) \neq \{0\} , \end{equation*}

and \(0\) is an eigenvalue.

(ii) \(\lambda \neq 0\): the second equation gives \(z_2 = 0\), then the first gives \(z_1 = 0\), so an eigenvector is \((0,0,z_3)\) with \(z_3 \neq 0\) and the third equation forces \(\lambda = 5\). Conversely \(T(0,0,z_3) = 5(0,0,z_3)\), so

\begin{equation*} \operatorname{null}(T - 5I) = \operatorname{span}\bigl( (0,0,1) \bigr) . \end{equation*}

Apply \(P\) to the eigenvalue equation. Take \(v \neq 0\) with \(Pv = \lambda v\) (5.5); then

\begin{equation*} \lambda v = Pv = P^2 v = P(\lambda v) = \lambda^2 v , \end{equation*}

so \(\lambda(\lambda - 1) v = 0\). Since \(v \neq 0\) we get \(\lambda(\lambda - 1) = 0\), and \(\mathbf{F}\) has no zero divisors, so \(\lambda = 0\) or \(\lambda = 1\).

The only eigenvalue is \(0\), whose eigenvectors are the nonzero constant polynomials.

Degree settles it. Suppose \(p \neq 0\) has degree \(m\) and \(p^{\prime} = \lambda p\) with \(\lambda \neq 0\). If \(m = 0\) then \(p^{\prime} = 0\) while \(\lambda p\) is the nonzero constant \(\lambda a_0\); if \(m \geq 1\) then

\begin{equation*} p^{\prime}(x) = a_1 + 2 a_2 x + \cdots + m a_m x^{m-1} \end{equation*}

has degree \(m - 1\) (its leading coefficient \(m a_m\) is nonzero, \(\mathbf{R}\) having characteristic \(0\)) whereas \(\deg(\lambda p) = m\). Either way \(p^{\prime} \neq \lambda p\), so no nonzero \(\lambda\) is an eigenvalue.

For \(\lambda = 0\) the condition \(p^{\prime} = 0\) says \(a_1 = 2a_2 = \cdots = m a_m = 0\), i.e. \(p\) is constant; the constant \(1\) is a nonzero such \(p\), so \(0\) is an eigenvalue.

The eigenvalues are \(0, 1, 2, 3, 4\), the eigenvectors for \(k\) being the nonzero multiples of \(x^k\); this is immediate from \(T x^j = x \cdot j x^{j-1} = j x^j\).

For completeness, write \(p = \sum_{j=0}^{4} a_j x^j\), so that \(Tp = \sum_{j=0}^{4} j \, a_j x^j\). Since \(1, x, x^2, x^3, x^4\) is a basis of \(\mathcal{P}_4(\mathbf{R})\), the equation \(Tp = \lambda p\) is equivalent to

\begin{equation*} (j - \lambda) a_j = 0 \qquad \text{for } j = 0, 1, 2, 3, 4 . \end{equation*}

(i) \(\lambda \notin \{0,1,2,3,4\}\): every \(a_j = 0\), so \(p = 0\) and \(\lambda\) is not an eigenvalue.

(ii) \(\lambda = k \in \{0,1,2,3,4\}\): \(a_j = 0\) for \(j \neq k\), so \(p(x) = a_k x^k\).

Take \(\delta\) to be the distance from \(\alpha\) to the nearest eigenvalue of \(T\) other than \(\alpha\) (and \(\delta = 1\) if there is none). By 5.12 the set

\begin{equation*} E = \{\mu \in \mathbf{F} : \mu \text{ is an eigenvalue of } T,\ \mu \neq \alpha\} \end{equation*}

is finite, say \(E = \{\lambda_1, \dots, \lambda_m\}\), so

\begin{equation*} \delta = \min\{|\alpha - \lambda_1|, \dots, |\alpha - \lambda_m|\} > 0 , \end{equation*}

each \(|\alpha - \lambda_j|\) being positive.

If \(0 < |\alpha - \lambda| < \delta\), then \(\lambda \neq \alpha\) and \(\lambda \notin E\) (otherwise \(|\alpha - \lambda| \geq \delta\)), so \(\lambda\) is not an eigenvalue of \(T\). As \(V\) is finite-dimensional, 5.7 then gives \(T - \lambda I\) invertible.

The eigenvalues are \(0\) and \(1\), with eigenvectors the nonzero vectors of \(W\) and of \(U\) respectively; both \(U\) and \(W\) are nonzero, so both are genuinely eigenvalues.

Since \(Pu = u\) for \(u \in U\), we have \(P^2 = P\), so Exercise 5A.8 confines the eigenvalues to \(\{0, 1\}\). Now decompose \(v = u + w\) with \(u \in U\), \(w \in W\) (uniquely, by 1.41), so that \(Pv = u\).

(i) \(Pv = v\) reads \(u = u + w\), i.e. \(w = 0\); thus \(\operatorname{null}(P - I) = U\).

(ii) \(Pv = 0\) reads \(u = 0\); thus \(\operatorname{null} P = W\).

(a) Conjugation respects the shift by \(\lambda I\): for each \(\lambda \in \mathbf{F}\),

\begin{equation*} S^{-1} T S - \lambda I = S^{-1} (T - \lambda I) S , \end{equation*}

since \(S^{-1}(\lambda I)S = \lambda I\). As \(S\) and \(S^{-1}\) are bijective, the left side is injective exactly when \(T - \lambda I\) is, and \(\lambda\) is an eigenvalue of an operator exactly when that operator minus \(\lambda I\) fails to be injective (5.5). Hence \(T\) and \(S^{-1}TS\) have the same eigenvalues.

(b) \(S\) carries eigenvectors of \(S^{-1}TS\) to eigenvectors of \(T\) for the same eigenvalue. Indeed, for \(w \neq 0\) (so \(Sw \neq 0\), \(S\) being injective), applying the invertible \(S\) turns \((S^{-1}TS)w = \lambda w\) into the equivalent equation

\begin{equation*} T(Sw) = \lambda (Sw) . \end{equation*}

Equivalently, \(\operatorname{null}(S^{-1}TS - \lambda I) = S^{-1}\bigl(\operatorname{null}(T - \lambda I)\bigr)\) for every \(\lambda\).

Take the quarter turn in each coordinate plane:

\begin{equation*} T(x_1, x_2, x_3, x_4) = (-x_2,\; x_1,\; -x_4,\; x_3) , \end{equation*}

which is linear since each output coordinate is linear in the input. If \(Tv = \lambda v\) with \(\lambda \in \mathbf{R}\), comparing coordinates gives

\begin{equation*} -x_2 = \lambda x_1, \qquad x_1 = \lambda x_2, \qquad -x_4 = \lambda x_3, \qquad x_3 = \lambda x_4 . \end{equation*}

Substituting the second into the first yields \((\lambda^2 + 1) x_2 = 0\), and \(\lambda^2 + 1 > 0\) for real \(\lambda\), so \(x_2 = 0\) and then \(x_1 = \lambda x_2 = 0\); the last two equations give \(x_4 = x_3 = 0\) the same way. Thus \(v = 0\), so \(T\) has no eigenvalue (5.5).

Dualizing turns surjectivity of \(T - \lambda I\) into injectivity of \(T^{\prime} - \lambda I_{V^{\prime}}\). Since \(I^{\prime}(\varphi) = \varphi \circ I = \varphi\), additivity and homogeneity of the dual map (3.120) give

\begin{equation*} (T - \lambda I)^{\prime} = T^{\prime} - \lambda I^{\prime} = T^{\prime} - \lambda I_{V^{\prime}} . \end{equation*}

Also \(\dim V^{\prime} = \dim V < \infty\) (3.111), so 5.7 applies to both \(T\) and \(T^{\prime}\):

\begin{equation*} \begin{aligned} \lambda \text{ is an eigenvalue of } T &\iff T - \lambda I \text{ is not surjective} \\ &\iff (T - \lambda I)^{\prime} \text{ is not injective} \\ &\iff T^{\prime} - \lambda I_{V^{\prime}} \text{ is not injective} \\ &\iff \lambda \text{ is an eigenvalue of } T^{\prime} , \end{aligned} \end{equation*}

the first and last steps by 5.7 and the second by 3.129 (which needs \(V\) finite-dimensional), applied to \(T - \lambda I\).

Test the eigenvector equation in the coordinate of largest modulus. Write \(A = \mathcal{M}(T)\), so \(Tv_k = \sum_{j} A_{j,k} v_j\) (3.31), and put \(C = \max_{j,k} |A_{j,k}|\). Let \(v = \sum_k c_k v_k\) be an eigenvector for \(\lambda\), and choose \(m\) with \(|c_m| = \max_k |c_k|\); since \(v \neq 0\) we have \(|c_m| > 0\). Expanding \(Tv = \lambda v\) in the basis and comparing the (unique) coefficients of \(v_j\),

\begin{equation*} \lambda c_j = \sum_{k=1}^{n} A_{j,k} c_k \qquad \text{for } j = 1, \dots, n . \end{equation*}

Taking \(j = m\), the triangle inequality with \(|A_{m,k}| \leq C\) and \(|c_k| \leq |c_m|\) gives

\begin{equation*} |\lambda| \, |c_m| = \Bigl| \sum_{k=1}^{n} A_{m,k} c_k \Bigr| \leq \sum_{k=1}^{n} |A_{m,k}| \, |c_k| \leq n C |c_m| , \end{equation*}

and dividing by \(|c_m| > 0\) yields \(|\lambda| \leq nC\).

The eigenvector equation splits into its two components. Since \(\lambda\) is real, \(\lambda(u + iv) = \lambda u + i \lambda v\), while \(T_{\mathbf{C}}(u+iv) = Tu + iTv\); as elements of \(V_{\mathbf{C}} = V \times V\) are equal exactly when both components agree,

\begin{equation*} T_{\mathbf{C}}(u+iv) = \lambda(u+iv) \iff Tu = \lambda u \ \text{ and } \ Tv = \lambda v . \end{equation*}

(i) If \(Tv = \lambda v\) with \(v \neq 0\), then \(v + i0 \neq 0\) in \(V_{\mathbf{C}}\) and the equivalence gives \(T_{\mathbf{C}}(v + i0) = \lambda(v + i0)\), so \(\lambda\) is an eigenvalue of \(T_{\mathbf{C}}\).

(ii) If \(T_{\mathbf{C}}(u+iv) = \lambda(u+iv)\) with \(u + iv \neq 0\), then \(u, v\) are not both \(0\), and whichever is nonzero is an eigenvector of \(T\) corresponding to \(\lambda\).

Conjugation \(\sigma(u + iv) = u - iv\) carries eigenvectors for \(\lambda\) to eigenvectors for \(\overline{\lambda}\). Since the two directions are symmetric (\(\sigma \circ \sigma\) is the identity, and \(\overline{\overline{\lambda}} = \lambda\)), it suffices to treat one. Three properties of \(\sigma\), all read off from \(T_{\mathbf{C}}(u+iv) = Tu + iTv\) and \((a+bi)(u+iv) = (au - bv) + i(av+bu)\):

(i) \(\sigma\) is injective, being the map \((u,v) \mapsto (u,-v)\) on \(V_{\mathbf{C}} = V \times V\); in particular \(\sigma(w) = 0\) only for \(w = 0\).

(ii) \(\sigma\) is conjugate-homogeneous: with \(\mu = a + bi\), both \(\sigma(\mu(u+iv))\) and \(\overline{\mu}\,\sigma(u+iv)\) equal \((au - bv) + i(-av - bu)\). (Check!)

(iii) \(\sigma\) commutes with \(T_{\mathbf{C}}\), since \(T(-v) = -Tv\):

\begin{equation*} \sigma\bigl(T_{\mathbf{C}}(u+iv)\bigr) = Tu + i\,T(-v) = T_{\mathbf{C}}\bigl(\sigma(u+iv)\bigr) . \end{equation*}

So if \(T_{\mathbf{C}} w = \lambda w\) with \(w \neq 0\), then \(\sigma(w) \neq 0\) and

\begin{equation*} T_{\mathbf{C}}\bigl(\sigma(w)\bigr) = \sigma(T_{\mathbf{C}} w) = \sigma(\lambda w) = \overline{\lambda}\, \sigma(w) , \end{equation*}

so \(\overline{\lambda}\) is an eigenvalue of \(T_{\mathbf{C}}\).

For every \(\lambda\), the equation \(Tz = \lambda z\) forces \(z = 0\). Comparing coordinates in

\begin{equation*} (0, z_1, z_2, \dots) = (\lambda z_1, \lambda z_2, \lambda z_3, \dots) \end{equation*}

gives \(\lambda z_1 = 0\) and \(z_k = \lambda z_{k+1}\) for every \(k \geq 1\).

(i) \(\lambda = 0\): the second family reads \(z_k = 0\) for every \(k\).

(ii) \(\lambda \neq 0\): then \(z_1 = 0\), and \(z_{k+1} = z_k / \lambda\) propagates this up the sequence by induction, so again every \(z_k = 0\).

An eigenvector must be nonzero, so \(T\) has no eigenvalues.

(a) For \(\lambda \in \mathbf{F}\) take the geometric sequence

\begin{equation*} w_\lambda = (1, \lambda, \lambda^2, \lambda^3, \dots) \qquad (\lambda^0 = 1) , \end{equation*}

which is nonzero (first coordinate \(1\)) and satisfies \(S w_\lambda = (\lambda, \lambda^2, \dots) = \lambda w_\lambda\), the \(k\)-th coordinates being \(\lambda^k = \lambda \cdot \lambda^{k-1}\). So every \(\lambda \in \mathbf{F}\) is an eigenvalue.

(b) The eigenvectors are exactly the vectors \(c\, w_\lambda\) with \(\lambda \in \mathbf{F}\) and \(c \neq 0\), such a vector corresponding to the eigenvalue \(\lambda\). Each is nonzero and satisfies \(S(c w_\lambda) = \lambda (c w_\lambda)\) by (a). Conversely, if \(Sz = \lambda z\) with \(z \neq 0\), comparing \(k\)-th coordinates in \((z_2, z_3, \dots) = (\lambda z_1, \lambda z_2, \dots)\) gives \(z_{k+1} = \lambda z_k\), hence by induction

\begin{equation*} z_k = \lambda^{k-1} z_1 \qquad \text{for every positive integer } k . \end{equation*}

Here \(z_1 \neq 0\), since \(z_1 = 0\) would force \(z = 0\); so \(z = z_1 w_\lambda\).

Both parts rest on one computation: if \(\lambda \neq 0\) and \(Tv = \lambda v\), then applying \(T^{-1}\) gives \(v = T^{-1}(\lambda v) = \lambda\, T^{-1}v\), so

\begin{equation*} T^{-1} v = \frac{1}{\lambda}\, v , \end{equation*}

with the very same vector \(v\). Since \(T^{-1}\) is invertible with \((T^{-1})^{-1} = T\), the same computation run with \(T^{-1}, \frac{1}{\lambda}\) in place of \(T, \lambda\) turns \(T^{-1}v = \frac{1}{\lambda}v\) back into \(Tv = \lambda v\).

(a) If \(\lambda \neq 0\) is an eigenvalue of \(T\) with eigenvector \(v\), the display exhibits the nonzero \(v\) as an eigenvector of \(T^{-1}\) for \(\frac{1}{\lambda}\); the reversed computation gives the converse.

(b) If \(v\) is an eigenvector of \(T\), say \(Tv = \lambda v\) with \(v \neq 0\), then \(\lambda \neq 0\) (an invertible \(T\) is injective, so \(\operatorname{null} T = \{0\}\)), and the display makes \(v\) an eigenvector of \(T^{-1}\). Applying this to \(T^{-1}\), whose inverse is \(T\), gives the reverse inclusion. Hence \(T\) and \(T^{-1}\) have the same eigenvectors.

The candidate eigenvectors are \(u + w\) and \(u - w\):

\begin{equation*} \begin{aligned} T(u + w) &= Tu + Tw = 3w + 3u = 3(u + w), \\ T(u - w) &= Tu - Tw = 3w - 3u = -3(u - w). \end{aligned} \end{equation*}

They are not both zero, since \((u+w) + (u-w) = 2u \neq 0\). So:

(i) if \(u + w \neq 0\), the first line makes \(3\) an eigenvalue of \(T\);

(ii) if \(u + w = 0\), then \(u - w = 2u \neq 0\) and the second line makes \(-3\) an eigenvalue of \(T\).

By symmetry in \(S\) and \(T\), it suffices to show every eigenvalue \(\lambda\) of \(ST\) is an eigenvalue of \(TS\). Fix \(v \neq 0\) with \((ST)v = \lambda v\).

(i) \(\lambda \neq 0\). Then \(w := Tv \neq 0\), since \(Tv = 0\) would give \(\lambda v = S0 = 0\) and hence \(v = 0\); and

\begin{equation*} (TS)w = T\bigl((ST)v\bigr) = T(\lambda v) = \lambda\, Tv = \lambda w . \end{equation*}

(ii) \(\lambda = 0\). Then \(ST\) is not invertible by 5.7 (\(V\) finite-dimensional), so \(S\) or \(T\) is not invertible. Were \(TS\) invertible, \(S\) would be injective (\(Sv = 0\) forces \((TS)v = 0\), so \(v = 0\)) hence invertible by 3.65, and then \(T = (TS)S^{-1}\) would be invertible too, a contradiction. So \(TS\) is not invertible, and \(0\) is an eigenvalue of \(TS\) by 5.7.

(a) The eigenvector is \(x = (1, \dots, 1)^{\mathsf{T}} \neq 0\): its image has \(j^{\text{th}}\) entry \(\sum_{k=1}^{n} A_{j,k} \cdot 1 = 1\) by the row hypothesis, so \(Tx = x = 1 \cdot x\).

(b) Here \(T - I\) fails to be surjective. Put \(B = A - I_n\), so \((T - I)x = Bx\); each column of \(B\) sums to \(0\), since subtracting \(I_n\) removes exactly \(1\) from the diagonal entry of column \(k\). Hence for every \(x \in \mathbf{F}^n\), swapping the order of summation,

\begin{equation*} \sum_{j=1}^{n} (Bx)_j = \sum_{k=1}^{n} x_k \Bigl( \sum_{j=1}^{n} B_{j,k} \Bigr) = 0 , \end{equation*}

so \(\operatorname{range}(T - I) \subseteq U := \{y : y_1 + \dots + y_n = 0\}\). Now \(U\) is the null space of the surjective linear functional \(y \mapsto y_1 + \dots + y_n\), so \(\dim U = n - 1\) by 3.21 and \(U \neq \mathbf{F}^n\). Thus \(T - I\) is not surjective, and \(1\) is an eigenvalue of \(T\) by 5.7.

Let \(Tu = \alpha u\), \(Tw = \beta w\), \(T(u+w) = \lambda(u+w)\) with \(u, w, u+w\) nonzero; we show \(\alpha = \beta\). Linearity gives

\begin{equation*} \lambda u + \lambda w = Tu + Tw = \alpha u + \beta w , \end{equation*}

that is, \((\alpha - \lambda)u + (\beta - \lambda)w = 0\). If \(\alpha \neq \beta\), then \(u, w\) is linearly independent by 5.11 (eigenvectors for distinct eigenvalues), so both coefficients vanish and \(\alpha = \lambda = \beta\) – contradicting \(\alpha \neq \beta\). Hence \(\alpha = \beta\).

If \(V = \{0\}\) then \(T = 0 = 0 \cdot I\), so assume \(V \neq \{0\}\). By hypothesis each nonzero \(v \in V\) has a scalar \(a_v\) with \(Tv = a_v v\), and \(a_v\) is unique because \(v \neq 0\). It suffices to show \(a_u = a_w\) for all nonzero \(u, w\).

(i) \(u, w\) linearly dependent. Then \(w = cu\) with \(c \neq 0\) (as \(u, w \neq 0\)), so

\begin{equation*} a_w w = Tw = c\,Tu = c\,a_u u = a_u w , \end{equation*}

whence \(a_w = a_u\).

(ii) \(u, w\) linearly independent. Then \(u + w \neq 0\) and

\begin{equation*} a_{u+w} u + a_{u+w} w = T(u+w) = a_u u + a_w w , \end{equation*}

so independence forces \(a_u = a_{u+w} = a_w\).

Writing \(\lambda\) for the common value, \(Tv = \lambda v\) for all nonzero \(v\), and \(T0 = 0 = \lambda \cdot 0\). Hence \(T = \lambda I\).

It suffices to show every nonzero \(v \in V\) is an eigenvector of \(T\), for then 5A.26 gives \(T = \lambda I\). Set \(n = \dim V\), so \(n \geq 2\) because \(k \in \{1, \dots, n-1\}\).

Suppose instead \(v \neq 0\) with \(Tv \notin \operatorname{span}(v)\), so that \(v, Tv\) is linearly independent; extend it by 2.32 to a basis \(v, Tv, w_3, \dots, w_n\) of \(V\) and put

\begin{equation*} U = \operatorname{span}(v, w_3, \dots, w_{k+1}) , \end{equation*}

legitimate since \(k + 1 \leq n\) (the list \(w_3, \dots, w_{k+1}\) is empty when \(k = 1\)). Being a sublist of a basis, \(v, w_3, \dots, w_{k+1}\) is linearly independent of length \(k\), so \(\dim U = k\) and \(U\) is invariant under \(T\) by hypothesis. Then \(Tv \in U\) expresses \(Tv\) in the basis \(v, Tv, w_3, \dots, w_n\) with coefficient \(0\) on \(Tv\), while the obvious expression has coefficient \(1\); uniqueness of basis representations (2.28) gives \(1 = 0\), a contradiction.

Hence \(Tv \in \operatorname{span}(v)\) for every nonzero \(v\), which is exactly the hypothesis of 5A.26. So \(T\) is a scalar multiple of the identity.

Let \(\lambda_1, \dots, \lambda_m\) be distinct eigenvalues of \(T\) with eigenvectors \(v_1, \dots, v_m\); this list is linearly independent by 5.11. At most one \(\lambda_j\) is \(0\), so relabel to make \(\lambda_1, \dots, \lambda_{m-1}\) nonzero. Then for \(j \leq m - 1\),

\begin{equation*} v_j = \frac{1}{\lambda_j}\, T v_j = T\Bigl( \frac{v_j}{\lambda_j} \Bigr) \in \operatorname{range} T , \end{equation*}

so \(v_1, \dots, v_{m-1}\) is a linearly independent list inside \(\operatorname{range} T\), a subspace of the finite-dimensional \(V\). Comparing it with a basis of \(\operatorname{range} T\), which spans, 2.22 gives \(m - 1 \leq \dim \operatorname{range} T\), that is, \(m \leq 1 + \dim \operatorname{range} T\).

It is enough that \(T - 9I\) be surjective. Since \(-4, 5, \sqrt{7}\) are three distinct eigenvalues and \(\dim \mathbf{R}^3 = 3\), by 5.12 the operator \(T\) has no other eigenvalue; as \(9 \notin \{-4, 5, \sqrt{7}\}\) (note \(\sqrt{7} < 3\)), the number \(9\) is not an eigenvalue of \(T\). So \(T - 9I\) is injective by 5.7, hence surjective by 3.65. Thus some \(x \in \mathbf{R}^3\) has \((T - 9I)x = (-4, 5, \sqrt{7})\).

Take \(v \neq 0\) with \(Tv = \lambda v\). Since \((T - \mu I)v = (\lambda - \mu)v\) for every \(\mu \in \mathbf{F}\), applying the three factors to \(v\) from the right gives

\begin{equation*} \begin{aligned} (T - 4I)v &= (\lambda - 4)v , \\ (T - 3I)(T - 4I)v &= (\lambda - 4)(\lambda - 3)v , \\ (T - 2I)(T - 3I)(T - 4I)v &= (\lambda - 2)(\lambda - 3)(\lambda - 4)v . \end{aligned} \end{equation*}

The left side of the last line is \(0\) by hypothesis, and \(v \neq 0\), so \((\lambda - 2)(\lambda - 3)(\lambda - 4) = 0\). As \(\mathbf{F}\) has no zero divisors, \(\lambda = 2\) or \(\lambda = 3\) or \(\lambda = 4\).

Take \(T\) to be counterclockwise rotation by \(45\) degrees, that is,

\begin{equation*} T(x,y) = \left( \frac{x - y}{\sqrt{2}},\ \frac{x + y}{\sqrt{2}} \right) , \end{equation*}

which is linear (Check!). Squaring,

\begin{equation*} T^2(x,y) = \left( \frac{-2y}{2},\ \frac{2x}{2} \right) = (-y, x) , \end{equation*}

so \(T^4(x,y) = T^2(-y,x) = (-x,-y)\). Hence \(T^4 = -I\).

Because polynomials in \(T\) factor as polynomials do (5.17) and \(z^4 - 1 = (z-1)(z+1)(z^2+1)\), the hypothesis \(T^4 = I\) gives

\begin{equation*} (T - I)(T + I)(T^2 + I) = T^4 - I = 0 . \end{equation*}

Suppose \(T^2 + I \neq 0\), say \(w := (T^2 + I)v \neq 0\); then \((T-I)(T+I)w = 0\). Put \(u = (T+I)w\).

(i) \(u = 0\). Then \(Tw = -w\) with \(w \neq 0\), so \(-1\) is an eigenvalue of \(T\).

(ii) \(u \neq 0\). Then \((T - I)u = 0\), so \(Tu = u\) and \(1\) is an eigenvalue of \(T\).

Both contradict the hypothesis that \(T\) has no eigenvalues. Hence \(T^2 + I = 0\), that is, \(T^2 = -I\).

Both directions are direct, since \(V\) is not assumed finite-dimensional; throughout \(T^m = T^{m-1}T = T\,T^{m-1}\) (5.13).

(a) By 3.15, injectivity means null space \(\{0\}\). If \(T\) is injective, induct: assuming \(T^{m-1}\) injective, \(T^m v = T^{m-1}(Tv) = 0\) forces \(Tv = 0\), hence \(v = 0\). Conversely, if \(T^m\) is injective and \(Tv = 0\), then

\begin{equation*} T^m v = T^{m-1}(Tv) = 0 , \end{equation*}

so \(v = 0\) (for \(m = 1\) this reads \(Tv = 0\) directly).

(b) Surjectivity means \(T(V) = V\). If \(T\) is surjective, induct: \(T^m(V) = T^{m-1}(T(V)) = T^{m-1}(V) = V\). Conversely, if \(T^m(V) = V\), then

\begin{equation*} V = T^m(V) = T\bigl(T^{m-1}(V)\bigr) \subseteq T(V) \subseteq V , \end{equation*}

the first inclusion because \(T^{m-1}(V) \subseteq V\). Hence \(T(V) = V\).

One direction is immediate from 5.11: eigenvectors corresponding to distinct eigenvalues form a linearly independent list.

For the other, suppose \(v_1, \ldots, v_m\) is linearly independent and extend it by 2.32 to a basis \(v_1, \ldots, v_n\) of \(V\), where \(n = \dim V\). The linear map lemma 3.4, applied to this basis, supplies \(T \in \mathcal{L}(V)\) with

\begin{equation*} T v_k = k\, v_k \qquad \text{for } k = 1, \ldots, n . \end{equation*}

Each \(v_k\) is nonzero (a linearly independent list contains no \(0\)), so \(v_1, \ldots, v_m\) are eigenvectors of \(T\) for the eigenvalues \(1, \ldots, m\), which are distinct in \(\mathbf{F}\).

Write \(f_k\) for the function \(x \mapsto e^{\lambda_k x}\) and set \(V = \operatorname{span}(f_1, \ldots, f_n)\), a subspace of the space of real-valued functions on \(\mathbf{R}\). Differentiation \(Df = f^{\prime}\) is linear and, since

\begin{equation*} f_k^{\prime} = \lambda_k f_k \in V \qquad \text{for } k = 1, \ldots, n , \end{equation*}

it maps each spanning vector of \(V\) into \(V\), hence maps \(V\) into \(V\); so \(D \in \mathcal{L}(V)\). Each \(f_k\) is nonzero in \(V\) because \(f_k(0) = 1\), so the display makes \(f_k\) an eigenvector of \(D\) for the eigenvalue \(\lambda_k\), and the \(\lambda_k\) are distinct by hypothesis. By 5.11 the list \(f_1, \ldots, f_n\) is linearly independent in \(V\), hence in the larger space, linear independence being a statement about vanishing linear combinations only.

Run the device of 5A.35 with the second derivative. Write \(c_k\) for \(x \mapsto \cos(\lambda_k x)\), set \(V = \operatorname{span}(c_1, \dots, c_n)\), and define \(Tf = f^{\prime\prime}\), which is linear and satisfies

\begin{equation*} T c_k = -\lambda_k^{\,2}\, c_k \in V \qquad \text{for } k = 1, \dots, n , \end{equation*}

so \(T\) maps a spanning list of \(V\) into \(V\) and hence \(T \in \mathcal{L}(V)\). Each \(c_k\) is nonzero since \(c_k(0) = 1\), so \(c_k\) is an eigenvector for the eigenvalue \(-\lambda_k^{\,2}\). These eigenvalues are distinct: \(\lambda_j^{\,2} = \lambda_k^{\,2}\) gives \((\lambda_j - \lambda_k)(\lambda_j + \lambda_k) = 0\), and \(\lambda_j + \lambda_k > 0\) by positivity, so \(\lambda_j = \lambda_k\) and \(j = k\). By 5.11 the list is linearly independent in \(V\), hence in the space of all real-valued functions on \(\mathbb{R}\).

If \(V = \{0\}\) then \(\mathcal{L}(V) = \{0\}\) too and both operators have no eigenvalues, so assume \(n = \dim V \geq 1\).

If \(\lambda\) is an eigenvalue of \(\mathcal{A}\), choose \(S \neq 0\) with \(TS = \lambda S\) and \(v\) with \(Sv \neq 0\); applying the equation to \(v\) gives \(T(Sv) = \lambda(Sv)\), so \(\lambda\) is an eigenvalue of \(T\).

Conversely, let \(Tv = \lambda v\) with \(v \neq 0\), fix a basis \(v_1, \dots, v_n\) of \(V\), and use 3.4 on that basis to get \(S \in \mathcal{L}(V)\) with

\begin{equation*} S v_1 = v, \qquad S v_j = 0 \ \text{ for } j = 2, \dots, n , \end{equation*}

so \(S \neq 0\). For \(w = a_1 v_1 + \cdots + a_n v_n\) we have \(Sw = a_1 v\), whence

\begin{equation*} (TS)w = T(a_1 v) = \lambda(a_1 v) = \lambda(Sw) . \end{equation*}

Thus \(\mathcal{A}(S) = TS = \lambda S\) with \(S \neq 0\), making \(\lambda\) an eigenvalue of \(\mathcal{A}\).

(a) Independence of the representative is exactly where invariance enters: if \(v + U = w + U\), then \(v - w \in U\), so \(T(v-w) \in U\) by invariance, that is, \(Tv - Tw \in U\), which says

\begin{equation*} Tv + U = Tw + U . \end{equation*}

So \(T/U\) is a well-defined function on \(V/U\), and it is linear because

\begin{equation*} \begin{aligned} (T/U)\bigl((v+U) + (w+U)\bigr) &= T(v+w) + U = (Tv + U) + (Tw + U) , \\ (T/U)\bigl(\alpha(v+U)\bigr) &= T(\alpha v) + U = \alpha\bigl((T/U)(v+U)\bigr) . \end{aligned} \end{equation*}

Hence \(T/U \in \mathcal{L}(V/U)\).

(b) Let \(\lambda\) be an eigenvalue of \(T/U\), so some \(v \notin U\) has \(Tv + U = \lambda v + U\), i.e. \(u := (T - \lambda I)v \in U\). Since \(U\) is invariant under \(T\) and under \(I\), it is invariant under \(T - \lambda I\), so \((T - \lambda I)|_U \in \mathcal{L}(U)\) with \(U\) finite-dimensional.

(i) \((T - \lambda I)|_U\) not injective. Some nonzero \(u^{\prime} \in U\) has \(Tu^{\prime} = \lambda u^{\prime}\).

(ii) \((T - \lambda I)|_U\) injective, hence surjective onto \(U\) by 3.65. Pick \(u^{\prime} \in U\) with \((T - \lambda I)u^{\prime} = u\); then

\begin{equation*} (T - \lambda I)(v - u^{\prime}) = u - u = 0 , \end{equation*}

and \(v - u^{\prime} \neq 0\) since \(v \notin U\).

Either way \(\lambda\) is an eigenvalue of \(T\).

For \(V = \{0\}\) both sides fail (no nonzero vectors; no subspace of dimension \(-1\)), and in each direction below the assumed hypothesis forces \(\dim V \geq 1\).

Suppose \(T\) has an eigenvalue \(\lambda\) and put \(W = \operatorname{range}(T - \lambda I)\). Then \(T - \lambda I\) is not injective by 5.7, hence not surjective by 3.65, so \(W \neq V\) and \(\dim W \leq n - 1\) where \(n = \dim V\). Extend a basis \(w_1, \dots, w_m\) of \(W\) to a basis \(w_1, \dots, w_m, u_{m+1}, \dots, u_n\) of \(V\) (2.32) and let \(U\) be the span of its first \(n - 1\) vectors, so \(\dim U = n - 1\) and \(W \subseteq U\) (as \(m \leq n-1\)). Then \(U\) is invariant:

\begin{equation*} Tu = (T - \lambda I)u + \lambda u \in W + U \subseteq U \qquad (u \in U) . \end{equation*}

Conversely, let \(U\) be invariant under \(T\) with \(\dim U = \dim V - 1\), so \(\dim V/U = 1\) by 3.105 and \(T/U \in \mathcal{L}(V/U)\) by 5A.38(a). Any nonzero \(w \in V/U\) spans \(V/U\), so \((T/U)w = \lambda w\) for some \(\lambda \in \mathbf{F}\); thus \(\lambda\) is an eigenvalue of \(T/U\), hence of \(T\) by 5A.38(b).

Everything follows from \((STS^{-1})^k = S T^k S^{-1}\) for \(k \geq 0\). This holds for \(k = 0\) (both sides are \(I\), by the convention \(R^0 = I\) of 5.13), and inductively

\begin{equation*} (STS^{-1})^{k+1} = (S T^k S^{-1})(S T S^{-1}) = S T^k (S^{-1}S) T S^{-1} = S T^{k+1} S^{-1} . \end{equation*}

Now write \(p(z) = a_0 + a_1 z + \cdots + a_m z^m\). By 5.14 and the previous display,

\begin{equation*} p(STS^{-1}) = a_0 S I S^{-1} + a_1 S T S^{-1} + \cdots + a_m S T^m S^{-1} , \end{equation*}

and \(R \mapsto SRS^{-1}\) is linear (composition distributes over addition and commutes with scalars), so the sum collapses to \(S\,p(T)\,S^{-1}\).

First, \(U\) is invariant under every power \(T^k\): true for \(k = 0\) since \(T^0 = I\) (5.13), and if \(T^k u \in U\) for all \(u \in U\), then \(T^{k+1}u = T(T^k u) \in U\) by invariance under \(T\). So for \(p(z) = a_0 + a_1 z + \cdots + a_m z^m\) and \(u \in U\), by 5.14

\begin{equation*} p(T)u = a_0 u + a_1 (Tu) + \cdots + a_m (T^m u) \in U , \end{equation*}

each term lying in \(U\) and \(U\) being closed under scalar multiples and sums.

With \(e_1, \dots, e_n\) the standard basis, the definition of \(T\) says \(Te_j = j\,e_j\).

(a) The eigenvalues are exactly \(1, 2, \dots, n\), with \(E(j, T) = \operatorname{span}(e_j)\); the eigenvectors for \(j\) are the nonzero multiples of \(e_j\). Indeed, comparing coordinates, \(Tx = \lambda x\) says

\begin{equation*} (k - \lambda)\, x_k = 0 \qquad \text{for } k = 1, \dots, n . \end{equation*}

If \(\lambda \notin \{1, \dots, n\}\) every factor \(k - \lambda\) is nonzero, forcing \(x = 0\); if \(\lambda = j\), the equations with \(k \neq j\) force \(x_k = 0\), leaving \(x \in \operatorname{span}(e_j)\).

(b) The invariant subspaces are exactly the \(2^n\) coordinate subspaces

\begin{equation*} U_\Omega = \operatorname{span}(e_j : j \in \Omega), \qquad \Omega \subseteq \{1, \dots, n\} . \end{equation*}

Each is invariant, since \(Te_j = j e_j \in U_\Omega\) on a spanning list. Conversely, let \(U\) be invariant and set

\begin{equation*} p_j(z) = \prod_{k \neq j,\ 1 \leq k \leq n} \frac{z - k}{j - k} \in \mathcal{P}(\mathbf{F}) , \end{equation*}

legitimate as \(j - k \neq 0\) in \(\mathbf{F}\), with \(p_j(j) = 1\) and \(p_j(k) = 0\) for \(k \neq j\). From \(T^m e_k = k^m e_k\) and 5.14 we get \(p(T)e_k = p(k) e_k\) for every polynomial \(p\), so for \(x = x_1 e_1 + \cdots + x_n e_n \in U\), invariance of \(U\) under \(p_j(T)\) (5A.41) gives

\begin{equation*} p_j(T)x = \sum_{k=1}^{n} x_k\, p_j(k)\, e_k = x_j\, e_j \in U . \end{equation*}

Hence \(x_j \neq 0\) implies \(e_j \in U\). With \(\Omega = \{j : e_j \in U\}\), every \(x \in U\) has \(x_j = 0\) off \(\Omega\), so \(U \subseteq U_\Omega\), while \(U_\Omega \subseteq U\) by definition of \(\Omega\). These \(2^n\) subspaces are distinct, since \(e_j \in U_\Omega\) if and only if \(j \in \Omega\) by independence of \(e_1, \dots, e_n\).

The set \(\mathcal{E} = \{p(T) : p \in \mathcal{P}(\mathbf{F})\}\) is commutative, since \(p(T)q(T) = q(T)p(T)\) by 5.17(b), whereas \(\mathcal{L}(V)\) is not. Indeed, let \(n = \dim V \geq 2\), take a basis \(v_1, \dots, v_n\) of \(V\) (2.31), and use 3.4 to get \(R, S \in \mathcal{L}(V)\) with

\begin{equation*} R v_1 = v_2, \qquad S v_1 = v_1, \qquad R v_j = S v_j = 0 \ \text{ for } j \geq 2 . \end{equation*}

Then

\begin{equation*} (RS)v_1 = R v_1 = v_2 \neq 0 = S v_2 = (SR)v_1 , \end{equation*}

so \(RS \neq SR\). Were \(\mathcal{E} = \mathcal{L}(V)\), these \(R, S\) would lie in \(\mathcal{E}\) and hence commute. So \(\mathcal{E} \neq \mathcal{L}(V)\).

Method (2): suppose \(\mathcal{E} = \mathcal{L}(V)\). Taking \(q(z) = z\) in 5.17(b), every element of \(\mathcal{E}\), hence every operator on \(V\), commutes with \(T\). Then every nonzero \(v\) is an eigenvector of \(T\): otherwise \(v, Tv\) is linearly independent, extends to a basis \(v, Tv, u_3, \dots, u_n\) (2.32), and 3.4 supplies \(S\) with \(Sv = v\), \(S(Tv) = 0\), \(Su_j = 0\), giving \((ST)v = 0 \neq Tv = (TS)v\) (that \(Tv \neq 0\) is part of the independence). So \(T = \lambda I\) by 5A.26, whence \(p(T) = p(\lambda)I\) for every \(p\) and \(\dim \mathcal{E} = \dim \operatorname{span}(I) = 1\). But \(\dim \mathcal{L}(V) = n^2 \geq 4\) by 3.72, a contradiction.

If \(Tv = \pm 3v\) with \(v \neq 0\), then

\begin{equation*} T^2 v = \pm 3\, Tv = (\pm 3)^2 v = 9v , \end{equation*}

so \(9\) is an eigenvalue of \(T^2\).

Conversely, take \(v \neq 0\) with \(T^2 v = 9v\). Since polynomials in \(T\) factor as polynomials do (5.17), \(T^2 - 9I = (T - 3I)(T + 3I)\), so with \(w = (T + 3I)v\) we get \((T - 3I)w = 0\).

(i) \(w = 0\): then \(Tv = -3v\) with \(v \neq 0\), so \(-3\) is an eigenvalue of \(T\).

(ii) \(w \neq 0\): then \(Tw = 3w\), so \(3\) is an eigenvalue of \(T\).

Suppose some invariant \(U\) were nonzero and finite-dimensional. Invariance makes \(T|_U \in \mathcal{L}(U)\), and \(U\) is a nonzero finite-dimensional complex vector space, so by 5.19 there is \(u \in U\) with \(u \neq 0\) and

\begin{equation*} Tu = (T|_U)u = \lambda u \end{equation*}

for some \(\lambda \in \mathbf{C}\). Then \(\lambda\) is an eigenvalue of \(T\), contradicting the hypothesis.

(a) The eigenvalues are \(0\) and \(n\); the eigenvectors for \(0\) are the nonzero \(x\) with \(x_1 + \cdots + x_n = 0\), and those for \(n\) are the nonzero multiples of \(\mathbf{1} = (1, \dots, 1)\).

Writing \(s(x) = x_1 + \cdots + x_n\), the definition of \(T\) says \(Tx = s(x)\mathbf{1}\). Hence \(Tx = 0\) iff \(s(x) = 0\), and since \(s\) is a surjective linear functional, 3.21 gives

\begin{equation*} \dim \operatorname{null} T = n - 1 \geq 1 , \end{equation*}

so \(0\) is an eigenvalue with the stated eigenvectors. Also \(T\mathbf{1} = n\mathbf{1}\) with \(\mathbf{1} \neq 0\). If \(Tx = \lambda x\) with \(x \neq 0\) and \(\lambda \neq 0\), then \(x = \lambda^{-1}s(x)\mathbf{1} = c\mathbf{1}\) with \(c \neq 0\), and applying \(T\) gives \(\lambda c \mathbf{1} = c n \mathbf{1}\), so \(\lambda = n\).

(b) The minimal polynomial is \(z^2 - nz\). For every \(x \in \mathbf{F}^n\),

\begin{equation*} T^2 x = s(x)\, T\mathbf{1} = n\, s(x)\, \mathbf{1} = n\, Tx , \end{equation*}

so \(T^2 - nT = 0\) and the minimal polynomial divides \(z(z-n)\) by 5.29. By 5.27(a) its zeros are the eigenvalues \(0\) and \(n\), which are distinct because \(n > 1\); hence its degree is at least \(2\), and being monic of degree \(2\) it equals \(z^2 - nz\).

Suppose first \(\alpha = p(\lambda)\) with \(Tv = \lambda v\), \(v \neq 0\). Then \(T^k v = \lambda^k v\) for every \(k \geq 0\), so writing \(p(z) = a_0 + a_1 z + \cdots + a_m z^m\),

\begin{equation*} p(T)v = \sum_{k=0}^{m} a_k T^k v = \Bigl(\sum_{k=0}^{m} a_k \lambda^k\Bigr) v = p(\lambda) v = \alpha v , \end{equation*}

and \(v \neq 0\), so \(\alpha\) is an eigenvalue of \(p(T)\).

Conversely, suppose \(\alpha\) is an eigenvalue of \(p(T)\), and set \(q(z) = p(z) - \alpha\). Then \(q\) is nonconstant, say \(\deg q = \deg p = m \geq 1\), so 4.13 factors it as \(q(z) = c(z-\lambda_1)\cdots(z-\lambda_m)\) with \(c \neq 0\). Because \(r \mapsto r(T)\) preserves products (5.17),

\begin{equation*} p(T) - \alpha I = q(T) = c\,(T - \lambda_1 I)\cdots(T - \lambda_m I). \end{equation*}

The left side is not injective, and a composition of injective operators is injective, so \(T - \lambda_j I\) fails to be injective for some \(j\); that is, \(\lambda_j\) is an eigenvalue of \(T\). Since \(q(\lambda_j) = 0\), we get \(\alpha = p(\lambda_j)\).

Take \(T(x,y) = (-y, x)\) on \(\mathbf{R}^2\), \(p(z) = z^2\), and \(\alpha = -1\).

Then \(T^2(x,y) = T(-y,x) = -(x,y)\), so \(p(T) = T^2 = -I\) and \(\alpha = -1\) is an eigenvalue of \(p(T)\). But \(T\) has no eigenvalues: \(T(x,y) = \lambda(x,y)\) says \(-y = \lambda x\) and \(x = \lambda y\), whence \((\lambda^2+1)y = 0\), so \(y = 0\) and then \(x = 0\) (as \(\lambda^2 + 1 > 0\) for real \(\lambda\)). Hence no eigenvalue \(\lambda\) of \(T\) satisfies \(\alpha = p(\lambda)\), and Exercise 4 fails over \(\mathbf{R}\).

The minimal polynomial of \(T\) is \(z^2 + 1\), for \(\mathbf{F} = \mathbf{R}\) and for \(\mathbf{F} = \mathbf{C}\). Indeed

\begin{equation*} T^2(w,z) = T(-z, w) = -(w,z), \end{equation*}

so \(T^2 + I = 0\) and the minimal polynomial divides \(z^2+1\) by 5.29. It is not \(1\) (which gives \(I \neq 0\)) nor of the form \(z - c\), since \(T(1,0) = (0,1)\) is not a multiple of \((1,0)\), so \(T \neq cI\). Hence its degree is \(2\) and it equals \(z^2 + 1\).

(a) Take \(S(w,z) = (z,0)\) and \(T(w,z) = (w,0)\). Then

\begin{equation*} (ST)(w,z) = S(w,0) = (0,0), \qquad (TS)(w,z) = T(z,0) = (z,0), \end{equation*}

so \(ST = 0\), with minimal polynomial \(z\), while \(TS \neq 0\) (it sends \((0,1)\) to \((1,0)\)) satisfies \((TS)^2(w,z) = (TS)(z,0) = 0\). Thus \(z^2\) annihilates \(TS\) and no degree-\(1\) monic polynomial does (\(TS = cI\) applied to \((1,0)\) forces \(c = 0\), hence \(TS = 0\)), so the minimal polynomial of \(TS\) is \(z^2 \neq z\).

(b) The statement is symmetric in \(S\) and \(T\), so assume \(S\) is invertible. For every integer \(k \geq 0\),

\begin{equation*} S^{-1}(ST)^k S = (S^{-1}S)(TS)^k = (TS)^k , \end{equation*}

by regrouping the \(2k\) alternating factors (the case \(k = 0\) being \(S^{-1}IS = I\)). Hence, for \(p(z) = a_0 + \cdots + a_m z^m\), linearity of \(A \mapsto S^{-1}AS\) gives the hint:

\begin{equation*} p(TS) = \sum_{k=0}^{m} a_k S^{-1}(ST)^k S = S^{-1} p(ST) S . \end{equation*}

Since \(S\) is invertible, \(p(TS) = 0\) if and only if \(p(ST) = 0\); so \(ST\) and \(TS\) are annihilated by exactly the same polynomials. As \(V\) is finite-dimensional, each has a minimal polynomial, namely the unique monic polynomial of smallest degree in that common set (5.22), so the two minimal polynomials coincide.

The minimal polynomial of \(T\) is

\begin{equation*} z^2 - \left(2\cos\tfrac{\pi}{180}\right) z + 1 . \end{equation*}

Write \(\theta = \pi/180\). With respect to the standard basis of \(\mathbb{R}^2\),

\begin{equation*} \mathcal{M}(T) = \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}, \end{equation*}

which in the notation of Exercise 11 has \(a = d = \cos\theta\), \(b = \sin\theta\), and \(c = -\sin\theta\). Here \(b = \sin\frac{\pi}{180} \neq 0\), so Exercise 11(b) puts us in the second case and the minimal polynomial is

\begin{equation*} z^2 - (a+d)z + (ad - bc) = z^2 - (2\cos\theta) z + (\cos^2\theta + \sin^2\theta), \end{equation*}

which is the polynomial displayed above.

Let \(n = \dim V\) and let \(A = \mathcal{M}(T)\) with respect to the given basis, so that \(A\) and all its powers have rational entries and, since \(\mathcal{M}\) is injective, linear, and multiplicative (3.43), a polynomial \(q\) satisfies \(q(T) = 0\) if and only if \(q(A) = 0\). Let \(p\) be the minimal polynomial of \(T\) and \(m = \deg p\). The point is that \(\mathbf{F}\) is \(\mathbb{R}\) or \(\mathbb{C}\), not \(\mathbb{Q}\), so the following bridge is needed.

Lemma. If \(B\) is a rational matrix with \(k\) columns and \(Bx = 0\) has a nonzero solution in \(\mathbf{F}^k\), then it has one in \(\mathbb{Q}^k\). Indeed, row reducing \(B\) uses only rational scalars, so its reduced row echelon form \(R\) is rational, and row operations are reversible and hence preserve the solution set over \(\mathbf{F}\) and over \(\mathbb{Q}\) alike. Were every column of \(R\) a pivot column, \(Rx = 0\) would force \(x = 0\) over \(\mathbf{F}\); so some column is free, and setting that free variable to \(1\), the other free variables to \(0\), and solving for the pivot variables gives a nonzero solution in \(\mathbb{Q}^k\).

The rational \(n\)-by-\(n\) matrices form a \(\mathbb{Q}\)-vector space of dimension \(n^2\), so \(I, A, \dots, A^{n^2}\) are linearly dependent over \(\mathbb{Q}\); let \(m^{\prime} \leq n^2\) be smallest with \(I, A, \dots, A^{m^{\prime}}\) dependent over \(\mathbb{Q}\), say

\begin{equation*} b_0 I + b_1 A + \cdots + b_{m^{\prime}} A^{m^{\prime}} = 0 \end{equation*}

with \(b_j \in \mathbb{Q}\) not all \(0\). Minimality of \(m^{\prime}\) forces \(b_{m^{\prime}} \neq 0\), so dividing by it produces a monic \(q \in \mathcal{P}(\mathbb{Q})\) of degree \(m^{\prime}\) with \(q(A) = 0\) and hence \(q(T) = 0\). By 5.29 the nonzero polynomial \(q\) is a polynomial multiple of \(p\), so \(m^{\prime} \geq m\).

For the reverse inequality, \(p(A) = 0\) is an \(\mathbf{F}\)-linear dependence among \(I, A, \dots, A^m\); listing the entries of these \(m+1\) rational matrices as the columns of a rational matrix \(B\), it is a nonzero solution of \(Bx = 0\) in \(\mathbf{F}^{m+1}\), so the Lemma yields a rational dependence among \(I, A, \dots, A^m\) and thus \(m^{\prime} \leq m\).

Hence \(m^{\prime} = m\), and \(q\) is monic of degree \(m\) with \(q(T) = 0\), so \(q = p\) by the uniqueness in 5.22. All coefficients of \(p\) are therefore rational.

Let \(n = \dim V\) and \(U = \operatorname{span}(v, Tv, \ldots, T^{n-1}v)\). By Exercise 23, \(U\) is invariant under \(T\); since \(v \in U\), induction on \(k\) gives

\begin{equation*} T^k v \in U \qquad \text{for every integer } k \geq 0 . \end{equation*}

Now fix an integer \(m \geq n - 1\). Then

\begin{equation*} U \subseteq \operatorname{span}(v, Tv, \ldots, T^m v) \subseteq U, \end{equation*}

the first inclusion because \(v, Tv, \ldots, T^{n-1}v\) is a sublist of \(v, Tv, \ldots, T^m v\), the second because \(U\) is a subspace containing each \(T^k v\). Hence the two spans are equal.

(a) Because \(\mathcal{M}\) is linear, injective, and multiplicative (3.43), it suffices to check the identity for \(A = \mathcal{M}(T)\), and multiplying out gives

\begin{equation*} A^2 = \begin{pmatrix} a^2 + cb & ac + cd \\ ab + db & bc + d^2 \end{pmatrix} = (a+d)A - (ad - bc) I \end{equation*}

(compare the four entries). Hence \(T^2 - (a+d)T + (ad-bc)I = 0\).

(b) Let \(p\) be the minimal polynomial of \(T\) and \(g(z) = z^2 - (a+d)z + (ad-bc)\), so \(g\) is monic of degree \(2\) with \(g(T) = 0\) by (a). Then \(\deg p \geq 1\), since the constant polynomial \(1\) gives \(I \neq 0\) (as \(\dim V = 2\)). Because \(\mathcal{M}\) is injective, \(T = \lambda I\) holds if and only if \(b = c = 0\) and \(a = d = \lambda\).

(i) If \(b = c = 0\) and \(a = d\), then \(T = aI\), so the monic degree-\(1\) polynomial \(z - a\) annihilates \(T\), and \(p(z) = z - a\) by the uniqueness in 5.22.

(ii) Otherwise \(T\) is not a scalar multiple of \(I\), so no monic polynomial of degree \(1\) annihilates \(T\), forcing \(\deg p = 2\); since \(g\) is monic of degree \(2\) with \(g(T) = 0\), 5.22 gives

\begin{equation*} p(z) = z^2 - (a+d) z + (ad - bc). \end{equation*}

The minimal polynomial of \(T\) is

\begin{equation*} (z-1)(z-2)(z-3)\cdots(z-n) = \prod_{j=1}^{n}(z - j). \end{equation*}

For the standard basis \(e_1, \dots, e_n\) of \(\mathbf{F}^n\) the definition of \(T\) reads \(Te_j = j\,e_j\), so \(T^k e_j = j^k e_j\) and hence \(q(T)e_j = q(j)e_j\) for every \(q \in \mathcal{P}(\mathbf{F})\). As \(e_1, \dots, e_n\) is a basis,

\begin{equation*} q(T) = 0 \iff q(j) = 0 \text{ for each } j = 1, \dots, n . \end{equation*}

Thus \(g(z) = (z-1)\cdots(z-n)\), monic of degree \(n\), satisfies \(g(T) = 0\). The minimal polynomial \(p\) has the \(n\) numbers \(1, \dots, n\) among its zeros, which are distinct in \(\mathbf{F}\) (\(= \mathbb{R}\) or \(\mathbb{C}\)), so \(\deg p \geq n\) by 4.8. Hence \(\deg p = n\), and \(p = g\) by the uniqueness in 5.22.

Take \(r\) to be the remainder on dividing \(p\) by the minimal polynomial \(q\) of \(T\) (which exists by 5.22), and set \(m = \deg q\).

Existence: by 4.9 there are \(s, r \in \mathcal{P}(\mathbf{F})\) with \(p = qs + r\) and \(\deg r < m\), so applying this to \(T\) and using \(q(T) = 0\),

\begin{equation*} p(T) = q(T)s(T) + r(T) = r(T). \end{equation*}

Uniqueness: if \(r_1, r_2\) both qualify, then \(u = r_1 - r_2\) satisfies \(u(T) = 0\) and \(\deg u < m\). By 5.29, \(u = qs\) for some \(s \in \mathcal{P}(\mathbf{F})\); were \(s \neq 0\) we would have \(\deg u = m + \deg s \geq m\), a contradiction. Hence \(s = 0\), so \(u = 0\) and \(r_1 = r_2\).

The minimal polynomial of \(T^{-1}\) is the reversal of \(p(z) = 4 + 5z - 6z^2 - 7z^3 + 2z^4 + z^5\), made monic:

\begin{equation*} \tfrac{1}{4} + \tfrac{1}{2}z - \tfrac{7}{4}z^2 - \tfrac{3}{2}z^3 + \tfrac{5}{4}z^4 + z^5. \end{equation*}

(Here \(T\) is invertible by 5.32, the constant term \(4\) of \(p\) being nonzero.)

Reversal lemma: if \(S\) is invertible, \(u(z) = b_0 + \cdots + b_k z^k\) satisfies \(u(S) = 0\), and \(\hat u(z) = b_k + b_{k-1}z + \cdots + b_0 z^k\), then

\begin{equation*} \begin{aligned} \hat u(S^{-1}) &= \sum_{j=0}^{k} b_{k-j} S^{-j} = S^{-k}\sum_{j=0}^{k} b_{k-j} S^{\,k-j} \\ &= S^{-k}\sum_{i=0}^{k} b_i S^{\,i} = S^{-k}u(S) = 0 . \end{aligned} \end{equation*}

Applied with \(S = T\) and \(u = p\), it gives \(\hat p(T^{-1}) = 0\) for \(\hat p(z) = 1 + 2z - 7z^2 - 6z^3 + 5z^4 + 4z^5\); dividing by \(4\), the monic degree-\(5\) polynomial \(g = \tfrac14 \hat p\) displayed above satisfies \(g(T^{-1}) = 0\), so the minimal polynomial \(\tilde q\) of \(T^{-1}\) has degree \(k \leq 5\).

Conversely, \(T^{-1}\) is invertible, so the constant term of \(\tilde q\) is nonzero by 5.32; hence \(\hat{\tilde q}\) has degree exactly \(k\), and the lemma with \(S = T^{-1}\) gives \(\hat{\tilde q}(T) = 0\). By 5.29 this nonzero polynomial is a multiple of \(p\), so \(k \geq \deg p = 5\).

Thus \(k = 5\), and \(g\) is monic of degree \(5\) with \(g(T^{-1}) = 0\), so \(g = \tilde q\) by the uniqueness in 5.22.

The function \(f\) fails to be continuous at any eigenvalue \(\lambda_0\) of \(T\), one of which exists by 5.19 because \(V\) is a nonzero finite-dimensional complex vector space.

Write \(n = \dim V\). By 3.21,

\begin{equation*} f(\lambda) = n - \dim \operatorname{null}(T - \lambda I), \end{equation*}

so \(f(\lambda) = n\) when \(\lambda\) is not an eigenvalue (then \(T - \lambda I\) is injective) while \(f(\lambda_0) \leq n - 1\). By 5.12 the set of eigenvalues of \(T\) is finite, so for each \(k \geq 1\) the infinite punctured disk \(\{\lambda : 0 < |\lambda - \lambda_0| < 1/k\}\) contains some \(\lambda_k\) that is not an eigenvalue. Then \(\lambda_k \to \lambda_0\) and

\begin{equation*} \lim_{k \to \infty} f(\lambda_k) = n > n - 1 \geq f(\lambda_0), \end{equation*}

so \(f\) is not continuous.

Write \(p(z) = a_0 + a_1 z + \cdots + a_{n-1}z^{n-1} + z^n\). Reading the columns of the displayed matrix against the standard basis \(e_1, \dots, e_n\),

\begin{equation*} Te_j = e_{j+1} \ \ (1 \leq j \leq n-1), \qquad Te_n = -a_0 e_1 - \cdots - a_{n-1}e_n , \end{equation*}

so induction gives \(T^k e_1 = e_{k+1}\) for \(k = 0, 1, \dots, n-1\).

Hence, rewriting each \(e_{k+1}\) as \(T^k e_1\) in the formula for \(Te_n\),

\begin{equation*} T^n e_1 = Te_n = -\bigl(a_0 e_1 + a_1 Te_1 + \cdots + a_{n-1}T^{n-1}e_1\bigr), \end{equation*}

that is, \(p(T)e_1 = 0\). Since \(e_j = T^{j-1}e_1\) and powers of \(T\) commute with \(p(T)\),

\begin{equation*} p(T)e_j = T^{j-1}\bigl(p(T)e_1\bigr) = 0 \qquad (j = 1, \dots, n), \end{equation*}

so \(p(T) = 0\).

No monic \(q\) of degree \(m < n\) annihilates \(T\): writing \(q(z) = b_0 + \cdots + b_{m-1}z^{m-1} + z^m\) and using \(T^k e_1 = e_{k+1}\) for \(k \leq m \leq n-1\),

\begin{equation*} q(T)e_1 = b_0 e_1 + b_1 e_2 + \cdots + b_{m-1}e_m + e_{m+1} \neq 0, \end{equation*}

since the coefficient of \(e_{m+1}\) in this combination of the linearly independent list \(e_1, \dots, e_{m+1}\) is \(1\).

So the minimal polynomial has degree \(n\), and being monic of degree \(n\) with \(p(T) = 0\), it equals \(p\) by the uniqueness in 5.22.

Everything follows from one shift lemma: for \(A \in \mathcal{L}(V)\), \(\mu \in \mathbf{F}\), and \(r \in \mathcal{P}(\mathbf{F})\) monic of degree \(m\), the polynomial \(r_\mu(z) = r(z+\mu)\) is again monic of degree \(m\), and \(r_\mu(A) = r(A + \mu I)\).

Indeed, write \(r(z) = c_0 + \cdots + c_m z^m\) with \(c_m = 1\). Expanding \(r_\mu = \sum_k c_k (z+\mu)^k\) by the binomial theorem, the only degree-\(m\) contribution is \(c_m z^m = z^m\), which gives the first assertion. For the second, \(r_\mu = \sum_k c_k s^k\) where \(s(z) = z + \mu\), and since \(r \mapsto r(A)\) is linear (5.14) and multiplicative (5.17(a)), so that \((s^k)(A) = (A + \mu I)^k\),

\begin{equation*} r_\mu(A) = \sum_{k=0}^{m} c_k (A + \mu I)^k = r(A + \mu I). \end{equation*}

Now put \(m = \deg p\) and \(S = T - \lambda I\). Since \(q = p_\lambda\), the lemma makes \(q\) monic of degree \(m\) with

\begin{equation*} q(S) = p(S + \lambda I) = p(T) = 0 . \end{equation*}

Hence the minimal polynomial \(\tilde q\) of \(S\) has \(m^{\prime} = \deg \tilde q \leq m\). Conversely, \(\tilde q_{-\lambda}\) is monic of degree \(m^{\prime}\) and

\begin{equation*} \tilde q_{-\lambda}(T) = \tilde q(T - \lambda I) = \tilde q(S) = 0, \end{equation*}

so minimality of \(\deg p\) (5.22) gives \(m \leq m^{\prime}\). Thus \(m^{\prime} = m\), and \(q\) and \(\tilde q\) are monic of the same degree \(m\) annihilating \(S\), so \(q = \tilde q\) by the uniqueness in 5.22.

As in Exercise 17, one scaling lemma does all the work: for \(A \in \mathcal{L}(V)\), \(\mu \in \mathbf{F}\setminus\{0\}\), and \(r \in \mathcal{P}(\mathbf{F})\) monic of degree \(m\), the polynomial \(r^{[\mu]}(z) = \mu^m r(z/\mu)\) is monic of degree \(m\) and \(r^{[\mu]}(\mu A) = \mu^m r(A)\).

Indeed, with \(r(z) = c_0 + \cdots + c_m z^m\) and \(c_m = 1\),

\begin{equation*} r^{[\mu]}(z) = \mu^m \sum_{k=0}^{m} c_k \mu^{-k} z^k = \sum_{k=0}^{m} c_k \mu^{\,m-k} z^k , \end{equation*}

whose coefficient of \(z^m\) is \(c_m = 1\); and since \((\mu A)^k = \mu^k A^k\),

\begin{equation*} r^{[\mu]}(\mu A) = \sum_{k=0}^{m} c_k \mu^{\,m-k}\mu^{k} A^k = \mu^m r(A). \end{equation*}

Let \(m = \deg p\), so the exercise’s \(q\) is \(p^{[\lambda]}\), monic of degree \(m\) with \(q(\lambda T) = \lambda^m p(T) = 0\). Hence the minimal polynomial \(\tilde q\) of \(\lambda T\) has \(m^{\prime} = \deg \tilde q \leq m\) by 5.22. Conversely, applying the lemma with \(\mu = 1/\lambda\) and \(A = \lambda T\), the polynomial \(\tilde q^{[1/\lambda]}\) is monic of degree \(m^{\prime}\) and, since \(\tfrac{1}{\lambda}(\lambda T) = T\),

\begin{equation*} \tilde q^{[1/\lambda]}(T) = \lambda^{-m^{\prime}}\, \tilde q(\lambda T) = 0, \end{equation*}

so minimality of \(\deg p\) gives \(m \leq m^{\prime}\). Thus \(m^{\prime} = m\), and \(q = \tilde q\) by the uniqueness in 5.22.

The list \(I, T, \dots, T^{m-1}\) is a basis of \(\mathcal{E}\), where \(p\) is the minimal polynomial of \(T\) and \(m = \deg p\); hence \(\dim \mathcal{E} = m\).

Each \(T^k\) lies in \(\mathcal{E}\) (take \(q(z) = z^k\)). The list spans: given \(q \in \mathcal{P}(\mathbf{F})\), divide by \(p\) using 4.9 to get \(q = ps + r\) with \(\deg r < m\), so by 5.17(a) and \(p(T) = 0\),

\begin{equation*} q(T) = p(T)s(T) + r(T) = r(T) = b_0 I + b_1 T + \cdots + b_{m-1}T^{m-1}, \end{equation*}

writing \(r(z) = b_0 + \cdots + b_{m-1}z^{m-1}\).

The list is linearly independent: if \(c_0 I + \cdots + c_{m-1}T^{m-1} = 0\) with some \(c_k \neq 0\), let \(j\) be the largest index with \(c_j \neq 0\); dividing by \(c_j\) gives

\begin{equation*} r(z) = \frac{c_0}{c_j} + \cdots + \frac{c_{j-1}}{c_j}z^{j-1} + z^{j}, \end{equation*}

a monic polynomial of degree \(j \leq m-1\) with \(r(T) = 0\), contradicting the minimality of \(\deg p\) in 5.22.

It suffices to show that \(r(z) = (z-3)^2(z-5)^2(z-8)^2\) is a polynomial multiple of the minimal polynomial \(p\) of \(T\), since then \(r(T) = 0\) by 5.29, and \(r(T) = (T-3I)^2(T-5I)^2(T-8I)^2\) by 5.17(a).

By 5.22, \(\deg p \leq \dim \mathbf{F}^4 = 4\), and by 5.27(a) the zeros of \(p\) are exactly the eigenvalues \(3, 5, 8\). Factoring these out one at a time by 4.6 (legitimate because each successive monic quotient is nonzero and vanishes at the next of the three numbers: writing \(p(z) = (z-3)q(z)\), the equation \(0 = p(5) = (5-3)q(5)\) forces \(q(5) = 0\), and similarly at \(8\)) gives

\begin{equation*} p(z) = (z-3)(z-5)(z-8)\,t(z) \end{equation*}

with \(t\) monic and \(\deg t = \deg p - 3 \leq 1\).

(i) \(\deg t = 0\): then \(p(z) = (z-3)(z-5)(z-8)\) and \(r(z) = p(z)(z-3)(z-5)(z-8)\).

(ii) \(\deg t = 1\): then \(t(z) = z - c\) with \(c\) a zero of \(p\), so \(c \in \{3,5,8\}\) by 5.27(a). Writing \(d, e\) for the other two members of \(\{3,5,8\}\),

\begin{equation*} p(z) = (z-c)^2(z-d)(z-e), \qquad r(z) = p(z)\,(z-d)(z-e). \end{equation*}

Put \(U = \operatorname{range} T\), which is invariant under \(T\) (if \(u \in U\) then \(Tu \in \operatorname{range}T = U\)), and let \(q\) be the minimal polynomial of \(T|_U \in \mathcal{L}(U)\), so that \(q\) is monic with \(\deg q \leq \dim U\) by 5.22.

Because \(U\) is invariant, induction on \(k\) gives \(T^k u = (T|_U)^k u\) for all \(u \in U\) and \(k \geq 0\), hence

\begin{equation*} q(T)u = q(T|_U)u = 0 \qquad \text{for all } u \in U . \end{equation*}

Now \(s(z) = z\,q(z)\) is monic with \(\deg s = 1 + \deg q \leq 1 + \dim U\), and \(s(T) = q(T)\,T\) by 5.17(a), so for every \(v \in V\),

\begin{equation*} s(T)v = q(T)(Tv) = 0 , \end{equation*}

the last step by the previous display, since \(Tv \in U\). Thus \(s\) is a monic polynomial annihilating \(T\), so by the minimality in 5.22 the minimal polynomial \(p\) of \(T\) satisfies

\begin{equation*} \deg p \leq \deg s \leq 1 + \dim \operatorname{range} T . \end{equation*}

Let \(n = \dim V\).

Suppose \(T\) is invertible, and let \(p(z) = c_0 + c_1 z + \dots + c_{m-1}z^{m-1} + z^m\) be its minimal polynomial, so \(m \leq n\) by 5.22 and \(c_0 \neq 0\) by 5.32. Dividing \(p(T) = 0\) by \(c_0\) and rearranging,

\begin{equation*} I = -\frac{c_1}{c_0} T - \dots - \frac{c_{m-1}}{c_0} T^{m-1} - \frac{1}{c_0} T^m , \end{equation*}

which lies in \(\operatorname{span}(T, T^2, \dots, T^n)\) because \(m \leq n\). (If \(m = 0\) then \(p = 1\) forces \(I = 0\) and \(V = \{0\}\), where \(I\) lies in every span.)

Conversely, suppose \(I = a_1 T + a_2 T^2 + \dots + a_n T^n\) and set \(S = a_1 I + a_2 T + \dots + a_n T^{n-1}\). Powers of \(T\) commute, so factoring \(T\) out on either side gives

\begin{equation*} TS = ST = a_1 T + a_2 T^2 + \dots + a_n T^n = I , \end{equation*}

so \(S\) is an inverse of \(T\).

Put \(U = \operatorname{span}(v, Tv, \dots, T^{n-1}v)\). Since \(T(T^k v) = T^{k+1}v \in U\) for \(k \leq n - 2\), the only point to prove is \(T^n v \in U\).

Let \(m\) be the smallest nonnegative integer with \(T^m v \in W\), where

\begin{equation*} W = \operatorname{span}(v, Tv, \dots, T^{m-1}v) . \end{equation*}

Such an \(m \leq n\) exists: the list \(v, Tv, \dots, T^n v\) has length \(n+1 > \dim V\), so it is linearly dependent by 2.22, and then 2.19 puts one of its terms in the span of the preceding terms.

Now \(W\) is invariant under \(T\): for \(j \leq m-2\) we have \(T(T^j v) = T^{j+1}v \in W\), while \(T(T^{m-1}v) = T^m v \in W\) by the choice of \(m\). Since \(v \in W\) (when \(m = 0\), \(W = \{0\}\) and \(v = 0\)), induction gives \(T^k v \in W\) for every \(k \geq 0\).

(i) \(m \leq n-1\): then \(W \subseteq U\) because the spanning list of \(W\) is a sublist of that of \(U\), and \(U \subseteq W\) by the previous display, so \(U = W\) is invariant under \(T\).

(ii) \(m = n\): minimality of \(m\) together with 2.19 makes \(v, Tv, \dots, T^{n-1}v\) linearly independent, hence a basis of \(V\) by 2.38 (its length is \(n = \dim V\)), so \(U = V\) is invariant under \(T\).

Let \(n = \dim V\) and let \(p\) be the minimal polynomial of \(T\), so \(\deg p \leq n\) by 5.22. Since \(\mathbf{F} = \mathbf{C}\), part (b) of 5.27 factors \(p\) into linear factors, and by part (a) its zeros are exactly the eigenvalues \(5\) and \(6\); hence

\begin{equation*} p(z) = (z-5)^j (z-6)^k, \qquad j, k \geq 1, \quad j + k = \deg p \leq n . \end{equation*}

Therefore \(j \leq n-1\) and \(k \leq n-1\), so the exponents below are nonnegative and

\begin{equation*} (z-5)^{n-1}(z-6)^{n-1} = p(z)\,(z-5)^{n-1-j}(z-6)^{n-1-k} . \end{equation*}

Being a polynomial multiple of \(p\), this polynomial evaluates to \(0\) at \(T\) by 5.29, that is, \((T-5I)^{n-1}(T-6I)^{n-1} = 0\).

Two computations drive both parts. Since \((T/U)(v+U) = Tv + U\) (Exercise 38 in Section 5A), induction gives \((T/U)^k(v+U) = T^k v + U\), so by linearity of the quotient map (3.104),

\begin{equation*} q(T/U)(v + U) = q(T)v + U \qquad (q \in \mathcal{P}(\mathbf{F}),\ v \in V). \quad (\ast) \end{equation*}

Likewise \(T^k u = (T|_U)^k u\) for \(u \in U\) (induction, using invariance of \(U\)), so

\begin{equation*} q(T)u = q(T|_U)u \qquad (q \in \mathcal{P}(\mathbf{F}),\ u \in U). \quad (\ast\ast) \end{equation*}

Both \(T|_U\) and \(T/U\) have minimal polynomials by 5.22, since \(U\) and \(V/U\) are finite-dimensional (3.105).

(a) Let \(p\) be the minimal polynomial of \(T\). By \((\ast)\), for every \(v \in V\),

\begin{equation*} p(T/U)(v + U) = p(T)v + U = 0 + U, \end{equation*}

so \(p(T/U) = 0\) and 5.29, applied to \(T/U\), makes \(p\) a polynomial multiple of the minimal polynomial of \(T/U\).

(b) Let \(r\) and \(s\) be the minimal polynomials of \(T|_U\) and \(T/U\). By 5.29 it suffices to prove \((rs)(T) = 0\). Fix \(v \in V\). Applying \((\ast)\) with \(q = s\) and using \(s(T/U) = 0\) gives \(s(T)v \in U\), so \((\ast\ast)\) yields

\begin{equation*} (rs)(T)v = r(T)\bigl(s(T)v\bigr) = r(T|_U)\bigl(s(T)v\bigr) = 0, \end{equation*}

the first equality by 5.17 and the last because \(r(T|_U) = 0\).

Let \(p\), \(r\), \(s\) be the minimal polynomials of \(T\), \(T|_U\), \(T/U\) (all exist by 5.22, as \(V\), \(U\), \(V/U\) are finite-dimensional), and write \(Z(q)\) for the set of zeros of \(q\) in \(\mathbf{F}\). By 5.27(a) these zero sets are the respective sets of eigenvalues, so it suffices to prove \(Z(p) = Z( r) \cup Z(s)\).

By 5.31 and by Exercise 25(a), \(p\) is a polynomial multiple of \(r\) and of \(s\); hence every zero of \(r\) or of \(s\) is a zero of \(p\), giving \(Z( r) \cup Z(s) \subseteq Z(p)\).

Conversely, Exercise 25(b) gives \(rs = pk\) for some \(k \in \mathcal{P}(\mathbf{F})\), so if \(\lambda \in Z(p)\) then

\begin{equation*} r(\lambda)s(\lambda) = p(\lambda)k(\lambda) = 0, \end{equation*}

and \(\mathbf{F}\) has no zero divisors, so \(r(\lambda) = 0\) or \(s(\lambda) = 0\). Hence \(Z(p) \subseteq Z( r) \cup Z(s)\).

Let \(p \in \mathcal{P}(\mathbf{R})\) be the minimal polynomial of \(T\). A basis \(v_1, \dots, v_n\) of \(V\) over \(\mathbf{R}\) is also a basis of \(V_{\mathbf{C}}\) over \(\mathbf{C}\) (Check!, using \(u + iv = \sum_j (a_j + ib_j)v_j\) and coordinatewise equality), so \(V_{\mathbf{C}}\) is finite-dimensional and \(T_{\mathbf{C}}\) has a minimal polynomial \(q \in \mathcal{P}(\mathbf{C})\) by 5.22. We show \(p\) and \(q\) are multiples of each other.

For \(g \in \mathcal{P}(\mathbf{R})\),

\begin{equation*} g(T_{\mathbf{C}})(u + iv) = g(T)u + i\, g(T)v \qquad (u, v \in V), \quad (\dagger) \end{equation*}

since \((T_{\mathbf{C}})^k(u+iv) = T^k u + iT^k v\) by induction and \(a(x+iy) = ax + iay\) for real \(a\).

Taking \(g = p\) in \((\dagger)\) gives \(p(T_{\mathbf{C}}) = 0\), so \(p\) is a polynomial multiple of \(q\) by 5.29.

Conversely, write \(q = q_1 + iq_2\) with \(q_1, q_2 \in \mathcal{P}(\mathbf{R})\) (real and imaginary parts of the coefficients). For \(u \in V\), \((\dagger)\) applied to \(q_1\) and \(q_2\) gives

\begin{equation*} 0 = q(T_{\mathbf{C}})(u + i0) = q_1(T)u + i\, q_2(T)u , \end{equation*}

and equality in \(V_{\mathbf{C}} = V \times V\) is coordinatewise, so \(q_1(T) = q_2(T) = 0\). By 5.29 each \(q_j\) is a polynomial multiple of \(p\), say \(q_j = pg_j\), whence \(q = p(g_1 + ig_2)\).

Thus \(q = ph\) and \(p = qg\), so \(p = phg\); as \(p \neq 0\), \(h\) is a nonzero constant, and comparing leading coefficients of the monic \(p\) and \(q\) gives \(h = 1\). Hence \(q = p\).

The two minimal polynomials agree because \(T\) and \(T^{\prime}\) are annihilated by exactly the same polynomials. (Both exist: \(\dim V^{\prime} = \dim V < \infty\) by 3.111, so 5.22 applies to each.)

First, \((T^k)^{\prime} = (T^{\prime})^k\) for every integer \(k \geq 0\): for \(k = 0\) this is \(I^{\prime}(\varphi) = \varphi \circ I = \varphi\), and inductively 3.120(c) gives

\begin{equation*} \begin{aligned} (T^{k+1})^{\prime} &= (T^{k} T)^{\prime} = T^{\prime}\,(T^{k})^{\prime} \\ &= T^{\prime} (T^{\prime})^{k} = (T^{\prime})^{k+1}. \end{aligned} \end{equation*}

Since \(S \mapsto S^{\prime}\) is linear by 3.120(a) and 3.120(b), applying this termwise to \(g(z) = a_0 + a_1 z + \dots + a_d z^d\) yields \(\bigl(g(T)\bigr)^{\prime} = g(T^{\prime})\) for every \(g \in \mathcal{P}(\mathbf{F})\).

Next, \(S = 0\) if and only if \(S^{\prime} = 0\), because \(\dim \operatorname{range} S^{\prime} = \dim \operatorname{range} S\) by 3.130(a) and an operator vanishes exactly when its range has dimension \(0\). Combining the two facts,

\begin{equation*} g(T) = 0 \iff \bigl(g(T)\bigr)^{\prime} = 0 \iff g(T^{\prime}) = 0 . \end{equation*}

So if \(p\) and \(\tilde p\) are the minimal polynomials of \(T\) and \(T^{\prime}\), then \(p(T^{\prime}) = 0\) gives \(\deg \tilde p \leq \deg p\) and \(\tilde p(T) = 0\) gives \(\deg p \leq \deg \tilde p\). Thus \(\tilde p\) is monic of degree \(\deg p\) with \(\tilde p(T) = 0\), so \(\tilde p = p\) by the uniqueness in 5.22.

Take \(W = \operatorname{span}(w, Tw)\), where \(w = h(T)v\) for the vector \(v\) and the factorization \(q = gh\) produced below.

(i) If \(T = \lambda I\), every subspace of \(V\) is invariant under \(T\), and \(\dim V \geq 2\) supplies a two-dimensional one.

(ii) Otherwise some \(v \in V\) has \(v, Tv\) linearly independent. For if not, then every nonzero \(u \in V\) satisfies \(Tu = \lambda_u u\) for some \(\lambda_u \in \mathbf{F}\) (in a dependence \(a u + b\,Tu = 0\) we must have \(b \neq 0\), else \(u = 0\)), and all these scalars agree: if \(w = tu\) with \(t \neq 0\), then \(\lambda_w w = t\lambda_u u = \lambda_u w\), while if \(u, w\) is linearly independent, then comparing coefficients in \(\lambda_{u+w} u + \lambda_{u+w} w = \lambda_u u + \lambda_w w\) gives \(\lambda_u = \lambda_{u+w} = \lambda_w\). Thus \(T = \lambda I\), a contradiction.

Fix such a \(v\) and let \(q\) be the monic polynomial of smallest degree \(m\) with \(q(T)v = 0\), which exists by Exercise 7(a) in Section 5C. Minimality gives the consequence, used twice below, that \(r(T)v = 0\) with \(\deg r < m\) forces \(r = 0\) (otherwise divide \(r\) by its leading coefficient). Also \(m \geq 2\), since \(m = 0\) would give \(v = Iv = 0\) and \(m = 1\) would give \(Tv = \lambda v\), contradicting the independence of \(v, Tv\).

Any monic \(q\) with \(\deg q = m \geq 2\) factors as \(q = gh\) with \(g, h\) monic, \(\deg g = 2\) and \(\deg h = m - 2\): over \(\mathbf{C}\) let \(g\) be the product of the first two linear factors supplied by 4.13, and over \(\mathbf{R}\) let \(g\) be a quadratic factor \(x^2 + b_1 x + c_1\) supplied by 4.16 if one occurs there and otherwise the product of two of the \(m\) linear factors. Write \(g(z) = z^2 + bz + c\).

Then \(w = h(T)v \neq 0\), since \(h\) is monic of degree \(m - 2 < m\), hence nonzero. By 5.17,

\begin{equation*} g(T)w = (gh)(T)v = q(T)v = 0, \qquad \text{so} \qquad T^2 w = -b\,Tw - c\,w , \end{equation*}

whence \(W = \operatorname{span}(w, Tw)\) is invariant under \(T\), because

\begin{equation*} T(\alpha w + \beta\, Tw) = (-\beta c) w + (\alpha - \beta b)\, Tw \in W . \end{equation*}

Finally \(\dim W = 2\): otherwise \(W = \operatorname{span}(w)\) with \(w \neq 0\), so \(Tw = \lambda w\) for some \(\lambda \in \mathbf{F}\), and then \(r(z) = (z - \lambda)h(z)\) is monic of degree \(m - 1\) with \(r(T)v = (T - \lambda I)w = 0\) by 5.17, contradicting the minimality of \(m\).

False: take \(\mathbf{F} = \mathbf{R}\), \(V = \mathbf{R}^2\), and \(T(x,y) = (-y, x)\), rotation by \(90\) degrees.

Then \(T^2(x,y) = T(-y,x) = (-x,-y)\), so \(T^2 = -I\), whose matrix with respect to every basis of \(\mathbf{R}^2\) is

\begin{equation*} \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix}, \end{equation*}

which is upper triangular.

But \(T\) has no eigenvalue: if \(T(x,y) = \lambda(x,y)\), then \(-y = \lambda x\) and \(x = \lambda y\), so \(x = -\lambda^2 x\), hence \((1 + \lambda^2) x = 0\) and \(x = 0\), and then \(y = -\lambda x = 0\) too. Were the matrix of \(T\) with respect to some basis upper triangular, its diagonal entries would be eigenvalues of \(T\) by 5.41, a contradiction.

Both claims are entrywise consequences of \(A_{j,k} = B_{j,k} = 0\) whenever \(j > k\) (5.38), with \(A_{k,k} = \alpha_k\) and \(B_{k,k} = \beta_k\).

(a) \((A+B)_{j,k} = A_{j,k} + B_{j,k}\), which is \(0\) when \(j > k\) and equals \(\alpha_k + \beta_k\) when \(j = k\).

(b) By 3.41, \((AB)_{j,k} = \sum_{r=1}^{n} A_{j,r} B_{r,k}\), and a nonzero term forces \(A_{j,r} \neq 0\) and \(B_{r,k} \neq 0\), hence

\begin{equation*} j \leq r \leq k . \end{equation*}

If \(j > k\), no such \(r\) exists, so \((AB)_{j,k} = 0\); if \(j = k\), the only such \(r\) is \(k\), so \((AB)_{k,k} = A_{k,k}B_{k,k} = \alpha_k \beta_k\).

Each \(\lambda_k \neq 0\), because \(T\) is invertible and hence injective, so \(0\) is not an eigenvalue of \(T\), while by 5.41 the eigenvalues of \(T\) are exactly \(\lambda_1, \ldots, \lambda_n\).

Fix \(k\) and set \(U_k = \operatorname{span}(v_1, \ldots, v_k)\), which is invariant under \(T\) by 5.39. Injectivity of \(T\) makes the operator \(T|_{U_k}\) on the finite-dimensional space \(U_k\) injective, hence surjective by 3.65, so \(T(U_k) = U_k\). Applying \(T^{-1}\) to \(v_k \in U_k = T(U_k)\) therefore gives

\begin{equation*} T^{-1}v_k \in \operatorname{span}(v_1, \ldots, v_k) \qquad \text{for each } k, \end{equation*}

so the matrix of \(T^{-1}\) with respect to \(v_1, \ldots, v_n\) is upper triangular by the implication (c) \(\Longrightarrow\) (a) of 5.39.

Let \(\mu_1, \ldots, \mu_n\) be its diagonal entries. By 3.43 (all three bases equal to \(v_1, \ldots, v_n\)),

\begin{equation*} \mathcal{M}(T)\,\mathcal{M}(T^{-1}) = \mathcal{M}(TT^{-1}) = \mathcal{M}(I), \end{equation*}

whose diagonal entries all equal \(1\), while Exercise 2(b) says that diagonal is \(\lambda_1 \mu_1, \ldots, \lambda_n \mu_n\). Hence \(\lambda_k \mu_k = 1\), that is, \(\mu_k = 1/\lambda_k\).

Take \(T \in \mathcal{L}(\mathbf{F}^2)\) with \(T(x,y) = (y,x)\). With respect to the standard basis \(e_1, e_2\),

\begin{equation*} \mathcal{M}(T) = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}, \end{equation*}

whose diagonal entries are both \(0\), yet \(T\bigl(T(x,y)\bigr) = T(y,x) = (x,y)\), so \(T^2 = I\) and \(T\) is invertible with \(T^{-1} = T\).

Take \(T \in \mathcal{L}(\mathbf{F}^2)\) with \(T(x,y) = (x+y,\; x+y)\). With respect to the standard basis \(e_1, e_2\),

\begin{equation*} \mathcal{M}(T) = \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}, \end{equation*}

whose diagonal entries both equal \(1\), yet \(T(1,-1) = (0,0)\) with \((1,-1) \neq 0\), so \(T\) is not injective and hence not invertible.

Take \(U = \operatorname{span}(v_1, \ldots, v_k)\), where \(v_1, \ldots, v_n\) is a basis of \(V\) with respect to which \(T\) has an upper-triangular matrix, supplied by 5.47 because \(V\) is a finite-dimensional complex vector space.

Then \(U\) is invariant under \(T\) by the implication (a) \(\Longrightarrow\) (b) of 5.39, and \(v_1, \ldots, v_k\) is a linearly independent spanning list of \(U\), being a sublist of a basis; hence \(\dim U = k\).

(a) Take \(p_v\) to be a member of smallest degree \(m\) of

\begin{equation*} S = \{\, q \in \mathcal{P}(\mathbf{F}) : q \text{ is monic and } q(T)v = 0 \,\} , \end{equation*}

which exists because \(S\) contains the minimal polynomial \(p\) of \(T\) (as \(p(T) = 0\) by 5.22, \(V\) being finite-dimensional), so \(\{\deg q : q \in S\}\) is a nonempty set of nonnegative integers.

For uniqueness, suppose \(q \in S\) also has degree \(m\). The leading terms of the two monic polynomials cancel, so \(p_v - q = 0\) or \(\deg(p_v - q) < m\), and \((p_v - q)(T)v = 0\) by the linearity of \(r \mapsto r(T)\) (5.14). Were \(p_v - q \neq 0\) with leading coefficient \(c\), the polynomial \(c^{-1}(p_v - q)\) would lie in \(S\) with degree less than \(m\), contradicting minimality. Hence \(q = p_v\).

(b) Since \(p_v \neq 0\), the division algorithm 4.9 gives \(p = s\,p_v + r\) with \(\deg r < \deg p_v\). Applying both sides to \(v\), using 5.14 and 5.17(a) together with \(p(T) = 0\) and \(p_v(T)v = 0\),

\begin{equation*} 0 = p(T)v = s(T)\bigl(p_v(T)v\bigr) + r(T)v = r(T)v . \end{equation*}

If \(r \neq 0\) with leading coefficient \(c\), then \(c^{-1}r\) is monic of degree \(\deg r < \deg p_v\) with \((c^{-1}r)(T)v = 0\), contradicting (a). So \(r = 0\) and \(p = s\,p_v\).

Both parts run through the nonzero \(T\)-invariant subspace \(U = \operatorname{span}(v, Tv)\), on which \(S = T|_U\) satisfies \(S^2 + 2S + 2I_U = 0\).

The hypothesis reads \((T^2 + 2T + 2I)v = 0\) with \(v \neq 0\), so \(T(Tv) = -2Tv - 2v \in U\) and \(U\) is invariant under \(T\); moreover \(T^2 + 2T + 2I\) also kills \(Tv\) (it commutes with \(T\)), hence kills all of \(U\), which is the displayed identity. So if \(\mu \in \mathbb{F}\) is an eigenvalue of \(S\), with eigenvector \(u \neq 0\), then

\begin{equation*} 0 = (S^2 + 2S + 2I_U)u = (\mu^2 + 2\mu + 2)u , \end{equation*}

so \((\mu + 1)^2 = -1\).

In either part, an upper-triangular matrix for \(T\) with diagonal \(\lambda_1, \dots, \lambda_n\) gives \((T - \lambda_1 I) \cdots (T - \lambda_n I) = 0\) by 5.40; since \(U\) is invariant under \(T\) and hence under each \(T - \lambda_k I\), restricting to \(U\) yields

\begin{equation*} (S - \lambda_1 I_U) \cdots (S - \lambda_n I_U) = 0 . \end{equation*}

(a) \(\mathbb{F} = \mathbb{R}\): no real \(\mu\) satisfies \((\mu+1)^2 = -1\), so \(S\) has no eigenvalue and every factor \(S - \lambda_k I_U\) is injective by 5.7. Then the composition is injective, so the display forces \(U = \{0\}\), contradicting \(v \neq 0\).

(b) \(\mathbb{F} = \mathbb{C}\): here \((\mu+1)^2 = -1\) forces \(\mu = -1 + i\) or \(\mu = -1 - i\). Since \(U \neq \{0\}\), the composition in the display is not injective, so some factor \(S - \lambda_k I_U\) is not injective; by 5.7 that \(\lambda_k\) is an eigenvalue of \(S\), hence equals \(-1 + i\) or \(-1 - i\), and it lies on the diagonal of \(A\).

Take \(A = \mathcal{M}\bigl(I, (u_1, \dots, u_n), (e_1, \dots, e_n)\bigr)\), the change-of-basis matrix described below.

Say \(B\) is \(n\)-by-\(n\), let \(e_1, \dots, e_n\) be the standard basis of \(\mathbb{C}^n\), and let \(T \in \mathcal{L}(\mathbb{C}^n)\) be the operator with \(Te_k = \sum_{j=1}^{n} B_{j,k}\,e_j\), so that \(\mathcal{M}\bigl(T, (e_1, \dots, e_n)\bigr) = B\). Because \(\mathbb{C}^n\) is a finite-dimensional complex vector space, 5.47 provides a basis \(u_1, \dots, u_n\) of \(\mathbb{C}^n\) for which

\begin{equation*} C = \mathcal{M}\bigl(T, (u_1, \dots, u_n)\bigr) \end{equation*}

is upper triangular.

This \(A\) is invertible: with \(A^{\prime} = \mathcal{M}\bigl(I, (e_1, \dots, e_n), (u_1, \dots, u_n)\bigr)\), the product formula 3.81 gives \(A A^{\prime} = \mathcal{M}(I, (e), (e))\) and \(A^{\prime} A = \mathcal{M}(I, (u), (u))\), both the identity matrix. The change-of-basis formula 3.84 then gives

\begin{equation*} A^{-1} B A = C , \end{equation*}

which is upper triangular.

Write \(A = \mathcal{M}\bigl(T, (v_1, \dots, v_n)\bigr)\), so that \(T v_k = \sum_{i=1}^{n} A_{i,k} v_i\), and prove (a) \(\Rightarrow\) (b) \(\Rightarrow\) (c) \(\Rightarrow\) (a).

(a) \(\Rightarrow\) (b). Lower triangularity says \(A_{i,j} = 0\) whenever \(i < j\), so

\begin{equation*} \begin{aligned} T v_j &\in \operatorname{span}(v_j, \dots, v_n) \\ &\subseteq \operatorname{span}(v_k, \dots, v_n) \qquad \text{for } j \geq k . \end{aligned} \end{equation*}

Thus \(T\) maps each vector of the spanning list \(v_k, \dots, v_n\) into \(\operatorname{span}(v_k, \dots, v_n)\), hence maps that subspace into itself.

(b) \(\Rightarrow\) (c). \(v_k\) lies in \(\operatorname{span}(v_k, \dots, v_n)\), which is invariant under \(T\).

(c) \(\Rightarrow\) (a). Writing \(T v_k\) as a combination of \(v_k, \dots, v_n\) alone says \(A_{i,k} = 0\) for all \(i < k\), so every entry of \(A\) above the diagonal is \(0\).

Reverse a triangularizing basis: let \(v_1, \dots, v_n\) be a basis of \(V\) with respect to which \(T\) has an upper-triangular matrix, supplied by 5.47 because \(V\) is a finite-dimensional complex vector space, and put \(w_j = v_{n+1-j}\).

Then \(w_1, \dots, w_n\) is again a basis of \(V\), being the same vectors in a different order, and 5.39 gives \(T v_m \in \operatorname{span}(v_1, \dots, v_m)\) for each \(m\). Fix \(k\) and set \(m = n+1-k\), so \(w_k = v_m\); since span does not depend on the order of the list,

\begin{equation*} \begin{aligned} \operatorname{span}(w_k, \dots, w_n) &= \operatorname{span}(v_{n+1-k}, \dots, v_1) \\ &= \operatorname{span}(v_1, \dots, v_m) . \end{aligned} \end{equation*}

Hence \(T w_k = T v_m \in \operatorname{span}(w_k, \dots, w_n)\), which is condition (c) of Exercise 10; so by Exercise 10 the matrix of \(T\) with respect to \(w_1, \dots, w_n\) is lower triangular.

Both parts come from the flag \(V_k = \operatorname{span}(v_1, \dots, v_k)\) of a basis \(v_1, \dots, v_n\) of \(V\) with respect to which \(T\) has an upper-triangular matrix: each \(V_k\) is invariant under \(T\) by 5.39 and has \(\dim V_k = k\), with \(V_0 = \{0\}\) and \(V_n = V\). Part (a) intersects this chain with \(U\); part (b) pushes it into \(V/U\).

(a) Set \(U_k = U \cap V_k\), invariant under \(T\) because \(U\) and \(V_k\) both are, with \(U_0 = \{0\}\), \(U_n = U\), and \(U_{k-1} \subseteq U_k\). Since \(U_k \cap V_{k-1} = U \cap V_{k-1} = U_{k-1}\) while \(U_k + V_{k-1} \subseteq V_k\), the dimension formula 2.43 gives

\begin{equation*} \dim U_k - \dim U_{k-1} + (k-1) = \dim(U_k + V_{k-1}) \leq k , \end{equation*}

so each step raises the dimension by \(0\) or \(1\). Let \(m = \dim U\) and let \(k_1 < \cdots < k_m\) be the indices at which it rises; counting the rises among \(1, \dots, k_j\) gives \(\dim U_{k_j} = j\).

Choose \(u_j \in U_{k_j}\) with \(u_j \notin U_{k_j - 1}\), possible since that inclusion is strict. Then \(\operatorname{span}(u_1, \dots, u_j) = U_{k_j}\) for each \(j\), by induction: \(U_{k_{j-1}} \subseteq U_{k_j - 1}\) and \(u_j \notin U_{k_j - 1}\) make \(u_1, \dots, u_j\) linearly independent, and a linearly independent list of length \(j\) in the \(j\)-dimensional space \(U_{k_j}\) is a basis of it (2.38). Taking \(j = m\) makes \(u_1, \dots, u_m\) a basis of \(U\), and each \(\operatorname{span}(u_1, \dots, u_j) = U_{k_j}\) is contained in \(U\) and invariant under \(T\), hence under \(T|_U\); so the implication (b) \(\Longrightarrow\) (a) of 5.39, applied to \(T|_U\), makes the matrix of \(T|_U\) with respect to \(u_1, \dots, u_m\) upper triangular.

(b) Let \(\pi \colon V \to V/U\) be the quotient map and \(W_k = \pi(V_k) = \operatorname{span}\bigl(\pi(v_1), \dots, \pi(v_k)\bigr)\). Each \(W_k\) is invariant under \(T/U\), since \((T/U)\bigl(\pi(v_i)\bigr) = \pi(T v_i) \in \pi(V_k) = W_k\) for \(i \leq k\). As \(\pi\) is surjective, \(\pi(v_1), \dots, \pi(v_n)\) spans \(V/U\); reduce it to a basis by the procedure of 2.30, deleting \(\pi(v_k)\) whenever it lies in the span of its predecessors, and let \(k_1 < \cdots < k_p\) be the surviving indices, \(w_j = \pi(v_{k_j})\).

Each deletion leaves the span of every initial segment unchanged, so induction on \(k\) gives

\begin{equation*} \begin{aligned} \operatorname{span}\bigl(w_j : k_j \leq k\bigr) &= \operatorname{span}\bigl(\pi(v_1), \dots, \pi(v_k)\bigr) = W_k . \end{aligned} \end{equation*}

Taking \(k = k_j\) gives \(\operatorname{span}(w_1, \dots, w_j) = W_{k_j}\), which is invariant under \(T/U\); so 5.39, applied to \(T/U\) and the basis \(w_1, \dots, w_p\) of \(V/U\), makes that matrix upper triangular.

Take the concatenated list \(x_1, \dots, x_{m+p}\) equal to \(u_1, \dots, u_m, w_1, \dots, w_p\), where \(u_1, \dots, u_m\) is a basis of \(U\) triangularizing \(T|_U\), the list \(\widetilde{w}_1, \dots, \widetilde{w}_p\) is a basis of \(V/U\) triangularizing \(T/U\), and \(\pi(w_j) = \widetilde{w}_j\) for the quotient map \(\pi \colon V \to V/U\), which is surjective with \(\operatorname{null} \pi = U\) and \((T/U) \circ \pi = \pi \circ T\).

This list is a basis of \(V\). It spans \(V\): given \(v \in V\), write \(\pi(v) = c_1 \widetilde{w}_1 + \cdots + c_p \widetilde{w}_p\); then \(\pi(v - c_1 w_1 - \cdots - c_p w_p) = 0\), so that vector lies in \(\operatorname{null} \pi = U = \operatorname{span}(u_1, \dots, u_m)\). It is linearly independent: applying \(\pi\) to \(a_1 u_1 + \cdots + a_m u_m + b_1 w_1 + \cdots + b_p w_p = 0\) kills the \(u_i\) and leaves \(b_1 \widetilde{w}_1 + \cdots + b_p \widetilde{w}_p = 0\), forcing all \(b_j = 0\) and then all \(a_i = 0\).

By 5.39 it now suffices to check that \(T x_k \in \operatorname{span}(x_1, \dots, x_k)\) for each \(k\).

(i) \(k = i \leq m\). Applying 5.39 to \(T|_U\) gives

\begin{equation*} \begin{aligned} T u_i = (T|_U)(u_i) &\in \operatorname{span}(u_1, \dots, u_i) \\ &= \operatorname{span}(x_1, \dots, x_i) . \end{aligned} \end{equation*}

(ii) \(k = m + j\) with \(1 \leq j \leq p\). Applying 5.39 to \(T/U\) gives \(c_1, \dots, c_j \in \mathbb{F}\) with \((T/U)(\widetilde{w}_j) = c_1 \widetilde{w}_1 + \cdots + c_j \widetilde{w}_j\), so

\begin{equation*} \pi(T w_j) = (T/U)\bigl(\pi(w_j)\bigr) = \pi(c_1 w_1 + \cdots + c_j w_j) , \end{equation*}

whence \(T w_j - (c_1 w_1 + \cdots + c_j w_j) \in \operatorname{null} \pi = U\) and therefore

\begin{equation*} \begin{aligned} T x_{m+j} &\in \operatorname{span}(u_1, \dots, u_m, w_1, \dots, w_j) \\ &= \operatorname{span}(x_1, \dots, x_{m+j}) . \end{aligned} \end{equation*}

Each side is equivalent, by 5.44, to the minimal polynomial \(p\) of \(T\) having the form

\begin{equation*} p(z) = (z - \lambda_1) \cdots (z - \lambda_m) \end{equation*}

for some \(\lambda_1, \dots, \lambda_m \in \mathbb{F}\).

For \(T\) itself this is 5.44 applied on \(V\). For \(T^{\prime}\), the space \(V^{\prime}\) is finite-dimensional with \(\dim V^{\prime} = \dim V\) (3.111), so 5.44 applies on \(V^{\prime}\) as well, and the minimal polynomial of \(T^{\prime}\) is again \(p\) by Exercise 28 in Section 5B. Hence the two triangularizability statements are equivalent.

Parts (a) and (b) both follow from this: if \(q \in \mathcal{P}(\mathbf{C})\) has no repeated zero and \(q(T) = 0\), then \(T\) is diagonalizable.

Indeed, the minimal polynomial \(p\) of \(T\) then divides \(q\) by 5.29, say \(q = ps\), and 4.13 lets us write

\begin{equation*} p(z) = (z - \alpha_1)^{m_1} \cdots (z - \alpha_k)^{m_k} \end{equation*}

with \(\alpha_1, \dots, \alpha_k \in \mathbf{C}\) distinct and each \(m_j \geq 1\) (with \(k = 0\) if \(p = 1\)). If some \(m_j \geq 2\), then \((z - \alpha_j)^2\) divides \(q\), so factoring \(q\) into linear factors repeats \(z - \alpha_j\), contradicting the uniqueness in 4.13 since the zeros of \(q\) are distinct. Hence every \(m_j = 1\), and 5.62 makes \(T\) diagonalizable.

(a) \(T^4 = I\) gives \(q(T) = 0\) for \(q(z) = z^4 - 1 = (z-1)(z+1)(z-i)(z+i)\), whose zeros \(1, -1, i, -i\) are distinct.

(b) \(T^4 = T\) gives \(q(T) = 0\) for

\begin{equation*} q(z) = z^4 - z = z(z-1)(z - \omega)(z - \omega^2), \qquad \omega = e^{2\pi i/3} , \end{equation*}

whose zeros \(0, 1, \omega, \omega^2\) are distinct, since \(\omega\) is not real, so \(\omega^2 = \overline{\omega} \neq \omega\), and neither is \(0\) or \(1\).

(c) Take \(T(w,z) = (z,0)\) on \(\mathbf{C}^2\). Then \(T^2 = 0\), so \(T^4 = T^2\), while \(T(0,1) = (1,0)\) shows \(T \neq 0\); hence the minimal polynomial of \(T\) is \(z^2\), which is not a product of distinct linear factors, so \(T\) is not diagonalizable by 5.62.

Everything follows from the identity \(E(\lambda, T) = \operatorname{span}(v_k : k \in S)\), where \(v_1, \dots, v_n\) is the basis giving the diagonal matrix \(A\), the numbers \(\lambda_1, \dots, \lambda_n\) are its diagonal entries, so that \(T v_k = \lambda_k v_k\), and

\begin{equation*} S = \{ k \in \{1, \dots, n\} : \lambda_k = \lambda \} . \end{equation*}

The inclusion \(\supseteq\) holds because \(T v_k = \lambda v_k\) for \(k \in S\) and \(E(\lambda, T)\) is a subspace (5.52). For \(\subseteq\), let \(v = a_1 v_1 + \cdots + a_n v_n\) lie in \(E(\lambda, T)\); then

\begin{equation*} \begin{aligned} a_1 \lambda_1 v_1 + \cdots + a_n \lambda_n v_n &= Tv = \lambda v \\ &= a_1 \lambda v_1 + \cdots + a_n \lambda v_n , \end{aligned} \end{equation*}

and uniqueness of the expansion in the basis \(v_1, \dots, v_n\) gives \(a_k(\lambda_k - \lambda) = 0\), so \(a_k = 0\) whenever \(k \notin S\).

Being a sublist of a basis, \((v_k)_{k \in S}\) is linearly independent, hence a basis of \(E(\lambda, T)\). Therefore \(\dim E(\lambda, T)\) is the number of elements of \(S\), that is, the number of times \(\lambda\) appears on the diagonal of \(A\).

Split a basis of eigenvectors according to whether its eigenvalue vanishes: by 5.55(b) there is a basis \(v_1, \dots, v_n\) of \(V\) with \(T v_k = \lambda_k v_k\), and with \(S = \{k : \lambda_k = 0\}\) and \(S^{\prime} = \{k : \lambda_k \neq 0\}\) we put

\begin{equation*} \begin{aligned} U &= \operatorname{span}(v_k : k \in S), \\ W &= \operatorname{span}(v_k : k \in S^{\prime}) . \end{aligned} \end{equation*}

Then \(V = U \oplus W\): the sum contains every \(v_k\), hence equals \(V\), and if \(u + w = 0\) with \(u \in U\) and \(w \in W\), then expanding both in \(v_1, \dots, v_n\) and using linear independence gives \(u = w = 0\), so the sum is direct by 1.45.

Moreover \(U = E(0, T) = \operatorname{null} T\), by the identity proved in Exercise 2 of this section taken with \(\lambda = 0\), and \(W = \operatorname{range} T\) because

\begin{equation*} \begin{aligned} \operatorname{range} T &= \operatorname{span}(T v_1, \dots, T v_n) \\ &= \operatorname{span}(\lambda_1 v_1, \dots, \lambda_n v_n) = W , \end{aligned} \end{equation*}

the terms with \(k \in S\) being \(0\) and \(\operatorname{span}(\lambda_k v_k) = \operatorname{span}(v_k)\) for \(k \in S^{\prime}\). Hence \(V = \operatorname{null} T \oplus \operatorname{range} T\).

Write \(N = \operatorname{null} T\) and \(R = \operatorname{range} T\), so that \(\dim N + \dim R = \dim V\) by 3.21, and prove (a) \(\Rightarrow\) (b) \(\Rightarrow\) (c) \(\Rightarrow\) (a).

(a) \(\Rightarrow\) (b). The statement \(V = N \oplus R\) asserts in particular that \(V = N + R\).

(b) \(\Rightarrow\) (c). By 2.43,

\begin{equation*} \dim V = \dim(N + R) = \dim N + \dim R - \dim(N \cap R) , \end{equation*}

whose right side is \(\dim V - \dim(N \cap R)\) by 3.21. Hence \(\dim(N \cap R) = 0\), that is, \(N \cap R = \{0\}\).

(c) \(\Rightarrow\) (a). The sum \(N + R\) is direct by 1.46, and by 2.43 and 3.21,

\begin{equation*} \dim(N \oplus R) = \dim N + \dim R - \dim(N \cap R) = \dim V . \end{equation*}

A subspace of \(V\) with the same dimension as \(V\) equals \(V\) (2.39), so \(V = N \oplus R\).

If \(T\) is diagonalizable, a basis \(v_1, \dots, v_n\) of eigenvectors of \(T\) (5.55(b)) consists of eigenvectors of \(T - \lambda I\) as well, since \((T - \lambda I)v_k = (\lambda_k - \lambda)v_k\). So \(T - \lambda I\) is diagonalizable by 5.55, and Exercise 3 of this section applied to \(T - \lambda I\) gives the displayed decomposition for every \(\lambda\).

Conversely, assume that decomposition for every \(\lambda \in \mathbf{C}\), so 1.46 gives

\begin{equation*} \operatorname{null}(T - \lambda I) \cap \operatorname{range}(T - \lambda I) = \{0\} \end{equation*}

for every \(\lambda \in \mathbf{C}\). Because \(\mathbf{F} = \mathbf{C}\), the factorization 4.13 writes the minimal polynomial of \(T\) as \(p(z) = (z - \lambda_1)^{m_1} \cdots (z - \lambda_k)^{m_k}\) with \(\lambda_1, \dots, \lambda_k\) distinct and each \(m_j \geq 1\) (with \(k = 0\) if \(p = 1\)). Suppose some \(m_j \geq 2\); relabel so \(j = 1\) and set

\begin{equation*} r(z) = \frac{p(z)}{(z - \lambda_1)^2} , \qquad q(z) = (z - \lambda_1) r(z) , \end{equation*}

both genuine polynomials since \(m_1 \geq 2\). As \(q \neq 0\) has degree less than \(\deg p\), it is not a polynomial multiple of \(p\), so \(q(T) \neq 0\) by 5.29; choose \(v\) with \(q(T)v \neq 0\) and set \(u = (T - \lambda_1 I) r(T) v\). By 5.17,

\begin{equation*} u = q(T)v \neq 0 , \qquad (T - \lambda_1 I)u = p(T)v = 0 , \end{equation*}

so \(u\) is a nonzero vector of \(\operatorname{null}(T - \lambda_1 I) \cap \operatorname{range}(T - \lambda_1 I)\), contradicting the display at \(\lambda = \lambda_1\). Hence every \(m_j = 1\), and 5.62 makes \(T\) diagonalizable.

Suppose neither \(T - 2I\) nor \(T - 6I\) is invertible. Then neither is injective, by 3.65 applied on the finite-dimensional space \(\mathbf{F}^5\), so \(E(2,T)\) and \(E(6,T)\) are nonzero; together with \(\dim E(8,T) = 4\) this makes \(2, 6, 8\) three distinct eigenvalues of \(T\). Then 5.54 gives

\begin{equation*} 6 \leq \dim E(2,T) + \dim E(6,T) + \dim E(8,T) \leq \dim \mathbf{F}^5 = 5 , \end{equation*}

a contradiction. Hence \(T - 2I\) or \(T - 6I\) is invertible.

For \(v \in V\) the conditions \(Tv = \lambda v\) and \(T^{-1}v = \frac{1}{\lambda} v\) are equivalent, which is the assertion, since by 5.52 these conditions describe membership in \(E(\lambda, T)\) and in \(E\!\left(\frac{1}{\lambda}, T^{-1}\right)\) respectively.

Applying \(T^{-1}\) to \(Tv = \lambda v\) gives \(v = \lambda\,T^{-1}v\), hence \(T^{-1}v = \frac{1}{\lambda} v\) as \(\lambda \neq 0\); applying \(T\) to \(T^{-1}v = \frac{1}{\lambda} v\) gives \(v = \frac{1}{\lambda}\,Tv\), hence \(Tv = \lambda v\).

Each \(E(\lambda_k, T)\) lies inside \(\operatorname{range} T\), since \(Tv = \lambda_k v\) with \(\lambda_k \neq 0\) gives \(v = T\!\left(\frac{1}{\lambda_k} v\right)\); hence so does their sum \(U = E(\lambda_1, T) + \cdots + E(\lambda_m, T)\).

That sum is direct, the \(\lambda_k\) being distinct eigenvalues (5.54), so 3.94 identifies \(\dim U\) with the sum of the dimensions and

\begin{equation*} \begin{aligned} \dim E(\lambda_1, T) + \cdots + \dim E(\lambda_m, T) &= \dim U \\ &\leq \dim \operatorname{range} T , \end{aligned} \end{equation*}

the inequality by 2.37, since \(U\) is a subspace of the finite-dimensional space \(\operatorname{range} T\).

Take \(S \in \mathcal{L}(\mathbf{F}^3)\) with \(S u_k = v_k\) for \(k = 1, 2, 3\), where \(u_1, u_2, u_3\) and \(v_1, v_2, v_3\) are bases of \(\mathbf{F}^3\) with

\begin{equation*} R u_k = \lambda_k u_k \qquad \text{and} \qquad T v_k = \lambda_k v_k \end{equation*}

for \(\lambda_1 = 2\), \(\lambda_2 = 6\), \(\lambda_3 = 7\).

Such bases exist because \(R\) and \(T\) each have \(3 = \dim \mathbf{F}^3\) distinct eigenvalues, so each is diagonalizable by 5.58; picking an eigenvector for each of \(2, 6, 7\) in that order gives the matching labels. The map \(S\) exists by 3.4, and its range contains the spanning list \(v_1, v_2, v_3\), so \(S\) is surjective and hence invertible by 3.65. Finally,

\begin{equation*} (S^{-1} T S) u_k = S^{-1} T v_k = \lambda_k S^{-1} v_k = \lambda_k u_k = R u_k \end{equation*}

for each \(k\), so \(S^{-1} T S\) and \(R\) agree on a basis of \(\mathbf{F}^3\) and are therefore equal.

Define \(R, T \in \mathcal{L}(\mathbf{F}^4)\) by

\begin{equation*} R(z_1, z_2, z_3, z_4) = (2z_1,\; 6z_2,\; 7z_3,\; 7z_4), \end{equation*}

\begin{equation*} T(z_1, z_2, z_3, z_4) = (2z_1,\; 6z_2,\; 7z_3 + z_4,\; 7z_4). \end{equation*}

With respect to the standard basis \(e_1, e_2, e_3, e_4\) both matrices are upper triangular with diagonal entries \(2, 6, 7, 7\), so by 5.41 each of \(R\) and \(T\) has eigenvalues exactly \(2, 6, 7\).

The \(7\)-eigenspaces differ in dimension: \(E(7, R) = \operatorname{span}(e_3, e_4)\), whereas \(Tz = 7z\) forces \(z_1 = z_2 = 0\) (from \(2z_1 = 7z_1\) and \(6z_2 = 7z_2\)) and \(z_4 = 0\) (from \(7z_3 + z_4 = 7z_3\)), so \(E(7, T) = \operatorname{span}(e_3)\). Thus \(\dim E(7,R) = 2\) and \(\dim E(7,T) = 1\).

Suppose \(S\) were invertible with \(R = S^{-1} T S\), that is, \(SR = TS\). For \(v \in E(7, R)\),

\begin{equation*} T(Sv) = (TS)v = (SR)v = S(7v) = 7(Sv), \end{equation*}

so \(S\) maps \(E(7, R)\) into \(E(7, T)\), injectively since \(S\) is injective; hence \(2 = \dim E(7,R) \le \dim E(7,T) = 1\) by 3.21, a contradiction. So no such \(S\) exists.

Define \(T \in \mathcal{L}(\mathbf{C}^3)\) by

\begin{equation*} T(z_1, z_2, z_3) = (6z_1 + z_2,\; 6z_2,\; 7z_3). \end{equation*}

Its matrix with respect to the standard basis \(e_1, e_2, e_3\) is upper triangular with diagonal entries \(6, 6, 7\), so by 5.41 the eigenvalues of \(T\) are exactly \(6\) and \(7\).

Both eigenspaces are one-dimensional: \(Tz = 6z\) reads \(6z_1 + z_2 = 6z_1\), \(7z_3 = 6z_3\), forcing \(z_2 = z_3 = 0\), so \(E(6,T) = \operatorname{span}(e_1)\); and \(Tz = 7z\) reads \(6z_2 = 7z_2\), \(6z_1 + z_2 = 7z_1\), forcing \(z_2 = 0\) and then \(z_1 = 0\), so \(E(7,T) = \operatorname{span}(e_3)\). Hence

\begin{equation*} \dim E(6, T) + \dim E(7, T) = 1 + 1 = 2 < 3 = \dim \mathbf{C}^3 , \end{equation*}

so condition (d) of 5.55 fails and \(T\) is diagonalizable with respect to no basis of \(\mathbf{C}^3\).

Method (2): were \(T\) diagonalizable, 5.62 together with 5.27 (the zeros of the minimal polynomial are the eigenvalues \(6, 7\)) would make the minimal polynomial \((z-6)(z-7)\). But \(Te_2 = e_1 + 6e_2\) gives

\begin{equation*} (T - 6I)(T - 7I) e_2 = (T - 6I)(e_1 - e_2) = 0 - e_1 = -e_1 \ne 0 . \end{equation*}

The desired equation says exactly that \((T - 8I)z = (6, 7, 13)\), so it suffices to show that \(T - 8I\) is surjective.

Were \(8\) an eigenvalue of \(T\), then \(6, 7, 8\) would be \(3 = \dim \mathbf{C}^3\) distinct eigenvalues of \(T\), making \(T\) diagonalizable by 5.58 and contradicting the hypothesis. Hence \(\operatorname{null}(T - 8I) = \{0\}\), so \(T - 8I\) is injective (3.15) and therefore surjective, injectivity and surjectivity being equivalent for an operator on the finite-dimensional space \(\mathbf{C}^3\) (3.65). Thus some \(z = (z_1, z_2, z_3)\) satisfies \((T - 8I)z = (6, 7, 13)\), which is to say

\begin{equation*} T(z_1, z_2, z_3) = (6 + 8z_1,\; 7 + 8z_2,\; 13 + 8z_3). \end{equation*}

Say \(A, B\) are \(n\)-by-\(n\) with \(A_{j,j} = \lambda_j\) and \(A_{j,k} = 0\) for \(j \ne k\), the \(\lambda_j\) distinct. Because all but one term of each sum vanishes, the entry formula 3.47 gives

\begin{equation*} (AB)_{j,k} = \sum_{r=1}^{n} A_{j,r} B_{r,k} = \lambda_j B_{j,k}, \qquad (BA)_{j,k} = \sum_{r=1}^{n} B_{j,r} A_{r,k} = \lambda_k B_{j,k}, \end{equation*}

so, comparing entries,

\begin{equation*} AB = BA \iff (\lambda_j - \lambda_k) B_{j,k} = 0 \ \text{ for all } j, k . \end{equation*}

(i) If \(AB = BA\) and \(j \ne k\), then \(\lambda_j - \lambda_k \ne 0\) by distinctness, so \(B_{j,k} = 0\); hence \(B\) is diagonal.

(ii) If \(B\) is diagonal, then \(B_{j,k} = 0\) when \(j \ne k\) and \(\lambda_j - \lambda_k = 0\) when \(j = k\), so the displayed condition holds and \(AB = BA\).

(a) Take \(V = \mathbf{C}^2\) and define \(T \in \mathcal{L}(\mathbf{C}^2)\) by

\begin{equation*} T(z_1, z_2) = (z_2, 0). \end{equation*}

Then \(T^2 = 0\), whose matrix is diagonal with respect to every basis. But the matrix of \(T\) with respect to the standard basis is upper triangular with both diagonal entries \(0\), so \(0\) is the only eigenvalue of \(T\) by 5.41, and

\begin{equation*} \begin{aligned} E(0, T) = \operatorname{null} T &= \{(z_1, z_2) \in \mathbf{C}^2 : z_2 = 0\} \\ &= \operatorname{span}\big((1,0)\big) \end{aligned} \end{equation*}

has dimension \(1 < 2 = \dim \mathbf{C}^2\); so condition (d) of 5.55 fails and \(T\) is not diagonalizable.

(b) If \(T\) is diagonalizable, take a basis \(v_1, \ldots, v_n\) of eigenvectors, \(Tv_j = \lambda_j v_j\); then \(T^k v_j = \lambda_j^{\,k} v_j\), so the same basis diagonalizes \(T^k\) (5.55(b)).

Conversely, suppose \(T^k\) is diagonalizable. By 5.62 the minimal polynomial of \(T^k\) is \((z - \mu_1) \cdots (z - \mu_m)\) with \(\mu_1, \ldots, \mu_m \in \mathbf{C}\) distinct, and each \(\mu_j \ne 0\), since \(T^k\) is invertible and the zeros of the minimal polynomial are the eigenvalues (5.27). Set

\begin{equation*} p(z) = (z^k - \mu_1)(z^k - \mu_2) \cdots (z^k - \mu_m) . \end{equation*}

Multiplicativity of \(q \mapsto q(T)\) (5.17) gives

\begin{equation*} p(T) = (T^k - \mu_1 I) \cdots (T^k - \mu_m I) = 0 , \end{equation*}

this being the minimal polynomial of \(T^k\) evaluated at \(T^k\). Each nonzero \(\mu_j\) has exactly \(k\) distinct \(k\)-th roots in \(\mathbf{C}\), and a zero shared by \(z^k - \mu_j\) and \(z^k - \mu_{j^{\prime}}\) would force \(\mu_j = \mu_{j^{\prime}}\); so \(p\), monic of degree \(km\), has \(km\) distinct zeros and its factorization 4.13 is into \(km\) distinct linear factors. Hence no \((z - \alpha)^2\) divides \(p\).

Let \(q\) be the minimal polynomial of \(T\), so \(p = qs\) for some \(s \in \mathcal{P}(\mathbf{C})\) by 5.29. Since \(\mathbf{F} = \mathbf{C}\), 4.13 gives \(q = (z - \lambda_1)^{d_1} \cdots (z - \lambda_r)^{d_r}\) with \(\lambda_1, \ldots, \lambda_r\) distinct. Any \(d_i \ge 2\) would make \((z - \lambda_i)^2\) a divisor of \(q\) and hence of \(p\), which is impossible; so every \(d_i = 1\) and \(T\) is diagonalizable by 5.62.

We prove (a) \(\Leftrightarrow\) (b) \(\Leftrightarrow\) (c) \(\Leftrightarrow\) (d), each by contraposition. (If \(V = \{0\}\) then \(p = 1\) and all four statements hold trivially, so assume \(V \ne \{0\}\), whence \(\deg p \ge 1\).)

(a) \(\Leftrightarrow\) (b). If (b) fails, say \(p = (z - \lambda)^2 r\), and \(T\) were diagonalizable, then 5.62 would give \(p = (z - \mu_1) \cdots (z - \mu_m)\) with \(\mu_1, \dots, \mu_m\) distinct; from \(p(\lambda) = 0\) we get \(\lambda = \mu_j\) for exactly one \(j\), so \(p = (z - \lambda)s\) with \(s = \prod_{k \ne j}(z - \mu_k)\) satisfying \(s(\lambda) \ne 0\). Comparing with \(p = (z - \lambda)\bigl((z - \lambda)r\bigr)\) and cancelling \(z - \lambda\) (legitimate since \(\mathcal{P}(\mathbf{C})\) has no zero divisors) gives \(s = (z - \lambda)r\) and hence \(s(\lambda) = 0\), a contradiction; so (a) fails. Conversely, if (b) holds then 5.27(b) writes \(p = (z - \lambda_1) \cdots (z - \lambda_n)\) over \(\mathbf{C}\), and a repetition \(\lambda_j = \lambda_k\) with \(j \ne k\) would exhibit \(p\) as a multiple of \((z - \lambda_j)^2\); so the \(\lambda_i\) are distinct and 5.62 makes \(T\) diagonalizable.

(b) \(\Leftrightarrow\) (c). If \(p = (z - \lambda)^2 r\), then

\begin{equation*} p^{\prime} = 2 (z - \lambda) r + (z - \lambda)^2 r^{\prime} , \end{equation*}

so \(p(\lambda) = p^{\prime}(\lambda) = 0\) and (c) fails. Conversely, if \(p(\lambda) = p^{\prime}(\lambda) = 0\), the factor theorem 4.6 gives \(p = (z - \lambda)q\), whence \(p^{\prime} = q + (z - \lambda)q^{\prime}\) and \(q(\lambda) = p^{\prime}(\lambda) = 0\); as \(q \ne 0\) (because \(p \ne 0\)), 4.6 applied to \(q\) gives \(q = (z - \lambda)r\) and so \(p = (z - \lambda)^2 r\), so (b) fails.

(c) \(\Leftrightarrow\) (d). Here \(p^{\prime} \ne 0\), its leading coefficient being \(\deg p \ne 0\) in \(\mathbf{C}\), so the greatest common divisor exists. If \(p(\lambda) = p^{\prime}(\lambda) = 0\), then \(z - \lambda\) divides both \(p\) and \(p^{\prime}\) by 4.6, so their greatest common divisor has degree at least \(1\) and (d) fails. Conversely, if that greatest common divisor \(q\) has \(\deg q \ge 1\), then 4.12 gives it a zero \(\lambda\), and \(p, p^{\prime}\) being multiples of \(q\) yields \(p(\lambda) = p^{\prime}(\lambda) = 0\), so (c) fails.

In the forward direction the subspaces to take are \(U_k = U \cap E(\lambda_k, T)\). Note \(V\) is finite-dimensional and, by 5.55(c),

\begin{equation*} V = E(\lambda_1, T) \oplus \cdots \oplus E(\lambda_m, T) . \end{equation*}

(i) Suppose \(U_k \subseteq E(\lambda_k, T)\) for each \(k\) and \(U = U_1 \oplus \cdots \oplus U_m\). Writing \(u \in U\) as \(u = u_1 + \cdots + u_m\) with \(u_k \in U_k\), each \(T u_k = \lambda_k u_k\) lies in the subspace \(U_k\), so

\begin{equation*} T u = \lambda_1 u_1 + \cdots + \lambda_m u_m \in U_1 + \cdots + U_m = U , \end{equation*}

and \(U\) is invariant under \(T\).

(ii) Suppose \(U\) is invariant under \(T\) and put \(U_k = U \cap E(\lambda_k, T)\), a subspace of \(E(\lambda_k, T)\) contained in \(U\). The sum \(U_1 + \cdots + U_m\) is direct: if \(u_1 + \cdots + u_m = 0\) with \(u_k \in U_k \subseteq E(\lambda_k, T)\), then each \(u_k = 0\) because the sum of the eigenspaces is direct (5.54), so 1.45 applies. Since \(U\) is invariant under the diagonalizable \(T\), the restriction \(T|_U\) is diagonalizable (5.65) on the finite-dimensional space \(U\) (2.25), so by 5.55(b) \(U\) has a basis of eigenvectors of \(T|_U\); each such vector is an eigenvector of \(T\), hence lies in \(E(\lambda_k, T)\) for some \(k\) and so in \(U_k\). Therefore

\begin{equation*} U \subseteq U_1 + \cdots + U_m \subseteq U , \end{equation*}

giving \(U = U_1 \oplus \cdots \oplus U_m\).

Fix a basis \(v_1, \dots, v_n\) of \(V\) and let \(E_{j,k} \in \mathcal{L}(V)\) be the operator with \(E_{j,k} v_k = v_j\) and \(E_{j,k} v_l = 0\) for \(l \ne k\) (3.4); then the list of \(n^2\) operators

\begin{equation*} E_{j,j} \quad (j = 1, \dots, n) \qquad\text{and}\qquad E_{j,j} + E_{j,k} \quad (j \neq k) \end{equation*}

is a basis of \(\mathcal{L}(V)\) consisting of diagonalizable operators. (For \(n = 0\) the empty list serves.)

It is a basis: the \(E_{j,k}\) have as matrices the \(n^2\) standard matrix units, which form a basis of \(\mathbf{F}^{n,n}\), and \(\mathcal{M}\) is an isomorphism of \(\mathcal{L}(V)\) onto \(\mathbf{F}^{n,n}\) (3.71), so the \(E_{j,k}\) form a basis of \(\mathcal{L}(V)\) and \(\dim \mathcal{L}(V) = n^2\); our list spans them all, since \(E_{j,k} = (E_{j,j} + E_{j,k}) - E_{j,j}\) for \(j \ne k\), and a spanning list of length \(n^2\) is a basis (2.42).

Each \(E_{j,j}\) is diagonalizable, since \(v_1, \dots, v_n\) is a basis of eigenvectors of it (eigenvalue \(1\) for \(v_j\), eigenvalue \(0\) for the rest), so 5.55(b) applies. For \(j \ne k\) put \(S = E_{j,j} + E_{j,k}\), so that \(S v_j = v_j\), \(S v_k = v_j\), and \(S v_l = 0\) for \(l \notin \{j,k\}\). Replacing \(v_k\) by \(v_k - v_j\) gives a list of length \(n\) still spanning \(V\), hence a basis (2.42), and it consists of eigenvectors of \(S\):

\begin{equation*} S v_j = 1 \cdot v_j, \qquad S(v_k - v_j) = 0 \cdot (v_k - v_j), \qquad S v_l = 0 \cdot v_l . \end{equation*}

So \(S\) too is diagonalizable by 5.55(b).

Push a basis of eigenvectors of \(T\) down to \(V/U\). Since \(T\) is diagonalizable, \(V\) is finite-dimensional and 5.55(b) gives a basis \(v_1, \dots, v_n\) of \(V\) with \(T v_i = \lambda_i v_i\). Let \(\pi \colon V \to V/U\) be the quotient map \(\pi v = v + U\), which is linear and surjective and satisfies \((T/U)(\pi v) = Tv + U = \pi(Tv)\) by the definition of the quotient operator (Exercise 38 in Section 5A). Hence

\begin{equation*} (T/U)(\pi v_i) = \pi(T v_i) = \lambda_i \, \pi v_i , \end{equation*}

so every nonzero \(\pi v_i\) is an eigenvector of \(T/U\); and \(\pi v_1, \dots, \pi v_n\) spans \(V/U\) because \(\pi\) is linear and surjective. Discarding the zero entries leaves a spanning list of eigenvectors of \(T/U\), which contains a basis of \(V/U\) (2.30). By 5.55(b), \(T/U\) is diagonalizable.

False: take \(V = \mathbf{F}^2\) with \(T(x,y) = (y, 0)\) and \(U = \operatorname{span}\bigl((1,0)\bigr)\).

Then \(U\) is invariant under \(T\), since \(T(x,0) = (0,0)\), and \(T|_U = 0\) is diagonalizable on the one-dimensional space \(U\). Also \(T/U = 0\) on \(V/U\), since for every \((x,y) \in \mathbf{F}^2\),

\begin{equation*} (T/U)\bigl( (x,y) + U \bigr) = (y, 0) + U = 0 + U , \end{equation*}

so \(T/U\) is diagonalizable as well.

But \(T \ne 0\) and \(T^2 = 0\), so the minimal polynomial of \(T\) is \(z^2\), which is not a product of distinct linear factors; hence \(T\) is not diagonalizable by 5.62.

The operators \(T\) and \(T^{\prime}\) have the same minimal polynomial, after which 5.62 settles both directions at once.

First, \(\bigl(q(T)\bigr)^{\prime} = q(T^{\prime})\) for every \(q \in \mathcal{P}(\mathbf{F})\). Indeed \(I^{\prime} = I\), since \(I^{\prime}\varphi = \varphi \circ I = \varphi\), and \((T^k)^{\prime} = (T^{\prime})^k\) by induction from 3.120(c):

\begin{equation*} \begin{aligned} (T^k)^{\prime} &= (T \, T^{k-1})^{\prime} = (T^{k-1})^{\prime} \, T^{\prime} \\ &= (T^{\prime})^{k-1} T^{\prime} = (T^{\prime})^k . \end{aligned} \end{equation*}

So for \(q(z) = a_0 + a_1 z + \cdots + a_d z^d\), linearity of \(S \mapsto S^{\prime}\) (3.120(a) and (b)) gives

\begin{equation*} \bigl( q(T) \bigr)^{\prime} = a_0 I + a_1 T^{\prime} + \cdots + a_d (T^{\prime})^d = q(T^{\prime}) . \end{equation*}

Second, \(S^{\prime} = 0\) forces \(S = 0\): if \(Sv \ne 0\), extend the linearly independent list \(Sv\) to a basis of \(V\) (2.32) and use 3.4 to get \(\varphi \in V^{\prime}\) with \(\varphi(Sv) = 1\) and \(\varphi\) zero on the other basis vectors; then \((S^{\prime}\varphi)(v) = \varphi(Sv) = 1 \ne 0\).

Combining, \(q(T) = 0 \iff q(T^{\prime}) = 0\) for every \(q\), so the monic polynomial of smallest degree annihilating \(T\) is the one annihilating \(T^{\prime}\) (5.24); call it \(p\). Since \(\dim V^{\prime} = \dim V < \infty\) (3.111), 5.62 applies to each of \(T\) and \(T^{\prime}\): each is diagonalizable exactly when \(p\) is a product of distinct linear factors. Hence \(T\) is diagonalizable if and only if \(T^{\prime}\) is.

Throughout, write

\begin{equation*} \varphi = \frac{1 + \sqrt{5}}{2}, \qquad \psi = \frac{1 - \sqrt{5}}{2} . \end{equation*}

Note \(\varphi - \psi = \sqrt{5}\) and \(\lambda^2 = \lambda + 1\) for \(\lambda \in \{\varphi, \psi\}\).

(a) Induct on \(n\). For \(n = 0\), \(T^0(0,1) = (0,1) = (F_0, F_1)\); and if \(T^n(0,1) = (F_n, F_{n+1})\), then the recurrence gives

\begin{equation*} T^{n+1}(0,1) = T(F_n, F_{n+1}) = (F_{n+1}, \, F_n + F_{n+1}) = (F_{n+1}, F_{n+2}) . \end{equation*}

(b) The eigenvalues are \(\varphi\) and \(\psi\). Indeed \(T(x,y) = \lambda(x,y)\) says \(y = \lambda x\) and \(x + y = \lambda y\), so \((\lambda^2 - \lambda - 1)x = 0\); here \(x = 0\) would force \(y = \lambda x = 0\), so \(\lambda^2 - \lambda - 1 = 0\) and \(\lambda = \frac{1 \pm \sqrt{5}}{2}\). Conversely each such \(\lambda\) is an eigenvalue, since

\begin{equation*} T(1, \lambda) = (\lambda, 1 + \lambda) = (\lambda, \lambda^2) = \lambda (1, \lambda) . \end{equation*}

(c) The basis is \((1, \varphi), (1, \psi)\): by (b) these are eigenvectors for the distinct eigenvalues \(\varphi \ne \psi\), hence linearly independent (5.11), and a linearly independent list of length \(2 = \dim \mathbf{R}^2\) is a basis (2.38).

(d) Writing \((0,1) = a(1,\varphi) + b(1,\psi)\) gives \(a + b = 0\) and \(a\varphi + b\psi = 1\), so \(a(\varphi - \psi) = 1\) and \(a = 1/\sqrt{5} = -b\). Applying \(T^n\),

\begin{equation*} T^n(0,1) = \frac{\varphi^n}{\sqrt{5}} (1, \varphi) - \frac{\psi^n}{\sqrt{5}} (1, \psi) , \end{equation*}

whose first coordinate is \((\varphi^n - \psi^n)/\sqrt{5}\) and equals \(F_n\) by (a). That is the asserted formula.

(e) Since \(2 < \sqrt{5} < 3\) we have \(-1 < \psi < 0\), so \(|\psi|^n \le 1\) and (d) gives

\begin{equation*} \begin{aligned} \left| F_n - \frac{1}{\sqrt{5}} \varphi^n \right| &= \frac{|\psi|^n}{\sqrt{5}} \leq \frac{1}{\sqrt{5}} < \frac{1}{2} . \end{aligned} \end{equation*}

An integer \(m\) with \(|t - m| < \frac12\) is the unique closest integer to \(t\), since any other \(m^{\prime} \in \mathbf{Z}\) has \(|t - m^{\prime}| \ge |m - m^{\prime}| - |t - m| > \frac12 > |t - m|\). Hence \(F_n\) is the integer closest to \(\varphi^n / \sqrt{5}\).

Apply the Gershgorin disk theorem 5.67 to the basis \(v_1, \dots, v_n\) of \(V\) with respect to which \(A = \mathcal{M}(T)\); note \(\dim V = n < \infty\), so \(T\) is invertible if and only if it is injective (3.63 with 3.65), that is, exactly when \(0\) is not an eigenvalue of \(T\).

If \(T\) were not invertible, then \(0\) would be an eigenvalue of \(T\) and hence, by 5.67, lie in some Gershgorin disk: for some \(j\),

\begin{equation*} \lvert A_{j,j} \rvert = \lvert 0 - A_{j,j} \rvert \leq \sum_{\substack{k=1 \\ k \neq j}}^{n} |A_{j,k}| , \end{equation*}

contradicting the strict inequality assumed for every \(j\). Hence \(T\) is invertible.

Write \(A\) for the matrix of \(T\) with respect to a basis \(v_1, \dots, v_n\) of \(V\); the changed \(k^{\text{th}}\) disk is

\begin{equation*} \begin{aligned} D_k = \left\{ z \in \mathbf{F} : |z - A_{k,k}| \leq r_k \right\}, \qquad r_k = \sum_{\substack{j=1 \\ j \neq k}}^{n} |A_{j,k}| , \end{aligned} \end{equation*}

and the claim is that every eigenvalue of \(T\) lies in some \(D_k\).

Let \(Tw = \lambda w\) with \(w = c_1 v_1 + \cdots + c_n v_n \neq 0\). Exactly as in the proof of 5.67, expanding \(T v_k = \sum_{j} A_{j,k} v_j\) and using uniqueness of the basis representation of \(\lambda w\) gives \(\lambda c_j = \sum_{k} A_{j,k} c_k\) for each \(j\), hence by the triangle inequality

\begin{equation*} \lvert \lambda - A_{j,j} \rvert \, |c_j| \leq \sum_{\substack{k=1 \\ k \neq j}}^{n} |A_{j,k}| \, |c_k| . \end{equation*}

Rather than single out a largest \(|c_j|\), sum over \(j\) and interchange the order of summation in the resulting finite double sum over \(\{(j,k) : j \neq k\}\):

\begin{equation*} \begin{aligned} \sum_{j=1}^{n} |\lambda - A_{j,j}| \, |c_j| &\leq \sum_{k=1}^{n} |c_k| \sum_{\substack{j=1 \\ j \neq k}}^{n} |A_{j,k}| \\ &= \sum_{k=1}^{n} |c_k| \, r_k . \end{aligned} \end{equation*}

If \(\lambda\) lay in no \(D_k\), then \(|\lambda - A_{k,k}| > r_k\) for every \(k\); multiplying by \(|c_k| \ge 0\) and summing gives a strict inequality, since \(c_k \neq 0\) for at least one \(k\), and hence

\begin{equation*} \begin{aligned} \sum_{k=1}^{n} |c_k| \, |\lambda - A_{k,k}| &\leq \sum_{k=1}^{n} |c_k| \, r_k \\ &< \sum_{k=1}^{n} |c_k| \, |\lambda - A_{k,k}| , \end{aligned} \end{equation*}

a finite number strictly less than itself. Hence \(\lambda \in D_k\) for some \(k\).

With \(e_1, e_2, e_3, e_4\) the standard basis of \(\mathbf{F}^4\), take

\begin{equation*} S(x_1, x_2, x_3, x_4) = (x_2, 0, 0, 0), \qquad T(x_1, x_2, x_3, x_4) = (0, 0, x_4, 0), \end{equation*}

so \(S e_2 = e_1\) and \(T e_4 = e_3\), while each kills the remaining standard basis vectors.

These commute, since \(ST = 0 = TS\): the second coordinate of \(T(x_1,x_2,x_3,x_4) = (0,0,x_4,0)\) is \(0\), and the fourth coordinate of \(S(x_1,x_2,x_3,x_4) = (x_2,0,0,0)\) is \(0\).

(i) \(U = \operatorname{span}(e_1, e_4)\) is invariant under \(S\), as \(S e_1 = S e_4 = 0\), but not under \(T\), as \(T e_4 = e_3 \notin U\).

(ii) \(W = \operatorname{span}(e_2, e_3)\) is invariant under \(T\), as \(T e_2 = T e_3 = 0\), but not under \(S\), as \(S e_2 = e_1 \notin W\).

Here \(V\) is finite-dimensional, since a diagonalizable operator has a matrix with respect to a (finite) basis.

(\(\Rightarrow\)) If every element of \(\mathcal{E}\) has a diagonal matrix with respect to one basis, then for \(S, T \in \mathcal{E}\) the matrices \(\mathcal{M}(S)\) and \(\mathcal{M}(T)\) are diagonal of the same size and so commute, whence \(ST = TS\) by 5.74.

(\(\Leftarrow\)) Suppose every pair in \(\mathcal{E}\) commutes. We show by strong induction on \(\dim V\) that \(V\) has a basis of vectors that are eigenvectors of every element of \(\mathcal{E}\); with respect to such a basis each \(T \in \mathcal{E}\) has a diagonal matrix, since \(T v_k = \mu_k v_k\) makes the \(k^{\text{th}}\) column \(\mu_k\) times the \(k^{\text{th}}\) standard column. If \(\dim V \le 1\), every operator on \(V\) is a scalar multiple of \(I\) and any basis works. So let \(\dim V = n \ge 2\).

(i) If every element of \(\mathcal{E}\) is a scalar multiple of \(I\), any basis of \(V\) consists of common eigenvectors.

(ii) Otherwise choose \(S \in \mathcal{E}\) that is not a scalar multiple of \(I\), with distinct eigenvalues \(\lambda_1, \dots, \lambda_m\). Since \(S\) is diagonalizable, 5.55(c) gives

\begin{equation*} V = E(\lambda_1, S) \oplus \cdots \oplus E(\lambda_m, S) , \end{equation*}

where \(m \ge 2\) (else \(V = E(\lambda_1, S)\), i.e. \(S = \lambda_1 I\)); the summands being nonzero, each \(\dim E(\lambda_k, S) < n\). Fix \(k\) and set \(U = E(\lambda_k, S)\). Every \(T \in \mathcal{E}\) commutes with \(S\), so \(U\) is invariant under \(T\) (5.75) and \(T|_U\) is diagonalizable (5.65); any two such restrictions commute, since

\begin{equation*} (T|_U)(R|_U) u = (TR)u = (RT)u = (R|_U)(T|_U) u \qquad (u \in U) . \end{equation*}

By the induction hypothesis applied to \(U\) and \(\{T|_U : T \in \mathcal{E}\}\), the eigenspace \(E(\lambda_k, S)\) has a basis of common eigenvectors of the restrictions, and each such vector \(u\) is an eigenvector of every \(T \in \mathcal{E}\) itself, as \(u \ne 0\) and \(Tu = (T|_U)u\) is a scalar multiple of \(u\).

Concatenating these bases over \(k = 1, \dots, m\) gives a list spanning \(V\) of length \(\sum_k \dim E(\lambda_k, S) = \dim V\), hence a basis of \(V\), consisting of common eigenvectors of every element of \(\mathcal{E}\). The induction is on \(\dim V\), so \(\mathcal{E}\) may be infinite.

Both parts follow from \(p(S) T = T p(S)\). Indeed \(S^k T = T S^k\) for every \(k \ge 0\), by induction from \(ST = TS\):

\begin{equation*} S^{k+1} T = S(S^k T) = S(T S^k) = (ST) S^k = (TS) S^k = T S^{k+1} , \end{equation*}

so writing \(p(S) = a_0 I + a_1 S + \cdots + a_N S^N\) gives

\begin{equation*} p(S) T = \sum_{k=0}^{N} a_k S^k T = \sum_{k=0}^{N} a_k T S^k = T p(S) . \end{equation*}

(a) If \(p(S) v = 0\), then \(p(S)(Tv) = \bigl(T p(S)\bigr) v = T 0 = 0\), so \(Tv \in \operatorname{null} p(S)\).

(b) If \(v = p(S) u\), then \(Tv = \bigl(T p(S)\bigr) u = p(S)(Tu) \in \operatorname{range} p(S)\).

The statement is false. Here is a counterexample of size \(2\)-by-\(2\):

\begin{equation*} A = \begin{pmatrix} 1 & 0 \\ 0 & 2 \end{pmatrix}, \qquad B = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}. \end{equation*}

Here \(A\) is diagonal and \(B\) is upper triangular of the same size, yet

\begin{equation*} AB = \begin{pmatrix} 1 & 0 \\ 0 & 2 \end{pmatrix}\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}, \qquad BA = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}\begin{pmatrix} 1 & 0 \\ 0 & 2 \end{pmatrix} = \begin{pmatrix} 0 & 2 \\ 0 & 0 \end{pmatrix}. \end{equation*}

so \(AB \neq BA\).

Both directions rest on the order reversal \((RT)^{\prime} = T^{\prime} R^{\prime}\) of 3.120(c). Let \(S, T \in \mathcal{L}(V)\) with \(V\) finite-dimensional.

If \(ST = TS\), then

\begin{equation*} T^{\prime} S^{\prime} = (ST)^{\prime} = (TS)^{\prime} = S^{\prime} T^{\prime} . \end{equation*}

Conversely, if \(S^{\prime}T^{\prime} = T^{\prime}S^{\prime}\), then \((TS)^{\prime} = (ST)^{\prime}\), so \(R = ST - TS\) satisfies \(R^{\prime} = 0\) by linearity of \(R \mapsto R^{\prime}\) (3.120(a) and (b)). Hence \((\operatorname{range} R)^0 = \operatorname{null} R^{\prime} = V^{\prime}\) by 3.128(a), and 3.127(b), applicable since \(V\) is finite-dimensional, gives \(\operatorname{range} R = \{0\}\). Thus \(R = 0\), that is, \(ST = TS\).

Method (2): with \(A = \mathcal{M}(S)\), \(B = \mathcal{M}(T)\) with respect to a basis of \(V\) and the dual basis, 3.132 makes the matrices of \(S^{\prime}, T^{\prime}\) equal to \(A^{\mathrm{t}}, B^{\mathrm{t}}\), so by 5.74 twice, \((XY)^{\mathrm{t}} = Y^{\mathrm{t}}X^{\mathrm{t}}\), and injectivity of transposition,

\begin{equation*} S^{\prime} T^{\prime} = T^{\prime} S^{\prime} \iff (BA)^{\mathrm{t}} = (AB)^{\mathrm{t}} \iff BA = AB \iff ST = TS . \end{equation*}

Take \(\alpha = A_{n,n}\) and \(\lambda = B_{n,n}\), where (using that \(V\) is a nonzero finite-dimensional complex vector space and \(S, T\) commute) 5.80 supplies a basis \(v_1, \dots, v_n\) of \(V\) making both \(A = \mathcal{M}(S)\) and \(B = \mathcal{M}(T)\) upper triangular. Both ranges then lie in \(U = \operatorname{span}(v_1, \dots, v_{n-1})\), a subspace with \(\dim U = n - 1 < n\).

Indeed, for \(k \le n-1\), upper-triangularity of \(A\) gives \(S v_k \in \operatorname{span}(v_1, \dots, v_k) \subseteq U\) by 5.39, and \(\alpha v_k \in U\), so \((S - \alpha I)v_k \in U\); while for \(k = n\),

\begin{equation*} (S - \alpha I) v_n = \sum_{j=1}^{n} A_{j,n} v_j - A_{n,n} v_n = \sum_{j=1}^{n-1} A_{j,n} v_j \in U . \end{equation*}

Since \(v_1, \dots, v_n\) spans \(V\), the vectors \((S - \alpha I)v_k\) span \(\operatorname{range}(S - \alpha I)\), so that range lies in \(U\); the identical argument with \(B, \lambda\) puts \(\operatorname{range}(T - \lambda I)\) in \(U\). Hence

\begin{equation*} \operatorname{range}(S - \alpha I) + \operatorname{range}(T - \lambda I) \subseteq U \neq V . \end{equation*}

Concatenate, over the distinct eigenvalues \(\lambda_1, \dots, \lambda_m\) of \(S\), bases of the eigenspaces \(E(\lambda_k, S)\) with respect to which \(T\) is upper triangular there.

Since \(S\) is diagonalizable, \(V\) is finite-dimensional and 5.55(c) gives

\begin{equation*} V = E(\lambda_1, S) \oplus \cdots \oplus E(\lambda_m, S). \end{equation*}

As \(S\) and \(T\) commute, each \(E(\lambda_k, S)\) is invariant under \(T\) by 5.75, so \(T|_{E(\lambda_k,S)}\) is an operator on a finite-dimensional complex vector space and 5.47 gives a basis \(u_1^{(k)}, \dots, u_{d_k}^{(k)}\) of \(E(\lambda_k, S)\) making it upper triangular; equivalently (5.39),

\begin{equation*} T u_j^{(k)} \in \operatorname{span}\bigl(u_1^{(k)}, \dots, u_j^{(k)}\bigr) \qquad (1 \le j \le d_k) . \end{equation*}

Concatenating these lists in the order \(k = 1, \dots, m\) gives a list \(v_1, \dots, v_n\) spanning \(V\) of length \(\sum_k \dim E(\lambda_k, S) = \dim V\), hence a basis of \(V\).

Each \(v_i\) lies in some \(E(\lambda_k, S)\), so \(S v_i = \lambda_k v_i\) and \(\mathcal{M}(S, (v_1, \dots, v_n))\) is diagonal. And if \(v_i = u_j^{(k)}\), then \(u_1^{(k)}, \dots, u_j^{(k)}\) occupy positions at most \(i\) in the concatenation, so the display above gives \(T v_i \in \operatorname{span}(v_1, \dots, v_i)\); by 5.39 the matrix of \(T\) is upper triangular.

Take the monomials \(w^j z^k\) with \(j + k \le 3\), listed in order of increasing total degree:

\begin{equation*} \begin{aligned} &v_1 = 1; \quad v_2 = w, \ v_3 = z; \quad v_4 = w^2, \ v_5 = wz, \ v_6 = z^2; \\ &v_7 = w^3, \ v_8 = w^2 z, \ v_9 = w z^2, \ v_{10} = z^3 . \end{aligned} \end{equation*}

These span \(\mathcal{P}_3(\mathbf{C}^2, \mathbf{C})\) by 5.73 and are linearly independent, since a polynomial in two variables vanishing on all of \(\mathbf{C}^2\) has every coefficient \(0\) (Check!); so they form a basis.

Both operators strictly lower total degree: for \(j + k \le 3\),

\begin{equation*} D_w(w^j z^k) = j\, w^{j-1} z^k, \qquad D_z(w^j z^k) = k\, w^j z^{k-1} , \end{equation*}

each a scalar multiple of a monomial of total degree \(j + k - 1\), or \(0\) (when \(j = 0\), respectively \(k = 0\)). Since the list places every monomial of total degree \(d-1\) before every monomial of total degree \(d\),

\begin{equation*} D_w v_i, \; D_z v_i \in \operatorname{span}(v_1, \ldots, v_{i-1}) \end{equation*}

for each \(i\), which lies in \(\operatorname{span}(v_1, \ldots, v_i)\); so by 5.39 both \(D_w\) and \(D_z\) have upper-triangular matrices with respect to \(v_1, \ldots, v_{10}\).

(a) Induct on \(n = \dim V \ge 1\). For \(n = 1\), any nonzero \(v\) spans \(V\), so \(Tv \in \operatorname{span}(v)\) makes \(v\) an eigenvector of every \(T \in \mathcal{E}\). Let \(n > 1\).

(i) If every element of \(\mathcal{E}\) is a scalar multiple of \(I\), any nonzero \(v \in V\) serves.

(ii) Otherwise choose \(S \in \mathcal{E}\) that is not a scalar multiple of \(I\). Since \(V\) is a nonzero finite-dimensional complex vector space, \(S\) has an eigenvalue \(\lambda\) (5.19); put \(U = E(\lambda, S)\), so \(U \ne \{0\}\), and \(U \ne V\) since \(U = V\) would mean \(S = \lambda I\). Hence \(1 \le \dim U \le n-1\). Every \(T \in \mathcal{E}\) commutes with \(S\), so \(U\) is invariant under \(T\) (5.75), and the restrictions again commute pairwise, since

\begin{equation*} (T_1|_U)(T_2|_U)u = T_1 T_2 u = T_2 T_1 u = (T_2|_U)(T_1|_U)u \qquad (u \in U) . \end{equation*}

By the induction hypothesis some nonzero \(v \in U\) is an eigenvector of every \(T|_U\), hence of every \(T \in \mathcal{E}\), as \(Tv = (T|_U)v\). (The eigenvalues depend on \(T\); the vector does not.)

(b) Induct on \(n = \dim V \ge 1\); for \(n = 1\) every \(1\)-by-\(1\) matrix is upper triangular. Let \(n > 1\). By (a) there is a nonzero \(v_1\) that is an eigenvector of every element of \(\mathcal{E}\), so \(U = \operatorname{span}(v_1)\) is invariant under every element of \(\mathcal{E}\) and the quotient operators \(T/U\), \((T/U)(v + U) = Tv + U\) (Exercise 38 in Section 5A), are defined on \(V/U\), where \(\dim V/U = n - 1 \ge 1\) by 3.105. They commute pairwise:

\begin{equation*} (T_1/U)(T_2/U)(v + U) = T_1T_2v + U = T_2T_1v + U = (T_2/U)(T_1/U)(v + U). \end{equation*}

So the induction hypothesis gives a basis \(u_2 + U, \ldots, u_n + U\) of \(V/U\) with respect to which every \(T/U\) is upper triangular.

Then \(v_1, u_2, \ldots, u_n\) is a basis of \(V\): it has length \(n\), and if \(a_1 v_1 + a_2 u_2 + \cdots + a_n u_n = 0\), applying the quotient map (which sends \(v_1\) to \(0 + U\)) forces \(a_2 = \cdots = a_n = 0\) by independence in \(V/U\), whence \(a_1 v_1 = 0\) and \(a_1 = 0\); now 2.38 applies.

Fix \(T \in \mathcal{E}\). Then \(T v_1 \in \operatorname{span}(v_1)\), and for \(k \ge 2\) upper-triangularity of \(\mathcal{M}(T/U)\) gives, by 5.39, scalars \(a_2, \ldots, a_k\) with

\begin{equation*} Tu_k + U = \sum_{j=2}^{k} a_j (u_j + U) = \Big(\sum_{j=2}^{k} a_j u_j\Big) + U , \end{equation*}

so \(Tu_k - \sum_{j=2}^{k} a_j u_j \in U = \operatorname{span}(v_1)\) and \(Tu_k \in \operatorname{span}(v_1, u_2, \ldots, u_k)\). By 5.39 again, every \(T \in \mathcal{E}\) has an upper-triangular matrix with respect to \(v_1, u_2, \ldots, u_n\). Both inductions are on \(\dim V\), so \(\mathcal{E}\) may be infinite.

Take \(V = \mathbf{R}^2\) and define \(S, T \in \mathcal{L}(\mathbf{R}^2)\) by

\begin{equation*} S(x, y) = (-y, x), \qquad T(x, y) = (y, -x). \end{equation*}

Thus \(T = -S\), and \(S, T\) commute: \(ST = -S^2 = TS\).

Neither has an eigenvalue: \(S(x,y) = \lambda(x,y)\) says \(-y = \lambda x\) and \(x = \lambda y\), so \((\lambda^2 + 1)y = 0\), forcing \(y = 0\) (as \(\lambda \in \mathbf{R}\)) and then \(x = \lambda y = 0\); and \(T(x,y) = \lambda(x,y)\) is the equation \(S(x,y) = (-\lambda)(x,y)\).

But \(S + T = 0\) has eigenvalue \(0\), and \(ST = I\) has eigenvalue \(1\), since

\begin{equation*} (ST)(x,y) = S(y, -x) = (x, y) . \end{equation*}

Since \(S\) and \(T\) have no eigenvalues at all, no number whatsoever is an eigenvalue of \(S\) plus an eigenvalue of \(T\), or an eigenvalue of \(S\) times an eigenvalue of \(T\); in particular \(0\) and \(1\) are not.

True: the double sum equals \(\|v_1 + \cdots + v_m\|^2\).

Put \(v = v_1 + \cdots + v_m\). Additivity in the second slot (6.6(d)) gives \(\sum_{k} \langle v_j, v_k \rangle = \langle v_j, v \rangle\), and additivity in the first slot (6.2) then gives

\begin{equation*} \sum_{j=1}^{m} \sum_{k=1}^{m} \langle v_j, v_k \rangle = \sum_{j=1}^{m} \langle v_j, v \rangle = \langle v, v \rangle = \|v\|^2 \geq 0 , \end{equation*}

the norm by 6.7 and the inequality by positivity in 6.2, which also makes the sum a nonnegative real number when \(\mathbf{F} = \mathbf{C}\).

Of the five properties in 6.2, all but definiteness hold for \(\langle \cdot, \cdot \rangle_1\) regardless of \(S\): positivity is \(\langle v, v \rangle_1 = \langle Sv, Sv \rangle \geq 0\), conjugate symmetry is inherited from \(\langle \cdot, \cdot \rangle\), and linearity of \(S\) turns additivity and homogeneity of \(\langle \cdot, \cdot \rangle\) in the first slot into the same for \(\langle \cdot, \cdot \rangle_1\), as in

\begin{equation*} \begin{aligned} \langle \lambda u + w, v \rangle_1 &= \langle \lambda Su + Sw, Sv \rangle \\ &= \lambda \langle u, v \rangle_1 + \langle w, v \rangle_1 . \end{aligned} \end{equation*}

So \(\langle \cdot, \cdot \rangle_1\) is an inner product if and only if it is definite. By definiteness of \(\langle \cdot, \cdot \rangle\),

\begin{equation*} \langle v, v \rangle_1 = 0 \iff \langle Sv, Sv \rangle = 0 \iff Sv = 0 , \end{equation*}

so \(\{v \in V : \langle v, v \rangle_1 = 0\} = \operatorname{null} S\). Hence \(\langle \cdot, \cdot \rangle_1\) is definite if and only if \(\operatorname{null} S = \{0\}\), which by 3.15 holds if and only if \(S\) is injective.

(a) Homogeneity in the first slot fails at \(\lambda = -1\), \(u = v = (1, 0)\): writing \(\varphi\) for the function,

\begin{equation*} \varphi(-u, v) = |(-1)(1)| + 0 = 1 \neq -1 = -\varphi(u, v) . \end{equation*}

So \(\varphi\) is not an inner product on \(\mathbf{R}^2\).

(b) Definiteness fails at \(v = (0, 1, 0) \neq 0\): writing \(\psi\) for the function,

\begin{equation*} \psi(v, v) = 0 \cdot 0 + 0 \cdot 0 = 0 . \end{equation*}

So \(\psi\) is not an inner product on \(\mathbf{R}^3\).

Method (2) for (b): \(\psi(u, v) = \langle Su, Sv \rangle\) for the Euclidean inner product and \(S \in \mathcal{L}(\mathbf{R}^3)\) given by \(S(x_1, x_2, x_3) = (x_1, 0, x_3)\), which is not injective; apply Exercise 6A.2.

By 3.15 it suffices to show \(\operatorname{null}(T - \sqrt{2}\, I) = \{0\}\). If \((T - \sqrt{2}\, I)v = 0\), then \(Tv = \sqrt{2}\, v\), so by 6.9(b) and the hypothesis

\begin{equation*} \sqrt{2}\, \|v\| = \|\sqrt{2}\, v\| = \|Tv\| \leq \|v\| , \end{equation*}

whence \((\sqrt{2} - 1)\|v\| \leq 0\). Since \(\sqrt{2} - 1 > 0\) and norms are nonnegative, \(\|v\| = 0\) and thus \(v = 0\) by 6.9(a).

(a) Expanding by additivity in each slot (6.2 and 6.6(d), with 6.6(e) at \(\lambda = -1\)),

\begin{equation*} \begin{aligned} \langle u + v, u - v \rangle &= \langle u, u \rangle - \langle u, v \rangle \\ &\qquad {} + \langle v, u \rangle - \langle v, v \rangle , \end{aligned} \end{equation*}

and the middle terms cancel because \(\mathbf{F} = \mathbf{R}\) makes conjugate symmetry plain symmetry. So \(\langle u + v, u - v \rangle = \|u\|^2 - \|v\|^2\) by 6.7.

(b) If \(\|u\| = \|v\|\), then \(\langle u + v, u - v \rangle = \|u\|^2 - \|v\|^2 = 0\) by (a), which is orthogonality by 6.10.

(c) Place the rhombus in \(\mathbf{R}^2\) with a vertex at the origin, and let \(u, v\) be the sides emanating from it; being a parallelogram, its vertices are \(0, u, u+v, v\), its four sides are \(\pm u, \pm v\) with lengths \(\|u\|, \|v\|, \|u\|, \|v\|\) by 6.9(b), and \(u, v\) are linearly independent. Equal side lengths thus say exactly \(\|u\| = \|v\|\), while the diagonals are the vectors \(u + v\) and \(u - v\), both nonzero by independence. By (b) they are orthogonal, hence perpendicular (Exercise 6A.15).

\(\Rightarrow\) If \(\langle u, v \rangle = 0\) and \(a \in \mathbf{F}\), then \(\langle u, av \rangle = \overline{a} \langle u, v \rangle = 0\) by 6.6(e), so \(u \perp av\) and the Pythagorean theorem 6.12 gives

\begin{equation*} \|u + av\|^2 = \|u\|^2 + \|av\|^2 \geq \|u\|^2 . \end{equation*}

\(\Leftarrow\) Suppose \(\|u\| \leq \|u + av\|\) for all \(a \in \mathbf{F}\). If \(v = 0\) then \(\langle u, v \rangle = 0\) by 6.6(c). Otherwise the orthogonal decomposition 6.13 gives \(u = cv + w\) with

\begin{equation*} c = \frac{\langle u, v \rangle}{\|v\|^2}, \quad w = u - cv, \quad \langle w, v \rangle = 0 , \end{equation*}

and then \(\langle cv, w \rangle = c \overline{\langle w, v \rangle} = 0\), so 6.12 and 6.9(b) give

\begin{equation*} \|u\|^2 = \|cv\|^2 + \|w\|^2 = |c|^2 \|v\|^2 + \|w\|^2 . \end{equation*}

The hypothesis at \(a = -c\) reads \(\|u\| \leq \|u - cv\| = \|w\|\); squaring and substituting yields \(|c|^2 \|v\|^2 \leq 0\). Since \(\|v\| \neq 0\) by 6.9(a), we get \(c = 0\) and hence \(\langle u, v \rangle = 0\).

Everything follows from the identity

\begin{equation*} \|au + bv\|^2 - \|bu + av\|^2 = (a^2 - b^2)\big(\|u\|^2 - \|v\|^2\big) \qquad (a, b \in \mathbf{R}) . \end{equation*}

To prove it, expand by additivity in both slots (6.2, 6.6(d)) and use \(\overline{a} = a\), \(\overline{b} = b\) together with 6.6(e), 6.9(b) and conjugate symmetry, so that the cross terms \(ab\langle u, v \rangle + ab\overline{\langle u, v \rangle}\) combine:

\begin{equation*} \begin{aligned} \|au + bv\|^2 = a^2 \|u\|^2 + b^2 \|v\|^2 {} + 2ab \operatorname{Re}\langle u, v \rangle . \end{aligned} \end{equation*}

Swapping \(a\) and \(b\) fixes the cross term and exchanges the first two, giving the identity.

If \(\|u\| = \|v\|\), the right side vanishes for all real \(a, b\), so \(\|au + bv\| = \|bu + av\|\). Conversely, \(a = 1, b = 0\) gives \(\|u\| = \|v\|\) directly.

Apply Cauchy–Schwarz (6.14) in \(\mathbb{R}^5\) with its Euclidean inner product to

\begin{equation*} u = (a, b, c, x, y), \qquad w = (1, 1, 1, 4, 9) . \end{equation*}

Here \(\|w\| = \sqrt{1 + 1 + 1 + 16 + 81} = 10\), while the hypothesis says \(\|u\|^2 \leq 1\), so \(\|u\| \leq 1\). Hence

\begin{equation*} a + b + c + 4x + 9y = \langle u, w \rangle \leq \|u\| \, \|w\| \leq 10 . \end{equation*}

Conjugate symmetry gives \(\langle v, u \rangle = \overline{\langle u, v \rangle} = 1\), so expanding \(\|u - v\|^2\) by 6.7 and 6.6(a), (d), (e),

\begin{equation*} \begin{aligned} \|u - v\|^2 &= \langle u - v, u - v \rangle \\ &= \langle u, u \rangle - \langle u, v \rangle {} - \langle v, u \rangle + \langle v, v \rangle \\ &= \|u\|^2 - \langle u, v \rangle - \langle v, u \rangle + \|v\|^2 \\ &= 1 - 1 - 1 + 1 \\ &= 0 . \end{aligned} \end{equation*}

By 6.9(a), \(\|u - v\| = 0\) implies \(u - v = 0\), that is, \(u = v\).

Write \(a = \|u\| \in [0,1]\) and \(b = \|v\| \in [0,1]\). Cauchy–Schwarz (6.14) gives \(|\langle u, v \rangle| \leq ab\), so \(1 - ab \leq 1 - |\langle u, v \rangle|\) and it suffices to prove the numerical inequality

\begin{equation*} \sqrt{1 - a^2} \, \sqrt{1 - b^2} + ab \leq 1 . \end{equation*}

Apply 6.14 again, in \(\mathbb{R}^2\) with the Euclidean inner product, to

\begin{equation*} s = \bigl( \sqrt{1 - a^2},\, a \bigr), \quad t = \bigl( \sqrt{1 - b^2},\, b \bigr), \end{equation*}

both of which are unit vectors, since \(\|s\| = \sqrt{(1 - a^2) + a^2} = 1\) and likewise \(\|t\| = 1\). Hence

\begin{equation*} \sqrt{1 - a^2} \, \sqrt{1 - b^2} + ab = \langle s, t \rangle \leq \|s\|\,\|t\| = 1 . \end{equation*}

Method (2) for the numerical step: both sides of \(\sqrt{1 - a^2} \sqrt{1 - b^2} \leq 1 - ab\) are nonnegative since \(ab \leq 1\), so squaring is equivalent, and \((1 - a^2)(1 - b^2) \leq (1 - ab)^2\) reduces to \(0 \leq (a - b)^2\).

Take

\begin{equation*} u = \left( \tfrac{7}{10}, \tfrac{21}{10} \right), \qquad v = \left( \tfrac{3}{10}, -\tfrac{1}{10} \right). \end{equation*}

These come from the orthogonal decomposition 6.13 with \(c = \langle (1,2), (1,3) \rangle / \|(1,3)\|^2 = 7/10\), so that \(u = c(1,3)\) and \(v = (1,2) - u\). Then \(u\) is a multiple of \((1,3)\), \(\langle v, (1,3) \rangle = \tfrac{3}{10} - \tfrac{3}{10} = 0\), and \(u + v = (1,2)\). (Check!)

Apply Cauchy–Schwarz (6.14) in \(\mathbb{R}^4\) to

\begin{equation*} \begin{aligned} s &= \left( \sqrt{a}, \sqrt{b}, \sqrt{c}, \sqrt{d} \right), \\ t &= \left( \tfrac{1}{\sqrt{a}}, \tfrac{1}{\sqrt{b}}, \tfrac{1}{\sqrt{c}}, \tfrac{1}{\sqrt{d}} \right), \end{aligned} \end{equation*}

which are defined because \(a, b, c, d > 0\), and satisfy \(\langle s, t \rangle = 4\), \(\|s\|^2 = a + b + c + d\), and \(\|t\|^2 = \frac1a + \frac1b + \frac1c + \frac1d\).

(a) Squaring \(|\langle s, t \rangle| \leq \|s\| \, \|t\|\) gives

\begin{equation*} 16 = \langle s, t \rangle^2 \leq \|s\|^2 \, \|t\|^2 = (a + b + c + d)\left( \tfrac1a + \tfrac1b + \tfrac1c + \tfrac1d \right). \end{equation*}

(b) Exactly when \(a = b = c = d\). If so, the product is \((4a)(4/a) = 16\). Conversely, equality forces equality in 6.14, so by its equality condition one of \(s, t\) is a scalar multiple of the other; both are nonzero, so \(t = \lambda s\) for some \(\lambda \in \mathbb{R}\). Comparing coordinates,

\begin{equation*} \lambda = \tfrac{1}{a} = \tfrac{1}{b} = \tfrac{1}{c} = \tfrac{1}{d} , \end{equation*}

and positivity gives \(a = b = c = d\).

Apply Cauchy–Schwarz (6.14) in \(\mathbb{R}^n\) to \(a = (a_1, \dots, a_n)\) and \(e = (1, \dots, 1)\), for which \(\langle a, e \rangle = a_1 + \dots + a_n\), \(\|a\|^2 = a_1^{\,2} + \dots + a_n^{\,2}\), and \(\|e\|^2 = n\). Squaring \(|\langle a, e \rangle| \leq \|a\| \, \|e\|\) gives

\begin{equation*} (a_1 + \dots + a_n)^2 \leq n \, (a_1^{\,2} + \dots + a_n^{\,2}) , \end{equation*}

and dividing by \(n^2 > 0\) gives

\begin{equation*} \left( \frac{a_1 + \dots + a_n}{n} \right)^{\!2} \leq \frac{a_1^{\,2} + \dots + a_n^{\,2}}{n} , \end{equation*}

the square of the average bounded by the average of the squares.

The left side equals \(\bigl| \, \|v\| - 1 \, \bigr|\): since \(\|v\| > 0\) by 6.9(a), 6.9(b) gives

\begin{equation*} \left\| v - \frac{v}{\|v\|} \right\| = \left| 1 - \frac{1}{\|v\|} \right| \|v\| = \bigl| \, \|v\| - 1 \, \bigr| . \end{equation*}

Now let \(\|u\| = 1\). Expanding by 6.7 and using conjugate symmetry to write the cross terms as \(2 \operatorname{Re}\langle v, u \rangle\),

\begin{equation*} \|v - u\|^2 = \|v\|^2 - 2 \operatorname{Re} \langle v, u \rangle + 1 , \end{equation*}

while \(\operatorname{Re} \langle v, u \rangle \leq |\langle v, u \rangle| \leq \|v\| \, \|u\| = \|v\|\) by Cauchy–Schwarz (6.14). Hence

\begin{equation*} \|v - u\|^2 \geq \|v\|^2 - 2\|v\| + 1 = \bigl( \|v\| - 1 \bigr)^2 , \end{equation*}

and taking nonnegative square roots gives the asserted inequality.

Equality forces \(\operatorname{Re} \langle v, u \rangle = \|v\| = \|v\| \, \|u\|\), so both inequalities above are equalities. In particular Cauchy–Schwarz is an equality, so by the equality condition in 6.14 one of \(v, u\) is a multiple of the other; as both are nonzero, \(v = \lambda u\) with \(\lambda \neq 0\). Then \(\langle v, u \rangle = \lambda\|u\|^2 = \lambda\) and \(\|v\| = |\lambda|\), so \(\operatorname{Re}\lambda = |\lambda|\) forces \(\lambda > 0\), whence \(\lambda = \|v\|\) and

\begin{equation*} u = \frac{v}{\lambda} = \frac{v}{\|v\|} . \end{equation*}

Expanding in the (symmetric, since real) Euclidean inner product gives the identity

\begin{equation*} \|u - v\|^2 = \|u\|^2 + \|v\|^2 - 2\langle u, v\rangle , \end{equation*}

in which \(\|u\|, \|v\|, \|u - v\|\) are the ordinary Euclidean lengths. Let \(\theta \in [0, \pi]\) be the angle between \(u\) and \(v\).

(i) \(u, v\) not scalar multiples of each other. Then \(0, u, v\) form a genuine triangle, with sides \(\|u\|, \|v\|\) at the vertex \(0\), included angle \(\theta\), and opposite side \(\|u - v\|\), so the law of cosines gives

\begin{equation*} \|u - v\|^2 = \|u\|^2 + \|v\|^2 - 2\|u\|\,\|v\| \cos\theta . \end{equation*}

Comparing with the identity yields \(\langle u, v\rangle = \|u\|\,\|v\|\cos\theta\).

(ii) \(v = tu\) with \(t \neq 0\) (both vectors being nonzero). Then \(\|v\| = |t|\,\|u\|\) by 6.9(b) and \(\cos\theta = \operatorname{sign}(t)\), since \(\theta = 0\) for \(t > 0\) and \(\theta = \pi\) for \(t < 0\); hence

\begin{equation*} \langle u, v\rangle = t\|u\|^2 = \operatorname{sign}(t)\,\|u\|\,|t|\,\|u\| = \|u\|\,\|v\|\cos\theta . \end{equation*}

Because \(\arccos\) has domain exactly \([-1, 1]\), being the inverse of \(\cos\) restricted to \([0,\pi]\), the definition means nothing unless

\begin{equation*} \frac{\langle x, y\rangle}{\|x\|\,\|y\|} \in [-1, 1] . \end{equation*}

The denominator is positive because \(x, y \neq 0\) gives \(\|x\|, \|y\| \neq 0\) by 6.9(a); and it is exactly the Cauchy–Schwarz inequality (6.14), \(|\langle x, y\rangle| \le \|x\|\,\|y\|\), that puts the quotient in \([-1,1]\) after dividing by \(\|x\|\,\|y\| > 0\). Nothing in the definitions of \(\langle\cdot,\cdot\rangle\) and \(\|\cdot\|\) alone gives that bound, so without 6.14 the arccosine could be undefined.

Split \(a_k b_k = (\sqrt{k}\, a_k)(b_k/\sqrt{k})\), legitimate since \(k \ge 1 > 0\), and apply Cauchy–Schwarz in coordinate form 6.16(a) to

\begin{equation*} x = \left( \sqrt{k}\,a_k \right)_{k=1}^{n}, \qquad y = \left( b_k/\sqrt{k} \right)_{k=1}^{n} , \end{equation*}

for which \(x_k y_k = a_k b_k\), \(x_k^2 = k a_k^2\), and \(y_k^2 = b_k^2/k\). Substituting into \((x_1y_1 + \cdots + x_ny_n)^2 \le (x_1^2 + \cdots + x_n^2)(y_1^2 + \cdots + y_n^2)\) gives

\begin{equation*} \left( \sum_{k=1}^{n} a_k b_k \right)^{\!2} \le \left( \sum_{k=1}^{n} k a_k^2 \right)\left( \sum_{k=1}^{n} \frac{b_k^2}{k} \right). \end{equation*}

Write \(F(b) = \int_1^b f\) and \(G(b) = \int_1^b x^2 f(x)^2\,dx\); both integrands are continuous and nonnegative, so \(F, G\) are nondecreasing and the improper integrals \(L = \lim_{b\to\infty} F(b)\) and \(M = \lim_{b\to\infty} G(b)\) exist in \([0,\infty]\).

(a) Apply Cauchy–Schwarz in the integral form 6.16(b) on \([1,b]\) to \(g(x) = 1/x\) and \(h(x) = x f(x)\), both continuous there, with \(gh = f\):

\begin{equation*} F(b)^2 \le \left( \int_1^b \frac{dx}{x^2} \right) G(b) = \left( 1 - \tfrac{1}{b} \right) G(b) \le G(b) \le M . \end{equation*}

Letting \(b \to \infty\) gives \(L^2 \le M\), which is the claim (trivially so if \(M = \infty\)).

(b) Exactly the functions \(f(x) = c/x^2\) with \(c \ge 0\) constant. For such \(f\),

\begin{equation*} \int_1^{\infty} f = c, \qquad \int_1^{\infty} x^2 f(x)^2\,dx = c^2 \int_1^{\infty} \frac{dx}{x^2} = c^2 , \end{equation*}

both finite and equal after squaring the first. Conversely suppose \(L^2 = M < \infty\), and set \(c = L \ge 0\). Expanding the square in \(\Phi(b) = \int_1^b \bigl( x f(x) - \tfrac{c}{x} \bigr)^2 dx\), using \(xf(x) \cdot \tfrac{c}{x} = c f(x)\),

\begin{equation*} \Phi(b) = G(b) - 2c F(b) + c^2\left(1 - \tfrac1b\right) \longrightarrow M - 2L^2 + L^2 = 0 . \end{equation*}

Since \(\Phi\) is nondecreasing and nonnegative with limit \(0\), it vanishes identically. A continuous nonnegative function with zero integral over \([1,b]\) is identically \(0\) there, so \(x f(x) = c/x\), that is, \(f(x) = c/x^2\), for all \(x \ge 1\).

Write \(A_{j,k} = \mathcal{M}(T)_{j,k}\), so \(T v_k = \sum_{j} A_{j,k} v_j\) by 3.31, and let \(v = \sum_k c_k v_k\) be an eigenvector for \(\lambda\), so that \(\sum_k |c_k|^2 > 0\) since \(v \neq 0\). Expanding \(Tv = \lambda v\) in the basis and comparing coefficients (the representation being unique) gives

\begin{equation*} \sum_{k=1}^{n} A_{j,k} c_k = \lambda c_j \qquad\text{for } j = 1, \ldots, n . \end{equation*}

Fix \(j\) and apply Cauchy–Schwarz (6.14) in \(\mathbf{F}^n\) with \(\langle x, y\rangle = \sum_k x_k \overline{y_k}\) to \(x = (c_1, \ldots, c_n)\) and \(y = \bigl( \overline{A_{j,1}}, \ldots, \overline{A_{j,n}} \bigr)\), for which \(\langle x, y \rangle = \sum_k A_{j,k} c_k\) and \(\|y\|^2 = \sum_k |A_{j,k}|^2\):

\begin{equation*} |\lambda|^2 |c_j|^2 \le \left( \sum_{k=1}^{n} |A_{j,k}|^2 \right) \left( \sum_{k=1}^{n} |c_k|^2 \right). \end{equation*}

Summing over \(j\) and dividing by \(\sum_k |c_k|^2 > 0\) gives \(|\lambda|^2 \le \sum_{j,k} |A_{j,k}|^2\).

It suffices to bound \(\pm(\|u\| - \|v\|)\) by \(\|u - v\|\). The triangle inequality (6.17) applied to \(u = (u - v) + v\) gives

\begin{equation*} \|u\| \le \|u - v\| + \|v\| , \end{equation*}

so \(\|u\| - \|v\| \le \|u - v\|\); interchanging \(u\) and \(v\) gives \(\|v\| - \|u\| \le \|v - u\| = \|u - v\|\), the last equality by 6.9(b) with \(\lambda = -1\). Hence \(\bigl| \|u\| - \|v\| \bigr| \le \|u - v\|\).

\(\|v\| = \sqrt{17}\). The parallelogram equality (6.21) gives

\begin{equation*} 4^2 + 6^2 = 2\bigl( 3^2 + \|v\|^2 \bigr) , \end{equation*}

so \(2\|v\|^2 = 52 - 18 = 34\) and \(\|v\|^2 = 17\).

Combine \(ab \le \frac{a^2 + b^2}{2}\) (from \(0 \le (a-b)^2\)) at \(a = \|u + v\|\), \(b = \|u - v\|\) with the parallelogram equality (6.21):

\begin{equation*} \|u + v\| \, \|u - v\| \le \frac{\|u + v\|^2 + \|u - v\|^2}{2} = \|u\|^2 + \|v\|^2 . \end{equation*}

Average over all \(2^m\) sign vectors. Put \(A = \{1, -1\}^m\) and \(N(a) = \|a_1 v_1 + \cdots + a_m v_m\|^2\) for \(a \in A\). Since each \(a_k\) is real, expanding by 6.6 in each slot gives

\begin{equation*} N(a) = \sum_{j=1}^{m} \sum_{k=1}^{m} a_j a_k \langle v_j, v_k \rangle , \end{equation*}

so summing over \(a \in A\) and interchanging the order of summation,

\begin{equation*} \sum_{a \in A} N(a) = \sum_{j=1}^{m} \sum_{k=1}^{m} \Bigl( \sum_{a \in A} a_j a_k \Bigr) \langle v_j, v_k \rangle . \end{equation*}

The inner sum is \(2^m\) when \(j = k\) (each term is \(a_j^2 = 1\)) and \(0\) when \(j \ne k\), because flipping the \(j^{\text{th}}\) coordinate is a fixed-point-free involution of \(A\) that negates \(a_j a_k\). Hence

\begin{equation*} \sum_{a \in A} N(a) = 2^m \sum_{k=1}^{m} \|v_k\|^2 \le 2^m m . \end{equation*}

So some \(N(a) \le m\), since otherwise the sum of the \(2^m\) terms would exceed \(2^m m\); for that \(a\), \(\|a_1 v_1 + \cdots + a_m v_m\| \le \sqrt{m}\).

True: no inner product on \(\mathbb{R}^2\) has \(\max\{|x|, |y|\}\) as its norm, since that function violates the parallelogram equality (6.21) at \(u = (1, 0)\), \(v = (0, 1)\). Indeed \(u + v = (1,1)\) and \(u - v = (1,-1)\), so all four of \(\|u\|, \|v\|, \|u+v\|, \|u-v\|\) equal \(1\) and

\begin{equation*} \|u + v\|^2 + \|u - v\|^2 = 2 \neq 4 = 2\bigl( \|u\|^2 + \|v\|^2 \bigr) . \end{equation*}

Hence for any inner product norm on \(\mathbb{R}^2\), at least one of these four vectors has norm different from the maximum of its coordinates absolute values.

Write \(N_p(x, y) = (|x|^p + |y|^p)^{1/p}\).

\(\Leftarrow\) For \(p = 2\), the Euclidean inner product on \(\mathbb{R}^2\) (an inner product by 6.3(a)) has associated norm \(\sqrt{x^2 + y^2} = N_2(x, y)\) by 6.7.

\(\Rightarrow\) If \(N_p\) comes from an inner product, the parallelogram equality (6.21) applies to \(u = (1, 0)\) and \(v = (0, 1)\), for which \(\|u\| = \|v\| = 1\) and \(\|u + v\| = \|u - v\| = 2^{1/p}\):

\begin{equation*} 2 \cdot 2^{2/p} = 2^{1 + 2/p} = 2(1 + 1) = 2^2 . \end{equation*}

Injectivity of \(t \mapsto 2^t\) gives \(1 + 2/p = 2\), so \(p = 2\).

Expanding by additivity in both slots (6.2, 6.6(d), 6.6(e)) and using \(\langle v, u \rangle = \langle u, v \rangle\), which is conjugate symmetry over \(\mathbf{R}\),

\begin{equation*} \|u \pm v\|^2 = \|u\|^2 + \|v\|^2 \pm 2 \langle u, v \rangle . \end{equation*}

Subtracting cancels \(\|u\|^2 + \|v\|^2\) and leaves \(\|u + v\|^2 - \|u - v\|^2 = 4 \langle u, v \rangle\).

Everything follows from the expansion

\begin{equation*} \|u + w\|^2 = \|u\|^2 + \|w\|^2 + \langle u, w \rangle + \overline{\langle u, w \rangle} = \|u\|^2 + \|w\|^2 + 2 \operatorname{Re} \langle u, w \rangle , \end{equation*}

valid for all \(u, w \in V\) by conjugate symmetry and \(z + \overline{z} = 2 \operatorname{Re} z\). Taking \(w = v\) and \(w = -v\) (using \(\|-v\| = \|v\|\) and \(\langle u, -v \rangle = -\langle u, v \rangle\), by 6.6(e)) gives

\begin{equation*} \|u + v\|^2 - \|u - v\|^2 = 4 \operatorname{Re} \langle u, v \rangle . \end{equation*}

Taking \(w = iv\) and \(w = -iv\) instead, and using \(\|{\pm} iv\| = |{\pm} i| \, \|v\| = \|v\|\) together with

\begin{equation*} \begin{aligned} \langle u, \pm iv \rangle &= \overline{\pm i} \, \langle u, v \rangle = \mp i \langle u, v \rangle , \\ \operatorname{Re} \bigl( -i \langle u, v \rangle \bigr) &= \operatorname{Im} \langle u, v \rangle \end{aligned} \end{equation*}

(by 6.6(e), and writing \(\langle u, v \rangle = a + bi\), so \(-i(a+bi) = b - ai\)), gives

\begin{equation*} \|u + iv\|^2 - \|u - iv\|^2 = 4 \operatorname{Im} \langle u, v \rangle . \end{equation*}

Hence the numerator on the right side of the asserted formula is

\begin{equation*} 4 \operatorname{Re} \langle u, v \rangle + 4 \bigl( \operatorname{Im} \langle u, v \rangle \bigr) i = 4 \langle u, v \rangle , \end{equation*}

and dividing by \(4\) gives the identity.

Take

\begin{equation*} \varphi(u, v) = \frac{\|u + v\|^2 - \|u - v\|^2}{4} , \end{equation*}

which is the required inner product when \(\mathbb{F} = \mathbb{R}\); when \(\mathbb{F} = \mathbb{C}\), take \(\langle u, v \rangle = \varphi(u, v) + \varphi(u, iv) \, i\), the expression of Exercise 27. Write (P) for the parallelogram equality. Two consequences of the norm axioms are used repeatedly: \(\|-u\| = \|u\|\) (homogeneity with \(\alpha = -1\)), and the reverse triangle inequality

\begin{equation*} \bigl| \, \|a\| - \|b\| \, \bigr| \le \|a - b\| \tag{R} \end{equation*}

(from \(\|a\| \le \|a - b\| + \|b\|\) and the same with \(a, b\) interchanged).

(i) Symmetry, positivity, definiteness. Since \(\|v - u\| = \|u - v\|\) we get \(\varphi(v, u) = \varphi(u, v)\), and

\begin{equation*} \varphi(u, u) = \frac{\|2u\|^2 - \|0\|^2}{4} = \|u\|^2 \ge 0 , \end{equation*}

which vanishes exactly when \(u = 0\). Also \(\varphi(u, 0) = \varphi(0, u) = 0\) and \(\varphi(-u, v) = -\varphi(u, v)\). (Check!)

(ii) Additivity in the first slot. Apply (P) to the pair \(u + w, v + w\) and to the pair \(u - w, v - w\); both pairs have difference \(u - v\), so subtracting cancels \(\|u - v\|^2\) and leaves

\begin{equation*} \begin{aligned} \|u + v + 2w\|^2 - \|u + v - 2w\|^2 &= 2 \bigl( \|u + w\|^2 - \|u - w\|^2 \bigr) \\ &\qquad + 2 \bigl( \|v + w\|^2 - \|v - w\|^2 \bigr) , \end{aligned} \end{equation*}

that is, \(\varphi(u + v, 2w) = 2 \bigl( \varphi(u, w) + \varphi(v, w) \bigr)\). Setting \(v = 0\) here gives \(\varphi(u, 2w) = 2 \varphi(u, w)\); applying that with \(u\) replaced by \(u + v\) and comparing yields

\begin{equation*} \varphi(u + v, w) = \varphi(u, w) + \varphi(v, w) , \tag{4} \end{equation*}

and by symmetry \(\varphi\) is additive in the second slot too.

(iii) Real homogeneity. From (4) by induction, together with \(\varphi(0, v) = 0\) and \(\varphi(-u, v) = -\varphi(u, v)\), we get \(\varphi(kw, v) = k \varphi(w, v)\) for every \(k \in \mathbb{Z}\). Taking \(w = \tfrac{1}{n} u\) and \(k = n\) gives \(\varphi(\tfrac{1}{n} u, v) = \tfrac{1}{n} \varphi(u, v)\), so \(\varphi(\tfrac{m}{n} u, v) = \tfrac{m}{n} \varphi(u, v)\) for all \(m \in \mathbb{Z}\) and positive integers \(n\). By (R),

\begin{equation*} \bigl| \, \|tu \pm v\| - \|su \pm v\| \, \bigr| \le \|(t - s)u\| = |t - s| \, \|u\| , \end{equation*}

so \(t \mapsto \varphi(tu, v)\) is continuous; agreeing with the continuous function \(t \mapsto t \varphi(u, v)\) on the dense set \(\mathbb{Q}\), it agrees on all of \(\mathbb{R}\):

\begin{equation*} \varphi(tu, v) = t \, \varphi(u, v) \qquad \text{for all } t \in \mathbb{R} . \tag{5} \end{equation*}

For \(\mathbb{F} = \mathbb{R}\) this completes the proof: (i), (ii), (iii) are the five conditions of 6.2, and \(\varphi(u, u)^{1/2} = \|u\|\).

(iv) The complex case. Restricting scalars to \(\mathbb{R}\) leaves \(\| \cdot \|\) a norm satisfying (P), so (i), (ii), (iii) apply verbatim and \(\varphi\) is a symmetric \(\mathbb{R}\)-bilinear form with \(\varphi(u, u) = \|u\|^2\). Complex homogeneity of the norm gives

\begin{equation*} \varphi(iu, iv) = \frac{|i|^2 \|u + v\|^2 - |i|^2 \|u - v\|^2}{4} = \varphi(u, v), \tag{6} \end{equation*}

whence, using \(i \cdot iu = -u\), then (5) in the second slot, then symmetry,

\begin{equation*} \varphi(v, iu) = \varphi(iv, i \cdot iu) = \varphi(iv, -u) = -\varphi(iv, u) = -\varphi(u, iv). \tag{7} \end{equation*}

Since \(\|u \pm iu\| = |1 \pm i| \, \|u\| = \sqrt{2} \, \|u\|\), we get \(\varphi(u, iu) = 0\) and hence

\begin{equation*} \langle u, u \rangle = \varphi(u, u) = \|u\|^2 , \end{equation*}

giving positivity, definiteness, and the required identity \(\|u\| = \langle u, u \rangle^{1/2}\). Additivity in the first slot is immediate from (4) applied to both terms, and conjugate symmetry follows from symmetry of \(\varphi\), (7), and the fact that \(\varphi\) is real valued:

\begin{equation*} \langle v, u \rangle = \varphi(u, v) - \varphi(u, iv) \, i = \overline{\langle u, v \rangle} . \end{equation*}

Finally (5) gives \(\langle tu, v \rangle = t \langle u, v \rangle\) for \(t \in \mathbb{R}\), while (6), symmetry and (7) give

\begin{equation*} \langle iu, v \rangle = \varphi(iu, v) + \varphi(u, v) \, i = -\varphi(u, iv) + \varphi(u, v) \, i = i \langle u, v \rangle , \end{equation*}

so for \(\alpha = a + bi\) with \(a, b \in \mathbb{R}\), additivity yields

\begin{equation*} \langle \alpha u, v \rangle = a \langle u, v \rangle + b \langle iu, v \rangle = (a + bi) \langle u, v \rangle = \alpha \langle u, v \rangle . \end{equation*}

All five conditions of 6.2 hold.

Each of the five conditions of 6.2 holds coordinatewise. Write \(\langle \cdot, \cdot \rangle_k\) for the inner product on \(V_k\) (all over the same field \(\mathbf{F}\), as is implicit in forming the product), so that for \(u = (u_1, \ldots, u_m)\) and \(v = (v_1, \ldots, v_m)\) in \(V = V_1 \times \cdots \times V_m\) the asserted formula reads \(\langle u, v \rangle = \sum_{k=1}^{m} \langle u_k, v_k \rangle_k \in \mathbf{F}\).

Positivity and definiteness: \(\langle v, v \rangle = \sum_{k=1}^{m} \langle v_k, v_k \rangle_k\) is a finite sum of nonnegative real numbers, hence is nonnegative, and it equals \(0\) exactly when every term does, which by definiteness in each \(V_k\) happens exactly when every \(v_k = 0\), that is, when \(v = 0\).

Additivity and homogeneity in the first slot: because the operations on \(V\) are coordinatewise, \(u + v\) has \(k\)-th coordinate \(u_k + v_k\) and \(\lambda u\) has \(k\)-th coordinate \(\lambda u_k\), so for \(w = (w_1, \ldots, w_m)\) and \(\lambda \in \mathbf{F}\),

\begin{equation*} \begin{aligned} \langle u + v, w \rangle &= \sum_{k=1}^{m} \bigl( \langle u_k, w_k \rangle_k + \langle v_k, w_k \rangle_k \bigr) = \langle u, w \rangle + \langle v, w \rangle , \\ \langle \lambda u, v \rangle &= \sum_{k=1}^{m} \lambda \langle u_k, v_k \rangle_k = \lambda \langle u, v \rangle . \end{aligned} \end{equation*}

Conjugate symmetry: since conjugation is additive,

\begin{equation*} \langle u, v \rangle = \sum_{k=1}^{m} \overline{\langle v_k, u_k \rangle_k} = \overline{\sum_{k=1}^{m} \langle v_k, u_k \rangle_k} = \overline{\langle v, u \rangle} . \end{equation*}

(a) The five conditions of 6.2 all reduce to symmetry and bilinearity of \(\langle \cdot, \cdot \rangle\) on the real space \(V\). Throughout, elements of \(V_{\mathbf{C}}\) are written uniquely as \(u + iv\) with \(u, v \in V\) (Exercise 1B.8), with

\begin{equation*} (a + bi)(u + iv) = (au - bv) + i(av + bu) , \qquad a, b \in \mathbf{R} . \end{equation*}

Positivity and definiteness. For \(z = u + iv\), the defining formula with \(w = u\), \(x = v\) gives, since \(\langle v, u \rangle = \langle u, v \rangle\) kills the imaginary part,

\begin{equation*} \langle z, z \rangle_{\mathbf{C}} = \langle u, u \rangle + \langle v, v \rangle = \|u\|^2 + \|v\|^2 \ge 0 , \end{equation*}

which equals \(0\) if and only if \(u = v = 0\) (by 6.9(a)), that is, if and only if \(z = 0\).

Additivity in the first slot. Since \(z_1 + z_2 = (u_1 + u_2) + i(v_1 + v_2)\), applying additivity of \(\langle \cdot, \cdot \rangle\) to each of the four terms and regrouping gives \(\langle z_1 + z_2, z \rangle_{\mathbf{C}} = \langle z_1, z \rangle_{\mathbf{C}} + \langle z_2, z \rangle_{\mathbf{C}}\). (Check!)

Homogeneity in the first slot. For \(a \in \mathbf{R}\) we have \(az = (au) + i(av)\), so all four terms of the defining formula scale by \(a\) and \(\langle az, z^{\prime} \rangle_{\mathbf{C}} = a \langle z, z^{\prime} \rangle_{\mathbf{C}}\). For \(i\), note \(iz = (-v) + iu\), so with \(z^{\prime} = w + ix\),

\begin{equation*} \begin{aligned} \langle iz, z^{\prime} \rangle_{\mathbf{C}} &= \bigl( -\langle v, w \rangle + \langle u, x \rangle \bigr) + \bigl( \langle u, w \rangle + \langle v, x \rangle \bigr) i \\ &= i \langle z, z^{\prime} \rangle_{\mathbf{C}} , \end{aligned} \end{equation*}

since multiplying \(\alpha + \beta i\) by \(i\) gives \(-\beta + \alpha i\). Hence for \(\lambda = a + bi\), writing \(\lambda z = az + b(iz)\) and using the additivity just proved, \(\langle \lambda z, z^{\prime} \rangle_{\mathbf{C}} = \lambda \langle z, z^{\prime} \rangle_{\mathbf{C}}\).

Conjugate symmetry. Conjugating the defining formula for \(\langle z^{\prime}, z \rangle_{\mathbf{C}}\) flips the sign of \(\langle x, u \rangle - \langle w, v \rangle\), and symmetry of \(\langle \cdot, \cdot \rangle\) in all four terms then turns the result into \(\langle z, z^{\prime} \rangle_{\mathbf{C}}\). (Check!)

(b) Identifying \(u \in V\) with \(u + i0\), the defining formula and \(\langle 0, y \rangle = \langle y, 0 \rangle = 0\) (6.6(b), 6.6(c)) give

\begin{equation*} \langle u, v \rangle_{\mathbf{C}} = \langle u, v \rangle + \langle 0, 0 \rangle + \bigl( \langle 0, v \rangle - \langle u, 0 \rangle \bigr) i = \langle u, v \rangle , \end{equation*}

and the positivity computation in (a) is exactly \(\| u + iv \|_{\mathbf{C}}^{\,2} = \|u\|^2 + \|v\|^2\).

This is the parallelogram equality 6.21 applied to \(a = w - u\) and \(b = w - v\), for which

\begin{equation*} a + b = 2 \Bigl( w - \tfrac{1}{2}(u + v) \Bigr) \quad \text{and} \quad a - b = v - u . \end{equation*}

By homogeneity of the norm (6.9(b)), \(\|a + b\|^2 = 4 \bigl\| w - \tfrac{1}{2}(u + v) \bigr\|^2\) and \(\|a - b\|^2 = \|u - v\|^2\), so \(\|a + b\|^2 + \|a - b\|^2 = 2(\|a\|^2 + \|b\|^2)\) reads

\begin{equation*} 4 \Bigl\| w - \tfrac{1}{2}(u + v) \Bigr\|^2 + \|u - v\|^2 = 2 \bigl( \|w - u\|^2 + \|w - v\|^2 \bigr) . \end{equation*}

Dividing by \(4\) and rearranging gives the asserted identity.

Suppose \(u_1, u_2 \in E\) both minimize the distance to \(w\); the midpoint of \(u_1\) and \(u_2\) then beats them unless they coincide. Indeed, taking \(x = u_2\) in the minimizing property of \(u_1\) and \(x = u_1\) in that of \(u_2\) gives \(\|w - u_1\| = \|w - u_2\|\); call this common value \(d\). The hypothesis puts \(\tfrac{1}{2}(u_1 + u_2)\) in \(E\), so the minimizing property of \(u_1\) applied to it gives \(d^2 \le \bigl\| w - \tfrac{1}{2}(u_1 + u_2) \bigr\|^2\), while Exercise 31 (with \(u = u_1\), \(v = u_2\)) evaluates that right side:

\begin{equation*} d^2 \le \Bigl\| w - \tfrac{1}{2}(u_1 + u_2) \Bigr\|^2 = \frac{d^2 + d^2}{2} - \frac{\|u_1 - u_2\|^2}{4} = d^2 - \frac{\|u_1 - u_2\|^2}{4} . \end{equation*}

Hence \(\|u_1 - u_2\|^2 \le 0\), so \(\|u_1 - u_2\| = 0\) and \(u_1 = u_2\) by 6.9(a).

(a) This is the one-variable product rule applied coordinatewise. With the Euclidean inner product of 6.3(a) on \(\mathbf{R}^n\) we have \(\langle f(t), g(t) \rangle = \sum_{k=1}^{n} f_k(t) g_k(t)\), a finite sum of products of differentiable real-valued functions, so

\begin{equation*} \begin{aligned} \langle f(t), g(t) \rangle^{\prime} &= \sum_{k=1}^{n} f_k{}^{\prime}(t) \, g_k(t) + \sum_{k=1}^{n} f_k(t) \, g_k{}^{\prime}(t) \\ &= \langle f^{\prime}(t), g(t) \rangle + \langle f(t), g^{\prime}(t) \rangle , \end{aligned} \end{equation*}

by the definitions of \(f^{\prime}(t)\) and \(g^{\prime}(t)\).

(b) The function \(t \mapsto \langle f(t), f(t) \rangle = \| f(t) \|^2 = c^2\) is constant, so its derivative is \(0\); by (a) with \(g = f\), and the symmetry of the inner product on the real space \(\mathbf{R}^n\), that derivative equals \(2 \langle f^{\prime}(t), f(t) \rangle\). Hence \(\langle f^{\prime}(t), f(t) \rangle = 0\).

(c) The curve traced by \(f\) lies on the sphere of radius \(c\) centered at the origin, and (b) says its tangent vector \(f^{\prime}(t)\) is orthogonal (6.10) to the radius vector \(f(t)\) at every point.

Apollonius’s identity is Exercise 31 rewritten. Label the vertices \(u, v, w\) in \(\mathbf{R}^2\) (with the Euclidean inner product of 6.3(a)) so that the side of length \(c\) joins \(u\) and \(v\) and \(w\) is the opposite vertex:

\begin{equation*} c = \|u - v\| , \qquad a = \|w - v\| , \qquad b = \|w - u\| \end{equation*}

(the identity is symmetric in \(a\) and \(b\), so this naming loses no generality). The midpoint of the side of length \(c\) is \(\tfrac{1}{2}(u + v)\), since by 6.9(b) it lies on that segment at distance \(\tfrac{1}{2}\|u - v\|\) from each endpoint; hence \(d = \bigl\| w - \tfrac{1}{2}(u + v) \bigr\|\). Exercise 31 now reads

\begin{equation*} d^2 = \frac{b^2 + a^2}{2} - \frac{c^2}{4} , \end{equation*}

which rearranges to \(a^2 + b^2 = \tfrac{1}{2} c^2 + 2 d^2\).

Method (2): translate so the midpoint is the origin, that is, \(v = -u\); then \(c = \|2u\| = 2\|u\|\), \(d = \|w\|\), and the parallelogram equality 6.21 gives

\begin{equation*} a^2 + b^2 = \|w + u\|^2 + \|w - u\|^2 = 2 \bigl( \|w\|^2 + \|u\|^2 \bigr) = 2 d^2 + \tfrac{1}{2} c^2 . \end{equation*}

Take \(p = q + (1 - \|x\|^2) r\) with \(r\) produced below. Since \(1 - \|x\|^2 = 1 - x_1^2 - \cdots - x_n^2\) is itself a polynomial, \(p\) is a polynomial, and \(p = q\) on the unit sphere because that factor vanishes there; only \(\Delta p = 0\) needs work.

For \(k \ge 0\) let \(\mathcal{P}_k\) be the span of the monomials \(x_1^{m_1} \cdots x_n^{m_n}\) with \(m_1 + \cdots + m_n \le k\), and let \(\mathcal{P}_k = \{0\}\) for \(k < 0\); each \(\mathcal{P}_k\) is finite-dimensional (finitely many such monomials) and every polynomial lies in some \(\mathcal{P}_k\). Fix \(m\) with \(q \in \mathcal{P}_m\). Since

\begin{equation*} \frac{\partial^2}{\partial x_j^{\,2}} \bigl( x_1^{m_1} \cdots x_n^{m_n} \bigr) = m_j (m_j - 1) \, x_1^{m_1} \cdots x_j^{\,m_j - 2} \cdots x_n^{m_n} \end{equation*}

is \(0\) when \(m_j \le 1\) and otherwise a scalar times a monomial of total degree two lower, the linear map \(\Delta\) sends \(\mathcal{P}_k\) into \(\mathcal{P}_{k-2}\); and multiplication by \(1 - \|x\|^2\) sends \(\mathcal{P}_k\) into \(\mathcal{P}_{k+2}\). Hence

\begin{equation*} Tr = \Delta \bigl( (1 - \|x\|^2) r \bigr) \end{equation*}

defines an operator \(T\) on the finite-dimensional space \(\mathcal{P}_{m-2}\), linear as a composition of those two linear maps.

\(T\) is injective. If \(Tr = 0\), then \(h = (1 - \|x\|^2) r\) is a polynomial with \(\Delta h = 0\) and \(h = 0\) on the unit sphere, so \(h = 0\) by the fact quoted in the statement. Thus \(r(x) = 0\) whenever \(\|x\| \ne 1\), and \(r\), being a polynomial and hence continuous, vanishes on the unit sphere too (for \(\|x\| = 1\), the points \((1 + 1/k)x\) converge to \(x\) and have norm \(\ne 1\)). So \(\operatorname{null} T = \{0\}\), giving injectivity by 3.15 and therefore surjectivity by 3.65.

Since \(q \in \mathcal{P}_m\) gives \(-\Delta q \in \mathcal{P}_{m-2}\), surjectivity supplies \(r \in \mathcal{P}_{m-2}\) with \(Tr = -\Delta q\), and then linearity of \(\Delta\) gives

\begin{equation*} \Delta p = \Delta q + \Delta \bigl( (1 - \|x\|^2) r \bigr) = \Delta q - \Delta q = 0 . \end{equation*}

Apply the hypothesis with one, then two, nonzero coefficients. Taking \(a_k = 1\) and all other coefficients \(0\) gives \(\|e_k\|^2 = 1\) for each \(k\).

Now fix \(j \ne k\), let \(a \in \mathbb{F}\), and take coefficient \(1\) on \(e_j\), coefficient \(a\) on \(e_k\), and \(0\) elsewhere. The hypothesis gives \(\|e_j + a e_k\|^2 = 1 + |a|^2\), while expanding by 6.6 and using \(\|e_j\| = \|e_k\| = 1\) and \(\langle e_k, e_j \rangle = \overline{\langle e_j, e_k \rangle}\) gives

\begin{equation*} \|e_j + a e_k\|^2 = 1 + |a|^2 + \overline{a} \langle e_j, e_k \rangle + \overline{\overline{a} \langle e_j, e_k \rangle} = 1 + |a|^2 + 2 \operatorname{Re} \bigl( \overline{a} \langle e_j, e_k \rangle \bigr) . \end{equation*}

Hence \(\operatorname{Re} \bigl( \overline{a} \langle e_j, e_k \rangle \bigr) = 0\) for every \(a \in \mathbb{F}\).

(i) \(\mathbb{F} = \mathbb{R}\): taking \(a = 1\) gives \(\langle e_j, e_k \rangle = \operatorname{Re} \langle e_j, e_k \rangle = 0\).

(ii) \(\mathbb{F} = \mathbb{C}\): write \(\langle e_j, e_k \rangle = x + iy\) with \(x, y \in \mathbb{R}\). Then \(a = 1\) gives \(x = 0\), and \(a = i\) gives

\begin{equation*} 0 = \operatorname{Re} \bigl( -i(x + iy) \bigr) = \operatorname{Re}(y - ix) = y . \end{equation*}

So the list is orthonormal.

Throughout, \(\mathbb{R}^2\) carries the Euclidean inner product.

(a) By \(\cos^2\theta + \sin^2\theta = 1\) each of the three vectors has norm \(1\), and

\begin{equation*} \langle (\cos\theta, \sin\theta), (-\sin\theta, \cos\theta) \rangle = 0 = \langle (\cos\theta, \sin\theta), (\sin\theta, -\cos\theta) \rangle . \end{equation*}