Problems for Chapter 7

(a) Let $𝑇\in \mathrm{L}(𝑉)$, and let ${𝐯}_1,\dots ,{𝐯}_{𝑚}$ be eigenvectors of $𝑇$ with distinct eigenvalues ${𝜆}_1,\dots ,{𝜆}_{𝑚}$. Then ${𝐯}_1,\dots ,{𝐯}_{𝑚}$ is linearly independent (Theorem 7.2.21). No hypothesis on $dim𝑉$ is needed.

(b) Killing terms. Suppose ${𝑐}_1{𝐯}_1+$$\cdots +$${𝑐}_{𝑚}{𝐯}_{𝑚}=$$\mathrm{𝟎}$, and fix $𝑖$. Apply ${𝑃}_{𝑖}=$${\prod }_{𝑗\ne 𝑖}(𝑇-{𝜆}_{𝑗}\text{id})$. By Proposition 7.2.28, ${𝑃}_{𝑖}{𝐯}_{𝑘}=$${\prod }_{𝑗\ne 𝑖}({𝜆}_{𝑘}-{𝜆}_{𝑗}){𝐯}_{𝑘}$, which is $\mathrm{𝟎}$ for $𝑘\ne 𝑖$ (the factor $𝑗=𝑘$ vanishes). So ${𝑐}_{𝑖}{\prod }_{𝑗\ne 𝑖}({𝜆}_{𝑖}-{𝜆}_{𝑗}){𝐯}_{𝑖}=$$\mathrm{𝟎}$. The product is nonzero because the $𝜆$’s are distinct, and ${𝐯}_{𝑖}\ne \mathrm{𝟎}$, so ${𝑐}_{𝑖}=0$.

A shortest relation. Suppose the list is dependent, and among all nontrivial relations choose one, ${\sum }_{𝑖\in 𝑆}{𝑐}_{𝑖}{𝐯}_{𝑖}=\mathrm{𝟎}$ with every ${𝑐}_{𝑖}\ne 0$, for which $|𝑆|$ is smallest. Since eigenvectors are nonzero, $|𝑆|\ge 2$. Pick $𝑗\in 𝑆$ and apply $𝑇-{𝜆}_{𝑗}\text{id}$:

The coefficients are nonzero and the index set $𝑆\setminus \{𝑗\}$ is nonempty, so this is a nontrivial relation that is shorter, a contradiction.

(c) $𝑉$ is a subspace of ${ℝ}^{ℝ}$, and ${𝐷}^2$ is linear. Since ${𝐷}^{2}cos𝑘𝑥=$$-{𝑘}^{2}cos𝑘𝑥$, ${𝐷}^{2}sin𝑘𝑥=$$-{𝑘}^{2}sin𝑘𝑥$, and ${𝐷}^{2}1=0$, the functions $1,cos𝑥,\dots ,cos𝑛𝑥$ are eigenvectors of ${𝐷}^2$ with the distinct eigenvalues $0,-1,-4,\dots ,-{𝑛}^2$, so they are independent by (a). The same holds for $sin𝑥,\dots ,sin𝑛𝑥$.

For the full list, suppose ${𝑎}_0+$${\sum }_{𝑘=1}^{𝑛}({𝑎}_{𝑘}cos𝑘𝑥+{𝑏}_{𝑘}sin𝑘𝑥)=$$0$ for all $𝑥$. Let ${𝑓}_0={𝑎}_0$ and ${𝑓}_{𝑘}=$${𝑎}_{𝑘}cos𝑘𝑥+$${𝑏}_{𝑘}sin𝑘𝑥$. Each ${𝑓}_{𝑘}$ lies in the eigenspace ${𝐸}_{-{𝑘}^2}({𝐷}^2)$, and ${𝑓}_0+\cdots +{𝑓}_{𝑛}=\mathrm{𝟎}$. By Corollary 7.2.23 the eigenspaces are independent, so every ${𝑓}_{𝑘}=\mathrm{𝟎}$. So ${𝑎}_0=0$, and for $𝑘\ge 1$, evaluating ${𝑓}_{𝑘}$ at $𝑥=0$ gives ${𝑎}_{𝑘}=0$, and then at $𝑥=𝜋/(2𝑘)$ gives ${𝑏}_{𝑘}=0$.

The theorem separates eigenvalues and nothing more. Inside one eigenspace, here the plane spanned by $cos𝑘𝑥$ and $sin𝑘𝑥$, independence needs a separate argument, which the two evaluations supply.

(a) Let $dim𝑉=𝑛$, $𝑇\in \mathrm{L}(𝑉)$, with distinct eigenvalues ${𝜆}_1,\dots ,{𝜆}_{𝑚}$. Then $𝑇$ is diagonalizable if and only if $𝑉={𝐸}_{{𝜆}_1}\oplus \cdots \oplus {𝐸}_{{𝜆}_{𝑚}}$, if and only if ${\sum }_{𝑖}gm({𝜆}_{𝑖})=𝑛$, if and only if ${𝑝}_{𝑇}$ splits over $𝐹$ and $gm({𝜆}_{𝑖})=am({𝜆}_{𝑖})$ for all $𝑖$ (Theorem 7.4.16).

(b) The sum ${𝐸}_{{𝜆}_1}+\cdots +{𝐸}_{{𝜆}_{𝑚}}$ is direct (Corollary 7.2.23), so its dimension is ${\sum }_{𝑖}gm({𝜆}_{𝑖})$ (Theorem 4.4.13). If ${\sum }_{𝑖}gm({𝜆}_{𝑖})=𝑛$, the sum is all of $𝑉$, and a basis of each ${𝐸}_{{𝜆}_{𝑖}}$, glued together, is a basis of $𝑉$ (Theorem 4.4.13) consisting of eigenvectors, so $𝑇$ is diagonalizable. Conversely, if $\mathrm{ℬ︀}$ is an eigenbasis, every vector of $\mathrm{ℬ︀}$ lies in some ${𝐸}_{{𝜆}_{𝑖}}$, so $𝑉=$$span\mathrm{ℬ︀}\subseteq$${\sum }_{𝑖}{𝐸}_{{𝜆}_{𝑖}}$ and ${\sum }_{𝑖}gm({𝜆}_{𝑖})=𝑛$.

For the second equivalence, $gm({𝜆}_{𝑖})\le am({𝜆}_{𝑖})$ for each $𝑖$ (Theorem 7.4.15), and ${\sum }_{𝑖}am({𝜆}_{𝑖})\le$$deg{𝑝}_{𝑇}=$$𝑛$ (Proposition 7.1.27), with equality if and only if ${𝑝}_{𝑇}$ splits: if it splits, the multiplicities add up to $𝑛$ (Proposition 7.1.34); if they add up to $𝑛$, then $\prod {(𝑥-{𝜆}_{𝑖})}^{am({𝜆}_{𝑖})}$ divides ${𝑝}_{𝑇}$ and has the same degree, and both are monic, so they are equal. Hence

and $\sum gm=𝑛$ exactly when both inequalities are equalities, that is, when ${𝑝}_{𝑇}$ splits and every $gm({𝜆}_{𝑖})=am({𝜆}_{𝑖})$.

(c) By Proposition 7.3.15, ${𝑝}_{𝑀}={𝑝}_{𝐵}{𝑝}_{𝐷}$, which splits because ${𝑝}_{𝐵}$ and ${𝑝}_{𝐷}$ do (Theorem 7.4.16). Let $𝜆$ be an eigenvalue of $𝑀$; it is a root of exactly one of ${𝑝}_{𝐵}$, ${𝑝}_{𝐷}$, so ${am}_{𝑀}(𝜆)$ is ${am}_{𝐵}(𝜆)$ or ${am}_{𝐷}(𝜆)$.

If ${𝑝}_{𝐵}(𝜆)=0$, the map $𝐱\mapsto (𝐱,\mathrm{𝟎})$ is injective and sends ${𝐸}_{𝜆}(𝐵)$ into ${𝐸}_{𝜆}(𝑀)$, because $𝑀(𝐱,\mathrm{𝟎})=$$(𝐵𝐱,\mathrm{𝟎})$. So ${gm}_{𝑀}(𝜆)\ge$${gm}_{𝐵}(𝜆)=$${am}_{𝐵}(𝜆)=$${am}_{𝑀}(𝜆)$.

If ${𝑝}_{𝐷}(𝜆)=0$, then $𝜆$ is not an eigenvalue of $𝐵$, so $𝐵-𝜆{𝐼}_{𝑘}$ is invertible. For $𝐲\in {𝐸}_{𝜆}(𝐷)$ let $𝐱=$$-{(𝐵-𝜆{𝐼}_{𝑘})}^{-1}𝐶𝐲$. Then $𝐵𝐱+$$𝐶𝐲=$$(𝐵-𝜆{𝐼}_{𝑘})𝐱+$$𝜆𝐱+$$𝐶𝐲=$$𝜆𝐱$ and $𝐷𝐲=𝜆𝐲$, so $(𝐱,𝐲)\in {𝐸}_{𝜆}(𝑀)$. The map $𝐲\mapsto (𝐱,𝐲)$ is linear and injective, so ${gm}_{𝑀}(𝜆)\ge$${gm}_{𝐷}(𝜆)=$${am}_{𝑀}(𝜆)$.

With Theorem 7.4.15, ${gm}_{𝑀}={am}_{𝑀}$ at every eigenvalue, and $𝑀$ is diagonalizable by (b). Without the hypothesis, $𝑀=\left(\begin{array}{cc}1& 1\\ 0& 1\end{array}\right)$ ($𝐵=𝐷=(1)$, $𝐶=(1)$) is not diagonalizable (Example 7.4.8).

Another route: if $𝑋$ solves $𝐵𝑋-𝑋𝐷=-𝐶$, then ${\left(\begin{array}{cc}{𝐼}_{𝑘}& 𝑋\\ 0& {𝐼}_{𝑙}\end{array}\right)}^{-1}𝑀\left(\begin{array}{cc}{𝐼}_{𝑘}& 𝑋\\ 0& {𝐼}_{𝑙}\end{array}\right)=$$\left(\begin{array}{cc}𝐵& 0\\ 0& 𝐷\end{array}\right)$, and with no common eigenvalue such an $𝑋$ always exists (Exercise 7.7.11(b) and rank–nullity on $𝑋\mapsto 𝐵𝑋-𝑋𝐷$).

(a) Let $𝑇\in \mathrm{L}(𝑉)$, $𝜆\in 𝐹$, $𝑁=𝑇-𝜆\text{id}$, and suppose ${𝑁}^{𝑘}𝐯\ne \mathrm{𝟎}$ and ${𝑁}^{𝑘+1}𝐯=\mathrm{𝟎}$. Then $𝐯,𝑁𝐯,\dots ,{𝑁}^{𝑘}𝐯$ is linearly independent (Proposition 7.2.37). Moreover ${𝑁}^{𝑘}𝐯$ is an eigenvector of $𝑇$ with eigenvalue $𝜆$, and the span of the chain is the Krylov subspace ${𝐾}_{𝑇}(𝐯)$, of dimension $𝑘+1$ (Corollary 7.2.38).

(b) Suppose ${\sum }_{𝑗=0}^{𝑘}{𝑐}_{𝑗}{𝑁}^{𝑗}𝐯=$$\mathrm{𝟎}$, and suppose some ${𝑐}_{𝑗}$ is nonzero; let $𝑖$ be the smallest such index. Apply ${𝑁}^{𝑘-𝑖}$: every term with $𝑗>𝑖$ becomes ${𝑐}_{𝑗}{𝑁}^{𝑗+𝑘-𝑖}𝐯=$$\mathrm{𝟎}$, since $𝑗+𝑘-𝑖\ge 𝑘+1$. What remains is ${𝑐}_{𝑖}{𝑁}^{𝑘}𝐯=\mathrm{𝟎}$, so ${𝑐}_{𝑖}=0$, a contradiction. For the last vector, $𝑁({𝑁}^{𝑘}𝐯)=$${𝑁}^{𝑘+1}𝐯=$$\mathrm{𝟎}$ and ${𝑁}^{𝑘}𝐯\ne \mathrm{𝟎}$, so $𝑇{𝑁}^{𝑘}𝐯=𝜆{𝑁}^{𝑘}𝐯$.

(c) Let $𝑁=𝑇-𝜆\text{id}$. By Theorem 7.5.13, $𝑇$ has an upper-triangular matrix in some basis, and by Proposition 7.5.9 its diagonal entries are the roots of ${𝑝}_{𝑇}$, all equal to $𝜆$. So the matrix of $𝑁$ is strictly upper triangular, and ${𝑁}^{𝑛}=0$ (Lemma 7.6.21).

Next, for $𝑗\ge 0$, $𝑁$ maps $ker{𝑁}^{𝑗+1}$ into $ker{𝑁}^{𝑗}$, and the kernel of this restriction is $ker𝑁\cap$$ker{𝑁}^{𝑗+1}=$$ker𝑁$. By Theorem 4.2.11,

using $dimker𝑁=gm(𝜆)=1$. Starting from $dimker{𝑁}^0=0$, induction gives $dimker{𝑁}^{𝑗}\le 𝑗$. In particular $dimker{𝑁}^{𝑛-1}\le$$𝑛-$$1<$$𝑛$, so ${𝑁}^{𝑛-1}\ne 0$.

Choose $𝐯$ with ${𝑁}^{𝑛-1}𝐯\ne \mathrm{𝟎}$; also ${𝑁}^{𝑛}𝐯=\mathrm{𝟎}$. By (b), the chain $𝐯,$$𝑁𝐯,$$\dots ,$${𝑁}^{𝑛-1}𝐯$ is independent, and it has $𝑛$ vectors, so it is a basis of $𝑉$. By Corollary 7.2.38(c), in the basis ${𝑁}^{𝑛-1}𝐯,$$\dots ,$$𝑁𝐯,$$𝐯$ the matrix of $𝑇$ is as claimed.

Conversely, that matrix has $gm(𝜆)=$$𝑛-$$rank(𝑁)=$$1$. So for an operator with a single eigenvalue, “one eigenvector line” and “one chain” are the same condition. Over any field the same proof works once ${𝑁}^{𝑛}=0$ is known, which Cayley–Hamilton supplies (Theorem 7.6.20). This is the simplest case of the Jordan form (Section 10.3).

We have $𝐴-$$2{𝐼}_3=$$\left(\begin{array}{ccc}0& 1& 0\\ 0& 0& 0\\ 0& 0& 1\end{array}\right)$, ${(𝐴-2{𝐼}_3)}^2=$$\left(\begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ 0& 0& 1\end{array}\right)$, and $𝐴-$$3{𝐼}_3=$$\left(\begin{array}{ccc}-1& 1& 0\\ 0& -1& 0\\ 0& 0& 0\end{array}\right)$. So $𝐺=span({𝐞}_1,{𝐞}_2)$, $𝐸=span({𝐞}_3)$, and $(𝐴-3{𝐼}_3){(𝐴-2{𝐼}_3)}^2=$$0$, since the third column of $𝐴-3{𝐼}_3$ is zero.

The splitting. In $\mathbb{Q}[𝑥]$, ${(𝑥-2)}^2-$$(𝑥-3)(𝑥-1)=$$({𝑥}^2-4𝑥+4)-$$({𝑥}^2-4𝑥+3)=$$1$. Substituting $𝐴$ (Proposition 7.2.27), for every $𝐮\in {\mathbb{Q}}^3$,

The first term is killed by $𝐴-3{𝐼}_3$, so it lies in $𝐸$; the second is killed by ${(𝐴-2{𝐼}_3)}^2$, since ${(𝐴-2{𝐼}_3)}^2(𝐴-3{𝐼}_3)(𝐴-{𝐼}_3)=$$(𝐴-{𝐼}_3)(𝐴-3{𝐼}_3){(𝐴-2{𝐼}_3)}^2=$$0$, so it lies in $𝐺$. If $𝑈$ is invariant, it is invariant under every polynomial in $𝐴$, so for $𝐮\in 𝑈$ both terms lie in $𝑈$. Hence $𝑈=$$(𝑈\cap 𝐺)+$$(𝑈\cap 𝐸)$, and the sum is direct because $𝐺\cap 𝐸=\{\mathrm{𝟎}\}$.

The pieces. $𝑈\cap 𝐸$ is a subspace of the line $𝐸$, so it is $\{\mathrm{𝟎}\}$ or $𝐸$. $𝑈\cap 𝐺$ is an invariant subspace of $𝐺$, on which $𝐴$ acts by $𝐴{𝐞}_1=2{𝐞}_1$, $𝐴{𝐞}_2={𝐞}_1+2{𝐞}_2$. An invariant line is spanned by an eigenvector (Proposition 7.2.7), and the eigenvectors in $𝐺$ are the nonzero multiples of ${𝐞}_1$, since $(𝐴-2{𝐼}_3)(𝑎{𝐞}_1+𝑏{𝐞}_2)=$$𝑏{𝐞}_1$. So $𝑈\cap 𝐺$ is $\{\mathrm{𝟎}\}$, $span({𝐞}_1)$, or $𝐺$.

Conversely, each of the $2\cdot 3$ combinations is invariant, being a sum of invariant subspaces. The six invariant subspaces are

The same argument shows, for any $𝑇$ with $𝑝(𝑇)𝑞(𝑇)=0$ and $𝑝,𝑞$ coprime, that invariant subspaces split along $ker𝑝(𝑇)\oplus$$ker𝑞(𝑇)$ (compare Exercise 7.2.55). With three distinct eigenvalues there would be $2^3=8$ invariant subspaces (Exercise 7.2.48); the chain in $𝐺$ is what removes two of them.

(a) True. Write $𝐴=𝑆𝑅{𝑆}^{-1}$ with $𝑅$ upper triangular (Corollary 7.5.14). Then ${𝐴}^2=𝑆{𝑅}^2{𝑆}^{-1}$, and ${𝑅}^2$ is upper triangular with diagonal entries ${𝑟}_{𝑖𝑖}^2$. By Proposition 7.5.8, the eigenvalues of $𝐴$ are the ${𝑟}_{𝑖𝑖}$ and those of ${𝐴}^2$ are the ${𝑟}_{𝑖𝑖}^2$. Similar matrices have the same eigenvalues (Proposition 7.3.11), so the finite set $𝑆$ of eigenvalues of $𝐴$ equals $\{{𝜆}^2|𝜆\in 𝑆\}$. In particular $𝜆\in 𝑆$ implies ${𝜆}^2,{𝜆}^4,{𝜆}^8,\dots \in 𝑆$. Since $𝑆$ is finite, ${𝜆}^{2^{𝑖}}={𝜆}^{2^{𝑗}}$ for some $𝑖<𝑗$, that is, ${𝜆}^{2^{𝑖}}({𝜆}^{2^{𝑗}-2^{𝑖}}-1)=$$0$. So $𝜆=0$ or ${𝜆}^{2^{𝑗}-2^{𝑖}}=1$.

(b) False. With $𝐴={𝐼}_{𝑛}$ and $𝐵=-{𝐼}_{𝑛}$, ${𝐴}^2={𝐵}^2={𝐼}_{𝑛}$, but a matrix similar to ${𝐼}_{𝑛}$ is $𝑆{𝐼}_{𝑛}{𝑆}^{-1}={𝐼}_{𝑛}$, and $𝐵\ne {𝐼}_{𝑛}$.

(c) False. Let $𝐴=\left(\begin{array}{cc}1& 1\\ 0& 1\end{array}\right)$. Then ${𝐴}^{-1}=\left(\begin{array}{cc}1& -1\\ 0& 1\end{array}\right)$ and $𝐴+{𝐴}^{-1}=2{𝐼}_2$, which is diagonal, while $𝐴$ is not diagonalizable (Example 7.4.8). The map $𝜆\mapsto 𝜆+{𝜆}^{-1}$ has derivative $0$ at $𝜆=1$, and that is where the shear disappears.

(d) True. If ${𝑁}^{𝑘}=0$, the only eigenvalue of $𝑁$ is $0$ (Proposition 7.2.28). If $({𝐼}_{𝑛}+𝑁)𝐯=𝜇𝐯$ with $𝐯\ne \mathrm{𝟎}$, then $𝑁𝐯=(𝜇-1)𝐯$, so $𝜇=1$. A diagonalizable matrix with the single eigenvalue $1$ is ${𝐼}_{𝑛}$ (Proposition 7.4.7), so ${𝐼}_{𝑛}+𝑁={𝐼}_{𝑛}$ and $𝑁=0$. The same proof works over every field.

(a) Work in ${𝐹[𝑥]}^{2𝑛\times 2𝑛}$, where determinants are multiplicative and block triangular matrices have determinant the product of the diagonal blocks (Proposition 7.3.2). By Theorem 2.4.34,

The two outer factors have determinant $1$, so ${𝑝}_{𝑀}=$$det(𝑥{𝐼}_{𝑛}-2𝐴)det(𝑥{𝐼}_{𝑛})=$${𝑥}^{𝑛}{𝑝}_{2𝐴}$.

(b) Let $𝑆=$$\left(\begin{array}{cc}{𝐼}_{𝑛}& {𝐼}_{𝑛}\\ {𝐼}_{𝑛}& -{𝐼}_{𝑛}\end{array}\right)$. Then ${𝑆}^2=2{𝐼}_{2𝑛}$, so ${𝑆}^{-1}=(1/2)𝑆$, and

For $𝜇\in 𝐹$, $ker({𝑀}^{\prime }-𝜇{𝐼}_{2𝑛})=$$ker(2𝐴-𝜇{𝐼}_{𝑛})\times$$ker(-𝜇{𝐼}_{𝑛})$, so ${gm}_{{𝑀}^{\prime }}(𝜇)=$${gm}_{2𝐴}(𝜇)$ for $𝜇\ne 0$ and ${gm}_{{𝑀}^{\prime }}(0)=$${gm}_{2𝐴}(0)+$$𝑛$. By (a) applied to ${𝑀}^{\prime }$, or directly, ${𝑝}_{{𝑀}^{\prime }}={𝑥}^{𝑛}{𝑝}_{2𝐴}$, so ${am}_{{𝑀}^{\prime }}(𝜇)=$${am}_{2𝐴}(𝜇)$ for $𝜇\ne 0$, ${am}_{{𝑀}^{\prime }}(0)=$${am}_{2𝐴}(0)+$$𝑛$, and ${𝑝}_{{𝑀}^{\prime }}$ splits iff ${𝑝}_{2𝐴}$ does. By Theorem 7.4.16(IV), ${𝑀}^{\prime }$ is diagonalizable iff $2𝐴$ is. Finally $2𝐴={𝑆}_0𝐷{𝑆}_0^{-1}$ iff $𝐴={𝑆}_0(𝐷/2){𝑆}_0^{-1}$, so $2𝐴$ is diagonalizable iff $𝐴$ is, and $𝑀$ is similar to ${𝑀}^{\prime }$.

(c) By Theorem 2.4.34 every block of ${𝑀}^2$ is ${𝐴}^2+{𝐴}^2=2{𝐴}^2=0$ in characteristic $2$. So $𝑀$ is nilpotent, and its only eigenvalue is $0$. If $𝑀$ is diagonalizable, then $𝑀=0$ (Proposition 7.4.7), that is, $𝐴=0$; conversely $𝐴=0$ gives $𝑀=0$. This matches (a): in characteristic $2$, $2𝐴=0$ and ${𝑝}_{𝑀}={𝑥}^{2𝑛}$. The similarity of (b) fails because $𝑆$ is singular when $2=0$.

(a) $𝜆{𝐰}^{\text{𝖳}}𝐯=$${𝐰}^{\text{𝖳}}(𝐴𝐯)=$${({𝐴}^{\text{𝖳}}𝐰)}^{\text{𝖳}}𝐯=$$𝜇{𝐰}^{\text{𝖳}}𝐯$, using ${({𝐴}^{\text{𝖳}}𝐰)}^{\text{𝖳}}={𝐰}^{\text{𝖳}}𝐴$ (Proposition 2.4.23). So $(𝜆-𝜇){𝐰}^{\text{𝖳}}𝐯=0$, and $𝜆\ne 𝜇$ gives ${𝐰}^{\text{𝖳}}𝐯=0$.

(b) From $𝐴𝑆=𝑆𝐷$ we get ${𝑆}^{-1}𝐴=𝐷{𝑆}^{-1}$, and transposing, ${𝐴}^{\text{𝖳}}{({𝑆}^{-1})}^{\text{𝖳}}=$${({𝑆}^{-1})}^{\text{𝖳}}𝐷$. The columns of ${({𝑆}^{-1})}^{\text{𝖳}}$ are ${𝐰}_1,\dots ,{𝐰}_{𝑛}$, so column $𝑖$ of this equation reads ${𝐴}^{\text{𝖳}}{𝐰}_{𝑖}={𝜆}_{𝑖}{𝐰}_{𝑖}$. The $(𝑖,𝑗)$ entry of ${𝑆}^{-1}𝑆={𝐼}_{𝑛}$ is row $𝑖$ of ${𝑆}^{-1}$ times column $𝑗$ of $𝑆$, which is ${𝐰}_{𝑖}^{\text{𝖳}}{𝐯}_{𝑗}$. Finally, $𝑆𝐷$ has columns ${𝜆}_{𝑖}{𝐯}_{𝑖}$, and a product of a matrix with columns ${𝐚}_{𝑖}$ and a matrix with rows ${𝐛}_{𝑖}^{\text{𝖳}}$ is ${\sum }_{𝑖}{𝐚}_{𝑖}{𝐛}_{𝑖}^{\text{𝖳}}$ (Theorem 2.4.34, or the column–row expansion of Theorem 4.3.21). So $𝐴=$$(𝑆𝐷){𝑆}^{-1}=$${\sum }_{𝑖}{𝜆}_{𝑖}{𝐯}_{𝑖}{𝐰}_{𝑖}^{\text{𝖳}}$.

(c) By (b), ${𝑃}_{𝑖}{𝑃}_{𝑗}=$${𝐯}_{𝑖}({𝐰}_{𝑖}^{\text{𝖳}}{𝐯}_{𝑗}){𝐰}_{𝑗}^{\text{𝖳}}$, which is $0$ for $𝑖\ne 𝑗$ and ${𝑃}_{𝑖}$ for $𝑖=𝑗$. By the column–row expansion again, ${\sum }_{𝑖}{𝑃}_{𝑖}=$$𝑆{𝑆}^{-1}=$${𝐼}_{𝑛}$. And ${𝐴}^{𝑘}=$$𝑆{𝐷}^{𝑘}{𝑆}^{-1}=$${\sum }_{𝑖}{𝜆}_{𝑖}^{𝑘}{𝑃}_{𝑖}$ by the same expansion applied to $𝑆{𝐷}^{𝑘}$.

So a diagonalizable $𝐴$ is a combination of rank-one idempotents ${𝑃}_{𝑖}$, the projection onto $span({𝐯}_{𝑖})$ along the span of the other ${𝐯}_{𝑗}$; the left eigenvectors ${𝐰}_{𝑖}$ measure the ${𝐯}_{𝑖}$-coordinate. When the ${𝜆}_{𝑖}$ are distinct, ${𝑃}_{𝑖}$ is the Lagrange polynomial ${𝐿}_{𝑖}(𝐴)$ of Technique 7.6.13. Over $ℝ$ the ${𝐰}_{𝑖}$ equal the ${𝐯}_{𝑖}$ exactly when ${𝑆}^{-1}={𝑆}^{\text{𝖳}}$, which makes $𝐴=𝑆𝐷{𝑆}^{\text{𝖳}}$ symmetric; conversely every real symmetric matrix arises this way (Section 8.2).

(a) Let ${𝑣}_{𝑗}=sin𝑗𝜃$, and set ${𝑣}_0=sin0=0$. Row $𝑗$ of ${𝐴}_{𝑛}$ has ones in columns $𝑗-1$ and $𝑗+1$ (when they exist), so ${({𝐴}_{𝑛}𝐯)}_{𝑗}=$${𝑣}_{𝑗-1}+$${𝑣}_{𝑗+1}$ for $𝑗\le 𝑛-1$ and ${({𝐴}_{𝑛}𝐯)}_{𝑛}={𝑣}_{𝑛-1}$. By the sum formula $sin(𝑎-𝑏)+$$sin(𝑎+𝑏)=$$2sin𝑎cos𝑏$,

which gives the rows $1\le 𝑗\le 𝑛-1$ (row $1$ because ${𝑣}_0=0$). For row $𝑛$, ${𝑣}_{𝑛-1}=$$2cos𝜃sin𝑛𝜃-$$sin((𝑛+1)𝜃)$ by the same formula. This is the claim.

(b) Let ${𝜃}_{𝑘}=𝑘𝜋/(𝑛+1)$ for $𝑘=1,\dots ,𝑛$. Then $sin((𝑛+1){𝜃}_{𝑘})=$$sin𝑘𝜋=$$0$, so ${𝐴}_{𝑛}{𝐯}_{𝑘}=$$2cos{𝜃}_{𝑘}{𝐯}_{𝑘}$, and ${𝐯}_{𝑘}\ne \mathrm{𝟎}$ because its first entry is $sin{𝜃}_{𝑘}>0$. So each $2cos{𝜃}_{𝑘}$ is an eigenvalue. The ${𝜃}_{𝑘}$ lie in $(0,𝜋)$, where $cos$ is strictly decreasing, so these $𝑛$ eigenvalues are distinct. An $𝑛\times 𝑛$ matrix has at most $𝑛$ eigenvalues (Corollary 7.2.22), so these are all of them, and ${𝐴}_{𝑛}$ is diagonalizable (Corollary 7.4.5).

(c) Let ${𝑑}_{𝑚}=det{𝐴}_{𝑚}$. Directly, ${𝑑}_1=0$ and ${𝑑}_2=det\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)=-1$. By Proposition 5.5.18 with diagonal entries $0$ and off-diagonal entries $1$, ${𝑑}_{𝑚}=$$0\cdot$${𝑑}_{𝑚-1}-$$1\cdot$$1\cdot$${𝑑}_{𝑚-2}=$$-{𝑑}_{𝑚-2}$ for $𝑚\ge 3$. Hence $det{𝐴}_{𝑛}=0$ for odd $𝑛$ and $det{𝐴}_{𝑛}={(-1)}^{𝑛/2}$ for even $𝑛$. Since ${𝐴}_{𝑛}$ is diagonalizable with the eigenvalues of (b), its determinant is their product (Theorem 7.5.22), so

For odd $𝑛$ this is visible directly, since the factor with $𝑘=(𝑛+1)/2$ is $2cos(𝜋/2)=0$. In the same way, $tr{𝐴}_{𝑛}=0$ says ${\sum }_{𝑘}cos(𝑘𝜋/(𝑛+1))=$$0$, which also follows from $cos(𝜋-𝑥)=-cos𝑥$.

${𝐴}_{𝑛}$ is the adjacency matrix of a path with $𝑛$ vertices; Section 16.5 studies such matrices. The eigenvector was guessed from the recurrence ${𝑣}_{𝑗-1}+$${𝑣}_{𝑗+1}=$$𝑐{𝑣}_{𝑗}$, whose solutions are combinations of $sin𝑗𝜃$ and $cos𝑗𝜃$, and the boundary rows chose $sin$ and $𝜃$.

(a) If ${𝑝}_{𝐯}$ divides $𝑓$, say $𝑓=𝑞{𝑝}_{𝐯}$, then $𝑓(𝑇)𝐯=$$𝑞(𝑇){𝑝}_{𝐯}(𝑇)𝐯=$$\mathrm{𝟎}$ (Proposition 7.2.27). Conversely, suppose $𝑓(𝑇)𝐯=\mathrm{𝟎}$, and divide: $𝑓=𝑞{𝑝}_{𝐯}+𝑟$ with $𝑟=\mathrm{𝟎}$ or $deg𝑟<deg{𝑝}_{𝐯}$ (Theorem 7.1.14). Then $𝑟(𝑇)𝐯=$$𝑓(𝑇)𝐯-$$𝑞(𝑇){𝑝}_{𝐯}(𝑇)𝐯=$$\mathrm{𝟎}$. If $𝑟\ne \mathrm{𝟎}$, dividing $𝑟$ by its leading coefficient gives a monic polynomial of smaller degree than ${𝑝}_{𝐯}$ killing $𝐯$, which contradicts the choice of ${𝑝}_{𝐯}$. So $𝑟=\mathrm{𝟎}$.

(b) $𝑓(𝑇)=0$ if and only if $𝑓(𝑇){𝐛}_{𝑖}=\mathrm{𝟎}$ for every $𝑖$, since a linear map is determined by its values on a basis (Theorem 2.2.1). By (a), this holds if and only if every ${𝑝}_{{𝐛}_{𝑖}}$ divides $𝑓$. By definition, ${𝑚}_{𝑇}$ is the monic polynomial of least degree with ${𝑚}_{𝑇}(𝑇)=0$ (Theorem 7.6.2), so it is the monic polynomial of least degree divisible by all the ${𝑝}_{{𝐛}_{𝑖}}$.

(c) For ${𝐞}_1$: $𝐴{𝐞}_1=(3,0,1)$ and ${𝐴}^2{𝐞}_1=(8,0,4)$. The first two are independent, and ${𝐴}^2{𝐞}_1-$$4𝐴{𝐞}_1+$$4{𝐞}_1=$$(8-12+4,0,4-4)=$$\mathrm{𝟎}$, so ${𝑝}_{{𝐞}_1}=$${𝑥}^2-$$4𝑥+$$4=$${(𝑥-2)}^2$. For ${𝐞}_2$: $𝐴{𝐞}_2=(1,2,1)$ and ${𝐴}^2{𝐞}_2=$$(4,4,4)=$$4𝐴{𝐞}_2-$$4{𝐞}_2$, so again ${𝑝}_{{𝐞}_2}={(𝑥-2)}^2$. For ${𝐞}_3$: $𝐴{𝐞}_3=(-1,0,1)$ and ${𝐴}^2{𝐞}_3=$$(-4,0,0)=$$4𝐴{𝐞}_3-$$4{𝐞}_3$, so ${𝑝}_{{𝐞}_3}={(𝑥-2)}^2$. By (b), ${𝑚}_{𝐴}={(𝑥-2)}^2$. It has a repeated root, so $𝐴$ is not diagonalizable (Theorem 7.6.15).