Chapter 5 Eigenvalues and Eigenvectors

Section 5A Invariant Subspaces

5A.1

Suppose $𝑇 ∈ ℒ(𝑉)$ and $𝑈$ is a subspace of $𝑉$.

(a) Prove that if $𝑈 ⊆ \text{null } T$, then $𝑈$ is invariant under $𝑇$.

Proof:

Assume $u \in U ⊆ \text{null } T$， $T(u) = 0 \in U$

$\square$

(b) Prove that if $\text{range } T ⊆ 𝑈$, then $𝑈$ is invariant under $𝑇$.

Proof:

If $u \in U$, then $T(u) \in \text{range } T ⊆ U$.

$\square$

5A.2

Suppose that $𝑇 ∈ ℒ(𝑉)$ and $V_1, \cdots, V_m$ are subspaces of $𝑉$ invariant under $𝑇$. Prove that $𝑉_1 +⋯ + 𝑉_𝑚$ is invariant under $𝑇$.

Proof:

Assume $v \in 𝑉_1 +⋯ + 𝑉_𝑚$, then we have

$$v = v_1 + \cdots + v_m, v_i \in V_i$$

So

$$T(v) = T(v_1) + \cdots + T(v_m)$$

Since $T(v_i) \in V_i$, then $T(v) \in 𝑉_1 +⋯ + 𝑉_𝑚$.

$\square$

5A.3

Suppose $𝑇 ∈ ℒ(𝑉)$. Prove that the intersection of every collection of subspaces of $𝑉$ invariant under $𝑇$ is invariant under $𝑇$.

Proof:

Let

$$U = \cap_i U_i$$

where $U_i$ is invariant under $𝑇$.

If $v \in U$, then $v \in U_i$, then $T(v) \in U_i$, so $T(v) \in U$.

$\square$

5A.4

Prove or give a counterexample: If $𝑉$ is finite-dimensional and $𝑈$ is a subspace of $𝑉$ that is invariant under every operator on $𝑉$ , then $𝑈 = \{0\}$ or $𝑈 = 𝑉$.

Proof:

The statement is correct. We prove by contradition.

Assume $U$ is a subspace and $U \neq \{0\}$ and $U \neq V$.

Let $u_1, \cdots, u_m$ be a basis of $U$ and we can extends it to a basis of $V$ by adding $v_1, \cdots, v_n$.

Define $T$ such that $T(u_i) = v_1, T(v_j) = v_1$. So $T$ is an operator.

Then $U$ is not an invariant under $T$. We reach a contradition.

$\square$

5A.5

Suppose $𝑇 ∈ ℒ(𝐑^2)$ is defined by $𝑇(𝑥, 𝑦) = (−3𝑦, 𝑥)$. Find the eigenvalues of $𝑇$.

Solution:

$$\begin{align*} (−3𝑦, 𝑥) &= k(x, y) \\ & \Rightarrow \\ x = ky&, -3y = kx \\ & \Rightarrow \\ k^2 &= -3 \end{align*}$$

So $T$ does not have any eigenvalues.

$\square$

5A.6

Define $𝑇 ∈ ℒ(𝐅^2)$ by $𝑇(𝑤, 𝑧) = (𝑧, 𝑤)$. Find all eigenvalues and eigenvectors of $𝑇$.

Solution:

Note $F$ can either be $\mathbb{R}$ or $\mathbb{C}$.

$$\begin{align*} k(w,z) &= (z,w) \\ & \Rightarrow \\ z = kw &, kz=w \\ & \Rightarrow \\ k^2 &= 1 \\ & \Rightarrow \\ k = &\pm1 \end{align*}$$

If $k = 1$, then the eigenvectors are $(a,a)$. If $k = -1$, then the eigenvectors are $(a,-a)$.

$\square$

5A.7

Define $𝑇 ∈ ℒ(𝐅^3)$ by $𝑇(𝑧_1, 𝑧_2, 𝑧_3) = (2𝑧_2, 0, 5𝑧_3)$. Find all eigenvalues and eigenvectors of $𝑇$.

Proof:

Note $F$ can either be $\mathbb{R}$ or $\mathbb{C}$.

$$\begin{align*} k(𝑧_1, 𝑧_2, 𝑧_3) &= (2𝑧_2, 0, 5𝑧_3) \\ & \Rightarrow \\ k z_1 = 2 z_2, k z_2 &= 0, k z_3 = 5 z_3 \end{align*}$$

If $k = 0$, then $z_3 = 0$. Also we have $0 = k z_1 = 2 z_2$, so $z_2 = 0$, so the eigenvectors for $k = 0$ will be $(a, 0, 0)$ for $a \neq 0$.

If $k \neq 0$, then $z_2 = 0$, then $z_1 = 0$. Since the eigenvectors cannot be $0$, so $z_3 \neq 0$, so $k = 5$. And the eigenvectors will be $(0,0, a)$ for $a \neq 0$.

$\square$

5A.8

Suppose $𝑃 ∈ ℒ(𝑉)$ is such that $𝑃^2 = 𝑃$. Prove that if $\lambda$ is an eigenvalue of $𝑃$, then $\lambda = 0$ or $\lambda = 1$.

Proof:

First we have the assumption that $V$ is a vector space over $\mathbb{R}$ or $\mathbb{C}$.

If $\lambda$ is a eigenvalue and $u$ is a an eigenvector, then

$$\begin{align*} P^2(u) &= P(u) \\ & \Rightarrow \\ P(\lambda u) &= \lambda u \\ & \Rightarrow \\ \lambda^2 u &= \lambda u \\ \end{align*}$$

Since $u \neq 0$, then $\lambda^2 = \lambda$. Since $\lambda$ is a real or complex number, then $\lambda = 0$ or $\lambda = 1$.

$\square$

5A.9

Define $𝑇 ∶ 𝒫(𝐑) → 𝒫(𝐑)$ by $𝑇𝑝 = 𝑝'$. Find all eigenvalues and eigenvectors of $T$.

Proof:

Assume

$$p(x) = a_0 + a_1 x + \cdots + a_{m} x^{m}$$

Then

$$T(p) = p'(x) = a_1 + 2 a_2 x + \cdots + m a_{m} x^{m-1}$$

If $\lambda$ is an eigenvalue and $p$ is an eigenvector,

then

$$\begin{align*} & \lambda a_0 + \lambda a_1 x + \cdots + \lambda a_{m} x^{m} \\ &= a_1 + 2 a_2 x + \cdots + m a_{m} x^{m-1} \end{align*}$$

So we have

$$\begin{align*} \lambda a_m &= 0 \\ \lambda a_{m-1} &= m a_m \\ &\cdots \\ \lambda a_1 &= 2 a_2 \\ \lambda a_0 &= a_1 \\ \end{align*}$$

If $\lambda = 0$, then $a_m = \cdots = a_1 = 0$, so the eigenvectors are $p(x) = a \neq 0$.

If $\lambda \neq 0$, then $a_m = \cdots = a_0 = 0$, so there is no eigenvectors.

$\square$

5A.10

Define $𝑇 ∈ ℒ(𝒫_4(𝐑))$ by $(𝑇𝑝)(𝑥) = 𝑥𝑝'(𝑥)$ for all $𝑥 ∈ 𝐑$. Find all eigenvalues and eigenvectors of $𝑇$.

Proof:

Assume

$$p(x) = a_0 + a_1 x + \cdots + a_{m} x^{m}$$

Then

$$T(p) = xp'(x) = a_1x + 2 a_2 x^2 + \cdots + m a_{m} x^{m}$$

If $\lambda$ is an eigenvalue and $p$ is an eigenvector,

then

$$\begin{align*} & \lambda a_0 + \lambda a_1 x + \cdots + \lambda a_{m} x^{m} \\ &= a_1x + 2 a_2 x^2 + \cdots + m a_{m} x^{m} \end{align*}$$

So we have

$$\begin{align*} \lambda a_m &= m a_m \\ \lambda a_{m-1} &= (m-1) a_{m-1} \\ &\cdots \\ \lambda a_1 &= a_1 \\ \lambda a_0 &= 0 \\ \end{align*}$$

If $\lambda = 0$, then $a_m = \cdots = a_1 = 0$, so the eigenvectors are $p(x) = a \neq 0$.

If $\lambda \neq 0$, then we can have $\lambda$ to be any positive integer $n$, and then eigenvectors will be $ax^n$ for $a \neq 0$.

$\square$

5A.11

Suppose $𝑉$ is finite-dimensional, $𝑇 ∈ ℒ(𝑉)$, and $𝛼 ∈ 𝐅$. Prove that there exists $𝛿 > 0$ such that $𝑇− \lambda 𝐼$ is invertible for all $\lambda ∈ 𝐅$ such that $0 < |\alpha − \lambda| < \delta$.

Proof:

Note $𝐅$ is either $\mathbf{R}$ or $\mathbf{C}$.

Since $V$ is finite-dimensional, then from "5.12 operator cannot have more eigenvalues than dimension of vector space", $T$ has at most $\dim V$ eigenvalues.

Then we can find $\delta$ such that $|\alpha - \lambda_i| \geq \delta$ where $\lambda_i$ is an eigenvalue.

So if $0 < |\alpha − \lambda| < \delta$, then $\lambda$ is not an eigenvalue.

Then from "5.7 equivalent conditions to be an eigenvalue" we know $𝑇− \lambda 𝐼$ is invertible.

$\square$

5A.12

Suppose $𝑉 = 𝑈 ⊕ 𝑊$, where $𝑈$ and $𝑊$ are nonzero subspaces of $𝑉$. Define $𝑃 ∈ ℒ(𝑉)$ by $𝑃(𝑢 + 𝑤) = 𝑢$ for each $𝑢 ∈ 𝑈$ and each $𝑤 ∈ 𝑊$. Find all eigenvalues and eigenvectors of $𝑃$ .

Proof:

Assume $\lambda$ is an eigenvalue, then

$$\begin{align*} P(u+w) &= \lambda (u+w) = u \\ & \Rightarrow \\ (\lambda - 1) u + \lambda w &= 0 \\ & \Rightarrow \\ (\lambda - 1) u &= \lambda w = 0 & \text{because } 𝑉 = 𝑈 ⊕ 𝑊 \\ \end{align*}$$

If $\lambda \neq 1$ and $\lambda \neq 0$, then we must have $u = w = 0$, then $u+w$ is not an eigenvector, as a result, $\lambda$ is not an eigenvalue.

If $\lambda = 1$, then every nonzero vector in $U$ is an eigenvector. Indeed, given $u \in U, u \neq 0$, then $P(u) = u = 1 \cdot u$.

If $\lambda = 0$, then every nonzero vector in $W$ is an eigenvector. Indeed, given $w \in W, w \neq 0$, then $P(w) = 0 = 0 \cdot w$.

$\square$

5A.13

Suppose $𝑇 ∈ ℒ(𝑉)$. Suppose $𝑆 ∈ ℒ(𝑉)$ is invertible.

(a) Prove that $𝑇$ and $𝑆^{−1}𝑇𝑆$ have the same eigenvalues.

Proof:

If $\lambda$ is an eigenvalue of $T$, then we can find $v \neq 0$ such that $T(v) = \lambda v$. Since $S$ is invertible, we can find $u$ such that $u = S^{-1}(v)$. Since $S^{-1}$ is invertible and $v \neq 0$, then $u \neq 0$.

$$\begin{align*} 𝑆^{−1}𝑇𝑆(u) &= 𝑆^{−1}𝑇𝑆(S^{-1}(v)) \\ &= 𝑆^{−1}𝑇(v) \\ &= 𝑆^{−1}(\lambda v) \\ &= \lambda 𝑆^{−1}(v) \\ &= \lambda u \\ \end{align*}$$

On the other hand, if $\lambda$ is an eigenvalue of $𝑆^{−1}𝑇𝑆$, and assume $v \neq 0$ such that $𝑆^{−1}𝑇𝑆(v) = \lambda v$

Then we have

$$\begin{align*} S(𝑆^{−1}𝑇𝑆(v)) &= S(\lambda v) \\ TS(v) &= \lambda S(v) \\ \end{align*}$$

Since $v \neq 0$ and $S$ is invertible, then $S(v) \neq 0$. So $S(v)$ is an eigenvector of $T$.

$\square$

(b) What is the relationship between the eigenvectors of $𝑇$ and the eigenvectors of $𝑆^{−1}𝑇𝑆$?

Proof:

From (a), if $v$ is an eigenvector of $T$, then $S^{-1}(v)$ is an eigenvector of $𝑆^{−1}𝑇𝑆$.

Conversely, if $v$ is an eigenvector of $𝑆^{−1}𝑇𝑆$, then $S(v)$ is an eigenvector of $T$.

$\square$

5A.14

Give an example of an operator on $𝐑^4$ that has no (real) eigenvalues.

Solution:

Consider this operator (borrowed from 5.9(b))

$$T : (a, b, c, d) \rightarrow (-b, a, -d, c)$$

Assume $k$ is an real eigenvalue, then

$$T((a, b, c, d)) = k (a, b, c, d) = (-b, a, -d, c)$$

Then

$$\begin{align*} ka &= -b \\ kb &= a \\ kc &= -d \\ kd &= c \\ & \Rightarrow \\ k^2 b = -b &, k^2 d = -d \end{align*}$$

If $k \neq 0$, then $(k^2+1) > 0$ so $(k^2+1)b = 0$ means $b = 0$. Similarly $d = 0$, then $a = c = 0$, so $k$ is not an eigenvalue.

If $k = 0$, then $a = c = 0$, then $b = d = 0$, so $k$ is still not an eigenvalue.

$\square$

5A.15

Suppose $𝑉$ is finite-dimensional, $𝑇 ∈ ℒ(𝑉)$, and $\lambda ∈ 𝐅$. Show that $\lambda$ is an eigenvalue of $𝑇$ if and only if $\lambda$ is an eigenvalue of the dual operator $𝑇' ∈ ℒ(𝑉')$.

Proof

Given $v_1, \cdots, v_m$ be a basis of $V$ and $\varphi_1, \cdots, \varphi_m$ be its dual basis.

Let $A$ be the matrix of $T$ under basis $v_1, \cdots, v_m$ and $A'$ be the matrix of $T'$ under basis $\varphi_1, \cdots, \varphi_m$.

From "3.132 matrix of $𝑇'$ is transpose of matrix of $𝑇$", we know $A' = A^t$.

Let $E$ be the identity matrix, then the matrix for operator $T - \lambda I$ is $A - \lambda E$. Similarly, the matrix for operator $T' - \lambda I'$, is $A' - \lambda E$.

Note that $(A - \lambda E)^t = A^t - \lambda E^t = A' - \lambda E$.

Then from "5.7 equivalent conditions to be an eigenvalue" we have the following equivalent relation.

$$\begin{align*} & \lambda \text{ is an eigenvalue of } 𝑇 \\ & \Leftrightarrow \\ & T - \lambda I \text{ is not invertible} \\ & \Leftrightarrow \\ & \text{dim range } T - \lambda I < \dim V \\ & \Leftrightarrow \\ & \text{column rank of } A - \lambda E < \dim V \\ & \Leftrightarrow \\ & \text{row rank of } A - \lambda E < \dim V \\ & \Leftrightarrow \\ & \text{column rank of } A' - \lambda E < \dim V & \text{because } (A - \lambda E)^t = A' - \lambda E\\ & \Leftrightarrow \\ & \text{dim range } T' - \lambda I' < \dim V \\ & \Leftrightarrow \\ & T' - \lambda I' \text{ is not invertible} \\ & \Leftrightarrow \\ & \lambda \text{ is an eigenvalue of the dual operator } T' \end{align*}$$

$\square$

The following is my original thought, but it seems not working.

$\Rightarrow$

Assume $\lambda$ is an eigenvalue of $𝑇$, and let $v_1$ be an eigenvector so $T(v_1) = \lambda v_1 = w_1$.

Note $w_1 \in \text{range } T$, we can extend it to a basis of $\text{range } T$ by adding $w_2, \cdots, w_m$

Then we can find a group of linear independent vectors $v_1, \cdots, v_m$ such that $T(v_i) = w_i$.

Then we can find a basis of $\text{null } T$: $u_1, \cdots, u_n$.

Together $v_1, \cdots, v_m, u_1, \cdots, u_n$ forms a basis of $V$.

Furthermore, we can extend $w_1, \cdots, w_m$ to a basis of $V$ by adding vectors $x_1, \cdots, x_n$

Then consider the dual basis $\varphi_1, \cdots, \varphi_{m+n}$ of $w_1, \cdots, w_m, x_1, \cdots, x_n$.

Let

$$ $$\begin{align*} v &= a_1 v_1 + \cdots + a_m v_m + b_1 u_1 + \cdots + b_n u_n \\ &= a_1 v_1 + \cdots + a_m v_m + b_1 u_1 + \cdots + b_n u_n \end{align*}$$

$$

Then

$$\begin{align*} T'(\varphi_1)(v) &= \varphi_1(T(v)) \\ &= \varphi_1(a_1 Tv_1 + \cdots + a_m Tv_m) \\ &= \varphi_1(a_1 w_1 + \cdots + a_m w_m) \\ &= a_1 w_1 \end{align*}$$

$\square$

5A.16

Suppose $𝑣_1, …, 𝑣_𝑛$ is a basis of $𝑉$ and $𝑇 ∈ ℒ(𝑉)$. Prove that if $\lambda$ is an eigenvalue of $𝑇$, then

$$|\lambda| ≤ 𝑛 \max \{∣ℳ(𝑇)_{𝑗, 𝑘}∣ : 1 ≤ 𝑗, 𝑘 ≤ 𝑛 \}$$

, where $ℳ(𝑇)_{𝑗, 𝑘}$ denotes the entry in row $𝑗$, column $𝑘$ of the matrix of $𝑇$ with respect to the basis $𝑣_1, …, 𝑣_𝑛$.

Proof:

Let $A = ℳ(𝑇)$, and $M = \max \{∣ℳ(𝑇)_{𝑗, 𝑘}∣ : 1 ≤ 𝑗, 𝑘 ≤ 𝑛 \}$ .

Assume $v$ is an eigenvector and $T(v) = \lambda v$.

Also assume

$$v = a_1 v_1 + \cdots + a_n v_n$$

So we have

$$\begin{align*} \lambda v &= T(v) \\ &= a_1 T(v_1) + \cdots + a_n T(v_n) \end{align*}$$

On the other hand

$$\lambda v = \lambda a_1 v_1 + \cdots + \lambda a_n v_n$$

So we have

$$a_1 T(v_1) + \cdots + a_n T(v_n) = \lambda a_1 v_1 + \cdots + \lambda a_n v_n$$ Compare the coefficients of $v_1, \cdots, v_n$ on both sides, we have

$$\begin{align*} \lambda a_1 &= a_1 A_{1,1} + a_2 A_{1,2} + \cdots + a_n A_{1, n} \\ \lambda a_2 &= a_1 A_{2,1} + a_2 A_{2,2} + \cdots + a_n A_{2, n} \\ \cdots \\ \lambda a_n &= a_1 A_{n,1} + a_2 A_{n,2} + \cdots + a_n A_{n, n} \\ \end{align*}$$

Assume $|a_1| \geq |a_i|$, then we must have $|a_1| > 0$. This is because $v$ as an eigenvector cannot be $0$.

$$\begin{align*} |\lambda a_1| &\leq |a_1 A_{1,1}| + |a_2 A_{1,2}| + \cdots + |a_n A_{1, n}| \\ &\Rightarrow \\ |\lambda | &\leq |a_1 A_{1,1}|/|a_1| + |a_2 A_{1,2}|/|a_1| + \cdots + |a_n A_{1, n}|/|a_1| \\ &\leq |A_{1,1}| + |A_{1,2}| + \cdots + |A_{1, n}|\\ & \leq M + M + \cdots + M \\ & \leq nM \\ \end{align*}$$

$\square$

5A.17

Suppose $𝐅 = 𝐑, 𝑇 ∈ ℒ(𝑉)$, and $\lambda ∈ 𝐑$. Prove that $\lambda$ is an eigenvalue of 𝑇 if and only if $\lambda$ is an eigenvalue of the complexification $𝑇_𝐂$.

The complexification is defined in Exercise 1B.8 See Exercise 33 in Section 3B for the definition of $𝑇_𝐂$.

Proof:

$\Rightarrow$

Assume $\lambda$ is an eigenvalue of $T$ and $v$ is an eigenvector, then $T(v) = \lambda v$.

Now consider

$$\begin{align*} T_c (v + iv) &= T(v) + i T(v) \\ &= \lambda v + i (\lambda v) \\ &= \lambda (v + iv) \\ \end{align*}$$

Since $v \neq 0$, then $v + iv \neq 0$. Then $v + iv$ is an eigenvector and $\lambda$ is an eigenvalue.

$\Leftarrow$

Assume $\lambda$ is an eigenvalue of $T_C$ and $u + iv \neq 0$ is an eigenvector. Then $u \neq 0$ or $v \neq 0$.

$$\begin{align*} T_C(u+iv) &= \lambda (u+iv) \\ &\Rightarrow \\ T(u) + i T(v) &= \lambda u + i \lambda v \\ &\Rightarrow \\ T(u) = \lambda u &, T(v) = \lambda v \\ \end{align*}$$

So one of $u$ or $v$ is an eigenvector for sure. And $\lambda$ is an eigenvalue.

$\square$

5A.18

Suppose $𝐅 = 𝐑, 𝑇 ∈ ℒ(𝑉)$, and $\lambda ∈ 𝐂$. Prove that $\lambda$ is an eigenvalue of the complexification $𝑇_𝐂$ if and only if $\overline{\lambda}$ is an eigenvalue of $𝑇_𝐂$.

Proof:

$\Rightarrow$

$\lambda = a + bi$ is an eigenvalue of the complexification $𝑇_𝐂$, then there is a $u+iv \neq 0$ such that

So we have

$$\begin{align*} T_C(u+iv) &= T(u) + iT(v) \\ &= \lambda (u+iv) \\ &= (a+bi) (u+iv) \\ &= (au - bv) + i(bu+av) \\ &\Rightarrow \\ T(u) = (au - bv) &, T(v) = bu+av \\ \end{align*}$$

Now consider

$$\begin{align*} T_C(u-iv) &= T(u) - iT(v) \\ &= (au - bv) - i(bu+av) \\ &= (a-ib)(u-iv) \\ \end{align*}$$

So $\overline{\lambda} = a-bi$ is also an eigenvalue of $T_C$.

The other direction is similar.

$\square$

5A.19

Show that the forward shift operator $𝑇 ∈ ℒ(𝐅^∞)$ defined by

$$𝑇(𝑧_1, 𝑧_2, … ) = (0, 𝑧_1, 𝑧_2, … )$$

has no eigenvalues.

Proof:

We use contradition.

Assume $\lambda$ is an eigenvalue and $(𝑧_1, 𝑧_2, … )$ is an eigenvector So we can find $z_m \neq 0$.

Also

$$\begin{align*} \lambda z_1 &= 0 \\ \lambda z_2 &= z_1 \\ &\cdots \\ \lambda z_{m} &= z_{m-1} \\ \lambda z_{m+1} &= z_{m} \\ \end{align*}$$

If $\lambda = 0$, then $\lambda z_{m+1} = 0 \neq z_m$, so we have contradition.

If $\lambda \neq 0$, then since $\lambda z_1 = 0$, we have $z_1 = 0$, similarly, $z_2 = \cdots = z_m = 0$, we reach a contradition.

So $T$ has no eigenvalues.

$\square$

5A.20

Define the backward shift operator $𝑆 ∈ ℒ(𝐅^∞)$ by

$$𝑆(𝑧_1, 𝑧_2, 𝑧_3, … ) = (𝑧_2, 𝑧_3, … ).$$

(a) Show that every element of $𝐅$ is an eigenvalue of $𝑆$.

Proof:

Let $\lambda \in F$.

If $\lambda = 0$, then $(1, 0, 0, \cdots )$ is an eigenvector.

If $\lambda \neq 0$, then $(1, \lambda, \lambda ^2, \cdots )$ is an eigenvector.

$\square$

(b) Find all eigenvectors of $𝑆$.

Solution:

From part (a), the eigenvectors are $(a, 0, 0, \cdots )$ with $a \neq 0$ and $(1, \lambda, \lambda ^2, \cdots )$.

$\square$

5A.21

Suppose $𝑇 ∈ ℒ(𝑉)$ is invertible.

(a) Suppose $\lambda ∈ 𝐅$ with $\lambda ≠ 0$. Prove that $\lambda$ is an eigenvalue of $𝑇$ if and only if $1/\lambda$ is an eigenvalue of $𝑇^{−1}$.

Proof:

$\Rightarrow$

$$\begin{align*} &\lambda \text{ is an eigenvalue } \\ &\Rightarrow \\ \text{We can find } u &\neq 0, \text{ such that } T(u) = \lambda u \\ &\Rightarrow \\ u = T^{-1}(T(u)) &= T^{-1}(\lambda u) = \lambda T^{-1}(u)\\ &\Rightarrow \\ T^{-1}(u) &= 1/\lambda u & \text{because } \lambda \neq 0 \end{align*}$$

Since $u \neq 0$, then $1/\lambda$ an eigenvalue of $𝑇^{−1}$.

$\Leftarrow$

The reverse side is similar.

$\square$

(b) Prove that $𝑇$ and $𝑇^{−1}$ have the same eigenvectors.

Proof:

The proof in (a) already showed if $u$ is an eigenvector of $T$, it's also a eigenvector of $𝑇^{−1}$ and vice versa.

$\square$

5A.22

Suppose $𝑇 ∈ ℒ(𝑉)$ and there exist nonzero vectors $𝑢$ and $𝑤$ in $𝑉$ such that

$$Tu = 3w \text{ and } Tw = 3u$$

Prove that 3 or −3 is an eigenvalue of $𝑇$.

Proof:

If $u = -w$, then $Tu = 3w = -3u$, since $u \neq 0$, then $-3$ is an eigenvalue.

If $u \neq -w$, then consider

$$T(u+w) = Tu + Tw = 3w + 3u = 3 (u + w)$$

Since $u \neq -w$, then $u+w \neq 0$. So $3$ is an eigenvalue.

$\square$

5A.23

Suppose $𝑉$ is finite-dimensional and $𝑆, 𝑇 ∈ ℒ(𝑉)$. Prove that $𝑆𝑇$ and $𝑇𝑆$ have the same eigenvalues.

Proof:

If $\lambda$ is an eigenvalue of $ST$ and $u \neq 0$ is an eigenvector such that $ST(u) = \lambda u$.

Then we have $TS(T(u)) = T(ST(u)) = T(\lambda u) = \lambda T(u)$.

If $T(u) \neq 0$, then $\lambda$ is an eigenvalue of $TS$.

If $T(u) = 0$, then $\lambda = 0$, then we need to show $\text{null } TS \neq \{0\}$.

If $S$ is invertible, then we can find $v$ such that $S(v) = u$. Note $v$ can not be $0$ since $S(0) = 0 \neq u$. Then $TS(v) = T(u) = 0$. So $v \in \text{null } TS$.

If $S$ is not invertible, then we can find $v \neq 0$, such that $S(v) = 0$. Then $TS(v) = T(S(v)) = T(0) = 0$, so $v \in \text{null } TS$.

In any case, $\text{null } TS \neq \{0\}$. And $0$ is also an eigenvalue of $TS$.

The reverse side is similar.

$\square$

5A.24

Suppose $𝐴$ is an $𝑛$-by-$𝑛$ matrix with entries in $𝐅$. Define $𝑇 ∈ ℒ(𝐅^𝑛)$ by $𝑇𝑥 = 𝐴𝑥$, where elements of $𝐅^𝑛$ are thought of as $𝑛$-by-$1$ column vectors.

(a) Suppose the sum of the entries in each row of $𝐴$ equals $1$. Prove that $1$ is an eigenvalue of $𝑇$.

Proof:

Consider the vector

$$\begin{align*} v &= e_1 + \cdots + e_n \neq 0\\ &\Rightarrow \\ T(v) &= T(e_1) + \cdots + T(e_n) \\ &= Ae_1 + \cdots + Ae_n \\ &= (\sum_{i = 1}^{n} a_{1i}) e_1 + \cdots + (\sum_{i = 1}^{n} a_{ni}) e_n \\ &= e_1 + \cdots + e_n \\ &= v \end{align*}$$

So $T(v) = v$ and $1$ is an eigenvalue of $𝑇$.

$\square$

(b) Suppose the sum of the entries in each column of $𝐴$ equals $1$. Prove that $1$ is an eigenvalue of $𝑇$.

Proof:

Consider $T'$ is the dual map of $T$ in $(F^n)'$. Then its matrix under dual basis is $A^T$. Then the sum of the entries in each row of $A^T$ equals $1$. Then we can apply part (a) and know $1$ is an eigenvalue of $T'$.

No we use the Exercise 5A.15 in this section which states $𝜆$ is an eigenvalue of $𝑇$ if and only if $𝜆$ is an eigenvalue of the dual operator $𝑇' ∈ ℒ(𝑉')$. And then we know $1$ is an eigenvalue of $T$.

$\square$

5A.25

Suppose $𝑇 ∈ ℒ(𝑉)$ and $𝑢, 𝑤$ are eigenvectors of $𝑇$ such that $𝑢 + 𝑤$ is also an eigenvector of $𝑇$. Prove that $𝑢$ and $𝑤$ are eigenvectors of $𝑇$ corresponding to the same eigenvalue.

Proof:

Assume

$$\begin{align*} T(u) &= \lambda_1 u \\ T(w) &= \lambda_2 w \\ T(u+w) &= \lambda_3 (u + w) \\ \end{align*}$$

Then

$$\begin{align*} \lambda_3 (u + w) & = T(u+w) \\ &= T(u) + T(w) \\ &= \lambda_1 u + \lambda_2 w \\ & \Rightarrow \\ (\lambda_3 - \lambda_1) u + (\lambda_3 - \lambda_2) w &= 0 \end{align*}$$

If $\lambda_1 \neq \lambda_2$, then $u$ and $w$ is independent, so $\lambda_3 - \lambda_1 = \lambda_3 - \lambda_2 = 0$.

Then $\lambda_1 = \lambda_2 = \lambda_3$, we have an contradition.

So $\lambda_1 = \lambda_2$.

$\square$

5A.26

Suppose $𝑇 ∈ ℒ(𝑉)$ is such that every nonzero vector in $𝑉$ is an eigenvector of $𝑇$. Prove that $𝑇$ is a scalar multiple of the identity operator.

Proof:

Consider $v_1 \neq 0, v_2 \neq 0$, and their eigenvalues are $\lambda_1, \lambda_2$.

If $v_1 = -v_2$, then $T(v_1+v_2) = 0$, then $\lambda_1 v_1+ \lambda_2 v_2 = 0$. Then $(\lambda_1 - \lambda_2) v_1 = 0$.

Since $v_1 \neq 0$ then $\lambda_1 - \lambda_2 = 0$, i.e. $\lambda_1 = \lambda_2$.

If $v_1 \neq -v_2$, then $v_1 + v_2 \neq 0$ is also an eigenvector of $T$. From the last Exercise, we know $\lambda_1 = \lambda_2$.

So for any $v$, $T(v) = \lambda_1 v$. Then $𝑇$ is a scalar multiple of the identity operator.

$\square$

5A.27

Suppose that $𝑉$ is finite-dimensional and $𝑘 ∈ {1, …, \dim 𝑉 − 1}$. Suppose $𝑇 ∈ ℒ(𝑉)$ is such that every subspace of $𝑉$ of dimension $𝑘$ is invariant under $𝑇$. Prove that $𝑇$ is a scalar multiple of the identity operator.

Proof:

Let $\dim V = n$. Consider $v_1 \in V, v_1 \neq 0$, we can extend $v_1$ to a basis of $V$ by adding $v_2, \cdots, v_n$.

Assume

$$T(v_1) = a_1 v_1 + \cdots + a_n v_n$$

If $a_n \neq 0$, then let $U = \text{span}(𝑣_1, \cdots, 𝑣_k)$. $U$ is an invariant of $T$.

So $T(v_1) = a_1 v_1 + \cdots + a_k v_k$. Since $T(v_1)$ can be represented by $v_1, \cdots, v_n$ uniquely, we have $a_n = 0$.

Similarly we can show $a_2 = \cdots = a_n = 0$. So $v_1$ is an eigenvector.

That means every vector $v$ is an eigenvector of $T$.

From last Exercise, we know $𝑇$ is a scalar multiple of the identity operator.

$\square$

5A.28

Suppose $𝑉$ is finite-dimensional and $𝑇 ∈ ℒ(𝑉)$. Prove that $𝑇$ has at most $1 + \text{dim range } T$ distinct eigenvalues.

Proof:

We prove by contradition. Assume there are more than $1 + \text{dim range } T$, then there are a least $1 + \text{dim range } T$ nonzero eigenvalues.

Let $m = \text{dim range } T$, we can find $m+1$ eigenvectors with different eigenvalues, i.e.

$$\begin{align*} T(v_1) = \lambda_1 v_1, &\cdots, T(v_{m+1}) = \lambda_{m+1} v_{m+1} \\ & \Rightarrow \\ \lambda_1 v_1, &\cdots , \lambda_{m+1} v_{m+1} \in \text{range } T \\ & \Rightarrow \\ v_1, &\cdots , v_{m+1} \in \text{range } T & \text{Because } \lambda_i \neq 0 \\ & \Rightarrow \\ \text{dim range } T &\geq m+1 & \text{Because } v_i \text{ are independent} \\ \end{align*}$$

So we reached an contradition.

$\square$

5A.29

Suppose $𝑇 ∈ ℒ(𝐑^3)$ and $−4$, $5$, and $\sqrt{7}$ are eigenvalues of $𝑇$. Prove that there exists $𝑥 ∈ 𝐑^3$ such that $𝑇𝑥 − 9𝑥 = (−4, 5, \sqrt{7})$.

Proof:

$−4$, $5$, and $\sqrt{7}$ are eigenvalues of $𝑇$, and from 5.12, we know operator cannot have more eigenvalues than dimension of vector space. Since $\dim \mathbb{R}^3 = 3$, then $T$ cannot have other eigenvalues, so $9$ is not an eigenvalue of $T$.

Since $9$ is not an eigenvalue of $T$, then $T - 9I$ is invertible. Then we can find $x$ such that $(T - 9I) x = (−4, 5, \sqrt{7})$.

$\square$

5A.30

Suppose $𝑇 ∈ ℒ(𝑉)$ and $(𝑇 − 2𝐼)(𝑇 − 3𝐼)(𝑇 − 4𝐼) = 0$. Suppose 𝜆 is an eigenvalue of $𝑇$. Prove that 𝜆 = 2 or 𝜆 = 3 or 𝜆 = 4.

Proof:

We prove by contradition. Assume $k$ is an eigenvalue of $𝑇$, where $k \neq 2, 3, 4$. If $v \neq 0$ is an eigenvector and $T(v) = k v$, then

$$\begin{align*} (𝑇 − 2𝐼)(𝑇 − 3𝐼)(𝑇 − 4𝐼)(v) &= (k-4)(𝑇 − 2𝐼)(𝑇 − 3𝐼)(v) \\ &= (k-4) (k-3) (𝑇 − 2𝐼) (v) \\ &= (k-4) (k-3) (k-2) (v) \\ \end{align*}$$

Since $k \neq 2, 3, 4, v \neq 0$, then $(k-4) (k-3) (k-2) (v) \neq 0$.

But this contradicts to the fact $(𝑇 − 2𝐼)(𝑇 − 3𝐼)(𝑇 − 4𝐼) = 0$.

So $k = 2$ or $k = 3$ or $k = 4$.

$\square$

5A.31

Give an example of $𝑇 ∈ ℒ(𝐑^2)$ such that $𝑇^4 = −𝐼$.

Proof:

Consider the matrix of $T$ is

$$\mathcal{M}(T) = \begin{bmatrix} \sqrt[]{2}/2 & -\sqrt[]{2}/2 \\ \sqrt[]{2}/2 & \sqrt[]{2}/2 \\ \end{bmatrix}$$

Then since $\mathcal{M}(T)^4 = -I$ then $T^4 = -I$.

$\square$

5A.32

Suppose $𝑇 ∈ ℒ(𝑉)$ has no eigenvalues and $𝑇^4 = 𝐼$. Prove that $𝑇^2 = −𝐼$.

Proof:

$$\begin{align*} 0&=T^4 - I \\ &= (T^2 + I)(T + I)(T - I) \end{align*}$$

Since $T$ has no eigenvalues then $T-I, T+I$ are invertible. So is $(T + I)(T - I)$.

So for any $u \in V$, we can find $v \in V$, such that

$$u = (T + I)(T - I)(v)$$

So $(T^2 + I)(u) = (T^2 + I)(T + I)(T - I)(v)=0$.

So $T^2 + I = 0$, then $𝑇^2 = −𝐼$.

$\square$

5A.33

Suppose $𝑇 ∈ ℒ(𝑉)$ and $𝑚$ is a positive integer.

(a) Prove that $𝑇$ is injective if and only if $𝑇^𝑚$ is injective.

Proof:

$\Rightarrow$

Assume $T$ is injective and if $T^m(u) = 0$, then $0 = T^m(u) = T(T^{m-1}(u))$, we have $T^{m-1}(u) = 0$. Similarly $T^{m-2}(u) = \cdots = T^2(u) = T(u) = 0$, so $u = 0$, That means $T^{m}$ is injective.

$\Leftarrow$

Assume $T(u) = 0$ then $T^{m}(u) = T^{m-1}(T(u))=0$, so $u = 0$, then $T$ is injective.

$\square$

(b) Prove that $𝑇$ is surjective if and only if $𝑇^𝑚$ is surjective.

Proof:

$\Rightarrow$

We prove by induction. First, $T$ is surjective. Assume $T^k$ is surjective, and given $w \in V$.

Since $T^k$ is surjective, we can find $v \in V$ such that $T^k(v) = w$. Since $T$ is also surjective, we can find $u \in V$ such that $T(u) = v$, then $T^{k+1}(u) = T^k(T(u)) = T^k(v) = w$.

So $T^{k+1}$ is also surjective. Then we finish the induction.

$\Leftarrow$

$T^m$ is surjective, and given $w \in V$. We can find $v \in V$ such that $T^{m}(v) = w$, then we have $T(T^{m-1}(v)) = w$.

$\square$

5A.34

Suppose $𝑉$ is finite-dimensional and $𝑣_1, …, 𝑣_𝑚 ∈ 𝑉$. Prove that the list $𝑣_1, …, 𝑣_𝑚$ is linearly independent if and only if there exists $𝑇 ∈ ℒ(𝑉)$ such that $𝑣_1, …, 𝑣_𝑚$ are eigenvectors of $𝑇$ corresponding to distinct eigenvalues.

Proof:

$\Leftarrow$

It is an immediate results from "5.11 linearly independent eigenvectors".

$\Rightarrow$

Since $V$ is finite-dimensional, and $𝑣_1, …, 𝑣_𝑚$ is linearly independent， we can extend it to a basis of $V$ by adding $w_1, \cdots, w_n$.

Let $T(v_i) = i v_i$ and $T(w_i) = 0$, then $T$ is a linear map and $𝑣_1, …, 𝑣_𝑚$ are eigenvectors of $𝑇$ corresponding to distinct eigenvalues

$\square$

5A.35

Suppose that $\lambda_{1}, …, \lambda_{𝑛}$ is a list of distinct real numbers. Prove that the list $𝑒^{\lambda_{1}𝑥} , …, 𝑒^{\lambda_{𝑛}𝑥}$ is linearly independent in the vector space of real-valued functions on 𝐑.

Hint: Let $𝑉 = \text{span}(𝑒^{\lambda_{1}𝑥} , …, 𝑒^{\lambda_{𝑛}𝑥})$, and define an operator $𝐷 ∈ ℒ(𝑉)$ by $𝐷 𝑓 = 𝑓'$. Find eigenvalues and eigenvectors of $𝐷$.

Proof:

Based on the hint, $Df = f'$ is a linear operator of $ℒ(𝑉)$. Also note from calculus, we know

$$D 𝑒^{\lambda_{1}𝑥} = \lambda_1 𝑒^{\lambda_{1}𝑥}$$

Then $𝑒^{\lambda_{1}𝑥} , …, 𝑒^{\lambda_{𝑛}𝑥}$ are eigenvectors with different eigenvalues.

From "5.11 linearly independent eigenvectors", we know $𝑒^{\lambda_{1}𝑥} , …, 𝑒^{\lambda_{𝑛}𝑥}$ is linearly independent.

$\square$

5A.36

Suppose that $\lambda_1, …, \lambda_𝑛$ is a list of distinct positive numbers. Prove that the list $\cos(\lambda_1𝑥), …, \cos(\lambda_𝑛𝑥)$ is linearly independent in the vector space of real-valued functions on 𝐑.

Proof:

We know that $Df = f''$ is a linear operator of $ℒ(𝑉)$.

Also note from calculus, we know

$$D \cos(\lambda_i 𝑥) = -\lambda_i^2 \cos \lambda_i 𝑥$$

Since $\lambda_i$ are distinct positive numbers, so $-\lambda_i^2$ are also distinct.

From "5.11 linearly independent eigenvectors", we know $\cos(\lambda_1𝑥), …, \cos(\lambda_𝑛𝑥)$ is linearly independent.

$\square$

5A.37

Suppose $𝑉$ is finite-dimensional and $𝑇 ∈ ℒ(𝑉)$. Define $𝒜 ∈ ℒ(ℒ(𝑉))$ by $𝒜(𝑆) = 𝑇𝑆$ for each $𝑆 ∈ ℒ(𝑉)$. Prove that the set of eigenvalues of $𝑇$ equals the set of eigenvalues of $𝒜$.

Proof:

Assume $\lambda$ is an eigenvalue of $T$. Then we can find $u \neq 0$, such that $T(u) = \lambda u$. We need to fine $S$ such that $𝒜(S) = \lambda S$.

Let $v_1, \cdots, v_m$ be a basis of $V$, let $S(v_i) = u$, then $S \in ℒ(𝑉)$.

$$\begin{align*} TS(a_1 v_1 + \cdots + a_m v_m) &= T(a_1 Sv_1 + \cdots + a_m Sv_m) \\ &= T(a_1 u + \cdots + a_m u) \\ &= a_1 Tu + \cdots + a_m Tu \\ &= a_1 \lambda u + \cdots + a_m \lambda u \\ &= \lambda (a_1 u + \cdots + a_m u) \\ &= \lambda (a_1 Sv_1 + \cdots + a_m Sv_m) \\ &= \lambda S(a_1 v_1 + \cdots + a_m v_m) \end{align*}$$

So $𝒜(S) = \lambda S$.

Now assume $\lambda$ is an eigenvalue of $𝒜$, then we can find $S \neq 0$, such that $TS = \lambda S$.

Since $S \neq 0$, we can find $v \in \text{range } S$ such that $v \neq 0$. Let $S(u) = v$, then

$$\begin{align*} T(v) &= TS(u) \\ &= \lambda S(u) \\ &= \lambda v \end{align*}$$

So $\lambda$ is an eigenvalue of $T$.

$\square$

5A.38

Suppose $𝑉$ is finite-dimensional, $𝑇 ∈ ℒ(𝑉)$, and $𝑈$ is a subspace of $𝑉$ invariant under $𝑇$. The quotient operator $𝑇/𝑈 ∈ ℒ(𝑉/𝑈)$ is defined by

$$(T/U)(v + U) = Tv + U$$

for each $v \in V$.

(a) Show that the definition of $𝑇/𝑈$ makes sense (which requires using the condition that $𝑈$ is invariant under 𝑇) and show that $𝑇/𝑈$ is an operator on $𝑉/𝑈$.

Proof:

Assume $u+U = v+U$, then from "3.101 two translates of a subspace are equal or disjoint", we know $u-v \in U$.

$$\begin{align*} (T/U)(v + U) &= Tv + U \\ (T/U)(u + U) &= Tu + U \\ (Tv+U) - (Tu+U) &= (Tv - Tu) + U \\ &= T(v-u) + U \\ &= U & \text{ since } u-v \in U \text{ which is invariant under } T\\ \end{align*}$$

So $T/U$ is well defined.

Next, we will prove $T/U$ is linear map.

First

$$\begin{align*} (T/U)((u+U)+(v+U)) &=(T/U)((u+v)+U) \\ &= T(u+v) + U \\ &= (T(u) + T(v)) + U \\ &= (T(u) + U) + (T(v) + U) \\ &= (T/U)(u+U) + (T/U)(v+U) \\ \end{align*}$$

Second

$$\begin{align*} (T/U)(a(u+U)) &= (T/U)(au + U) \\ &= T(au) + U \\ &= aT(u) + U \\ &= a(T(u) + U) \\ &= a((T/U)(u+U)) \\ \end{align*}$$

$\square$

(b) Show that each eigenvalue of $𝑇/𝑈$ is an eigenvalue of $𝑇$.

Proof:

Assume $k$ is an eigenvalue of $T/U$. Then $T/U - kI$ is not injective, so we can find $u \not\in U$ such that $(T/U-kI)(u+U) = (T-kI)(u) + U = 0+U$, that means we can find $v \in U$ such that $(T-kI)(u) = v \in U$.

Since $U$ is invariant under $T$, so we can consider $T$ limited on $U$, i.e. $T|_U$.

If $(T|_U-kI)$ is not invertible, we can find $w \in U$ and $w \neq 0$, such that $(T|_U-kI)(w) = 0$, since $T|_U(w) = T(w)$. Then $(T-kI)(w) = 0$. So $k$ is eigenvalue of $T$.

If $(T|_U-kI)$ is invertible, we can find $w \in U$ such that $(T|_U-kI)(w) = v$, then we also have $(T-kI)(w) = v$.

Since $u \not\in U, w \in U$, we have $u-w \neq 0$. And $(T-kI)(u-w) = (T-kI)(u) - (T-kI)(w) = v - v = 0$.

Therefore, $k$ is an eigenvalue of $T$.

$\square$

5A.39

Suppose $𝑉$ is finite-dimensional and $𝑇 ∈ ℒ(𝑉)$. Prove that $𝑇$ has an eigenvalue if and only if there exists a subspace of $𝑉$ of dimension $\dim 𝑉 − 1$ that is invariant under $𝑇$.

Proof:

$\Leftarrow$

Let $U$ be a invariant space under $T$ and $\dim U = \dim V - 1$.

Then consider the quotient space $V/U$ and the operator $T/U$ on it defined in the last exercise.

Since $\dim V/U = \dim V - \dim U = 1$, then $T/U$ has an eigenvalue. From the previous exercise part (b), each eigenvalue of $𝑇/𝑈$ is an eigenvalue of $𝑇$. So $T$ has an eigenvalue.

$\Rightarrow$

We will use induction. Let $m = \dim V$. First if $m = 1$, then $\{0\}$ is invariant and it's dimension is $0 = m - 1$. The statement is true.

Now assume the statement holds for all $m \leq n$, and we will prove when $m = n + 1$.

Assume $k$ is an eigenvalue, consider the linear mapping $T-kI$.

Consider $\text{range } T- kI$. From "5.18 null space and range of $𝑝(𝑇)$ are invariant under $𝑇$", we know $\text{range } T- kI$ is invariant under $T$.

Since $T-kI$ is not surjective, then $\text{dim range } T-kI < n+1$

If $\text{dim range } T-kI = n$, then we are done.

If $\text{dim range } T-kI < n$, then from fundamental theorem of linear maps (3.21), we know $\text{dim null } T-kI \geq 2$.

Then we can find $v_1, v_2 \in \text{null } T-kI$ which are linear independent.

Let $U = \text{span}(𝑣_1)$

Now we can use exercise 5A.38 and consider the quotient space $V/U$ and its operator $T/U$.

Note that $k$ is an eigenvalue of $T/U$ and $v_2+U$ is an eigenvector since

$$\begin{align*} (T/U)(v_2+U) &= T(v_2) + U \\ &= k v_2 + U \\ &= k (v_2 + U) \end{align*}$$

Since $\dim U = 1$, then $\dim V/U = \dim V - \dim U = (n+1)-1 = n$. And $T/U$ has an eigenvalue $k$, so we can apply the induction and get an invariant subspace $W'$ and $\dim W' = n-1$.

Now assume $w_1+U, \cdots, w_{n-1}+U$ is a basis of $W'$.

Then we can cosider the space $W = \text{span}(w_1, \cdots, w_{n-1}, v_1)$.

Since $W'$ is invariant under $T/U$, then we have

$$\begin{align*} T(w_i) + U &= (T/U)(w_i+U) \\ &= a_1 (w_1+U) + \cdots + a_{n-1} (w_{n-1}+U) \\ &= (a_1 w_1 + \cdots + a_{n-1} w_{n-1}) + U \end{align*}$$

And since $U = \text{span}(𝑣_1)$, it means

$$T(w_i) = a_1 w_1 + \cdots + a_{n-1} w_{n-1} + a_n v_1 \in W$$

Also $T(v_1) = k v_1 \in W$. So $W$ is invariant under $T$.

Now if

$$\begin{align*} a_1 w_1 + \cdots + a_{n-1} w_{n-1} + a_n v_1 &= 0 \\ &\Rightarrow \\ a_1 w_1 + \cdots + a_{n-1} w_{n-1} &= -a_n v_1 \\ &\Rightarrow \\ (a_1 w_1 + \cdots + a_{n-1} w_{n-1}) + U &= -a_n v_1 + U \\ &\Rightarrow \\ a_1 (w_1+U) + \cdots + a_{n-1} (w_{n-1}+U) &= 0 + U \\ &\Rightarrow \\ a_1 = \cdots = a_{n-1} &= 0 \\ &\Rightarrow \\ a_n &= 0 \text{ as well} \end{align*}$$

So $w_1, \cdots, w_{n-1}, v_1$ are linear independent.

The $\dim W = n$, this completes the induction.

$\square$

Here is the anwser from AI:

Let $k$ be an eigenvalue of $T$ and let $W = range(T − kI)$. By 5.18, $W$ is invariant under $T$. Since $T − kI$ is not surjective ($k$ is an eigenvalue, so $T − kI$ is not injective, and $V$ is finite-dimensional), $\dim W < \dim V$.

Case 1: $\dim W = \dim V − 1$. Then $W$ is the desired invariant subspace.

Case 2: $\dim W < \dim V − 1$.

Since $W$ is invariant under $T$, the quotient operator $T/W ∈ L(V/W)$ is well-defined. For any $v ∈ V$:

$$\begin{align*} (T/W)(v+W) &= Tv + W \\ &= kv + (T-kI)v + W \\ &= k(v + W) \end{align*}$$

So $(T/W) = kI$.

Let $Z = π⁻¹(Z')$ be the preimage of $Z'$ under the quotient map $π : V → V/W$.

Then $Z$ is invariant under $T$, for $z \in Z$,

$$\begin{align*} π(T(z)) &= T(z) + W \\ &= (T/W)(z+W) \\ &= k(z+W) \\ &= k π(z) \\ &\in Z' \\ &\Rightarrow \\ T(z) &\in Z \end{align*}$$

Then note $\text{null } \pi = W$ and $\text{range } \pi = Z'$

So $\dim Z = \dim W + \dim Z' = n-1$.

$\square$

5A.40

Suppose $𝑆, 𝑇 ∈ ℒ(𝑉)$ and $𝑆$ is invertible. Suppose $𝑝 ∈ 𝒫(𝐅)$ is a polynomial. Prove that

$$𝑝(𝑆𝑇𝑆^{−1}) = 𝑆𝑝(𝑇)𝑆^{−1}$$

Proof:

Note since $𝑆$ is invertible, then $SS^{-1} = I$.

So $(STS^{-1})^n = (STS^{-1}) \cdots (STS^{-1}) = ST^nS^{-1}$.

If $p(z) = a_0 + a_1 z + \cdots + a_{m} z^{m}$

Then

$$\begin{align*} 𝑝(𝑆𝑇𝑆^{−1}) &= a_0 + a_1 (𝑆𝑇𝑆^{−1}) + \cdots + a_{m} (𝑆𝑇𝑆^{−1})^{m} \\ &= a_0 SS^{-1} + a_1 (𝑆𝑇𝑆^{−1}) + \cdots + a_{m} (𝑆𝑇^{m}𝑆^{−1}) \\ &= S(a_0 + a_1 T + \cdots + a_{m} T^{m})S^{-1} \\ &= 𝑆𝑝(𝑇)𝑆^{−1} \end{align*}$$

$\square$

5A.41

Suppose $𝑇 ∈ ℒ(𝑉)$ and $𝑈$ is a subspace of $𝑉$ invariant under $𝑇$. Prove that $𝑈$ is invariant under $𝑝(𝑇)$ for every polynomial $𝑝 ∈ 𝒫(𝐅)$.

Proof:

We use the induction to show $U$ is invariant under $T^n$. Since $U$ is invariant under $T$, assume $U$ is invariant under $T^n$.

For any $u \in U$, since $T^n(u) \in U$, then $T^{n+1}(u) = T(T^n(u)) \in U$, so $U$ is invariant under $T^{n+1}$.

Then given any $u \in U$, $a_nT^{n}(u) \in U$.

$$\begin{align*} p(T)(u) &= (a_0 + a_1 T + \cdots + a_{m} T^{m})(u) \\ &= a_0(u) + a_1 T(u) + \cdots + a_{m} T^{m}(u) \\ & \in U \end{align*}$$

So $𝑈$ is invariant under $𝑝(𝑇)$ for every polynomial $𝑝 ∈ 𝒫(𝐅)$.

$\square$

5A.42

Define $𝑇 ∈ ℒ(𝐅^𝑛)$ by $𝑇(𝑥_1, 𝑥_2, 𝑥_3, …, 𝑥_𝑛) = (𝑥_1, 2𝑥_2, 3𝑥_3, …, 𝑛𝑥_𝑛)$.

(a) Find all eigenvalues and eigenvectors of 𝑇.

Proof:

For any $i$ such that $1 \leq i \leq n$, $i$ is an eigenvalue, because let $e_i$ be the ith item of standard basis of $𝐅^𝑛$.

$$T(e_i) = i \cdot e_i$$

Since $\dim 𝐅^𝑛 = n$, then $T$ can only have $n$ distinct eigenvalues. Then we found all eigenvalue.

If $v = a_1 e_1 + \cdots + a_n e_n$ is an eigenvector with eigenvalue of $i$, then

$$\begin{align*} i v &= T(v) \\ &= a_1 Te_1 + \cdots + a_n Te_n \\ &= 1 a_1 e_1 + \cdots + n a_n e_n \\ & \Rightarrow \\ (1-i)a_1 e_1 + \cdots + (n-i) a_n e_n &= 0 \\ & \Rightarrow \\ (k-i) a_k &= 0 \\ & \Rightarrow \\ a_k &= \begin{cases} 0 &\text{if } k \neq i \\ \text{anything } &\text{if } k = i\\ \end{cases} \end{align*}$$

so $v = a_i e_i$ with $a_i \neq 0$. That's all eigenvectors of $T$.

$\square$

(b) Find all subspaces of $𝐅^𝑛$ that are invariant under $𝑇$.

Solution:

Assume $U$ is invariant under $T$ and $v \in U$ and $v \neq 0$, let

$$v = a_1 e_1 + \cdots + a_n e_n$$

If $a_i \neq 0$, we want to show $e_i \in U$. If $a_j = 0$ when $j \neq i$, then $a_i^{-1}v = e_i \in U$, we are done.

Other we have some $a_j \neq 0$ with $j \neq i$.

Since $U$ is invariant subspace, then $T(\frac{1}{j} v) \in U$, so we have $v - T(\frac{1}{j}v) \in U$. We also have

$$\begin{align*} v - T(\frac{1}{j}v) &= (a_1 e_1 + \cdots + a_n e_n) - \frac{1}{j}(a_1 Te_1 + \cdots + a_n Te_n) \\ &= \sum \limits_{k=1}^{n} (1-k/j)a_k e_k \\ &= \sum \limits_{k=1}^{n} b_k e_k \end{align*}$$

So, for $v - T(\frac{1}{j}v)$, $b_i \neq 0$ but $b_j = 0$. For other $k \neq i, j$, if $a_k \neq 0$ then $b_k = (1-k/j)a_k \neq 0$. If $a_k = 0$, then $b_k = (1-k/j)a_k = 0$.

So the number of non zero coefficients is reduced by $1$. By continuing this process, we can show $e_i \in U$.

That means $\text{span}(e_i, \cdots, e_j) \subseteq U$, on the other hand $u = a_1 e_1 + \cdots + a_n e_n$. So $U \subseteq \text{span}(e_i, \cdots, e_j)$.

That means all subspaces of $𝐅^𝑛$ that are invariant under $𝑇$ are the form of $\text{span}(e_i, \cdots, e_j)$.

$\square$

5A.43

Suppose that $𝑉$ is finite-dimensional, $\dim 𝑉 > 1$, and $𝑇 ∈ ℒ(𝑉)$. Prove that $\{𝑝(𝑇) ∶ 𝑝 ∈ 𝒫(𝐅)\} ≠ ℒ(𝑉)$.

Proof:

Assume $\dim V = n$.

One strategy is that: since $\dim L(V) = n^2$, we should prove $\dim \{𝑝(𝑇) ∶ 𝑝 ∈ 𝒫(𝐅)\} < n^2$.

We will use the results from section about next section, i.e. section 5.3.

Assume $p(z)$ is the minimal polynomial of $T$. Then from 5.22 we know $\deg p \leq n$.

Given any $q(z)$, from 4.9 division algorithm for polynomials, we know we can find $s(z), r(z)$ such that $q(z) = s(z)p(z) + r(z)$, where $\deg r < n$.

That means $q(T) = s(T)p(T) + r(T)$, since $p(T) = 0$, then $q(T) = r(T)$.

Since $\deg r < n$, then we have

$$\begin{align*} \{𝑝(𝑇) ∶ 𝑝 ∈ 𝒫(𝐅)\} &= \text{span}(I, T, \cdots, T^{n-1}) \\ &\Rightarrow \\ \dim \{𝑝(𝑇) ∶ 𝑝 ∈ 𝒫(𝐅)\} &\leq n \end{align*}$$

Since $n > 1$, then $n < n^2$.

$\square$

Another strategy (not working so far) is we should find some $p$ such that $\text{range } p(T)$ or $\text{null } p(T)$ is not trivial invariant subspace.

We consider the $U = \text{null } T$. Since we know from exercise 5A.41, $U$ is invariant under $p(T)$, we want to find a operator $S$ such that $U$ is not invariant under $S$.

Case 1: $U = V$, then $T = 0$, then $p(T) = kI$. Let $v_1, v_2 \in V$ be linear independent, then let $S(v_1) = v_2, S(v_2) = v_1$, then $S \neq kI = P(T)$.

Case 2: $U \subset V$, we can find a basis of $U$: $u_1, \cdots, u_m$. We can also extend it to a basis of $V$ by adding $v_1, \cdots, v_n$. Let $S(u_i) = v_1$, then $U$ is not invariant under $S$.

$\square$

Section 5B The Minimal Polynomial

5B.1

Suppose $𝑇 ∈ ℒ(𝑉)$. Prove that $9$ is an eigenvalue of $𝑇^2$ if and only if $3$ or $−3$ is an eigenvalue of $𝑇$.

Proof:

$\Rightarrow$

Assume $u \neq 0$ is an eigenvector and $(T^2 - 9I)(u) = 0$. Then we have $(T-3I)(T+3I)(u) = 0$.

If $(T+3I)(u) = 0$, then $-3$ is an eigenvalue and $u$ is an eigenvector.

If $(T+3I)(u) \neq 0$, let $w = (T+3I)(u)$. We have $(T-3I)(w) = 0$. Then $+3$ is an eigenvalue and $w$ is an eigenvector.

$\Leftarrow$

If $3$ is an eigenvalue and $u \neq 0$ is an eigenvector such that $(T-3I)(u) = 0$, then

$$\begin{align*} (T^2 - 9I)(u) &= (T+3I)(T-3I)(u) \\ &= (T+3I)(0) \\ &= 0 \end{align*}$$

So $9$ is an eigenvalue of $𝑇^2$ and $u$ is an eigenvector. Similarly for the case when $−3$ is an eigenvalue of $𝑇$.

$\square$

5B.2

Suppose $𝑉$ is a complex vector space and $𝑇 ∈ ℒ(𝑉)$ has no eigenvalues. Prove that every subspace of $𝑉$ invariant under $𝑇$ is either $\{0\}$ or infinite- dimensional.

Proof:

Assume $U \subseteq V$ is subspace that is invariant under $T$. Then we can restrict $T$ under $U$ to get $T|_U$. Then $T|_U$ is an operator on $U$.

Assume $U \neq \{0\}$ is a finite-dimensional space.

Then from "5.19 existence of eigenvalues", it must have an eigenvalue $\lambda$ and an eigenvector of $u \neq 0$ such that $T|_U(u) = \lambda u$.

But since $u \in U$, then we also have $T(u) = T|_U(u) = \lambda u$. This means $T$ has an eigenvalue $\lambda$ and an eigenvector of $u \neq 0$.

Then we reached an contradition. So if $U \neq \{0\}$, then $U$ is infinite-dimensional.

$\square$

5B.3

Suppose $𝑛$ is a positive integer and $𝑇 ∈ ℒ(𝐅^𝑛)$ is defined by

$$T (x_1, \cdots, x_n) = (𝑥_1 +⋯ + 𝑥_𝑛, …, 𝑥_1 +⋯ + 𝑥_𝑛)$$

anwser the following:

(a) Find all eigenvalues and eigenvectors of $𝑇$.

Solution:

Assume $k$ is an eigenvalue and $(x_1, \cdots, x_n) \neq 0$ is an eigenvector,

Then

$$\begin{align*} T (x_1, \cdots, x_n) &= (𝑥_1 +⋯ + 𝑥_𝑛, …, 𝑥_1 +⋯ + 𝑥_𝑛) \\ &= k (x_1, \cdots, x_n) \end{align*}$$

Then we have

$$k x_1 = k x_2 = \cdots = k x_n = 𝑥_1 +⋯ + 𝑥_𝑛$$

If $k = 0$, then any $(x_1, \cdots, x_n) \neq 0$ satisfy $𝑥_1 +⋯ + 𝑥_𝑛 = 0$, is an eigenvector. So if $n = 1$, then $x_1 = 0$, then $0$ is not an eigenvalue.

If $k \neq 0$, then $x_1 = x_2 = \cdots = x_n = a \neq 0$ and $k = n$.

So in summary, there are 2 possible eigenvalues when $n > 1$:

1. $k = 0$, and $(x_1, \cdots, x_n) \neq 0$ satisfy $𝑥_1 +⋯ + 𝑥_𝑛 = 0$
2. $k = n$, and $(x_1, \cdots, x_n) = (a, \cdots, a)$ with $a \neq 0$.

if $n = 1$, then only case 2 is valid.

$\square$

(b) Find the minimal polynomial of 𝑇.

Solution:

We first assume $n > 1$.

Let $p(z)$ be the minimal polynomial of $𝑇$.

From "5.27 eigenvalues are the zeros of the minimal polynomial", we know

$$p(z) = z(z-n)q(z)$$

This means $p(z)$ is a multiple of $z(z-n)$.

Now we consider $(T-nI)T$ applied on $e_i$ where $e_i$ is one of the standard basis.

$$\begin{align*} (T-nI)T(e_i) &= (T-nI) (1, \cdots , 1) \\ &= (n, \cdots , n) - (n, \cdots , n) \\ &= 0 \end{align*}$$

Then $(T-nI)T(v) = 0$ for any $v \in F^n$. That means $(T-nI)T = 0$.

Then from "5.29 $𝑞(𝑇) = 0 ⟺ 𝑞$ is a polynomial multiple of the minimal polynomial", we know $z(z-n)$ is a multiple $p(z)$.

Since $z(z-n)$ is a monic polynomial, $p(z) \mid z(z-n)$ and $z(z-n) \mid p(z)$ we must have $p(z) = z(z-n)$.

Now we assume $n = 1$. In this case $T-I = 0$ because $(T-I)(v) = T(v) - v = v - v = 0$. so $p(z) \mid z - 1$. On the other hand $1$ is a zero of $p(z)$ so $z - 1 \mid p(z)$. So $p(z) = z - 1$.

$\square$

5B.4

Suppose $𝐅 = 𝐂, 𝑇 ∈ ℒ(𝑉), 𝑝 ∈ 𝒫(𝐂)$, and 𝛼 ∈ 𝐂. Prove that 𝛼 is an eigenvalue of 𝑝(𝑇) if and only if 𝛼 = 𝑝(𝜆) for some eigenvalue 𝜆 of 𝑇.

Proof:

$\Rightarrow$

Consider the polynomial $p(z) - \alpha \in 𝒫(𝐂)$. And consider $U = \text{null } p(T) - \alpha I$. Since $𝛼$ is an eigenvalue of $𝑝(𝑇)$, $U \neq \{0\}$.

From "5.18 null space and range of $𝑝(𝑇)$ are invariant under $𝑇$", we know $U$ is invariant under $T$. And then we can restrict $T$ on $U$ to get $T|_U$.

Let $q(z)$ be the minimal polynomial of $T|_U$. Since $p(T|_U) - \alpha$ is $0$ on $U$, from "5.29 $𝑞(𝑇) = 0 ⟺ 𝑞$ is a polynomial multiple of the minimal polynomial", we have $q \mid p(z) - \alpha$

Since $q(z) \in 𝒫(𝐂)$, from fundamental theorem of algebra, we can find one of its zero, say $\lambda$.

So $q(z) = (z-\lambda)s(z)$. Since $\deg s < \deg q$, then $s(z)$ is not a multiple of $q$, then $s(T|_U)$ is not $0$ on $U$ again from 5.29. We can find $v \in U$, such that $s(T|_U)(v) \neq 0$. Let $w = s(T|_U)(v)$, then we have $(T|_U - \lambda I)(w) = 0$.

Since $w \in U$, then we have $T(w) = T|_U(w)$, so $(T-\lambda I)(w) = (T|_U - \lambda I ) (w) = 0$, i.e. $T(w) = \lambda w$. So $\lambda$ is an eigenvalue of $T$.

Since $q \mid p(z) - \alpha$ and $\lambda$ is a zeor of $q$, we have $p(\lambda) - \alpha = 0$.

$\Leftarrow$

If $𝛼 = 𝑝(𝜆)$, and $T(u) = \lambda u$, then

$$\begin{align*} p(T)(u) &= (a_0 I + a_1 T + \cdots + a_{m} T^{m})(u) \\ &= a_0 I(u) + a_1 T(u) + \cdots + a_{m} T^{m}(u) \\ &= a_0 u + a_1 \lambda(u) + \cdots + a_{m} \lambda^{m}(u) \\ &= (a_0 + a_1 \lambda + \cdots + a_{m} \lambda^{m})(u) \\ &= \alpha u \end{align*}$$

So $𝛼$ is an eigenvalue of $𝑝(𝑇)$.

$\square$

5B.5

Give an example of an operator on $𝐑^2$ that shows the result in Exercise 4 does not hold if 𝐂 is replaced with 𝐑.

Solution:

Consider $T$ is an operator, it rotates a vector $(x, y)$ by 90 degree counterclock-wise.

So $T(1,0) = (0,1), T(0,1) = (-1,0)$.

So $T^2(1,0) = T(0,1) = (-1,0)$, then if we reuse the symbol in exercise 4, we have $p(z) = z^2, \alpha = -1$, then it's not possible to find a real number $\lambda$ such that $p(\lambda) = \lambda^2 = -1$.

$\square$

5B.6

Suppose $𝑇 ∈ ℒ(𝐅^2)$ is defined by $𝑇(𝑤, 𝑧) = (−𝑧, 𝑤)$. Find the minimal polynomial of $𝑇$.

Solution:

Consider

$$\begin{align*} T(1,0) &= (0, 1) \\ T^2(1,0) &= T(0, 1) \\ &= (-1, 0) \end{align*}$$

And

$$\begin{align*} T(0,1) &= (-1, 0) \\ T^2(0,1) &= T(-1, 0) \\ &= (0, -1) \end{align*}$$

That means $T^2(e_1) + e_1 = 0$ and $T^2(e_2) + e_2 = 0$.

So $T^2 + 1 = 0$. so $T^2 + 1$ is a multiple of minimal polynomial.

On the other hand, if $T+\lambda I$ is $0$, then

$$\begin{align*} (T+\lambda I)(e_1) &= 0 \\ & \Rightarrow \\ (0, 1) &= (-\lambda, 0) \end{align*}$$

This is no possible, so $T+\lambda I \neq 0$.

So $T^2 + 1$ is the minimal polynomial.

$\square$

5B.7

(a) Give an example of $𝑆, 𝑇 ∈ ℒ(𝐅^2)$ such that the minimal polynomial of $𝑆𝑇$ does not equal the minimal polynomial of $𝑇𝑆$.

Solution:

Consider $S(x,y) = (x, x)$ and $T(x,y)=(0,x)$.

Then $ST(x,y)=S(0,x)=(0,0)$ so $ST = 0$, then the minimal polynomial of $ST$ is $z$.

On the other hand,

$TS(x,y) = T(x,x) = (0,x)$, so $(TS)^2(1,0) = (TS)(0,1) = (0,0)$, and $(TS)^2(0,1) = (TS)(0,0) = (0,0)$, so the minimal polynomial of $TS$ is $z^2$.

$\square$

(b) Suppose $𝑉$ is finite-dimensional and $𝑆, 𝑇 ∈ ℒ(𝑉)$. Prove that if at least one of $𝑆, 𝑇$ is invertible, then the minimal polynomial of $𝑆𝑇$ equals the minimal polynomial of $𝑇𝑆$.

Hint: Show that if 𝑆 is invertible and 𝑝 ∈ 𝒫(𝐅), then $𝑝(𝑇𝑆) = 𝑆^{−1}𝑝(𝑆𝑇)𝑆$.

Proof:

Assume $S$ is invertible. $p$ is the minimal polynomial of $𝑆𝑇$. $q$ is the minimal polynomial of $TS$.

We will show $p(TS) = 0$ and $q(ST) = 0$.

Just use exercise 5A.40,

$$\begin{align*} p(TS) &= p(S^{-1}STS) \\ &= 𝑆^{−1}𝑝(𝑆𝑇)𝑆 \\ &= 𝑆^{−1} 0 S \\ &= 0 \end{align*}$$

Then use "5.29 $𝑞(𝑇) = 0 ⟺ 𝑞$ is a polynomial multiple of the minimal polynomial", we know $p$ is an multiple of $q$.

On the other hand,

$$\begin{align*} q(ST) &= q(STSS^{-1}) \\ &= S q(TS) S^{-1} \\ & = S 0 S^{-1} \\ & = 0 \end{align*}$$

So $q$ is a multiple of $p$.

In summary, $p = q$.

$\square$

5B.8

Suppose $𝑇 ∈ ℒ(𝐑^2)$ is the operator of counterclockwise rotation by $1^∘$. Find the minimal polynomial of $𝑇$.

Because $\dim 𝐑^2 = 2$, the degree of the minimal polynomial of $𝑇$ is at most $2$. Thus the minimal polynomial of $𝑇$ is not the tempting polynomial $𝑥^{180} + 1$, even though $𝑇^{180} = −𝐼$.

Solution:

We write the standard basis in polar form

$$e_1 = (1, 0) = (\cos 0^\circ, \sin 0^\circ) \\ e_2 = (0, 1) = (\cos 90^\circ , \sin 90^\circ) \\ $$

Then note

$$\begin{align*} T^2(e_1) &= (\cos 2^\circ , \sin 2^\circ) \\ T^2(e_1) + e_1 &= (\cos 2^\circ , \sin 2^\circ) + (\cos 0^\circ, \sin 0^\circ)\\ &= (2 \cos 1 ^ \circ \cos 1 ^ \circ, 2 \sin 1 ^\circ \cos 1 ^ \circ ) \\ &= 2 \cos 1^ \circ (\cos 1^\circ , \sin 1^\circ) \\ &= 2 \cos 1^ \circ T(e_1) \end{align*}$$

Similarly,

$$\begin{align*} T^2(e_2) + e_2 &= (\cos 90^\circ , \sin 90^\circ) + (\cos 92^\circ, \sin 92^\circ) \\ &= (2 \cos 91 ^ \circ \cos 1 ^ \circ, 2 \sin 91 ^\circ \cos 1 ^ \circ ) \\ &= 2 \cos 1^ \circ (\cos 91^\circ , \sin 91^\circ) \\ &= 2 \cos 1^ \circ T(e_2) \\ \end{align*}$$

So $T^2 - 2 \cos 1^ \circ T + I = 0$.

On the other hand, since $T$ is a rotation, $T$ does not have eigenvalue, so the minimal polynomial cannot be degree 1.

Therefore, the minimal polynomial is $p(z) = z^2 - 2 \cos 1^ \circ z + 1$.

$\square$

5B.9

Suppose $𝑇 ∈ ℒ(𝑉)$ is such that with respect to some basis of $𝑉$, all entries of the matrix of $𝑇$ are rational numbers. Explain why all coefficients of the minimal polynomial of $𝑇$ are rational numbers.

Proof: unsolved.

5B.10

Suppose $𝑉$ is finite-dimensional, $𝑇 ∈ ℒ(𝑉)$, and $𝑣 ∈ 𝑉$. Prove that

$$\text{span}(𝑣, Tv, \cdots, T^𝑚v) = \text{span}(𝑣, Tv, \cdots, T^{\dim V - 1} v)$$

for all integers $𝑚 ≥ \dim 𝑉 − 1$.

Proof:

Let $p$ be the minimal polynomial of $T$, from the polynomial division algorithm, we know given $z^m$, we can find $s(z), r(z)$ such that

$$z^m = p(z)s(z) + r(z)$$

with $\deg r < \deg p \leq \dim V$. So $\deg r \leq \dim V - 1$. Let $\deg r = n$.

So we have

$$\begin{align*} T^m &= p(T)s(T) + r(T) \\ & = 0 + r(T) \\ & = r(T) \\ & = a_0 I + a_1 T + \cdots + a_{n} T^{n} \end{align*}$$

So $T^m(v) = (a_0 I + a_1 T + \cdots + a_{n} T^{n})(v) \in \text{span}(𝑣, Tv, \cdots, T^{\dim V - 1} v).$ That means

$$\text{span}(𝑣, Tv, \cdots, T^𝑚v) \subseteq \text{span}(𝑣, Tv, \cdots, T^{\dim V - 1} v)$$

On the other hand, since $𝑚 ≥ \dim 𝑉 − 1$, we have

$$\text{span}(𝑣, Tv, \cdots, T^𝑚v) \supseteq \text{span}(𝑣, Tv, \cdots, T^{\dim V - 1} v)$$

So we have

$$\text{span}(𝑣, Tv, \cdots, T^𝑚v) = \text{span}(𝑣, Tv, \cdots, T^{\dim V - 1} v)$$

$\square$

5B.11

Suppose $𝑉$ is a two-dimensional vector space, $𝑇 ∈ ℒ(𝑉)$, and the matrix of $𝑇$ with respect to some basis of $𝑉$ is $$\begin{pmatrix} a & b \\ c & d \\ \end{pmatrix}$$

Prove that

(a) Show that $𝑇^2 − (𝑎 + 𝑑)𝑇 + (𝑎𝑑 − 𝑏𝑐)𝐼 = 0$.

Proof:

$$\begin{align*} T^2 &= \begin{pmatrix} a & b \\ c & d \\ \end{pmatrix}^2 \\ &= \begin{pmatrix} a & b \\ c & d \\ \end{pmatrix} \begin{pmatrix} a & b \\ c & d \\ \end{pmatrix} \\ &= \begin{pmatrix} a^2 + bc & ab + bd \\ ac + cd & bc + d^2 \\ \end{pmatrix} \\ \end{align*}$$

$$\begin{align*} (a+d)T &= (a+d) \begin{pmatrix} a & b \\ c & d \\ \end{pmatrix} \\ &= \begin{pmatrix} a^2 + ad & ab + bd \\ ac + cd & ad + d^2 \\ \end{pmatrix} \end{align*}$$

$$\begin{align*} (𝑎𝑑 − 𝑏𝑐)𝐼 &= \begin{pmatrix} ad-bc & 0 \\ 0 & ad-bc \\ \end{pmatrix} \end{align*}$$

So $𝑇^2 − (𝑎 + 𝑑)𝑇 + (𝑎𝑑 − 𝑏𝑐)𝐼 = 0$.

(b) Show that the minimal polynomial of $𝑇$ equals

$$\begin{cases} z-a &\text{if } b=c=0 \text{ and } a=d\\ z^2 - (a+d)z + (ad - bc) &\text{otherwise}\\ \end{cases}$$

Proof:

Assume the minimal polynomial of $T$ is $p$.

If $\deg T = 1$ then assume $T + \lambda I = 0$.

$$\begin{align*} 0 &= \begin{pmatrix} a & b \\ c & d \\ \end{pmatrix} + \lambda I \\ &= \begin{pmatrix} a + \lambda & b \\ c & d + \lambda \\ \end{pmatrix} \end{align*}$$

So $a = d = -\lambda$, $b = c = 0$.

If $\deg T = 2$, then from part (2) we know $𝑇^2 − (𝑎 + 𝑑)𝑇 + (𝑎𝑑 − 𝑏𝑐)𝐼 = 0$, since minimal polynomial is unique, then $z^2 - (a+d)z + (ad - bc)$ is the minimal polynomial.

$\square$

5B.12

Define $𝑇 ∈ ℒ(𝐅^𝑛)$ by $𝑇(𝑥_1, 𝑥_2, 𝑥_3, …, 𝑥_𝑛) = (𝑥_1, 2𝑥_2, 3𝑥_3, …, 𝑛𝑥_𝑛)$. Find the minimal polynomial of $𝑇$.

Proof:

Let $p$ be the minimal polynomial of $T$.

Let $e_1, \cdots, e_n$ be the standard basis of $F^n$. Then note $T(e_i) = i e_i$, then $i$ is an eigenvalue of $T$.

So from "5.27 eigenvalues are the zeros of the minimal polynomial", part (b) $p$ has the form of $(z-1)^{i_1}\cdots(z-n)^{i_n}$. with $i_1, \cdots, i_n \geq 1$.

On the other hand, $\deg p \leq n$, so $p = (z-1)\cdots(z-n)$.

$\square$

5B.13

Suppose $𝑇 ∈ ℒ(𝑉)$ and $𝑝 ∈ 𝒫(𝐅)$. Prove that there exists a unique $𝑟 ∈ 𝒫(𝐅)$ such that $𝑝(𝑇) = 𝑟(𝑇)$ and $\deg 𝑟$ is less than the degree of the minimal polynomial of $𝑇$.

Proof:

Let $q$ be the minimal polynomial of $T$, from the division algorithm of polynomials, we know we can find unique $s(z), r(z)$ such that

$$p(z) = s(z)q(z) + r(z)$$

and $\deg r < \deg q$, then

$$\begin{align*} p(T) &= s(T)q(T) + r(T) \\ &= s(T) 0 + r(T) \\ &= r(T) \end{align*}$$

For the uniqueness, if $r'(T) = P(T) = r(T)$, and $\deg r' < \deg q$, then $(r-r')(T) = 0$. Furthermore, $\deg (r-r') < \deg q$, since $q$ is the minimal polynomial of $T$, then $r-r'$ has to be $0$.

$\square$

5B.14

Suppose $𝑉$ is finite-dimensional and $𝑇 ∈ ℒ(𝑉)$ has minimal polynomial $4 + 5𝑧 − 6𝑧^2 − 7𝑧^3 + 2𝑧^4 + 𝑧^5$ . Find the minimal polynomial of $𝑇^{−1}$.

Solution:

First note that

$$\begin{align*} 4I + 5T − 6T^2 − 7T^3 + 2T^4 + T^5 &= 0 \\ & \Rightarrow \\ -(5 − 6T − 7T^2 + 2T^3 + T^4)T &= 4I \\ & \Rightarrow \\ T &\text{ is invertible} \end{align*}$$

Then we have

$$\begin{align*} 4I + 5T − 6T^2 − 7T^3 + 2T^4 + T^5 &= 0 \\ & \Rightarrow \\ T^{5} (4T^{-5} + 5 T^{-4} - 6T^{-3} - 7 T^{-2} + 2T^{-1} + I) &= 0\\ \end{align*}$$

Since $T$ is invertible, so is $T^{5}$. Then we have

$4T^{-5} + 5 T^{-4} - 6T^{-3} - 7 T^{-2} + 2T^{-1} + I = 0$.

If we can find a polynomial with degree less than $5$ that can nullify $T^{-1}$, then we can use similar approach to find a polynomial with degree less than $5$ to nullify $T$ which contradicts the fact that minimal polynomial of $T$ has degree $5$.

$\square$

5B.15

Suppose $𝑉$ is a finite-dimensional complex vector space with $\dim 𝑉 > 0$ and $𝑇 ∈ ℒ(𝑉)$. Define $𝑓 ∶ 𝐂 → 𝐑$ by

$$𝑓 (\lambda) = \text{dim range } (𝑇 − \lambda𝐼).$$

Prove that $𝑓$ is not a continuous function.

Proof:

Let $\dim V = n$. Since $𝑉$ is a finite-dimensional complex vector space, then from "5.19 existence of eigenvalues", we know $T$ has an eigenvalue, say $\lambda$.

From "5.12 operator cannot have more eigenvalues than dimension of vector space", we know $T$ can have at most $n$ distinct eigenvalues, so we can find a neighborhood of $\lambda$, $V_{\delta }(\lambda)$, such that if $c \in V_{\delta }(\lambda)$ and $c \neq \lambda$, then $c$ is not an eigenvalue.

For these $c$, $T - cI$ is invertible, so $\text{dim range } T-cI = n$. On the other hand, $\text{dim range } T-\lambda I < n$ since $\text{dim range } T-\lambda I = n - \text{dim null } T-\lambda I$, and $\text{dim null } T-\lambda I > 0$.

So $𝑓 (\lambda)$ is not continuous at $\lambda$.

$\square$

5B.16

Suppose $𝑎_0, …, 𝑎_{𝑛 − 1} ∈ 𝐅$. Let $𝑇$ be the operator on $𝐅^𝑛$ whose matrix (with respect to the standard basis) is

$$\begin{pmatrix} 0 & & & & & & -a_0 \\ 1 & 0 & & & & & -a_1 \\ & 1 & \ddots & & & & -a_2 \\ & & \ddots & & & & \vdots \\ & & & & & 0 & -a_{n-2} \\ & & & & & 1 & -a_{n-1} \\ \end{pmatrix}$$

Here all entries of the matrix are $0$ except for all $1’s$ on the line under the diagonal and the entries in the last column (some of which might also be $0$). Show that the minimal polynomial of $𝑇$ is the polynomial.

$$𝑎_0 + 𝑎_1𝑧 +⋯ + 𝑎_{𝑛 − 1}𝑧^{𝑛 − 1} + 𝑧^𝑛$$

Proof:

Let $p(z) = 𝑎_0 + 𝑎_1𝑧 +⋯ + 𝑎_{𝑛 − 1}𝑧^{𝑛 − 1} + 𝑧^𝑛$

Consider $e_1 = (1, 0, \cdots, 0)$. And note

$$\begin{align*} T(e_1) &= e_2 \\ T^2(e_1) &= T(e_2) = e_3 \\ & \cdots \\ T^{n-1}(e_1) &= e_n \\ T^n(e_1) &= T(e_n) = -(a_0, a_1, a_2, \cdots, a_{n-1}) \end{align*}$$

So

$$𝑎_0(e_1) + 𝑎_1T(e_1) +⋯ + 𝑎_{𝑛 − 1}T^{𝑛 − 1}(e_1) + T^𝑛(e_1) = 0$$

i.e. $p(T)(e_1) = 0$.

Furthermore, notice for $1 \leq i \leq n-1$

$$\begin{align*} 0 &= T^i(0) \\ &=T^i (𝑎_0(e_1) + 𝑎_1T(e_1) +⋯ + 𝑎_{𝑛 − 1}T^{𝑛 − 1}(e_1) + T^𝑛(e_1)) \\ &= 𝑎_0(T^i(e_1)) + 𝑎_1T(T^i(e_1)) +⋯ + 𝑎_{𝑛 − 1}T^{𝑛 − 1}(T^i(e_1)) + T^𝑛(T^i(e_1)) \\ &= 𝑎_0(e_{i+1}) + 𝑎_1T(e_{i+1}) +⋯ + 𝑎_{𝑛 − 1}T^{𝑛 − 1}(e_{i+1}) + T^𝑛(e_{i+1}) \end{align*}$$

i.e. $p(T)(e_{i+1}) = 0$. This means $p(T) = 0$.

Also, since $T^{i}(e_1) = e_{i+1}$, so $e_1, T(e_1), \cdots, T^{n-1}(e_1)$ are independent, then it's not possible to find a polynomial $q$ with $\deg q < n$, such that $q(T) = 0$. Otherwise $q(T)(e_1) = 0$, which is not possible.

So the minimal polynomial of $T$ is $p$.

$\square$

5B.17

Suppose $𝑉$ is finite-dimensional, $𝑇 ∈ ℒ(𝑉)$, and $𝑝$ is the minimal polynomial of $𝑇$. Suppose $𝜆 ∈ 𝐅$. Show that the minimal polynomial of $𝑇 − 𝜆𝐼$ is the polynomial $𝑞$ defined by $𝑞(𝑧) = 𝑝(𝑧 + 𝜆)$.

Proof:

First

$$\begin{align*} q(T - \lambda I) &= p(T - \lambda I + \lambda I) \\ &= p(T) \\ &= 0 \end{align*}$$

Now assume there is another monic polynomial $q'$ with $\deg q' < \deg q = \deg p$ and $q'(T-\lambda I) = 0$

Then let $p'(z) = q'(z-\lambda)$, then $p'$ is also a monic polynomial such that $\deg p' = \deg q' < \deg q = \deg p$.

Also $p'(T) = q'(T - \lambda I) = 0$.

This is contradictary to the fact that $p$ is the minimal polynomial.

So $q$ is the minimal polynomial of $T-\lambda I$.

$\square$

5B.18

Suppose $𝑉$ is finite-dimensional, $𝑇 ∈ ℒ(𝑉)$, and $𝑝$ is the minimal polynomial of $𝑇$. Suppose $\lambda ∈ 𝐅 \backslash \{0\}$. Show that the minimal polynomial of $𝜆𝑇$ is the polynomial $𝑞$ defined by $𝑞(𝑧) = \lambda ^{\deg 𝑝} 𝑝( 𝑧 / \lambda )$.

Proof:

let $m = \deg 𝑝$, and

$$p(z) = a_0 + a_1 z + \cdots + z^{m}$$

Then

$$\begin{align*} q(\lambda T) &= \lambda ^{m} p(\frac{\lambda T}{\lambda}) \\ &= \lambda ^{m} p(T) \\ &= 0 \\ \end{align*}$$

Also

$$\begin{align*} q(z) &= \lambda ^m p( z / \lambda ) \\ &= \lambda ^m a_0 + \lambda ^{m-1} a_1 z + \cdots + z^m \end{align*}$$

So $q$ is monic.

Now assume $q'(z) = b_0 + b_1 z + \cdots + z^{k}$ satisfy $k < m$ and $q'(\lambda T) = 0$.

Then

$$\begin{align*} 0 &= q'(\lambda T) \\ &= b_0 + b_1 (\lambda T) + \cdots + (\lambda T)^{k} \\ &= (\lambda)^k (b_0 \lambda^{-k} + b_1 \lambda^{1-k}T + \cdots + T^k) \end{align*}$$

Since $\lambda \neq 0$, so $\lambda^k \neq 0$, then $b_0 \lambda^{-k} + b_1 \lambda^{1-k}T + \cdots + T^k = 0$,

Let $p'(z) = b_0 \lambda^{-k} + b_1 \lambda^{1-k}z + \cdots + z^k$.

So we found a monic polynomial $p'$ with $p'(T) = 0$ and $\deg p' < \deg p$.

This is contradictary to the fact that $p$ is the minimal polynomial.

So the minimal polynomial of $𝜆𝑇$ is the polynomial $𝑞$ defined by $𝑞(𝑧) = \lambda ^{\deg 𝑝} 𝑝( 𝑧 / \lambda )$.

$\square$

5B.19

Suppose $𝑉$ is finite-dimensional and $𝑇 ∈ ℒ(𝑉)$. Let $ℰ$ be the subspace of $ℒ(𝑉)$ defined by

$$ℰ = \{q(T): q \in 𝒫(𝐅)\}$$

Prove that $\dim ℰ$ equals the degree of the minimal polynomial of $𝑇$.

Proof:

Let $p(z)$ be the minimal polynomial of $𝑇$. And $\deg p = m$. Given any $q(z)$, using the "4.9 division algorithm for polynomials", we know we can find $s, r$ such that

$$q(z) = s(z) p(z) + r(z)$$

with $\deg r < \deg p$.

The use "5.17 multiplicative properties", we know

$$q(T) = s(T) p(T) + r(T)$$

So

$$\begin{align*} q(T)(v) &= (s(T) p(T) + r(T))(v) \\ &= s(T) p(T)(v) + r(T)(v) \\ &= r(T) (v) \end{align*}$$

So $q(T) = r(T)$.

Since $\deg r < \deg p$, then $r(z)$ can be represented by $1, z, z^2, \cdots, z^{m-1}$.

So $q(T) = r(T) \in \text{span}(I, T, T^2, \cdots, T^{m-1})$. so $ℰ \subseteq \text{span}(I, T, T^2, \cdots, T^{m-1})$. So $\dim ℰ \leq m$

On the other hand, $I, T, T^2, \cdots, T^{m-1}$ are linear independent. Otherwise, we can find $a_i$ such that they are not all 0 and

$$a_0 I + a_1 T + \cdots + a_{m-1} T^{m-1} = 0$$

That means we find another polynomial with degree less than $m$ such that it's 0, which contradicts the assumption that $p(z)$ is the minimal polynomial.

Since $I, T, T^2, \cdots, T^{m-1}$ are linear independent, then $\dim ℰ \geq m$.

In summary, $\dim ℰ = m$.

$\square$

5B.20

Suppose $𝑇 ∈ ℒ(𝐅^4)$ is such that the eigenvalues of 𝑇 are 3, 5, 8. Prove that $(𝑇 − 3𝐼)^2(𝑇 − 5𝐼)^2(𝑇 − 8𝐼)^2 = 0$.

Proof:

Let $p(z)$ be the minimal polynomial of $T$. From "5.22 existence, uniqueness, and degree of minimal polynomial", $\deg p \leq \dim F^4 = 4$.

From "5.27 eigenvalues are the zeros of the minimal polynomial" we know $p(3) = p(5) = p(8) = 0$.

From "4.6 each zero of a polynomial corresponds to a degree-one factor" we know $p(z) = (z-3)(z-5)(z-8)q(z)$.

Since $\deg p \leq 4$, $\deg q = 1$ or $\deg q = 0$.

Case 1: $\deg q = 0$, then $p(z) = (z-3)(z-5)(z-8)$. Then $(z-3)^2(z-5)^2(z-8)^2$ is a multiple of $p(z)$.

Case 2: $\deg q = 1$, then $q(z) = z- \lambda$ and $p(z) = (z-\lambda)(z-3)(z-5)(z-8)$.

Use 5.27 again, we know $\lambda$ is an eigenvalue of $T$, so $\lambda = 3, 5, 8$. In any case, $(z-3)^2(z-5)^2(z-8)^2$ is a multiple of $p(z)$.

Then we can use 5.29 to conclude $(𝑇 − 3𝐼)^2(𝑇 − 5𝐼)^2(𝑇 − 8𝐼)^2 = 0$.

$\square$

5B.21

Suppose $𝑉$ is finite-dimensional and $𝑇 ∈ ℒ(𝑉)$. Prove that the minimal polynomial of $𝑇$ has degree at most $1 + \text{dim range } 𝑇$.

If $\text{dim range } T < \dim 𝑉 − 1$, then this exercise gives a better upper bound than 5.22 for the degree of the minimal polynomial of $𝑇$.

Proof:

Let $m = \text{dim range } T$. Let $U = \text{range } T$, then consider $T|_U$. Since $U$ is invariant under $T$, so $T|_U$ is well defined.

Let $p(z)$ be the minimal polynomial of $T|_U$, then from 5.22, $\deg p \leq m$. Consider $q(z) = p(z)z$. We would like to show $q(T) = p(T)T = 0$.

Given $v \in V$,

case 1: $v \in \text{null } T$, then $p(T)T(v) = p(T)(T(v)) = p(T)(0) = 0$.

case 2: $v \not\in \text{null } T$, then $T(v) = u \in \text{range } T$, so $p(T|_U)(u) = 0$. Also, since $u \in \text{range } T$, then $p(T)(u) = p(T|_U)(u)$, so $0 = p(T|_U)(u) = p(T)(u) = p(T)(T(u)) = (p(T)T)(u)$.

So in summary, $p(T)T = 0$.

Then from 5.29, $p(T)T$ is a multiple of minimal polynomial. Then the minimal polynomial of $𝑇$ has degree at most $1 + \text{dim range } 𝑇$.

$\square$

5B.22

Suppose $𝑉$ is finite-dimensional and $𝑇 ∈ ℒ(𝑉)$. Prove that $𝑇$ is invertible if and only if $𝐼 ∈ \text{span}(𝑇, 𝑇^2 , …, 𝑇^{\dim 𝑉})$.

Proof:

Let $m = \dim V$

$\Leftarrow$

$$\begin{align*} I &= a_1 T + \cdots + a_m T^m \\ &= T(a_1 I + \cdots + a_m T^{m-1}) \\ &= (a_1 I + \cdots + a_m T^{m-1})T \\ \end{align*}$$

So $T$ has an inverse mapping: $a_1 I + \cdots + a_m T^{m-1}$.

$\Rightarrow$

From 5.32 $𝑇$ not invertible ⟺ constant term of minimal polynomial of $𝑇$ is 0 we know if $p(z)$ is the minimal polynomial, then

$$p(z) = a_0 + a_1 z + \cdots + a_{k} z^{k}$$

where $a_0 \neq 0, k \leq m$, so

$$I = -\frac{a_1}{a_0} T - \cdots -\frac{a_k}{a_0} T^{k} \in \text{span}(𝑇, 𝑇^2 , …, 𝑇^{\dim 𝑉})$$

$\square$

5B.23

Suppose $𝑉$ is finite-dimensional and $𝑇 ∈ ℒ(𝑉)$. Let $𝑛 = \dim 𝑉$. Prove that if $𝑣 ∈ 𝑉$, then $\text{span}(𝑣, 𝑇𝑣, …, 𝑇^{𝑛 − 1}𝑣)$ is invariant under $𝑇$.

Proof:

Let $U = \text{span}(𝑣, 𝑇𝑣, …, 𝑇^{𝑛 − 1}𝑣)$.

We only need to prove $T^nv \in U$.

Since $𝑣, 𝑇𝑣, …, 𝑇^{𝑛 − 1}𝑣, T^n v$ are linear dependent, we can find

$$a_0 v + a_1 Tv + \cdots + a_{n-1} T^{n-1}v + a_n T^n v = 0$$

and $a_0, \cdots a_n$ are not all $0$.

Let $m$ be the biggest integer such that $a_m \neq 0$. So we have

$$a_0 v + a_1 Tv + \cdots + a_{m-1} T^{m-1} v + a_m T^m v = 0 \\ \Rightarrow \\ T^m v = -\frac{a_0}{a_m} v -\frac{a_1}{a_m} Tv \cdots -\frac{a_{m-1}}{a_m} T^{m-1} v$$

Apply $T^{n-m}$ on both sides we have

$$T^{n-m}(T^m v) = T^{n-m}(-\frac{a_0}{a_m} v -\frac{a_1}{a_m} Tv - \cdots -\frac{a_{m-1}}{a_m} T^{m-1} v) \\ \Rightarrow \\ T^n v = -\frac{a_0}{a_m} T^{n-m}v -\frac{a_1}{a_m} T^{n-m+1}v - \cdots -\frac{a_{m-1}}{a_m} T^{n-1} v$$

$\square$

5B.24

Suppose $𝑉$ is a finite-dimensional complex vector space. Suppose $𝑇 ∈ ℒ(𝑉)$ is such that $5$ and $6$ are eigenvalues of $𝑇$ and that $𝑇$ has no other eigenvalues. Prove that $(𝑇 − 5𝐼)^{\dim 𝑉 − 1}(𝑇 − 6𝐼)^{\dim 𝑉 − 1} = 0$.

Proof:

Let $p(z)$ be the minimal polynomial of $T$. From 5.27 eigenvalues are the zeros of the minimal polynomial (b)

$$p(z) = (z-\lambda_1) \cdots (z-\lambda_m)$$

where $\lambda_1, …, \lambda_𝑚$ is a list of all eigenvalues of $𝑇$, possibly with repetitions.

So $\lambda_i = 5, 6$. So $p(z) = (z-5)^s(z-6)^t, s,t \geq 1$.

If $s > \dim V - 1$, then $\deg p > \dim V$ which is not possible, so $s \leq \dim V - 1$. Similarly $t \leq \dim V - 1$.

Then $(𝑇 − 5𝐼)^{\dim 𝑉 − 1}(𝑇 − 6𝐼)^{\dim 𝑉 − 1}$ is a multiple of $p$. So $(𝑇 − 5𝐼)^{\dim 𝑉 − 1}(𝑇 − 6𝐼)^{\dim 𝑉 − 1} = 0$.

$\square$

5B.25

Suppose $𝑉$ is finite-dimensional, $𝑇 ∈ ℒ(𝑉)$, and $𝑈$ is a subspace of $𝑉$ that is invariant under $𝑇$.

(a) Prove that the minimal polynomial of $𝑇$ is a polynomial multiple of the minimal polynomial of the quotient operator $𝑇/𝑈$.

Proof:

Assume the minimal polynomial of $𝑇$ is $p(z)$, given any $v \in V$, we need to prove $p(T/U)(v+U) = 0$.

Since $T/U(v+U) = T(v) + U$, then

$$\begin{align*} p(T/U)(v+U) &= p(T)(v) + U \\ &= 0 + U \end{align*}$$

Then $p(T/U)$ is a zero operator on $V/U$.

Then from 5.29 $𝑞(𝑇) = 0 ⟺ 𝑞$ is a polynomial multiple of the minimal polynomial, we can conclude $p$ is a polynomial multiple of the minimal polynomial of the quotient operator $𝑇/𝑈$.

$\square$

(b) Prove that

(minimal polynomial of $𝑇|_𝑈$) × (minimal polynomial of $𝑇/𝑈$)

is a polynomial multiple of the minimal polynomial of 𝑇.

Proof:

Assume minimal polynomial of $𝑇|_𝑈$ is $p(z)$ and minimal polynomial of $𝑇/𝑈$ is $q(z)$.

We need to prove, given any $v \in V$, $p(T)q(T)(v) = 0$.

First since the minimal polynomial of $𝑇/𝑈$ is $q(z)$. Then $q(T/U)(v+U) = 0 + U$. On the other hand, $q(T/U)(v+U) = q(T)(v) + U$. Thus it means $q(T)(v) \in U$.

Since the minimal polynomial of $𝑇|_𝑈$ is $p(z)$, then $p(𝑇|_𝑈)(q(T)(v)) = 0$.

Since $U$ is invariant under $T$, then from exercise 5A.41 we know $U$ is also invariant under $p(T)$. Thus for any $u \in U$, $p(T|_U)(u) = p(T)(u)$. So $p(T)(q(T)(v)) = p(𝑇|_𝑈)(q(T)(v)) = 0$.

Then again from 5.29 $𝑞(𝑇) = 0 ⟺ 𝑞$ is a polynomial multiple of the minimal polynomial, we can conclude the statement.

$\square$

5B.26

Suppose $𝑉$ is finite-dimensional, $𝑇 ∈ ℒ(𝑉)$, and $𝑈$ is a subspace of $𝑉$ that is invariant under $𝑇$. Prove that the set of eigenvalues of $𝑇$ equals the union of the set of eigenvalues of $𝑇|𝑈$ and the set of eigenvalues of $𝑇/𝑈$.

Proof:

Let the minimal polynomial of $T$ be $s(z)$, minimal polynomial of $𝑇|_𝑈$ is $p(z)$ and minimal polynomial of $𝑇/𝑈$ is $q(z)$.

$\Rightarrow$

Assume $\lambda$ is a eigenvalue of $T$. Then $\lambda$ is a zero of $s(z)$. From the part (b) of exercise 5B.25, $p(z)q(z)$ is a multiple of $s(z)$, then $p(\lambda)q(\lambda) = 0$, so $p(\lambda) = 0$ or $q(\lambda) = 0$.

Then from 5.27, $\lambda$ is either an eigenvalue of $𝑇|𝑈$ or an eigenvalue of $𝑇/𝑈$.

$\Leftarrow$

From exercise 5A.38, we know an eigenvalue of $T/U$ is an eigenvalue of $T$.

Assume $\lambda$ is an eigenvalue of $T|_U$, then we can find a $u \neq 0$ and $u \in U$ such that $T|_U(u) = \lambda u$. Since $u \in U$, then $T(u) = T|_U(u) = \lambda u$.

So $\lambda$ is an eigenvalue of $T$.

$\square$

5B.27

Suppose $𝐅 = 𝐑$, $𝑉$ is finite-dimensional, and $𝑇 ∈ ℒ(𝑉)$. Prove that the minimal polynomial of $𝑇_𝐂$ equals the minimal polynomial of $𝑇$. The complexification $𝑇_𝐂$ was defined in Exercise 33 of Section 3B.

Proof:

Let $p$ be the minimal polynomial of $T$ and $p'$ be the minimal polynomial of $T_C$.

Note that $T_C(u+iv) = T(u) + iT(v)$, so

$$\begin{align*} T_C^n(u+iv) &= T_C^{n-1}(T_C(u+iv)) \\ &= T_C^{n-1}(T(u) + iT(v)) \\ &= T_C^{n-2}(T_C(T(u) + iT(v))) \\ &= T_C^{n-2}(T^2(u) + iT^2(v)) \\ &\cdots \\ &= T^n(u) + iT^n(v) \\ \end{align*}$$

Then from the linearity of $T_C$, we can see

$$\begin{align*} q(T_C)(u+iv) &= q(T)(u) + iq(T)(v) \end{align*}$$

So given any $u+iv$

$$\begin{align*} 0 &= p'(T_C)(u+iv) \\ &= p'(T)(u) + ip'(T)(v) \\ & \Rightarrow \\ p'(T)(u) = 0,&\qquad p'(T)(v) = 0 \end{align*}$$

So $p'$ is a multiple of $p$.

On the other hand

$$\begin{align*} 0 &= p(T)(u) + ip(T)(v) \\ &=p(T_C)(u+iv) \\ \end{align*}$$

So $p$ is a multiple of $p'$.

Then $p = p'$

$\square$

5B.28

Suppose $𝑉$ is finite-dimensional and $𝑇 ∈ ℒ(𝑉)$. Prove that the minimal polynomial of $𝑇' ∈ ℒ(𝑉')$ equals the minimal polynomial of $𝑇$.

Proof:

One way to think this problem is consider the matrix of $T$ and $T'$. They are transpose to each other.

But we will do another way. Note that given any linear functional $\varphi_i \in V'$, we have

$$\begin{align*} (T')^n(\varphi_i) &= (T')^{n-1} (T'(\varphi_i)) \\ &= (T')^{n-1} (\varphi_i \circ T) \\ &= (T')^{n-2} (T'(\varphi_i \circ T)) \\ &= (T')^{n-2} ((\varphi_i \circ T) \circ T) \\ &= (T')^{n-2} (\varphi_i \circ (T^2)) \\ &= \cdots \\ &= \varphi_i \circ T^n \end{align*}$$

And then use the distribution law of the composition of the linear mapping, we have $p(T')(\varphi_i) = \varphi_i \circ p(T)$

First assume $p(z)$ is the minimal polynomial of $𝑇$. And assume $p'(z)$ is the minimal polynomial of $𝑇'$.

Then given any linear functional $\varphi$ and $v \in V$,

$$\begin{align*} (p(T')(\varphi))(v) &= (\varphi \circ p(T))(v) \\ &= \varphi (p(T)(v)) \\ &= \varphi (0) \\ &= 0 \end{align*}$$

So $p$ is a multiple of $p'$.

Now assume $v_1, \cdots, v_m$ is a basis of $v$ and $\varphi_1, \cdots, \varphi_m$ is the dual basis. And assume $$p'(T)(v) = a_1 v_1 + \cdots + a_m v_m$$

On one hand

$\varphi_i(a_1 v_1 + \cdots + a_m v_m) = a_1 \varphi_i(v_1) + \cdots + a_m \varphi_i(v_m) = a_i$

On the other hand

$$\begin{align*} \varphi_i((p'(T)) (v)) &= (\varphi_i \circ p'(T)) (v)\\ &= (p'(T')(\varphi_i))(v) \\ &= 0(v) \\ &= 0 \end{align*}$$

So $a_i = 0$, i.e. $p'(T)(v) = 0$.

So $p'$ is a multiple of $p$.

Based on this, $p = p'$.

$\square$

5B.29

Show that every operator on a finite-dimensional vector space of dimension at least two has an invariant subspace of dimension two.

Proof:

Consider $T$ is the operator, and $p(z)$ is its minimal polynomial.

We distinguish the following cases:

case $1$: If $p(z) = z-\lambda$, then every vector is a eigenvector of $\lambda$. Since $\dim V \geq 2$, then we can pick two linear independent vectors $u_1, u_2$. Then $\text{span}(u_1, u_2)$ is invariant with dimension $2$.

case $2$: $\deg p(z) \geq 2$, we can assume $p(z) = (z^2+bz+c)q(z)$. This is because if $\mathbb{F} = \mathbb{R}$, we can use "4.16 factorization of a polynomial over $\mathbb{R}$". If $\mathbb{F} = \mathbb{C}$, we can use "4.13 fundamental theorem of algebra, second version".

Consider $\text{range } q(T)$. It cannot be $\{0\}$ because otherwise $q(z)$ will be the minimal polynomial.

Then assume we can find $u \in \text{range } q(T)$, such that $u$ is not an eigenvector of $T$, then $u, T(u)$ are linearly independent. Furthermore, since $(T^2+bT+c)(u) = 0$, we have $T^2u = -bTu - cu$. So $\text{span}(u, Tu)$ is invariant with dimension $2$.

Now assume all $u \in \text{range } q(T)$ are eigenvectors. If for all $u \in \text{range } q(T)$, $T(u) = \lambda u$, then $(z- \lambda)q(z)$ will be the minimal polynomial, which is contradictary to our assumption that $p(z)$ is the minimal polynomial.

So we can find $\lambda_1, \lambda_2$ which are 2 different eigenvalues and we can find 2 eigenvectors $u_1, u_2$, such that $T(u_1) = \lambda_1 u_1, T(u_2) = \lambda_2 u_2$.

From 5.11 linearly independent eigenvectors, we know $u_1, u_2$ are linear independent, then $\text{span}(u_1, u_2)$ is invariant with dimension $2$.

$\square$

Section 5C Upper-Triangular Matrices

5C.1

Prove or give a counterexample: If $𝑇 ∈ ℒ(𝑉)$ and $𝑇^2$ has an upper-triangular matrix with respect to some basis of $𝑉$, then $𝑇$ has an upper-triangular matrix with respect to some basis of $𝑉$.

Solution:

Consider $V = \mathbb{R}^2$. And $T$ be the linear operator that rotates $V$ by $90$ degree counterclockwise.

Since $T$ does not have an eigenvalue, then $T$ does not have a 1-dimensional subspace that is invariant under $T$ w.r.t any basis of $V$. Therefore, it's not possible that $𝑇$ has an upper-triangular matrix w.r.t any basis of $V$.

On the other hand, the matrix of $T^2$ w.r.t the standard basis is

$$\begin{bmatrix} -1 & 0 \\ 0 & -1 \\ \end{bmatrix}$$

which is an upper-triangular matrix.

$\square$

5C.2

Suppose $𝐴$ and $𝐵$ are upper-triangular matrices of the same size, with $𝛼_1, …, 𝛼_𝑛$ on the diagonal of $𝐴$ and $𝛽_1, …, 𝛽_𝑛$ on the diagonal of $𝐵$.

(a) Show that $𝐴 + 𝐵$ is an upper-triangular matrix with $𝛼_1 + 𝛽_1, …, 𝛼_𝑛 + 𝛽_𝑛$ on the diagonal.

Proof:

Let $C = A + B$, then the diagonal of $C$ on the $i$th row and $i$th column is

$$\begin{align*} C_{i,i} &= A_{i,i} + B_{i,i} \\ &= 𝛼_i + 𝛽_i \ \end{align*}$$

$\square$

(b) Show that $𝐴𝐵$ is an upper-triangular matrix with $𝛼_1 𝛽_1, …, 𝛼_𝑛 𝛽_𝑛$ on the diagonal.

Proof

Let $C = AB$, then the diagonal of $C$ on the $i$th row and $i$th column is

$$\begin{align*} C_{i,i} &= \sum_{k = 1}^{n} a_{i, k} b_{k, i} \\ \end{align*}$$

Note that if $i > k, a_{i, k} = 0$, and if $i < k, b_{k, i} = 0$, so $C_{i,i} = a_{i,i} b_{i,i} = 𝛼_i 𝛽_i$.

$\square$

5C.3

Suppose $𝑇 ∈ ℒ(𝑉)$ is invertible and $𝑣_1, …, 𝑣_𝑛$ is a basis of $𝑉$ with respect to which the matrix of $𝑇$ is upper triangular, with $𝜆_1, …, 𝜆_𝑛$ on the diagonal. Show that the matrix of $𝑇^{−1}$ is also upper triangular with respect to the basis $𝑣_1, …, 𝑣_𝑛$, with

$$\frac{1}{𝜆_1} , …, \frac{1}{𝜆_𝑛}$$

on the diagonal.

Proof:

First, from "5.41 determination of eigenvalues from upper-triangular matrix", $𝜆_1, …, 𝜆_𝑛$ are the eigenvalues of $T$. Since $T$ is invertible, then $0$ is not an eigenvalue of $T$. So $𝜆_1, …, 𝜆_𝑛 \neq 0$.

Then $\frac{1}{𝜆_1} , …, \frac{1}{𝜆_𝑛}$ is well defined.

Next, we show if $U$ is invariant under $T$ and $T$ is invertible, then $U$ is invariant under $T^{-1}$.

Let $u \in U$, and consider $T|_U$. Since $T$ is invertible, then $T$ is injective, then $T|_U$ is also injective. So $T|_U$ is invertible. Then we can find $w \in U$ such that $T|_U(w) = u$. then we have $T(w) = u$.

So $T^{-1}(u) = T^{-1}(T(w)) = w \in U$. Then $U$ is invariant under $T^{-1}$.

Now let $U_i = \text{span}(𝑣_1, \cdots, 𝑣_i)$. From "5.39 conditions for upper-triangular matrix", $U_i$ is invariant under $T$, then $U_i$ is also invariant under $T^{-1}$, then again use 5.39, The matrix of $𝑇^{-1}$ with respect to $𝑣_1, …, 𝑣_𝑛$ is upper triangular.

Let $\eta_i$ on the diagonal. Note $v_1 = T^{-1} (T v_1) = T^{-1} (\lambda_1 v_1) = \eta_1 \lambda_1 v_1$, so $\eta_1 \lambda_1 = 1$, thne $\eta_1 = \frac{1}{\lambda_1}$.

For $i \geq 2$,

$$T(v_i) = \lambda_i v_i + u_{i-1}$$

Where $u_{i-1} \in U_{i-1}$, then

$$\begin{align*} v_i &= T^{-1} (T v_i) \\ &= T^{-1} (\lambda_i v_i) + T^{-1} (u_{i-1}) \\ &= \eta_i \lambda_i v_i + T^{-1} (u_{i-1}) \\ & \Rightarrow \\ 0 = (1 - \eta_i \lambda_i) v_i + T^{-1} (u_{i-1}) \\ \end{align*}$$

Since $U_i$ is invariant under $T^{-1}$, then $T^{-1} (u_{i-1}) \in U_{i-1} = \text{span}(𝑣_1, \cdots, 𝑣_{i-1})$. Then we must have $1 - \eta_i \lambda_i = 0$, i.e. $\eta_i = \frac{1}{\lambda_i}$.

$\square$

5C.4

Give an example of an operator whose matrix with respect to some basis contains only $0$'s on the diagonal, but the operator is invertible. This exercise and the exercise below show that 5.41 fails without the hypothesis that an upper-triangular matrix is under consideration.

Proof:

Consider the vector space is $\mathbb{R}^2$, and the operator $T$ is rotation counterclockwise by 90 degree. And the basis is the standard basis, $e_1, e_2$.

$T(e_1) = e_2, T(e_2) = -e_1$. So

$$\mathcal{M}(T) = \begin{bmatrix} 0 & -1 \\ 1 & 0 \\ \end{bmatrix}$$

It's invertible because its inverse is rotation clockwise by 90 degree.

$\square$

5C.5

Give an example of an operator whose matrix with respect to some basis contains only nonzero numbers on the diagonal, but the operator is not invertible.

Proof:

Consider the vector space is $\mathbb{R}^2$, and the operator $T$ is defined as

$$T(e_1) = (1, 1) \\ T(e_2) = (1, 1) \\ $$

Since it's not injective, it's not invertible.

Also the

$$\mathcal{M}(T) = \begin{bmatrix} 1 & 1 \\ 1 & 1 \\ \end{bmatrix}$$

$\square$

5C.6

Suppose $𝐅 = 𝐂$, $𝑉$ is finite-dimensional, and $𝑇 ∈ ℒ(𝑉)$. Prove that if $𝑘 ∈ \{1, …, \dim 𝑉\}$, then $𝑉$ has a $𝑘$-dimensional subspace invariant under $𝑇$.

Proof:

Use "5.47 if 𝐅 = 𝐂, then every operator on 𝑉 has an upper-triangular matrix", then for the operator $T$, we can find a basis $v_1, \cdots, v_n$, such that $\mathcal{M}(T)$ is upper-triangular.

Then use "5.39 conditions for upper-triangular matrix", we know $\text{span}(𝑣_1, \cdots, 𝑣_k)$ is invariant under $T$ and its dimension is $k$.

$\square$

5C.7

Suppose $𝑉$ is finite-dimensional, $𝑇 ∈ ℒ(𝑉)$, and $𝑣 ∈ 𝑉$.

(a) Prove that there exists a unique monic polynomial $𝑝_𝑣$ of smallest degree such that $𝑝_𝑣(𝑇)𝑣 = 0$.

Proof:

Let $n = \dim V$ and then the following $1+n$ vectors must be linear dependent

$$v, Tv, \cdots, T^{n}v$$

Then we can find the smallest $k$, such that $T^{k} v$ is the linear combination of $v, \cdots, T^{k-1}v$ and then we find the $𝑝_𝑣$.

(b) Prove that the minimal polynomial of $𝑇$ is a polynomial multiple of $𝑝_𝑣$.

Proof:

Let $p$ be the minimal polynomial of $𝑇$.

Let $U = \text{span}(v, Tv, \cdots, T^{k-1}v)$. Then $U$ is invariant under $T$. Then we can consider $T|_U$. Note that $p_v$ is the minimal polynomial of $T|_U$ from part (a).

Then from "5.21 minimal polynomial of a restriction operator", we know $p$ is multiple of $𝑝_𝑣$.

$\square$

5C.8

Suppose $𝑉$ is finite-dimensional, $𝑇 ∈ ℒ(𝑉)$, and there exists a nonzero vector $𝑣 ∈ 𝑉$ such that $𝑇^2𝑣 + 2𝑇𝑣 = −2𝑣$.

(a) Prove that if $𝐅 = 𝐑$, then there does not exist a basis of $𝑉$ with respect to which $𝑇$ has an upper-triangular matrix.

Proof:

Let $p$ the minimal polynomial of $T$ and $p_v(z) = z^2 + 2z + 2$. Note that $p_v(z)$ does not have a real zero, which means $v$ is not an eigenvector.

Otherwise, if $T(v) = \lambda v$, then $0 = 𝑇^2𝑣 + 2𝑇𝑣 + 2𝑣 = \lambda^2𝑣 + 2\lambda𝑣 + 2𝑣 = (\lambda^2 + 2\lambda + 2)𝑣$, so $\lambda^2 + 2\lambda + 2 = 0$. No real $\lambda$ can satisfy this equation.

Then $p_v(z)$ satisfy the 5C.7(a), so $p$ is a multiple of $p_v$. Since $p_v$ cannot be written in the form of $(z-a)(z-b), a,b \in \mathbb{R}$.

Then use "5.44 necessary and sufficient condition to have an upper-triangular matrix", we know there does not exist a basis of $𝑉$ with respect to which $𝑇$ has an upper-triangular matrix.

$\square$

(b) Prove that if $𝐅 = 𝐂$ and $𝐴$ is an upper-triangular matrix that equals the matrix of $𝑇$ with respect to some basis of $𝑉$ , then $−1 + 𝑖$ or $−1 − 𝑖$ appears on the diagonal of $𝐴$.

Proof:

Let $\lambda_1 = −1 + 𝑖$ and $\lambda_2 = −1 - 𝑖$, then it's possible that

$$\begin{align*} p_v(z) &= (z - \lambda_1) \\ p_v(z) &= (z - \lambda_2) \\ p_v(z) &= (z-\lambda_1)(z - \lambda_2) \\ \end{align*}$$ .

Since $p$ is a multiple of $p_v$, then

either $\lambda_1$ or $\lambda_2$ or both are zeros of the $p$. Then $\lambda_1$ or $\lambda_2$ or both are eigenvalues of $T$.

From "5.41 determination of eigenvalues from upper-triangular matrix", we know $\lambda_1$ or $\lambda_2$ appears on the diagonal of $𝐴$.

$\square$

5C.9

Suppose $𝐵$ is a square matrix with complex entries. Prove that there exists an invertible square matrix $𝐴$ with complex entries such that $𝐴^{−1}𝐵𝐴$ is an upper-triangular matrix.

Proof:

Let $B$ be an $n$ by $n$ matrix. Consider $V = \mathbb{C}^n$. $T$ is an operator in $L(V)$ such that $\mathcal{M}(T)$ under the standard basis is $B$, i.e. $B = \mathcal{M}(T, (e_1, \cdots, e_n))$.

Use "5.47 if $𝐅 = 𝐂$, then every operator on 𝑉 has an upper-triangular matrix", then we know we can find a basis $v_1, \cdots, v_n$ such that $D = \mathcal{M}(T, (v_1, \cdots, v_n))$ is a upper-triangular matrix.

Then we can directly apply "3.84 change-of-basis formula" and let

$$A = \mathcal{M}(I,(v_1, \cdots, v_n),(e_1, \cdots, e_n))$$

Then we have $D = A^{-1} B A$.

$\square$

5C.10

Suppose $𝑇 ∈ ℒ(𝑉)$ and $𝑣_1, …, 𝑣_𝑛$ is a basis of $𝑉$. Show that the following are equivalent.

(a) The matrix of $𝑇$ with respect to $𝑣_1, …, 𝑣_𝑛$ is lower triangular.

(b) $\text{span}(𝑣_k, \cdots, 𝑣_n)$ is invariant under $𝑇$ for each $𝑘 = 1, …, 𝑛$.

(c) $𝑇{𝑣_𝑘} ∈ \text{span}(𝑣_k, \cdots, 𝑣_n)$ for each $𝑘 = 1, …, 𝑛$.

A square matrix is called lower triangular if all entries above the diagonal are $0$.

Proof:

$(a) \Rightarrow (b)$

Fix any $𝑘 = 1, …, 𝑛$ and $k \leq j \leq n$.

Since The matrix of $𝑇$ with respect to $𝑣_1, …, 𝑣_𝑛$ is lower triangular, then $𝑇{𝑣_j}$ is the linear combination of $𝑣_j, \cdots, 𝑣_n$, so $T v_j \in \text{span}(𝑣_j, \cdots, 𝑣_n) \subseteq \text{span}(𝑣_k, \cdots, 𝑣_n)$.

So $\text{span}(𝑣_k, \cdots, 𝑣_n)$ is invariant under $𝑇$.

$(b) \Rightarrow (c)$

$\text{span}(𝑣_k, \cdots, 𝑣_n)$ is invariant under $𝑇$, then $𝑇{𝑣_𝑘} ∈ \text{span}(𝑣_k, \cdots, 𝑣_n)$.

$(c) \Rightarrow (a)$

$𝑇{𝑣_𝑘} ∈ \text{span}(𝑣_k, \cdots, 𝑣_n)$, then $𝑇{𝑣_𝑘}$ is a linear combination of $𝑣_k, \cdots, 𝑣_n$. Then the coefficients of $v_1, \cdots, v_{k-1}$ are all $0$. This means the matrix of $𝑇$ with respect to $𝑣_1, …, 𝑣_𝑛$ is lower triangular.

$\square$

5C.11

Suppose $𝐅 = 𝐂$ and $𝑉$ is finite-dimensional. Prove that if $𝑇 ∈ ℒ(𝑉)$, then there exists a basis of $𝑉$ with respect to which $𝑇$ has a lower-triangular matrix.

Proof:

Use "5.47 if $𝐅 = 𝐂$, then every operator on 𝑉 has an upper-triangular matrix", then we can find a basis $v_1, \cdots, v_n$ such that $\mathcal{M}(T)$ is a upper-triangular matrix.

Now use "5.39 conditions for upper-triangular matrix", we know $\text{span}(𝑣_1, \cdots, 𝑣_k)$ is invariant under $T$ for $k = 1, \cdots, n$.

Now let $w_n = v_1, \cdots, w_1 = v_n$. We can see

$\text{span}(w_k, \cdots, w_n) = \text{span}(v_1, \cdots, v_{n-k+1})$ is invariant under $𝑇$.

Then apply exercise 5C.10 above, we get $\mathcal{M}(T)$ is a lower-triangular matrix under the basis $w_1, \cdots, w_n$.

$\square$

5C.12

Suppose $𝑉$ is finite-dimensional, $𝑇 ∈ ℒ(𝑉)$ has an upper-triangular matrix with respect to some basis of $𝑉$, and $𝑈$ is a subspace of $𝑉$ that is invariant under $𝑇$.

(a) Prove that $T|_U$ has an upper-triangular matrix with respect to some basis of $𝑈$.

Proof:

Let $p(z)$ be the minimal polynomial of $T$, and $q(z)$ be the minimal polynomial of $T|_U$.

From "5.44 necessary and sufficient condition to have an upper-triangular matrix", we know

$$p(z) = (z-\lambda_1) \cdots (z-\lambda_m)$$

Then from "5.31 minimal polynomial of a restriction operator", $p(z)$ is a multiple of $q(z)$.

Then $q(z)$ has to be in the form of

$$q(z) = (z-\alpha_1) \cdots (z-\alpha_l)$$

Then again apply 5.44, we know $T|_U$ has an upper-triangular matrix with respect to some basis of $𝑈$.

$\square$

(b) Prove that the quotient operator $𝑇/𝑈$ has an upper-triangular matrix with respect to some basis of $𝑉/𝑈$.

Proof: The logic is very similar to part (a).

Let $p(z)$ be the minimal polynomial of $T$, and $q(z)$ be the minimal polynomial of $T/U$.

From exercise 5B.25 part (a), the minimal polynomial of $𝑇$ is a polynomial multiple of the minimal polynomial of the quotient operator $𝑇/𝑈$.

Then the other part is exactly the same as part (a).

$\square$

5C.13

Suppose $𝑉$ is finite-dimensional and $𝑇 ∈ ℒ(𝑉)$. Suppose there exists a subspace $𝑈$ of $𝑉$ that is invariant under $𝑇$ such that $𝑇|_𝑈$ has an upper-triangular matrix with respect to some basis of $𝑈$ and also $𝑇/𝑈$ has an upper-triangular matrix with respect to some basis of $𝑉/𝑈$. Prove that $𝑇$ has an upper-triangular matrix with respect to some basis of $𝑉$.

Proof:

Let $p(z)$ be the minimal polynomial of $T$, $q(z)$ be that of $T|_U$ and $s(z)$ be that of $T/U$.

From "5.44 necessary and sufficient condition to have an upper-triangular matrix",

we have

$$q(z) = (z-\alpha_1) \cdots (z-\alpha_l) \\ s(z) = (z-\beta_1) \cdots (z-\beta_m)$$

From the exercise 5B.25 (b) we know that (minimal polynomial of $𝑇|_𝑈$) $×$ (minimal polynomial of $𝑇/𝑈$) is a polynomial multiple of the minimal polynomial of 𝑇.

Then due to unique factorization in $F[z]$ we have

$$p(z) = (z-\gamma_1) \cdots (z-\gamma_k)$$

Then again use 5.44, we know $𝑇$ has an upper-triangular matrix with respect to some basis of $𝑉$.

$\square$

5C.14

Suppose $𝑉$ is finite-dimensional and $𝑇 ∈ ℒ(𝑉)$. Prove that $𝑇$ has an upper- triangular matrix with respect to some basis of $𝑉$ if and only if the dual operator $𝑇'$ has an upper-triangular matrix with respect to some basis of the dual space $𝑉'$

Proof:

We use exercise 5B.28, which states that the minimal polynomial of $𝑇' ∈ ℒ(𝑉')$ equals the minimal polynomial of $𝑇$.

And then we can use "5.44 necessary and sufficient condition to have an upper-triangular matrix" again.

$\square$

Section 5D Diagonalizable Operators

5D.1

Suppose $𝑉$ is a finite-dimensional complex vector space and $𝑇 ∈ ℒ(𝑉)$.

(a) Prove that if $𝑇^4 = 𝐼$, then $𝑇$ is diagonalizable.

Proof: Since $T^4 - I = 0$ then $z^4 - 1$ is a multiple of $p$, which is the minimal polynomial of $T$. Also note

$$z^4 - 1 = (z-1)(z+1)(z-i)(z+i)$$

From "4.13 fundamental theorem of algebra, second version", we know both $z^4-1$ and $p$ has a unique factorization. So the zeros of $p$ can only be distinct numbers. Then apply "5.62 necessary and sufficient condition for diagonalizability", we know $T$ is diagonalizable.

$\square$

(b) Prove that if $𝑇^4 = 𝑇$, then $𝑇$ is diagonalizable.

Proof: Since $T^4 - T = 0$ then $z^4 - z$ is a multiple of $p$, which is the minimal polynomial of $T$. Also note

$$z^4 - z = z(z-1)(z-\frac{-1+\sqrt[]{3}i}{2})(z-\frac{-1-\sqrt[]{3}i}{2})$$

From "4.13 fundamental theorem of algebra, second version", we know both $z^4-z$ and $p$ has a unique factorization. So the zeros of $p$ can only be distinct numbers. Then apply "5.62 necessary and sufficient condition for diagonalizability", we know $T$ is diagonalizable.

$\square$

(c) Give an example of an operator $𝑇 ∈ ℒ(𝐂2)$ such that $𝑇^4 = 𝑇^2$ and $𝑇$ is not diagonalizable.

Solution:

Consider $T(x,y) = (y,0)$, then $T^2 = 0$. Since $T(0,1) = (1,0) \neq \lambda (0,1)$, then the degree of the minimal polynomial of $T$, i.e. $\deg p \geq 2$. So $p = z^2$.

Then apply "5.62 necessary and sufficient condition for diagonalizability", we know $T$ is not diagonalizable.

Furthermore $T^4 = (T^2)^2 = 0$.

$\square$

5D.2

Suppose $𝑇 ∈ ℒ(𝑉)$ has a diagonal matrix $𝐴$ with respect to some basis of $𝑉$. Prove that if $𝜆 ∈ 𝐅$, then $𝜆$ appears on the diagonal of $𝐴$ precisely $\dim 𝐸(𝜆, 𝑇)$ times.

Proof:

Assume $T$ has a diagonal matrix $𝐴$ w.r.t. the basis $v_1, \cdots, v_m$. And the elements on the diagonal matrix are $\lambda_1, \cdots, \lambda_m$.

Since $T(v_i) = \lambda_i v_i$, then they are eigenvectors.

If $\lambda$ is not an eigenvalue of $T$, then $\lambda \neq \lambda_i$, so it does not appear on the diagonal. Also $\dim E(\lambda, T) = 0$. So the conclusion holds.

Now if $\lambda$ is an eigenvalue, and if it appears on the diagonal more than $k = \dim E(\lambda, T)$ times. It means we can find at least $k+1$ linearly independent vectors and they are all eigenvectors w.r.t. to eigenvalue $\lambda$. So they all belong to $E(\lambda, T)$. It means $\dim E(\lambda, T) > k$, which is impossible.

So $\lambda$ appears on the diagonal at most $\dim E(\lambda, T)$ times. This is true of all eigenvalues.

On the other hand, from "5.55 conditions equivalent to diagonalizability" we have

$$\dim V = \dim E(\lambda_1, T) + \cdots + \dim E(\lambda_m, T)$$

It means, $\lambda$ cannot appear on the diagonal less than $\dim E(\lambda, T)$ times. Otherwise the righ side of the above equation is less than the left side.

So the conclusion holds.

$\square$

5D.3

Suppose $𝑉$ is finite-dimensional and $𝑇 ∈ ℒ(𝑉)$. Prove that if the operator $𝑇$ is diagonalizable, then $𝑉 = \text{null } 𝑇 ⊕ \text{range } 𝑇$.

Proof:

If $\text{null } T = \{0\}$, then $T$ is invertible. So $V = \text{range } T$, So $𝑉 = \text{null } 𝑇 ⊕ \text{range } 𝑇$.

Now assume $\text{null } T \neq \{0\}$, then note $\text{null } T = E(0,T)$.

Since $T$ is diagonalizable, then from "5.55 conditions equivalent to diagonalizability", we know $𝑉$

$V = E(\lambda_0, T) \oplus E(\lambda_1, T) \oplus \cdots \oplus E(\lambda_m, T)$, where $\lambda_0 = 0, \lambda_i \neq 0$ for $i > 0$ .

Note $E(\lambda_0, T) = \text{null } T$, so we just need to prove $\text{range } T = E(\lambda_1, T) \oplus \cdots \oplus E(\lambda_m, T)$.

Let $v \in V$, then since $V = E(\lambda_0, T) \oplus E(\lambda_1, T) \oplus \cdots \oplus E(\lambda_m, T)$, we can pick $v_i \in E(\lambda_i, T)$ sucht that

$v = v_0 + v_1 + \cdots + v_m$ Then

$$\begin{align*} T v &= Tv_0 + Tv_1 + \cdots + Tv_m \\ &= \lambda_0 v_0 + \lambda_1 v_1 + \cdots + \lambda_m v_m \\ &= \lambda_1 v_1 + \cdots + \lambda_m v_m \\ & \in E(\lambda_1, T) \oplus \cdots \oplus E(\lambda_m, T) \end{align*}$$

Which means $\text{range } T \subseteq E(\lambda_1, T) \oplus \cdots \oplus E(\lambda_m, T)$

On the other hand, if $v = v_1 + \cdots + v_m \in E(\lambda_1, T) \oplus \cdots \oplus E(\lambda_m, T)$, then let $w = 1/ \lambda_1 \cdot v_1 + \cdots + 1/ \lambda_m \cdot v_m$, we have $T(w) = v$.

So $\text{range } T \supseteq E(\lambda_1, T) \oplus \cdots \oplus E(\lambda_m, T)$.

In summary, $\text{range } T = E(\lambda_1, T) \oplus \cdots \oplus E(\lambda_m, T)$

Then we have $𝑉 = \text{null } 𝑇 ⊕ \text{range } 𝑇$.

$\square$

5D.4

Suppose $𝑉$ is finite-dimensional and $𝑇 ∈ ℒ(𝑉)$. Prove that the following are equivalent.

(a) $V = \text{null } T \oplus \text{range } T$

(b) $V = \text{null } T + \text{range } T$

(c) $\text{null } T \cap \text{range } T = \{0\}$

Proof:

$(a) \Rightarrow (b)$

Since $\text{null } T \oplus \text{range } T = \text{null } T + \text{range } T$, so $V = \text{null } T + \text{range } T$.

$(b) \Rightarrow (c)$

From fundamental theorem of linear mapping, we know

$$\dim V = \dim \text{null } T + \dim \text{range } T$$

From "2.43 dimension of a sum", we know

$$\dim \text{null } T + \text{range } T = \dim \text{null } T + \dim \text{range } T - \dim \text{null } T \cap \text{range } T$$

Then we have $\text{null } T \cap \text{range } T = \{0\}$

$(c) \Rightarrow (a)$

From (c), we know $\text{null } T + \text{range } T$ is a direct sum.

Then from "2.43 dimension of a sum", fundamental theorem of linear mapping and (c), we know

$$\begin{align*} \dim \text{null } T \oplus \text{range } T &= \dim \text{null } T + \text{range } T \\ &= \dim \text{null } T + \dim \text{range } T - \dim \text{null } T \cap \text{range } T \\ &= \dim V \end{align*}$$

So $V = \text{null } T \oplus \text{range } T$

$\square$

5D.5

Suppose $𝑉$ is a finite-dimensional complex vector space and $𝑇 ∈ ℒ(𝑉)$. Prove that $𝑇$ is diagonalizable if and only if

$$V = \text{null } (T- \lambda I) \oplus \text{range } (T - \lambda I)$$

for every $\lambda \in \mathbb{C}$.

Proof:

$\Rightarrow$

We use the exercise 5D.3. If $T$ is diagonalizable, then using "5.55 conditions equivalent to diagonalizability"

we can find a basis $v_1, \cdots, v_m$ which are all eigenvectors of $T$.

In fact, for any $\lambda$, they are also eigenvectors of $T - \lambda I$. To see this, note

$$\begin{align*} (T - \lambda I)(v_j) = T v_j - \lambda I v_j = (\lambda_j - \lambda) v_j \end{align*}$$

So again use "5.55 conditions equivalent to diagonalizability", we know $T - \lambda I$ is also diagonalizable. Then use 5D.3, we have

$$V = \text{null } (T- \lambda I) \oplus \text{range } (T - \lambda I)$$

$\Leftarrow$

Strategy:

Assume $Tu \neq 0$ but $T^2 u = 0$, can we prove it's not possible to write

$$V = \text{null } T \oplus \text{range } T$$

Of course we can do it, because $Tu \in \text{null } T \cap \text{range } T$.

Put it formally, let $p(z)$ be the minimal polynomial of $T$. Since $V$ is a complex vector space, $p(z) = (z - \lambda_1) \cdots (z - \lambda_m)$

If $\lambda_i$ are not all different, then we can write $p(z)$ as

$$p(z) = (z-\lambda)^2 q(z)$$

We can find $u$ such that $w = (T - \lambda I)q(T)u \neq 0 \neq 0$, otherwise $(T - \lambda I)q(T) = 0$ and its degree is less than $p(z)$ which is contradictary to the assumption that $p(z)$ is the minimal polynomial.

Then $w \in \text{null } T - \lambda I$ since

$$(T - \lambda I)w = (T - \lambda I)^2q(T)u = p(T) u = 0$$

On other other hand $w = (T - \lambda I)q(T)u$, so $w \in \text{range } T - \lambda I$.

So $w \in \text{null } T - \lambda I \cap \text{range } T - \lambda I$.

Then $V \neq \text{null } (T- \lambda I) \oplus \text{range } (T - \lambda I)$

Which is contradictary to our assumption.

Then $\lambda_i$ are all distinct. Then we can use "5.62 necessary and sufficient condition for diagonalizability" To conclude $𝑇$ is diagonalizable.

$\square$

5D.6

Suppose $𝑇 ∈ ℒ(𝐅^5)$ and $\dim 𝐸(8, 𝑇) = 4$. Prove that $𝑇 − 2𝐼$ or $𝑇 − 6𝐼$ is invertible.

Proof: We prove by contradition.

Assume $𝑇 − 2𝐼$ and $𝑇 − 6𝐼$ are all non-invertible. Then $\text{null } T-2I \neq \{0\}$ and $\text{null } T-6I \neq \{0\}$.

Then $\dim E(2,T) \geq 1$ and $\dim E(6,T) \geq 1$.

So $\dim E(2,T) + \dim E(6,T) + \dim E(8,T) \geq 6 > 5 = \dim F^5$.

This is contradictary to "5.54 sum of eigenspaces is a direct sum".

So Either $\text{null } T-2I = \{0\}$ or $\text{null } T-6I = \{0\}$.

So either $𝑇 − 2𝐼$ or $𝑇 − 6𝐼$ is invertible.

$\square$

5D.7

Suppose $𝑇 ∈ ℒ(𝑉)$ is invertible. Prove that

$$E(\lambda,T) = E(\frac{1}{\lambda},T^{-1})$$

for every $𝜆 ∈ 𝐅$ with $𝜆 ≠ 0$.

Proof:

Let $u \in E(\lambda,T)$. Since $T$ is invertible, then $T$ is injective. Furthermore, $E(\lambda,T)$ is invariant under $T$, so $T|_{E(\lambda,T)}$ is well defined and is also injective, so $T|_{E(\lambda,T)}$ is invertible.

So we can find $v \in E(\lambda,T)$ such that $T(v) = u$. So $\lambda v = u$, i.e. $v = \frac{1}{\lambda}u$ .

So

$$\begin{align*} T^{-1}(u) &= T^{-1} (T(v)) \\ &= v \\ &= \frac{1}{\lambda} u \end{align*}$$

So $u \in E(\frac{1}{\lambda},T^{-1})$, then $E(\lambda,T) \subseteq E(\frac{1}{\lambda},T^{-1})$.

Then if we replace $T$ with $T^{-1}$, and replace $T^{-1}$ with $T$, and replace $\lambda$ with $\lambda^{-1}$, we can show $$E(\frac{1}{\lambda},T^{-1}) \subseteq E(\lambda,T)$$

So $E(\frac{1}{\lambda},T^{-1}) = E(\lambda,T)$

$\square$

5D.8

Suppose $𝑉$ is finite-dimensional and $𝑇 ∈ ℒ(𝑉)$. Let $𝜆_1, …, 𝜆_𝑚$ denote the distinct nonzero eigenvalues of $𝑇$. Prove that

$$\dim E(\lambda_1,T) + \cdots + \dim E(\lambda_m,T) \leq \text{dim range } T$$

Proof:

if $\lambda \neq 0$ and $u \in E(\lambda,T)$, then $T(u) = \lambda u$, so $u = \frac{1}{\lambda} T(u) = T (\frac{1}{\lambda}u) \in \text{range } T$.

Then it means $E(\lambda,T) \subseteq \text{range } T$. So we have

$$E(\lambda_1,T) + \cdots + E(\lambda_m,T) \subseteq \text{range } T$$

Then from "5.54 sum of eigenspaces is a direct sum", we can know

$$E(\lambda_1,T) \oplus \cdots \oplus E(\lambda_m,T) \subseteq \text{range } T$$

Then from "3.94 a sum is a direct sum if and only if dimensions add up", we know

$$\dim E(\lambda_1,T) + \cdots + \dim E(\lambda_m,T) = \dim E(\lambda_1,T) \oplus \cdots \oplus E(\lambda_m,T)$$

So, we have

$$\dim E(\lambda_1,T) + \cdots + \dim E(\lambda_m,T) \leq \text{dim range } T$$

$\square$

5D.9

Suppose $𝑅, 𝑇 ∈ ℒ(𝐅^3)$ each have $2, 6, 7$ as eigenvalues. Prove that there exists an invertible operator $𝑆 ∈ ℒ(𝐅^3)$ such that $𝑅 = S^{-1} 𝑇𝑆$.

Proof:

Let $V = F^{3}$, since $\dim V = 3$ and $R, T$ have 3 different eigenvalues. Then by "5.58 enough eigenvalues implies diagonalizability", we know both $R, T$ have diagonal matrices under some basis.

Assume $R$ has a basis $u_1, u_2, u_3$ and $$R(u_1) = 2 u_1, R(u_2) = 6 u_2, R(u_3) = 7 u_3$$

Similarly, assume $T$ has a basis $v_1, v_2, v_3$ and

$$T(v_1) = 2 v_1, T(v_2) = 6 v_2, T(v_3) = 7 v_3$$

Then define $S$ as to be

$$S(u_1) = v_1, S(u_2) = v_2, S(u_3) = v_3$$

$S$ is invertible since $S$ is injective. We can also see $S^{-1}$ satisfy:

$$S^{-1}(v_1) = u_1, S^{-1}(v_2) = u_2, S^{-1}(v_3) = u_3$$

Then note

$$\begin{align*} S^{-1} TS (u_1) &= S^{-1} T(v_1) \\ &= S^{-1} (2 v_1) \\ &= 2 S^{-1} (v_1) \\ &= 2 u_1 \\ &= R(u_1) \\ \end{align*}$$

The same calculation can be applied to $u_2, u_3$. Since $u_1, u_2, u_3$ is a basis of of $R$, and when $R$ and $S^{-1}TS$ are applied to them the results are the same. So $R = S^{-1}TS$.

$\square$

5D.10

Find $𝑅, 𝑇 ∈ ℒ(𝐅^4)$ such that $𝑅$ and $𝑇$ each have $2, 6, 7$ as eigenvalues, $𝑅$ and $𝑇$ have no other eigenvalues, and there does not exist an invertible operator $𝑆 ∈ ℒ(𝐅^4)$ such that $𝑅 = S^{-1}TS$.

Solution:

If $R = S^{-1}TS$, then we use exercise 5A.40 which states:

Suppose $𝑆, 𝑇 ∈ ℒ(𝑉)$ and $𝑆$ is invertible. Suppose $𝑝 ∈ 𝒫(𝐅)$ is a polynomial. Then

$$𝑝(𝑆𝑇𝑆^{−1}) = 𝑆𝑝(𝑇)𝑆^{−1}$$

This means the polynomial of $R$ and $T$ are actually the same.

So our goal is to find 2 operators sucht that they have $2, 6, 7$ as eigenvalues, but their minimal polynomials are not the same.

Let $R$ be an operator such that under standard basis, its matrix is

$$\mathcal{M}(R) = \begin{bmatrix} 2 & 0 & 0 & 0 \\ 0 & 6 & 0 & 0 \\ 0 & 0 & 7 & 0 \\ 0 & 0 & 0 & 7 \\ \end{bmatrix}$$

Then it's minimal polynomial is $p(z) = (z-2)(z-6)(z-7)$.

Let $T$ be an operator such that under standard basis, its matrix is

$$\mathcal{M}(T) = \begin{bmatrix} 2 & 0 & 0 & 0 \\ 0 & 6 & 0 & 0 \\ 0 & 0 & 7 & 1 \\ 0 & 0 & 0 & 7 \\ \end{bmatrix}$$

Then it's minimal polynomial is $q(z) = (z-2)(z-6)(z-7)^2$.

So $p \neq q$. Then it's impossible to find an invertible operator $𝑆 ∈ ℒ(𝐅^4)$ such that $𝑅 = S^{-1}TS$.

$\square$