# Chapter 8 Exercise C

Exercise 1

Because

$$4 = \operatorname{dim} \mathbb{C}^4 = \operatorname{dim} G(3, T) + \operatorname{dim} G(5, T) + \operatorname{dim} G(8, T),$$

it follows that the multiplicities of the eigenvalues of $T$ are at most $2$. Thus $p(z) = (z-3)^2(z-5)^2(z-8)^2$ is a polynomial multiple of the characteristic polynomial. Therefore, we must have $p(T) = 0$.

Exercise 2

We have

$$1 \le \dim G(5, T), \dim G(6, T) \le n – 1.$$

Hence $p(z) = (z – 5I)^{n-1}(z – 6I)^{n-1}$ is a polynomial multiple of the characteristic polynomial. Therefore, $p(T) = 0$.

Exercise 3

Define $T \in \mathcal{L}(\mathbb{C}^4)$ by

$$T(z_1, z_2, z_3, z_4) = (7z_1, 7z_2, 8z_3, 8z_4).$$

Since $E(7, T) \subset G(7, T), E(8, T) \subset G(8, T)$ and $\mathbb{C}^4 = E(7, T) \oplus E(8, T)$, it follows that $E(7, T) = G(7, T)$ and $E(8, T) = G(8, T)$. We have $\dim E(7, T) = \dim E(8, T) = 2$. Hence, the characteristic polynomial of $T$ is

$$(z – 7)^2(z – 8)^2.$$

Exercise 4

$$A = \begin{pmatrix} 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 \end{pmatrix}$$

Let

$$A = \begin{pmatrix} 1 & 1 & 0 & 0\\ 0 & 5 & 1 & 0\\ 0 & 0 & 5 & 0\\ 0 & 0 & 0 & 5 \end{pmatrix}.$$

Then

\begin{aligned} (A – 5I)^2(A – I) &= \begin{pmatrix} -4 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 \end{pmatrix}^2 \begin{pmatrix} 0 & 1 & 0 & 0\\ 0 & 4 & 1 & 0\\ 0 & 0 & 4 & 0\\ 0 & 0 & 0 & 4 \end{pmatrix}\\ &= \begin{pmatrix} -4 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 \end{pmatrix} \begin{pmatrix} 0 & 0 & 1 & 0\\ 0 & 0 & 4 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 \end{pmatrix}\\ &= \begin{pmatrix} 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 \end{pmatrix}. \end{aligned}

Define $T \in \mathcal{L}(\mathbb{C}^4)$ by

$$T(z_1, z_2, z_3, z_4) = (z_1 + z_2, 5z_2 + z_3, 5z_3, 5z_4).$$

Then $\mathcal{M}(T) = A$ and the eigenvalues of $T$ are thus $1$ and $5$ (the entries on the diagonal). Now 8.36 implies that the minimal polynomial of $T$ is a polynomial multiple of $(z – 5)(z – 1)$. The previous work shows that $(T – 5I)(T – I) \neq 0$ and $(T – 5I)^2(T – I) = 0$. Hence $(z – 5)^2(z – 1)$ is the minimal polynomial of $T$. By Exercise 11 in section 8B, the multiplicity of $1$ is $1$ and of $5$ is $3$. Thereby the characteristic polynomial of $T$ is $(z – 1)(z – 5)^3$.

Exercise 5

Let

$$A = \begin{pmatrix} 0 & 0 & 0 & 0\\ 0 & 1 & 1 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 3 \end{pmatrix}.$$

Then

\begin{aligned} (A – I)^2(A – 3)A &= \begin{pmatrix} -1 & 0 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 2 \end{pmatrix}^2 \begin{pmatrix} -3 & 0 & 0 & 0\\ 0 & -2 & 1 & 0\\ 0 & 0 & -2 & 0\\ 0 & 0 & 0 & 0 \end{pmatrix} \begin{pmatrix} 0 & 0 & 0 & 0\\ 0 & 1 & 1 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 3 \end{pmatrix}\\ &= \begin{pmatrix} -1 & 0 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 2 \end{pmatrix}^2 \begin{pmatrix} 0 & 0 & 0 & 0\\ 0 & -2 & -1 & 0\\ 0 & 0 & -2 & 0\\ 0 & 0 & 0 & 0 \end{pmatrix}\\ &= \begin{pmatrix} -1 & 0 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 2 \end{pmatrix} \begin{pmatrix} 0 & 0 & 0 & 0\\ 0 & 0 & -2 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 \end{pmatrix}\\ &= \begin{pmatrix} 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 \end{pmatrix}. \end{aligned}

Define $T \in \mathcal{L}(\mathbb{C}^4)$ by

$$T(z_1, z_2, z_3, z_4) = (0, z_2 + z_3, z_3, 3z_4).$$

Then $\mathcal{M}(T) = A$. The rest of this exercise is almost the same as the previous one.

Exercise 6

Define $T \in \mathcal{L}(\mathbb{C}^4)$ by

$$T(z_1, z_2, z_3, z_4) = (0, z_1, z_2, 3z_4).$$

Then, the standard basis of $\mathbb{C}^4$ consists of eigenvectors of $T$ corresponding to the eigenvalues $0, 1, 1, 3$. Applying $T(T – I)(T – 3)$ to each of these basis vectors shows that $T(T – I)(T – 3) = 0$. Hence $z(z-1)(z-3)$ is the minimal polynomial of $T$. We have

$$\mathcal{M}(T) = \begin{pmatrix} 0 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 3 \end{pmatrix}.$$

Thus, by Exercise 11 in section 8B, the characteristic polynomial of $T$ is $z(z-1)^2(z-3)$.

Exerise 7

By Exercise 4 in section 5B, we have

$$V = \operatorname{null} P \oplus \operatorname{range} P. \tag{1}$$

It is easy to check that if $v \in \operatorname{range} P$ then $Pv = v$. Thus

$$\operatorname{null} P \subset G(0, T) \text{ and } \operatorname{range} P \subset G(1, T).$$

$(1)$ and 8.26 then imply that these inclusions are actually equalities, which gives the desired result.

Exercise 8

Let $p$ denote the minimal polynomial of $T$. We have

\begin{aligned} T \text{ is invertible } &\iff 0 \text{ is not an eigenvalue of } T\\ &\iff p(0) \neq 0\\ &\iff \text{the consant term of p is nonzero}, \end{aligned}

where the equivalence between the first and second lines follows from 8.49.

Exercise 9

Since

$$4 + 5T – 6T^2 – 7T^3 + 2T^4 + T^5 = 0,$$

multiplying both sides by $T^{-5}$ we get

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

Hence $4z^5 + 5z^4 – 6z^3 – 7z^2 + 2z + 1$ is a polynomial multiple of the minimal polynomial of $T^{-1}$ (by 8.46). As it turns out, this is actually the minimal polynomial of $T^{-1}$ (with the coefficients multiplied by $4$). To see this, suppose by contradiction that it is not. Hence, the minimal polynomial of $T$ has degree at most $4$. This means that

$$a_0 I + a_1 T^{-1} + a_2 T^{-2} + a_3 T^{-3} + a_4 T^{-4} = 0$$

for some $a_0, a_1, a_2, a_3, a_4 \in \mathbb{F}$, not all equal $0$. Multiplying both sides of the equation above by $T^4$, we get

$$a_0 T^4 + a_1 T^3 + a_2 T^2 + a_3 T + a_4 I = 0,$$

which is a contradiction, because the minimal polynomial of $T$ has degree $5$.

Exercise 10

Let $\lambda_1, \dots, \lambda_m$ denote the distinct eigenvalues of $T$ and $d_1, \dots, d_m$ their corresponding multiplicities. Then

$$p(z) = (z – \lambda_1)^{d_1}\cdots(z – \lambda_m)^{d_m}.$$

The eigenvalues of $T^{-1}$ are $\frac{1}{\lambda_1}, \dots, \frac{1}{\lambda_m}$ and by Exercise 3 in section 8A they have multiplicities $d_1, \dots, d_m$. Thus

\begin{aligned} q(z) &= \left(z – \frac{1}{\lambda_1}\right)^{d_1}\cdots\left(z – \frac{1}{\lambda_m}\right)^{d_m}\\ &= z^{d_1}\left(1 – \frac{1}{\lambda_1 z}\right)^{d_1}\cdots z^{d_m}\left(1 – \frac{1}{\lambda_m z}\right)^{d_m}\\ &= z^{d_1 + \dots + d_m}\left(1 – \frac{1}{\lambda_1 z}\right)^{d_1}\cdots \left(1 – \frac{1}{\lambda_m z}\right)^{d_m}\\ &= z^{\dim V}\left(1 – \frac{1}{\lambda_1 z}\right)^{d_1}\cdots \left(1 – \frac{1}{\lambda_m z}\right)^{d_m}\\ &= z^{\dim V} \frac{1}{\lambda_1^{d_1}}\left(\lambda_1 – \frac{1}{z}\right)^{d_1}\cdots \frac{1}{\lambda_m^{d_m}}\left(\lambda_m – \frac{1}{z}\right)^{d_m}\\ &= z^{\dim V} \frac{1}{\lambda_1^{d_1}\cdots\lambda_m^{d_m}}\left(\lambda_1 – \frac{1}{z}\right)^{d_1}\cdots\left(\lambda_m – \frac{1}{z}\right)^{d_m}\\ &= z^{\dim V} \frac{(-1)^{d_1 + \cdots + d_m}}{\lambda_1^{d_1}\cdots\lambda_m^{d_m}}\left(\frac{1}{z} – \lambda_1\right)^{d_1}\cdots\left(\frac{1}{z} – \lambda_m\right)^{d_m}\\ &= z^{\dim V} \frac{(-1)^{d_1 + \cdots + d_m}}{\lambda_1^{d_1}\cdots\lambda_m^{d_m}}p\left(\frac{1}{z}\right)\\ &= z^{\dim V} \frac{1}{p(0)} p\left(\frac{1}{z}\right). \end{aligned}

Exercise 11

Let $z^m + \dots + a_1z + a_0$ be the minimal polynomial of $T$. Then

$$a_0I = -T^m – \dots – a_1T$$

Multiplying both sides of the equation above by $\frac{1}{a_0}T^{-1}$ gives the desired result (note that $a_0$ is nonzero by Exercise 8).

Exercise 12

Suppose $V$ has a basis consisting of eigenvectors of $T$. Let $\lambda_1, \dots, \lambda_m$ denote the distinct eigenvalues of $T$. Then, it is easy to see that

$$(T – \lambda_1 I) \cdots (T – \lambda_m I) = 0$$

by applying the left side of the equation to each of the basis vectors, because the parentheses commute. By 8.46, $(z – \lambda_1) \cdots (z – \lambda_m)$ is a polynomial multiple of the minimal polynomial of $T$. This polynomial has no repeated zeros. Hence the minimal polynomial has no repeated zeros.

Conversely, suppose the minimal polynomial of $T$, call it $p$, has no repeated zeros. By 8.23, $T$ has a basis of generalized eigenvectors. Let $v$ be one vector in this basis. Then $v \in G(\lambda, T)$ for some eigenvalue $\lambda$ of $T$. By 8.49, $\lambda$ is zero of $p$. Thus, we can write $p(z) = (z – \lambda)q(z)$ for some polynomial $q$ with $q(\lambda) \neq 0$. We have

$$0 = p(T)v = q(T)(T – \lambda)v.$$

Note that $(T – \alpha)\hat{v} \neq 0$ for all nonzero $\hat{v} \in G(\lambda, T)$ and all $\alpha \in \mathbb{F}$ with $\alpha \neq \lambda$. This implies that $(T – \lambda)v = 0$, since $G(\lambda, T)$ is invariant under every polynomial of $T$ and so $(T-\lambda I)v \neq 0$ implies $q(T)(T-\lambda I) \neq 0$ (because we can factor $q(T)$ and $\lambda$ is not a zero of $q$). Therefore $v$ is an eigenvector of $T$ and the basis of $V$ consisting of generalized eigenvectors of $T$ actually consists of eigenvectors of $T$.

Exercise 13

If $\mathbb{F} = \mathbb{C}$, this follows directly from the previous exercise from the Complex Spectral Theorem (7.24).

Exercise 14

As you can see in the solution to Exercise 10, the constant term in the characteristic polynomial, which equals $p(0)$, is is $1$ or $-1$ times the product of the eigenvalues of $S$ raised to the their respective multiplicities. The eigenvalues of $S$ all have absolute value $1$ (see 7.43 (b)), hence $|p(0)| = 1$.

Exercise 15

(a) Just repeat the proof of 8.40 replacing $I$ with $v$, $T^j$ with $T^jv$ and $n^2$ with $n$.

(b) This is essentially the same as the proof of 8.46.

Let $q$ denote the minimal polynomial of $T$. Then we also have $q(T)v = 0$. Furthermore, $\deg q \ge \deg p$. By the Division Algorithm for Polynomials (4.8), there exist $s, r \in \mathbb{P}(\mathbb{F})$ such that

$$q = sp + r$$

and $\deg r < \deg p$. This implies that

$$0 = q(T)v = s(T)p(T)v = r(T)v = r(T)v.$$

By the same reasoning used in the proof, it follows that $r = 0$.

Exercise 16

We have

$$a_0I + a_1T + a_2T^2 + \dots + a_{m-1}T^{m-1} + T^m = 0.$$

Taking the adjoint of it side yields

$$\overline{a_0}I + \overline{a_1} T^* + \overline{a_2} (T^*)^2 + \dots + \overline{a_{m-1}}(T^*)^{m-1} + (T^*)^m = 0$$

and we see that the minimal polynomial of $T^*$ is

$$\overline{a_0} + \overline{a_1} z + \overline{a_2} z^2 + \dots + \overline{a_{m-1}}z^{m-1} + z^m.$$

To see this, suppose by contradiction this is not the minimal polynomial of $T^*$. Let $p$ denote the minimal polynomial $T$. Then $\deg p < m$. Because $p(T^*) = 0$, taking the adjoing of each side as we did above shows that $\overline{p}(T) = 0$, where $\overline{p}$ equals $p$ with conjugated coefficients. But $\deg \overline{p} < m$, which is a contradiction because the minimal polynomial of $T$ has degree $m$.

Exercise 17

The characteristic polynomial of $T$, call it $q$, has degree $\dim V$ (see 8.36) and is a polynomial multiple of the minimal polynomial of $T$ (see 8.48), call it $p$. Because $q$ is a multiple of $q$, it follows that $q = ps$ for some polynomial $s$ and, because $\deg q = \deg p$, it follows that $\deg s = 0$. Hence $s$ is a constant. Since they’re both monic, $s = 1$ and they should be equal.

Exercise 19

This follows directly from Exercise 11 in section 8B.

Exercise 20

This is a bit obvious if you realize that

$$\dim G(\lambda, T) = \dim G(\lambda, T|_{V_1}) + \dots + G(\lambda, T|_{V_m}) \tag{2}$$

for every $\lambda \in \mathbb{F}$. This is true because the subspaces on the right side of the equation are contained in $G(\lambda, T)$ and because when we write each $V_j$ as a direct sum of generalized eigenspaces of $T|_{V_j}$ in the equation

$$V = V_1 \oplus \dots \oplus V_m$$

the only way the dimensions will fit is if $(2)$ is true.