1. Solution: 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 $p(T) = 0$.

2. Solution: 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, $p(T) = 0$.

3. Solution: 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

$$ (z – 7)^2(z – 8)^2. $$

4. Solution: $$ A = \begin{pmatrix} 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 \end{pmatrix} $$ 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 $(z – 1)(z – 5)^3$.

5. Solution: 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.

6. Solution: 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 $z(z-1)^2(z-3)$.

7. Solution: 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.

8. Solution: 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.

9. Solution: 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 $5$.

10. Solution: 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

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

11. Solution: 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).

12. Solution: 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 $T$.

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

14. Solution: 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 $|p(0)| = 1$.

15. Solution:

(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, $\deg p \ge \deg q$. By the Division Algorithm for Polynomials (4.8), there exist $s, r \in \mathbb{P}(\mathbb{F})$ such that $$ p = sq + r $$ and $\deg r < \deg q$. This implies that

$$ 0 = p(T)v = s(T)q(T)v + r(T)v = r(T)v. $$ By the same reasoning used in the proof, it follows that $r = 0$.

16. Solution: 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 $m$.

17. Solution: 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.

19. Solution: This follows directly from Exercise 11 in section 8B.

20. Solution: 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.