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.

## Zen

5 Jul 2020E.18 is interesting, just as the author remarks, every monic polynomial can be the characteristic polynomial of some operator (at least if we choose a basis (a space as well) and construct such a matrix as in E.18).

Suppose $P(z) = a_0+a_1z+a_2z^2+\cdots+a_{n-1}z^{n-1}+z^n$, we construct a linear transformation(let $e_1,...,e_n$ be an euclidean basis of the finite complex vector space $V$, $n = dimV$),

$$

\begin{aligned}

T(e_1) &=e_2\\

T(e_2) &=e_3\\

\vdots\\

T(e_{n-1}) &=e_n\\

T(e_n)&=-a_0e_1-a_1e_2-\cdots-a_{n-1}e_n

\end{aligned}

$$

Then, $$

\begin{aligned}

P(T)(e_1) &= a_0I(e_1)+a_1T(e_1)+...+a_{n-1}T^{n-1}(e_1)+T^{n}(e_1)\\

&=a_0e_1+a_1e_2+...+a_{n-1}e_n+T(e_n) = 0

\end{aligned}

$$

What we have now is a polynomial $P(z)$ with degree $n$ and a vector $e_1 \in \mathbf{C}^n$ satisfying $P(T)e_1 = 0$, and we want to prove that $P(z)$ is actually the characteristic polynomial of the operator we just contruct, which reminds us of Ex.15 and Ex.17.

Suppose there exists a polynomial $Q(z)$, let it be $Q(z) = b_0+b_1z+b_2z^2+\cdots+b_{m}z^m$ with degree $m$ less than $deg~P(z) = n$, and $Q(T)e_1 = 0$, then $b_0e_1+b_1e_2+b_2e_3+\cdots+b_{m}e_{m+1} = 0$ which implies $b_0 = b_1 =...=b_m = 0$ since $e_1,...,e_n$ is an euclidean basis.

Thus, $P(z) $ has the smallest degree, and it can divide the minimal polynomial of $T$ and also is unique by E.15.

Since the minimal polynomial of any operator over complex vector space has a degree $k \leq dim(V)$ by $8.36,8.37,8.40$, for our constructed $T$, $k \geq dim(V)$, thus $k = dim(V)$ and the minimal polynomial of $T$ must be $P(z)$ by E.15, as well as the characteristic polynomial of $T$ by Ex.17 therefore (we can even skip the minimal polynomial to prove the equivalence between the characteristic polynomial and $P(z)$) .

## KongMing

4 Jun 2020A typo in Q15(b)

if q is the min poly then it should be p = sq+r by definition of min poly

perhaps a little mistake was overlooked while typing

## Linearity

4 Jun 2020Thank you. $p$ and $q$ are completely messed up.

It is fixed.

## LLL

14 Apr 2020in exercise 13,if V is on R,how can we solve the question?

## Linearity

20 Apr 2020If you think a little bit, you can figure out that you can still enlarge $\mathbb R$ to $\mathbb C$ and then show the minimal polynomial over $\mathbb C$ is a real polynomial.

## Chi Yuan Lau

10 Aug 2019In 5.A,23, we've proved $ST,TS$ have the same egeinvalues , actually , their ( Algebraic) multiplicities are equal !But I don't know how to prove.

## Linearity

10 Aug 2019It can be done by computing determinants. See Do 𝐴𝐵 and 𝐵𝐴 have same minimal and characteristic polynomials?

## Chi Yuan Lau

11 Aug 2019thank you ! That seems likes the easiest way !

## Chi Yuan Lau

10 Aug 2019There is an intersting fact on $\mathrm{ex}7, 8.\mathrm{C} $, if the operator $P$ is of 2-periodic , then $P$ can be diagonalized.

If $P^2=P$ , ,then $P^{2+i}=P$, by ex4 in 5B , we have

$$V=\mathrm{Im}P\oplus\ker P$$

Also , it is easy to verify that $0,1$ are the only egeinvalues of $P$ , we denote $\mathrm{dim}\ker P=n, \mathrm{dim Im}P=m$, and $$\mathrm{dim}G(0,P)=\mathrm{dim}P^{n+m}=\mathrm{dim}E(0,P)=n$$

So what we need is to show

$$\mathrm{dim}G(1,P)=\mathrm{dim}E(1,P)$$

Just note that

$$(T-I)^2=-(T-I)$$

Thus , by the same way , we can gain the conclusion .

However , it may fail to exist for some others periodic operatora , because $(T-I)^k$ may have non businesses with $(T-I)$.

## Chayan Ghosh

12 Dec 2018Can anyone post a solution for 8C problem 20?

## Marcel Ackermann

4 Feb 20188C18) How do the eigenvalues look like for this operator?

Tv=lambda v

=>

-a_0=lambda v_1

v_1 -a_1 v_n = lambda v_2

v_2 - a_2 v_n = lambda v_3

...

subsituting yields: [0] a_0 + a_1 lambda + a_2 lambda^2 + ... + a_{n-1} lambda^{n-1}=0

all eigenvalues fulfill this equation, every polynomial in C has a factorization with lambda_i the eigenvalues, so we know that [0] is the characteristic polynomial.

The minimal polynomial depends on the a_i values.

## Marcel Ackermann

25 Nov 20178C1) Because the multiplicities of the eigenvalues sum up to the dimensionality of the operator, we know that the multiplicity of one of the eigenvalues has to be two. So there are three possibilities for the minimal polynomial: q_1(z)=(z-3)^2(z-5)(z-8) (etc.). Applying T to these q yields q(T)=0 according to the Cayley-Hamilton theorem (8.37). Multiplying 0 with any polynomial yields the 0 polynomial.