If you find any mistakes, please make a comment! Thank you.

## Chapter 8 Exercise D

1. Solution: By Exercise 11 in section 8B the characteristic polynomial is $z^4$ and by 8.46 this is a polynomial multiple of the minimal polynomial. Since $N^3 \neq 0$, it follows that the minimal polynomial of $N$ is $z^4$.

2. Solution: Similarly, the characteristic polynomial is $z^6$ and a quick computation shows that the matrix of $N$ equals $0$ when raised to the third power, hence the minimal polynomial of $N$ is $z^3$.

3. Solution: Let $\mathcal{M}(N)$ denote the matrix of $N$ with respect to some Jordan basis for $N$. Then $\mathcal{M}(N)$ is a block diagonal matrix of the form
$$\mathcal{M}(N) = \begin{pmatrix} A_1 & & 0\\ & \ddots &\\ 0 & & A_p \end{pmatrix}.$$ Because $N$ is nilpotent, $0$ is the only eigenvalue of $N$ (see Exercise 7 in section 8A). Hence the diagonal entries of $\mathcal{M}(N)$ are all $0$ and each $A_j$ has the following form
$$A_j = \begin{pmatrix} 0 & 1 & & 0\\ & \ddots & \ddots &\\ & & \ddots & 1\\ 0 & & & 0 \end{pmatrix}.$$ Thus every string of consecutive $1$’s corresponds to one of these blocks and its length is the same as the length of the side of the block minus $1$. It is easy to see that if $A_j$ is $n$-by-$n$, then $A_j^{n-1} \neq 0$ and $A_j^n = 0$ (think of it as the matrix of an operator on a $n$ dimensional vector space with respect some basis, each basis vector is mapped to the previous one, except the first one obviously, and to send the last one to $0$ we have to apply the operator $n$ times). Exercise 9 in section 8B now implies that $\mathcal{M}(N)^{m} \neq 0$ and $\mathcal{M}(N)^{m+1} = 0$. Hence the minimal polynomial of $N$ is $z^{m+1}$.

4. Solution: The difference is that the order of the blocks in the diagonal is reversed and the $1$’s appear on the line below the diagonal.

5. Solution: By Exercise 9 in section 8B we just need to square each block on the diagonal of the Jordan form of $T$. In other words, the matrix of $T^2$ is a block diagonal matrix where each block has the following form
$$\begin{pmatrix} \lambda_j & 1 & & 0\\ & \ddots & \ddots &\\ & & \ddots & 1\\ 0 & & & \lambda_j \end{pmatrix}^2 = \begin{pmatrix} \lambda_j^2 & 2\lambda_j & 1 & & 0\\ & \ddots & \ddots & & 0\\ & & \ddots & 2\lambda_j & 1\\ 0 & & & \lambda_j^2 & 2\lambda_j \\ 0 & & & 0 & \lambda_j ^2 \end{pmatrix}.$$

6. Solution: No vector in the span of
$$N^{m_1-1}v_1, \dots, Nv_1, v_1, \dots, N^{m_n-1}v_n, \dots, Nv_n, v_n$$ is in $\operatorname{null} N$, because applying $N$ to any such vector we get a linear of combination of
$$N^{m_1}v_1, \dots, Nv_1, \dots, N^{m_n}v_n, \dots, Nv_n,$$ which is linearly independent. The vectors above are all in $\operatorname{range} N$, therefore, by the Fundamental Theorem of Linear Maps (3.22), the dimension of $\operatorname{null} N$ is at most the dimension of $V$ minus the dimension of the span of the vectors above, that is, at most $n$. Since $N^{m_1}v_1, \dots, N^{m_n} \in \operatorname{null} N$ is linearly independent and has length $n$, it must be a basis of $\operatorname{null} N$.

7. Solution: Let $\lambda_1, \dots, \lambda_m$ denote the distinct zeros of $p$ and $q$. Then
$$p(z) = (z – \lambda_1)^{d_1} \cdots (z – \lambda_m)^{d_m}$$ and
$$q(z) = (z – \lambda_1)^{h_1} \cdots (z – \lambda_m)^{h_m}$$ for some positive integers $d_1, \dots, d_m, h_1, \dots, h_m$, where $h_j \ge d_j$ for each $j$ (because $q$ is a polynomial multiple of $p$). Let $A$ equal the following block diagonal matrix $$A = \begin{pmatrix} A_1 & & 0\\ & \ddots &\\ 0 & & A_m \end{pmatrix}$$ where each $A_j$ is the $h_j$-by-$h_j$ matrix defiend by
$$A_j = \begin{pmatrix} \lambda_j & 1 & & & & &\\ & \ddots & \ddots & & & &\\ & & \ddots & 1 & & &\\ & & & \lambda_j & 0 & &\\ & & & & \lambda_j & \ddots &\\ & & & & & \ddots & 0\\ & & & & & & \lambda_j \end{pmatrix}$$ where the $1$’s appear up to the $d_j$-th column and the $0$’s fill the rest. Note that $A$ is a $\deg q$-by-$\deg q$ matrix. Define $T \in \mathcal{L}(\mathbb{C}^{\deg q})$ such that the matrix of $T$ with respect to the standard basis is $A$. Then each $\lambda_j$ is an eigenvalue of $T$, because $A$ is upper triangular and $\lambda_j$ appears on the diagonal of $A$. Moreover, the multiplicity of $\lambda_j$ as an eigenvalue of $T$ is $h_j$, by Exercise 11 in section 8B. Hence, the characteristic polynomial of $T$ is $q$.

It is easy to see that $(A_j – \lambda_j I)^{d_j – 1} \neq 0$ but $(A_j – \lambda_j I)^{d_j} = 0$ (we can use a reasoning similar to that of Exercise 3 to show this). This, together with Exercise 9 from section 8B, shows that the $j$-th block is nonzero in $(A – \lambda_j I)^{d_j-1}$ and is zero in $(A – \lambda_j I)^{d_j}$ (pay attention to this fact). It follows that
$$(A – \lambda_1 I)^{d_1} \cdots (A – \lambda_m I)^{d_m} = 0.$$ Hence $p(T) = 0$. We claim now $p$ is the minimal polynomial of $T$. To see this, suppose that it is not. Then we can subtract $1$ from one of the exponents in the equation above and it will still hold. Thus, we can write
$$0 = p'(T)(T – \lambda_j)^{d_j – 1}$$ for some $j \in \{1, \dots, m\}$ and some polynomial $p’$, with $p'(\lambda_j) \neq 0$. Let $v \in G(\lambda_j, T)$ such that $(T – \lambda_j)^{d_j – 1}v \neq 0$ (this $v$ exists due to the fact we pinpointed before). We have
$$0 = p'(T)(T – \lambda_j)^{d_j – 1}v,$$ but this is contradiction because $(T – \lambda_j)^{d_j – 1}v \in G(\lambda_j, T)$ and $\lambda_j$ is not an zero of $p’$.

8. Solution: Suppose there does not exist a direct sum decomposition of $V$ into two proper subspaces invariant under $T$. Therefore, the block diagonal matrix of $T$ with respect to some Jordan basis for $T$ only has one block (the span of the basis vectors that correspond to each block is invariant under $T$). Therefore $T$ only has one eigenvalue, call it $\lambda$. Exercise 3 now implies that $(z – \lambda)^{\dim V}$ is minimal polynomial of $T$.

For the other direction, we will prove the contrapositive. Suppose we can decompose $V$ into two proper subspaces invariant under $T$. If $T$ has more than one eigenvalue the result is obvious. Assume $T$ has only one eigenvalue, call it $\lambda$. Then $V = G(\lambda, T)$ and there are proper subspaces $U_1, U_2$ of $G(\lambda, T)$ invariant under $T$ such that $G(\lambda, T) = U_1 \oplus U_2$ with $1 \le \dim U_1, \dim U_2 < \dim V$. We have that $(z – \lambda)^{\dim U_1}$ and $(z – \lambda)^{\dim U_2}$ are the characteristic polynomials of $T|_{U_1}$ and $T|_{U_2}$. Thus $(T – \lambda)^{\max\{\dim U_1, \dim U_2\}} = 0$ and so the minimal polynomial of $T$ cannot be $z^{\dim V}$.