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
$$ \begin{pmatrix} \lambda_j & 1 & & 0\\ & \ddots & \ddots &\\ & & \ddots & 1\\ 0 & & & \lambda_j \end{pmatrix}^2 = \begin{pmatrix} \lambda_j^2 & 2\lambda_j & & 0\\ & \ddots & \ddots &\\ & & \ddots & 2\lambda_j\\ 0 & & & \lambda_j^2 \end{pmatrix}. $$

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 $z^{\dim V}$.

About Linearity

This website is supposed to help you study Linear Algebras. Please only read these solutions after thinking about the problems carefully. Do not just copy these solutions.
This entry was posted in Chapter 8 and tagged .