Part 1

Since 0 is an eigenvalue, there exists an eigenvector $\mathbf{v}$ such that $L\mathbf{v} = 0 \mathbf{v} = 0$. But then $\mathbf{v} \in \ker(L)$, so $\dim\ker(L) \geq 1$. Since $\ker(L) \neq 0$, $L$ can not be injective.

By the rank-nullity theorem, we must also have $5 = \dim\ker(L) + \dim \operatorname{im}(L)$. But then $\dim \operatorname{im}(L) \leq 5 = \dim {\mathbf{R}}^5$, so $L$ can not be surjective either.

Part 2

Since all eigenvalues are roots of the minimal polynomial and complex roots occur in conjugate pairs, we must have $$\begin{align*} \operatorname{Spec}(L) = \left\{{0, 1 \pm i, 1\pm 2i}\right\}. \end{align*}$$

Moreover, since this is a $5\times 5$ matrix and we have 5 eigenvalues, this is all of them, and we have the characteristic polynomial $$\begin{align*} \chi_L(x) = x(x^2-2x+2)(x^2 - 2x + 5) \in {\mathbf{R}}[x] \end{align*}$$

Since the minimal polynomial $p_L(x)$ must divide the characteristic polynomial and have every eigenvalue as a root, this forces $$\begin{align*} p_L(x) = \chi_L(x). \end{align*}$$

Part 3

If $L\mathbf{x} = \mathbf{x}$, then $\mathbf{x}$ is an eigenvector with eigenvalue $\lambda = 1$. Since $1 \not\in \operatorname{Spec}(L)$, such an $\mathbf{x}$ can not exist, so $L$ has only one fixed point: namely $\mathbf{x} = \mathbf{0}$.

Problem 2

Let $M$ be an $n \times n$ matrix such that $M_{ij} = 1$ for all $i, j$, and consider the possible eigenvectors of $M$.

We have $$\begin{align*} M [1,1, \cdots, 1]^t = [n, n, \cdots, n]^t = n[1,1,\cdots, 1]^t, \end{align*}$$

which exhibits $\mathbf{x} = [1,1,\cdots, 1]$ as an eigenvector with eigenvalue $\lambda = n$.

Now consider $$\begin{align*} \mathbf{x}_j \coloneqq\mathbf{e}_1 - \mathbf{e}_j = [1, 0,0, \cdots, 0,-1,0,\cdots, 0] \end{align*}$$ which has a $1$ in the $1$st coordinate and a $-1$ in the $j$th coordinate.

Then

\begin{align*}
M \mathbf{x}_j = 
\left[\begin{array}{c} 
1 + 0 + \cdots + 0 + (-1) + 0 + \cdots + 0 \\ 
1 + 0 + \cdots + 0 + (-1) + 0 + \cdots + 0 \\ 
\vdots 
\end{array}\right]
= [0,0,\cdots, 0]^t
,\end{align*}

which exhibits each $\mathbf{x}_j$ as an eigenvector with eigenvalue $\lambda = 0$.

But the set $\left\{{ \mathbf{x}_j \mathrel{\Big|}2 \leq j \leq n}\right\}$ with eigenvalue $0$ contains $n-1$ distinct eigenvectors, and we have an additional 1 eigenvector with eigenvalue $1$, which yields $n$ distinct eigenvectors.

So $M$ is fact diagonalizable and given by $$\begin{align*} JCF(M) = (n-1)J_0^{1} \oplus J_n^1 = \left[\begin{array}{ccccc} 0 & 0 & 0 & \cdots & 0 \\ 0 & 0 & 0 & \cdots & 0 \\ 0 & 0 & 0 & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & n \\ \end{array}\right] \end{align*}$$

Problem 3

Part 1

Note that we can’t have $T^j = 0$ for any $j\leq 4$, since then $T^5 = T^{5-k}T^k = T^{5-k} 0 = 0$, contradicting $T^5 \neq 0$.

So in fact $p_T(x) = x^6$ is the minimal polynomial of $T$, and since $V$ is 6 dimensional, the degree of the characteristic polynomial $\chi_T(x)$ is 6. Since $p_T \divides \chi_T$, and both are monic polynomials of degree 6, we in fact have $$\begin{align*} p_T(x) = \chi_T(x) = x^6. \end{align*}$$

But this means $T$ has eigenvalue $\lambda = 0$ with multiplicity 6. This means

• The size of the largest Jordan block associated to $\lambda = 0$ is size 6, since $0$ has multiplicity 6 in $p_T$, and
• The sum of the sizes of all Jordan blocks associated to $\lambda = 0$ is 6, since $0$ has multiplicity 6 in $\chi_T$

which forces $JCF(T)$ to have a single Jordan block of size 6, i.e. $$\begin{align*} JCF(T) = J_0^6 = \left[\begin{array}{cccccc} 0 & 1 & 0 & 0 & 0 & 0\\ 0 & 0 & 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 0 & 0 & 1\\ 0 & 0 & 0 & 0 & 0 & 0\\ \end{array}\right] \end{align*}$$

Part 2

By part (1), we know that these conditions uniquely specify their Jordan forms, so we have $M\coloneqq JCF(T) = JCF(S)$.

Moreover, since $M = JCF(T)$, we know there is a matrix $P$ such that $T = PMP^{-1}$.

Similarly, we know there is a matrix $Q$ such that $S = QMQ^{-1}$.

But then $P^{-1}TP = M$, and so $$\begin{align*} S = QMQ^{-1}= Q(P^{-1}T P) Q^{-1}= (QP^{-1}) T (QP^{-1})^{-1}\coloneqq ATA^{-1} \end{align*}$$

where $A = QP^{-1}$ is a product of invertible matrices and thus invertible.