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.