Similarity with Jordan Cells

Question

Show that for each nonzero number $a$, $a J(k)$ is similar to $J(k)$.

Show that if $a_1, a_2, \ldots, a_d$ are $d$ nonzero complex numbers and if $k_1,k_2,\ldots,k_d$ are arbitrary positive integers, then the block matrices

\begin{bmatrix} a_1 J(k_1) & 0 & \cdots & 0\\ 0 & a_2 J(k_2) & \cdots & 0\\ \vdots & & \ddots & \vdots\\ 0 & 0 & \cdots & a_d J(k_d)\end{bmatrix}

and \begin{bmatrix} J(k_1) & 0 & \cdots & 0\\ 0 & J(k_2) & \cdots & 0\\ \vdots & & \ddots & \vdots\\ 0 & 0 & \cdots & J(k_d)\end{bmatrix} are similar.

So far I have $J(k)$ denoted as the Jordan cell $J(0,k)$, i.e., $J(k)=J(0,k)= \begin{bmatrix} 0 & 1 & 0 & \cdots & 0\\ 0 & 0 & 1 & \cdots & 0\\ \vdots & & & \ddots & \vdots\\ 0 & 0 & \cdots & \ddots & 1\\ 0 & 0 & 0 & \cdots & 0\end{bmatrix}$

I am clueless after that, please help!

Answer 1

It suffices to show that for, every $a\not=0$, $aJ(0,k)$ and $J(0,k)$ are similar. That is true because, for every integer $p$, $\ker((aJ(0,k))^p)=\ker ((J(0,k))^p)$. (cf. Jordan's theory).

Answer 2

It suffices to note that there can be only one Jordan form for a matrix that only has the eigenvalue $0$ with an eigenspace of dimension $1$.

For the second part, it suffices to show that if $A_1,\dots,A_d$ are each similar to $B_1,\dots,B_d$ respectively, then $$ \pmatrix{A_1\\&\ddots\\&&A_d} \text{ and } \pmatrix{B_1\\&\ddots\\&&B_d} $$ are similar to each other.