TITLE: 'Diagonalization' of Jordan block
QUESTION [5 upvotes]: We know the Jordan block 
$$J = 
\begin{bmatrix}
 \lambda & 1 & & \\
 & \lambda & \ddots & \\
 & & \ddots & 1\\
 & & & \lambda       
\end{bmatrix}
$$
Prove that there exists an invertible $\bf{S}$ such that
$$\bf{SJS^{-1}} = 
\begin{bmatrix}
 \lambda & \varepsilon & & \\
 & \lambda & \ddots & \\
 & & \ddots & \varepsilon\\
 & & & \lambda       
\end{bmatrix}
$$ 
with any nonzero $\varepsilon$.
I have no idea how to construct the $\bf{S}$, or how to 'diagonalize' the Jordan block. Could anyone give me some hints? Thanks in advance!

REPLY [2 votes]: By Jordan normal form theorem, if $A\in k^{n\times n}$ is triangulable and $B\in k^{n\times n}$, then $A$ and $B$ are similar if and only if $\dim\ker (A-\lambda I)^m=\dim\ker (B-\lambda I)^m$ for all $m\in\Bbb N$ and for all $\lambda\in k$.
This is easily applied to this case, because $J_\varepsilon-\lambda I=\varepsilon(J-\lambda I)$, while $J_\varepsilon-\mu I$ and $J-\mu I$ are both invertible when $\mu\ne \lambda$.
The same lemma and some calculations prove that an upper-triangular matrix is similar to $J$ if and only if, for all $i$, $a_{i,i}=\lambda$ and $a_{i,i+1}\ne 0$.