'Diagonalization' of Jordan block

Question

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!

Answer 1

Take$$S=\begin{bmatrix}1&0&0&\cdots&0\\0&\varepsilon^{-1}&0&\cdots&0\\0&0&\varepsilon^{-2}&\cdots&0\\\vdots&\vdots&\vdots&\ddots&\vdots\\0&0&0&\cdots&\varepsilon^{-(n-1)}\end{bmatrix}$$(where $n$ is such that $J$ is a $n\times n$ matrix). Then$$SJS^{-1}=\begin{bmatrix}\lambda&\varepsilon&0&\cdots&0\\0&\lambda&\varepsilon&\cdots&0\\0&0&\lambda&\cdots&0\\\vdots&\vdots&\vdots&\ddots&\vdots\\0&0&0&\cdots&\lambda\end{bmatrix}.$$

Answer 2

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 $a_{i,i+1}\ne 0$.