Let U be nilpotent nxn matrix. $U^{n} = 0, U^{k} \neq 0$ if $k<n$. Therefore, in some basis U is Jordan block, i.e. $U = J_{0,n}$. Let T be semisimple matrix that in some basis is $\begin{pmatrix} t^{\lambda_1} & 0 & \cdots & 0 \\ 0 & t^{\lambda_2} & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & t^{\lambda_n} \end{pmatrix}$
Find all solutions of equation $TUT^{-1} = t^{-m}U, t\in \mathbb C, m \in \mathbb Z$
I assumed that U and T have good form($U=J_{0,n}$, T is diagonal) in same basis. And this gave me $T = \begin{pmatrix} t^{s} & 0 & \cdots & 0 \\ 0 & t^{s+m} & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & t^{s+(n-1)m} \end{pmatrix}$
Seems that this is all solutions of this equations, but I cant prove that.
Any help is welcome.
We consider the equation $TUT^{-1}=t^{-m}U$ where $U=J$, the nilpotent Jordan block of dimension $n$ and $T=[t_{ij}]$. The relation $TU=t^{-m}UT$ implies $t_{i+1,j}=t^mt_{i,j-1}$. Then, for $n=3$, $T$ is in the form: $T=\begin{pmatrix}a&b&c\\d&at^m&bt^m\\e&dt^m&at^{2m}\end{pmatrix}$.
Let $(e_i)_i$ be the canonical basis and $Te_n=\sum_i a_ie_i$. Now $TU^{n-1}e_n=t^{-(n-1)m}U^{n-1}Te_n$, that is $Te_1=t^{-(n-1)m}a_ne_1$; thus, the first column of $T$ has only one non-zero element (on the diagonal); finally $T$ is upper triangular in the form (when $n=3$) $T=\begin{pmatrix}a&b&c\\0&at^m&bt^m\\0&0&at^{2m}\end{pmatrix}$. The converse is easy, that is, all the matrices in this form agree, with the condition $a\not= 0$.