# Solve matrix solution $TUT^{-1} = t^{m}U.$

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 $a\not= 0$.
• Thx, but I already solved this problem. I want to share my solution :) We can look at $TUT^{-1} = t^{-m}U$ as at $TU - UT = mU$. Where T - semisimple, U - nilpotent. $U^n = 0, U^{n-1} \neq 0$.Now we look at this equation in basis where $T = diag(t_1, ... , t_n)$. So we have $(t_{i} - t_{j})u_{ij} = mu_{ij}$. Lets build graph with verticies $t_i$ and $t_i \rightarrow t_j$ if $t_i - t_j = m$. No we change basis in the way $t_1, t_2, ..., t_n$ are topologicaly sorted(graph obviously has no cycles). In this basis $(t_{i} - t_{j})u_{ij} = mu_{ij}$ give us that $u_{ij} =0 for j \leq u$. – SaveMyLife May 3 '16 at 13:49
• So that basis makes U upper trianfular. The fact $U^{n-1}\neq 0$ gives us that $u_{ii+1} \neq 0$ (upper diagonal elements). This proves that $t_2 = t_1 +m, t_3 = t_2 + m, ...$. And finaly we have $u_{ij} \neq 0$ for non upper diagonal elements. So we have basis where T is diagonal,$t_2 = t_1 +m, t_3 = t_2 + m, ...$, U is $J_{0,n}$. – SaveMyLife May 3 '16 at 13:56