Matrices similarity using the jordan form

Question

Given $A \in M_{n}(\mathbb{C}),$ a matrix of size $n\times n$ with characteristic polynomial $\chi_{A}(X) = (X-1)^n$, I need to show that $A$ is similar to $A^k$ for every $k \in \mathbb{N}$. I've been guided to use the jordan normal form but I can't see how it helps me. Would be glad for some help

Answer 1

This is what I found on BaiduZhidao, and I translated it into English:

Assume $J$ is a $k$-order nilpotent Jordan block. Then for any $0\leq i\leq k$, $\text{rank}\,J^i= k-i$, and for all $i> k$, $\text{rank}\, J^i=0$.

Therefore, if $B$ is a nilpotent $n$-order Jordan canonical matrix, for all $0\leq i\leq n-1$, $$ \text{rank}\, B^i - \text{rank}\, B^{i+1} $$ is the number of Jordan blocks with order no less than $i+1$.

Therefore, the number of Jordan blocks of certain order is totally determined by $\text{rank}\, B^i$.

Furthermore, if $C$ is a $n$-order nilpotent matrix, let $T$ be the invertible matrix such that $B = T^{-1}CT$ is a nilpotent Jordan canonical matrix, then the equivalent class of similarities of $C$ is totally determined by the number of Jordan blocks of $B$ of certain order, which is also determined by $$ \text{rank}\, C^i = \text{rank}\, B^i. $$

Therefore, to show two nilpotent matrices $B,C$ are similar, it suffices to show that $B^k,C^k$ have the same rank for all $k$.

Back to your question, to show $A$ and $A^k$ have the same order, it is equivalent to show $A-I$ is similar to $A^k-I$, where $I$ is the identity matrix.

As all eigenvalues of $A$ is $1$, $A-I$ has only one eigenvalue $0$, and $A-I$ is nilpotent.

On the other hand, $A^k$ has also only one eigenvalue $1$, so $A^k-I$ is also nilpotent.

According to the discussion above, it suffices to show $$ \text{rank}\, (A-I)^i = \text{rank}\, (A^k-I)^i. $$

Note that $$ A^k-I = (A-I)(A^{k-1}+A^{k-2}+\dotsb+I). $$ Denote $P\,\hat{=}\, A^{k-1}+A^{k-2}+\dotsb+I$. As all eigenvalue of $P$ is $k$, $P$ is invertible. As we also have
$$ A^k-I = (A-I)P = P(A-I), $$ so $$ \text{rank}\,(A^k-I)^i = \text{rank}\,(A-I)^iP^i = \text{rank}(A-I)^i, $$ as desired.

Answer 2

A slightly less abstract and more calculative answer!

Given the characteristic polynomial$(x-1)^n$ we know all the eigenvalues are $1$ and the jordan normal form of $A$ is a block diagonal matrix with Jordan blocks of different order in the form.

$$ PAP^{-1}=diag(J_{p_1}(1),J_{p_2}(1)...) $$ Then the $A^k$ is similar to a block diagonal matrix with same block structure. $$ PA^kP^{-1}=diag(J_{p_1}(1)^k,J_{p_2}(1)^k...)\\ $$

Given each jordan block defines its own invariant subspace, we just need to show any $J_{p}(1)^k$ is similar to $J_{p}(1)$

take an order $p$ jordan block as example $$J_p(1)=I+N= \begin{bmatrix} 1 & 1 & 0 & 0&... &0\\ 0 & 1 & 1 & 0&... &0\\ 0 & 0 & 1 & 1&... &0\\ &&&&...\\ 0 & 0 & 0 & 0&... &1\\ \end{bmatrix} $$

Then its power $k$ is $$ J_p(1)^k = (I+N)^k= \begin{bmatrix} 1 & \binom{k}{1} & \binom{k}{2} & \cdots & \cdots & \binom{k}{p-1} \\ & 1 & \binom{k}{1} & \cdots & \cdots & \binom{k}{p-2} \\ & & \ddots & \ddots & \vdots & \vdots\\ & & & \ddots & \ddots & \vdots\\ & & & & 1 & \binom{k}{1}\\ & & & & & 1 \end{bmatrix} $$ we just need to show $(I+N)^k$ is similar to $I+N$.

As the only eigenvalues $(I+N)^k$ and $I+N$ have are $1$, to show they are similar / have the same jordan normal form, we just need to show $1$ has the same geometric multiplicity, i.e. $I+N-I$ and $(I+N)^k-I$ have the same kernel dimension / same rank.

We know $rank(N)=p-1$, $$ (I+N)^k-I=\sum_{t=1}^k\binom{k}{t}N^t=kN+\binom{k}{2}N^2+...N^k\\ =N[kI+\binom{k}{2}N+...N^{k-1}] $$ Since $[I+\binom{k}{2}N+...N^{k-1}]$ is full rank, then $rank((I+N)^k-I)=rank N=p-1$.