When $A$ is a $n \times n $ Jordan block in $M_n(F)$, what is the Jordan form of $A^m$? Assume that $F$ be a field ( in special case let $F=\mathbb C$). Assume that $A$  be a  $n \times n $ Jordan block in $M_n(F)$. When $m$ is a natural number,  what is the Jordan form of $A^m$?
 A: Denote by $J_n(\lambda)$ the Jordan block of size $n$ for an eigenvalue $\lambda$. Hence $A=J_n(\lambda)$ for some $\lambda$.
(I) When $A$ is not nilpotent and the field has characteristic zero, the Jordan form of $A^m$ is just $J_n(\lambda^m)$. That is, the Jordan form itself is a single Jordan block of size $n$.
(II) When $A$ is not nilpotent and the field has finite characteristic, things get messier. For instance, over $GF(2)$, consider the case where $n=3,m=2$ and
$$
A=\pmatrix{1&1\\ &1&1\\ &&1},\quad A^m=A^2=\pmatrix{1&0&1\\ 0&1&0\\ 0&0&1}
\sim \pmatrix{1&1\\ 0&1\\ &&1}.
$$
So, the $A^m$ in this case has Jordan blocks of sizes smaller than $n$. The general case depends on $n,m$, the actual value of the eigenvalue as well as the characteristic of the field. It looks complicated and I'm not sure if there is an easy answer.
(III) When $A$ is nilpotent, there are two possibilities:


*

*$1\le m<n$. Then $T=A^m$ is the matrix whose only nonzero entries are a line of ones extended from the $(1,m+1)$-th entry southeastward. Hence the only nontrivial cyclic subspace of $T$ is subspace spanned by the subset $\{e_n,e_{n-1},e_{n-2},\ldots,e_m\}$ of standard basis. As the dimension of this cyclic subspace is $(n+1-m)$, the Jordan form of $T$ is $J_{n+1-m}(0)\oplus0_{(m-1)\times(m-1)}$.

*$m\ge n$. Then $T=A^m=0$ and the Jordan form of $T$ is just itself, the zero matrix.

