Minimal polynomial for any power of Jordan block is same as the minimal polynomial of the Jordan block.

Question

Let $J$ be the $n \times n$ Jordan block corresponding to the eigen value $1$. For any natural number $r$ is it true that the minimal polynomial for $J^r$ is $(X-1)^n$ ?

Another way to think about it to produce a cyclic vector of $J^r$. I can’t prove it. I need some help. Thanks.

Answer 1

Hint: write $J=I+N$ where $N$ is the shift matrix. $N$ is nilpotent with index $n$. Now expand $J^r=(I+N)^r=...$ and find out what is the smallest $m$ we need in order to $(J^r-I)^m=0$.

Answer 2

As $r(J-I)=r(J^r-I)$, so geometric multiplicity is $1$in both case are same and hence same minimal polynomial. Here $r$ means rank of matrix.