How to verify the rank of the operator on a finite dimensional vector space V?

Question

let $U$ be any linear operator on a finite dimensional vector space. If rank $U^m = U^{m+1},$ for some positive integer m. Then Rank $U^m =$ Rank $U^k,$ for k $\geq$ m.

For Diagnol matrices, the result is obvious.

Is that neccessary that the characteristic polynomial of finite dimensional space always splits. If so, the problem reduces to prove for only jordan blocks.

Answer 1

Hint Since rank is invariant under similarity, it's enough to prove this for matrices in Jordan Canonical Form.

Additional hint First establish the claim for a Jordan block of arbitrary eigenvalue and size.