rank$({A-\lambda I_n})^k$=rank$(B-\lambda I_n)^k$ iff $A$ is similar to $B$

Question

Let $A,B \in M_n(\mathbb{C})$. Prove that rank$({A-\lambda I_n})^k$=rank$(B-\lambda I_n)^k$ for every $k \in \mathbb{N}$ and $\lambda \in \mathbb{C}$ iff they are similar. I know that every matrix has a Jordan form since it's over the field of complex numbers and that they are similar to their Jordan form, but I do not know where to go from there. Any hints on where to start?

Answer 1

Hints:

• Show that $(A-\lambda I_n)^n$ and $(B-\lambda I_n)^n$ are similar.
• Show that $\text{rank}(U)=\text{rank}(V)$ if $U$ and $V$ are similar.