Proving two matrices have equal rank

Question

Let $A$ be a diagonalizable matrix and $\lambda$ an eigenvalue of $A$. Prove that

$ rank(\lambda I - A) = rank((\lambda I - A)^2) $

Any tips how I should get started proving this statement?

I've tried starting from the definition of A being a diagonalizable matrix i.e. there exists an invertible matrix $P$ such that $P^{-1}AP$ is a diagonal matrix, but have no clue on how to proceed.

Answer 1

If $M$ and $N$ are similar matrices, i.e. if there exists an invertible matrix $P$ such that $N=P^{-1}MP$, then they have the same rank (why?) and the same eigenvalues (again, why?).

Spoilers below.

Let $D=P^{-1}AP$ be the diagonalisation of $A$. Then $$\text{rank}(\lambda I-A)=\text{rank}(P^{-1}(\lambda I-A)P)=\text{rank}(\lambda I-P^{-1}AP)=\text{rank}(\lambda I-D).$$ Similarly, because $P^{-1}(\lambda I - A)^2P = \lambda^2 I -2\lambda P^{-1}AP+P^{-1}APP^{-1}AP = (\lambda^2 I - D)^2$, the original problem now reduces to: given a diagonal matrix $D$ with $\lambda$ its eigenvalue, prove that $$\text{rank}(\lambda I-D)=\text{rank}((\lambda I-D)^2).$$ Now it is clear that $\lambda I - D$ and $(\lambda I-D)^2$, being diagonal matrices themselves, will have the same number of nonzero entries on the diagonal, i.e. the same rank.