Prove rank(λI−A)=rank((λI−A)^2). Given that A is a diagonalizable matrix and λ an eigenvalue of A.

Question

The question gave the hint that :For any invertible matrix B,rank(AB)=rank(A)and rank(BA)=rank(A).

I was thinking to prove that λI−A is invertible, hence by the property of the rank of matrices given, rank(λI−A)=rank((λI−A)^2).

However I realized that since λ is an eigenvalue of A, the det(λI−A) is zero hence is not invertible. Is there anyway to prove this?

Answer 1

It suffices to show the dimension of the null space is the same for both. I.e. eigenvectors and some generalized eigenvectors.

And for diagonalizable matrices there is a basis for $\mathbb R^n$ consisting of just eigenvectors. So all generalized eigenvectors are eigenvectors.