Let $A$ be a diagonalizable matrix and $\lambda$ be an eigenvalue of A. Prove that rank($\lambda I - A$) = rank(($\lambda I - A)^2$). [closed]

Since $A$ is diagonalizable there is a matrix $U$, with $U U^{-1}=Id$ such that $D=U^{-1}A U$, or $A=UDU^{-1}$, where $D=diag(\lambda,\lambda_1,...\lambda_{d-1})$ is diagoanal and $\lambda_i =$eigenvalues of A ($\lambda_i \neq \lambda_j$ if $i \neq j$) then $rank(\lambda Id-A)=rank(\lambda Id-UDU^{-1})=rank(U(\lambda Id - D)U^{-1})=rank(\lambda Id-D)$, where I used your hint twice. Then I can see that $D-\lambda Id=diag(0,\lambda_1-\lambda,...\lambda_{d-1}-\lambda) $ and $(D-\lambda Id)^2=diag(0,(\lambda_1-\lambda)^2,...(\lambda_{d-1}-\lambda)^2) $. I think that should help you, be careful, in my computation I assumed that the eigenspaces of every $\lambda_i$ is onedimensional!

One approach is as follows: note that the image of $A - \lambda I$ is spanned by eigenvectors of $A$ that do not correspond to $\lambda$. Thus, $\ker(A - \lambda I) \cap \operatorname{im}(A - \lambda I) = \{0\}$. Conclude that $$ \operatorname{rank}(A - \lambda I)^2 = \dim [(A - \lambda I)^2(\Bbb C^n)] \\= \dim[\operatorname{im}[(A - \lambda I)|_{\operatorname{im}(A - \lambda I)}]] \\= \dim\operatorname{im}(A - \lambda I) \\ = \operatorname{rank}(A - \lambda I) $$ as desired.

A method that uses the actual hint: suppose that we have $S \Lambda S^{-1}$, where $S$ is invertible and $\Lambda$ is diagonal with $$ \Lambda = \pmatrix{\lambda_1 I_{m_1} \\ & \lambda_2 I_{m_2}\\ && \ddots \\ &&& \lambda_k I_{m_k}} $$ where the $\lambda_i$ are distinct and $\lambda_1 = \lambda$. Let $B = SMS^{-1}$, where $$ M = \pmatrix{I_{m_1} \\ & (\lambda_2 - \lambda) I_{m_2}\\ && \ddots \\ &&& (\lambda_k- \lambda) I_{m_k}}. $$ Verify that $B$ is invertible and that $B(A - \lambda I) = (A - \lambda I)^2$.