I have been stuck on this problem for a really long time and it is due in a few hours.
Let $A$ be a diagonalizable matrix and $λ$ an eigenvalue of $A$. Prove that $$rank(λI - A) = rank((λI - A)^2)$$
Hint 1: For any invertible matrix $B$, $rank(AB) = rank(BA) = rank(A)$.
Hint 2: Diagonalize $λI - A$ and $(λI - A)^2$.
Any help would be greatly appreciated.
Thanks in advance!
EDIT:
Various people have pointed out that a similar thread was opened just recently. I went over and could only partially follow one of the suggested solutions - the others were beyond my level of understanding.
In particular, this is the solution:
Since $A$ is diagonalizable, A can be written in the form $A=PDP^{-1}$, then $rank(λI − A) = rank(λI−PDP^{−1}) = rank(P(λI − D)P^{−1}) = rank(λI − D)$.
The person who suggested this solution also said that he made use of Hint 1 twice. Here is where I am lost. I understand that $rank(λI − A) = rank(λI − PDP^{−1})$, but how does $rank(λI − PDP^{−1}) = rank(P(λI − D)P^{−1}) = rank(λI − D)$? I do not even know how he made use of Hint 1 ):
I apologise in advance to anyone who can take the trouble to explain this solution to me, if it is supposed to be trivial!
To address your edit: note that $$ \begin{align} \operatorname{rank}(\lambda I−PDP^{−1})&= \operatorname{rank}(P(\lambda I)P^{-1} − PDP^{−1}) \\ &= \operatorname{rank}(P[(\lambda I)P^{-1} − DP^{−1}]) \\ &= \operatorname{rank}(P[(\lambda I) − D]P^{−1}) \\ &= \operatorname{rank}(P[(\lambda I − D)P^{−1}]) =\operatorname{rank}((\lambda I − D)P^{−1}) = \operatorname{rank}(\lambda I − D). \end{align} $$