So the eigenspace is $Ker(A-λI)$ where $λ$ is an eigenvalue of A and the root subspace is $Ker(A-λI)^r$ where $r$ is the exponent of $(x-λ)$ in the minimal polynomial for $A$. My professor stated that both of these are invariant for $A$, but didn't provide a proof.
Them being invariant means that if $u \in Ker(A-λI)$ then $Au \in Ker(A-λI)$. The same applies for $Ker(A-λI)^r$. So far I've gotten this: $$(A-λI)v=0,$$ $$A(A-λI)v=0,$$ but to move on I'd have to prove that $A$ commutes with $(A-λI)$. Is that really true?
Of course: $$ A(A-\lambda I)=A^2-\lambda A $$ and $$ (A-\lambda I)A=A^2-\lambda A $$ and