Question I am trying to proof something for normal matrices over $\mathbb{C}$. I have stumbled upon this "if and only if" condition for commuting matrices. I think I have proven it, could someone verify whether my proof is correct?
Lemma Let $A,B\in M_{n\times n}(\mathbb{K})$ with $A$ diagonizable. Then $AB=BA$ if and only if all eigenspaces of $A$ are $B$-invariant.
Proof: Let $A$ and $B$ be as in the lemma. $AB=BA$ is equivalent to $(A-\lambda I_n)B=B(A-\lambda I_n)$. Let $E_\lambda$ be an eigenspace of $A$ for an eigenvalue $\lambda\in\sigma(A)$.
Assuming $A$ and $B$ commute, we find $$((A-\lambda I_n)B)E_\lambda = (B(A-\lambda I_n))E_\lambda = B(\{0\}) = \{0\}$$ which implies that $B(E_\lambda)\subseteq E_\lambda$ because $(A-\lambda I_n)|_{E_\mu}$ is injective for $\mu\neq\lambda$ and $\mu\in\sigma(A)$ and eigenvalue.
For the other direction we assume $B(E_\lambda)\subseteq E_\lambda$. Using the same equation from the first direction again, we find that $$((A-\lambda I_n)B)|_{E_\lambda}=(B(A-\lambda I_n))|_{E_\lambda}\equiv 0$$ for all $\lambda\in\sigma(A)$. Because $\mathbb{K}^n=\bigoplus_{\lambda\in\sigma(A)}E_\lambda$ we find that $(A-\lambda I_n)B=B(A-\lambda I_n)$. This proofs that $A$ and $B$ commute.