How does a normal matrix act on the orthogonal complement of its eigenspace? Let $M$ be a $n\times n$ matrix with complex entries. And let $\lambda$ be an eigenvalue of $M$, with corresponding eigenspace $E_M(\lambda)$.
It is easy to see that $M(E_M(\lambda))\subset E_M(\lambda)$, but it is generally not true, that $M(E_M(\lambda)^\bot)\subset E_M(\lambda)^\bot.
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My question: If $M$ is normal, is the latter inclusion true?
I´m not able to prove it, nor can I find a counter example, hence why I´m asking.
 A: This is true. One way to see it is by using the spectral theorem for normal matrices. Denote by $\lambda_1, \dots, \lambda_k$ the distinct eigenvalues of $A$. Then we have $\mathbb{C}^n = \bigoplus_{i=1}^k E_M(\lambda_i)$ and this is an orthogonal direct sum decomposition. Thus, $E_M(\lambda_i)^{\perp} = \bigoplus_{j \neq i} E_M(\lambda_j)$ and the result follows by the fact that $M(E_M(\lambda_j)) \subseteq E_M(\lambda_j)$. 
Alternatively, for a normal matrix it holds that $E_M(\lambda) = E_{M^{*}}(\overline{\lambda})$. Then, if $x \in E_M(\lambda)^{\perp}$ and $y \in E_M(\lambda)$ we have
$$ \left< Mx, y \right> = \left< x, M^{*}y \right> = \left<x, \overline{\lambda}y \right> = \lambda \left< x, y \right> = 0$$
and so $Mx \in E_{M}(\lambda)^{\perp}$.
In fact, one can show that $M$ is a normal matrix if and only if for all $M$-invariant subspaces $W$ (that is, $M(W) \subseteq W$) we have that $W^{\perp}$ is also $M$-invariant (that is, $M(W^{\perp}) \subseteq W^{\perp}$). In particular, this holds for the eigenspaces of $M$.
