Show that if $v\in (V_c)^{\perp}$ then $(Av)\in (V_c)^{\perp}$ for a normal matrix $A$ with an eigenvalue $c$

Question

Suppose $A \in M_{n\times n}(\mathbb C)$ is a normal matrix and $c$ is an eigenvalue of $A$.

I'm trying to show that if $v\in (V_c)^{\perp}$ then $(Av)\in (V_c)^{\perp}$.

I know that if we were talking about $v'\in V_{c}$ then we knew that $v'c = Av'$.

I also know that $\forall x\in V_c : \langle x,v \rangle = 0$.

And the last thing I was able to conclude is that $AA^*=A^*A$ because $A$ is a normal matrix.

Am I'm looking in the right direction? Nothing of these facts helps me to figure out how to prove it.

Answer 1

A simple and classic result (prove it): if two endomorphisms commute then the eigenspace of one is invariant by the other. Now since $A$ and $A^*$ commute then $V_c$ is invariant by $A^*$ hence for all $w\in V_c$ we have $A^*w\in V_c$ so for $v\in (V_c)^\perp$:

$$\langle Av,w \rangle =\langle v,A^*w\rangle=0$$ and the result follows

Answer 2

Assume that $W_c$ is the matrix with orthonormal columns forming a basis of $V_c$ and let $W_c^\perp$ be such that $W:=[W_c,W_c^\perp]$ is unitary ($W_c$ contains a basis of the orthogonal complement of $V_c$). We have $AW_c=cW_c$, so $$ A[W_c,W_c^\perp]=[W_c,W_c^\perp]\begin{bmatrix}cI & B_{12} \\ 0 & B_{22}\end{bmatrix}=:WB, $$ that is, $A$ is unitarily similar to a block triangular matrix $B$. Now since $A$ is normal, $B$ must be normal as well. Requiring $B^*B=BB^*$ gives $B_{22}$ is normal and $B_{12}=0$, that is, $B$ is block diagonal and hence $AW_c^\perp=W_c^\perp B_{22}$ which gives what you want to prove ($AV_c^\perp=\mathcal{R}(AW_c^\perp)\subset\mathcal{R}(W_c^\perp)=V_c^\perp$).