Eigenvalue of normal matrices

Question

$A$ is a (complex) normal matrix, i.e. $AA^*=A^*A$ and if $Av=\lambda v$, show that $A^* v=\overline{\lambda}v$. Is there any direct and direct proof which does not involve the fact that it is unitarily diagonalizable by just exploiting the condition on inner product or so? I could not find it.

Answer 1

Let $B$ be normal. Then $$ \|B^*x\|^2 = (B^*x,B^*x) = (BB^*x,x) = (B^*Bx,x) = (Bx,Bx) = \|Bx\|^2. $$ Hence, $\ker B^* = \ker B$.

Now, with $A$ being normal also $A-\lambda I$ is normal with adjoint $A^*-\overline\lambda I$.