TITLE: New proof about normal matrix is diagonalizable.
QUESTION [9 upvotes]: I try to prove normal matrix is diagonalizable.
I found that $A^*A$ is hermitian matrix. I know that hermitian matrix is diagonalizable.
I can not go more. I want to prove statement use only this fact.
I need you help.
(professor said that we can prove only use this fact.

REPLY [2 votes]: As you told, $$A^*A=U^*DU$$ for some unitary matrix $U$, and $D$ a diagonal matrix. Since $AA^*=A^*A$, we have: $$(U^*DU)A=AA^*A=A(U^*DU).$$Consider LHS and RHS of this equation, we multiply $U$ on the left, $U^*$ on the right and get: $$D(UAU^*)=(UAU^*)D.$$Now the matrix $UAU^*$ is interchangable with a diagonal matrix. 
If the diagonal entries of $D$ are mutually distinct, then it is easy to show that $UAU^*$ is diagonal as well. Hence we just need to consider the case when $D=\lambda I$. In this case we have $$A^*A=U^*(\lambda I)U=\lambda I.$$Let $v$ be an eigenvector of $A$ with respect to an eigenvalue $\mu$. Then 
\begin{align}
v^*A^*Av=\bar{\mu}v^*\mu v=|\mu|^2|v|^2\\
v^*A^*Av=v^*(\lambda I)v=\lambda|v|^2
\end{align}
Comparing these two equations we get $\lambda=|\mu|^2$. So the matrix $\frac{1}{\mu}A$ is a unitary matrix. Moreover, we know unitary matrices are diagonalizable, so we are done.
$\textbf{Note: }$Maybe this is somewhat the proof you asked for. However, I am afraid this is not particularly easier than the one of @Joh Hughes.