Proof that a Hermitian Matrix has orthogonal eigenvectors, real eigenvalues Here's my feeble attempt:

A = A* is the condition for a Hermitian matrix. So I try expanding both sides in terms of their SVD:

A = USV* and A* = (USV*)* = ... = VSU* 

So equating I get:

VSU* = USV* 

Now maybe through this I can conclude something about how V and U relate to each other but I don't know how... 

Thanks for any help.
The process is fairly straightforward given the fact that any square matrix has at least one eigenvalue and eigenvector.

Suppose $Ax = \lambda x$. Then $\langle x , Ax \rangle = \langle Ax , x \rangle = \overline{\langle x , Ax \rangle} = \lambda \|x\|^2$, hence $\lambda \in \mathbb{R}$.

The point about being Hermitian is that if $x$ is an eigenvector of $A$, then both $\text{sp} \{x\}$ and the subspace $\{x\}^\bot$ are invariant under $A$. To see the latter, suppose $v \bot x$, then $\langle x , Av \rangle = \langle Ax , v \rangle = \lambda \langle x , v \rangle = 0$, hence $A v \bot x$. Now let $v_k$ be a basis for $\{x\}^\bot$, then in the basis $x, v_1,...,v_k$, $A$ must have a block diagonal form:
$$ A \sim \begin{bmatrix} \lambda &amp; \\ &amp; \tilde{A}\end{bmatrix}$$
(By $\sim$ I mean that the two matrices are similar.)
Now find an eigenvalue of $\tilde{A}$ (which must also be Hermitian) and repeat the process until $A$ is diagonalized.
Use the eigenvalue decomposition of the hermitian matrix $A$.

$A=S \Lambda S^{-1}$

$ A^H = (S \Lambda S^{-1})^{H} $

$ A^H = ( S^{-1})^{H} \Lambda^H S^H$

Since all eigenvalues for a hermitian matrix are real, $\Lambda = \Lambda^H$, and since $A$ is hermitian we have

$S \Lambda S^{-1} =  ( S^{-1})^{H} \Lambda S^H$

This shows $S^{-1}=S^H$ implying $SS^H=I$ as required.