Let $A \in \mathbb{C}^{n \times n}$ be hermitian. Prove all eigenvalues of $A$ are real and that eigenvectors corresponding to different eigenvalues are orthogonal.
(1.) The eigenvalues of A are real.
proof
Let $\lambda$ be an arbitrary eigenvalue of $A$, then there exists an $x \in \mathbb{C}^n$ such that $Ax = \lambda x$. Now
$$x^*Ax = x^*(Ax) = x^*\lambda x = \lambda||x||^2 \,\,\,\,\,\,\,\,\text { and } \\ x^*Ax = x^*A^*x = (Ax)^*x = (\lambda x)^*x= \lambda ^*||x||^2$$
So
$$ \lambda ||x||^2 = \lambda^*||x||^2 \\ \lambda = \lambda^*$$
Which is only possible if $Im(\lambda) = 0$.
(2.) The eigenvectors corresponding to different eigenvalues are orthogonal.
proof
Let $\lambda_x, \lambda_y $ be distinct eigenvalues of $A$ corresponding to vectors $x,y$. Then
$$(Ax)^*(Ay) = {\lambda}_{x}^* {\lambda}_{y} x^*y = {\lambda}_{x} {\lambda}_{y} x^*y$$
But
$$ (Ax)^*(Ay) = x^*A^*Ay = x^*A^*A^*y = (AAx)^*y = (\lambda_x^2 x)^*y = \lambda_x^2 x^*y $$
which implies
$$\lambda_x^2 x^*y = {\lambda}_{x} {\lambda}_{y} x^*y$$
And since (by assumption) $\lambda_x \neq \lambda_y$ the equality only holds if $x^*y = 0$.
Better ways to do it?
As pointed out in comments, I did not account properly for the case (in part 2) of a zero eigenvalue. The better way to do it, as mentioned by arugula is
$$\lambda_xx^*y = \lambda^*_xx^*y = (Ax)^*y = x^*A^*y = x^*(Ay) = \lambda_yx^*y$$
So that
$$(\lambda_x - \lambda_y) x^*y = 0$$
And by assumption $\lambda_x \neq \lambda_y$ so it must be that $x^*y = 0$.