 Solution

If $Au=\lambda u$ and $Av=\mu v$, the Hermitian matrix identity givesbecause both eigenvalues are real. Thus eigenvectors belonging to distinct eigenvalues are automatically orthogonal. A simple spectrum guarantees this for an arbitrary choice of one eigenvector per eigenvalue.

Within a degenerate eigenspace, arbitrary independent eigenvectors need not be orthogonal. The Gram-Schmidt process always supplies an orthonormal basis of that eigenspace. Accordingly the existence of an orthonormal eigenbasis requires no additional condition, whereas automatic orthogonality of arbitrary independent eigenvectors is ensured by distinct eigenvalues. Scalar multiples of the same nonzero eigenvector are of course not mutually orthogonal.