= Solution
Let $Hv=\lambda v$ with $v\ne0$ and $H^*=H$. Then $v^*Hv=\lambda v^*v$, while taking the <complex conjugate> gives $v^*Hv=\overline\lambda v^*v$. Since $v^*v>0$, every <eigenvalue> of a <Hermitian matrix> is real.
If $Hv=\lambda v$ and $Hw=\mu w$, then
$$
\lambda\,v^*w=(Hv)^*w=v^*Hw=\mu\,v^*w.
$$
Distinct <eigenvalues> therefore force $v^*w=0$: their <eigenvectors> are orthogonal in the <Hermitian inner product>. A degenerate <eigenvalue> occurs more than once in the spectrum, counted with multiplicity. For a <Hermitian matrix>, its <algebraic multiplicity> equals its <geometric multiplicity>, so degeneracy gives an <eigenspace> of dimension greater than one; an <orthonormal basis> can be chosen within it.
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