 Solution

Let $Hv=\lambda v$ with $v\ne 0$ and $H^{\ast }=H$. Then $v^{\ast }Hv=\lambda v^{\ast }v$, while taking the complex conjugate gives $v^{\ast }Hv=\overline{\lambda }{v}^{\ast }v$. Since $v^{\ast }v>0$, every eigenvalue of a Hermitian matrix is real.

If $Hv=\lambda v$ and $Hw=\mu w$, thenDistinct eigenvalues therefore force $v^{\ast }w=0$: 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.