The eigenvalues and eigenvectors of Hermitian matrices have some special properties. First of all, the eigenvalues must be real! To see why this relationship holds, start with the eigenvector equation \begin{equation} M |v\rangle = \lambda |v\rangle \label{eigen} \end{equation} and multiply on the left by $\langle v|$ (that is, by $v^\dagger$): \begin{equation} \langle v | M | v \rangle = \langle v | \lambda | v \rangle = \lambda \langle v | v\rangle . \label{vleft} \end{equation} But we can also compute the Hermitian conjugate (that is, the conjugate transpose) of (\ref{eigen}), which is \begin{equation} \langle v| M^\dagger = \langle v| \lambda^* . \end{equation} Using the fact that $M^\dagger=M$, and multiplying by $|v\rangle$ on the right now yields \begin{equation} \langle v | M | v \rangle = \langle v | \lambda^* | v \rangle = \lambda^* \langle v | v \rangle . \label{vright} \end{equation} Comparing (\ref{vleft}) with (\ref{vright}) now shows that \begin{equation} \lambda^* = \lambda \end{equation} as claimed.

The second property is that eigenvectors corresponding to different eigenvalues must be orthogonal. The argument establishing this relationship is similar to the one above. Suppose that \begin{align} M |v\rangle &= \lambda |v\rangle ,\\ M |w\rangle &= \mu |w\rangle . \end{align} Then \begin{equation} \langle v | \lambda | w \rangle = \langle v | M | w \rangle = \langle v | \mu | w \rangle \end{equation} or equivalently \begin{equation} (\lambda - \mu) \langle v | w \rangle = 0 . \end{equation} Thus, if $\lambda\ne\mu$, $v$ must be orthogonal to $w$.

This argument can be extended to the case of repeated eigenvalues; it is always possible to find an orthonormal basis of eigenvectors for any Hermitian matrix.