\begin{equation}
\langle v | v \rangle
= \langle v | U^\dagger U | v \rangle
= \langle v | \lambda^* \lambda | v \rangle
= |\lambda|^2 \langle v | v \rangle\tag{5.5.3}
\end{equation}
Thus, the eigenvalues of a unitary matrix are unimodular, that is, they have norm 1, and hence can be written as \(e^{i\alpha}\) for some \(\alpha\text{.}\)
Eigenvectors corresponding to different eigenvalues must be orthogonal.
Just as for Hermitian matrices, eigenvectors of unitary matrices corresponding to different eigenvalues must be orthogonal. The argument is similar to the proof for Hermitian matrices in Section 5.2. Suppose that
\begin{align}
U |v\rangle \amp = e^{i\lambda} |v\rangle,\tag{5.5.5}\\
U |w\rangle \amp = e^{i\mu} |w\rangle\text{.}\tag{5.5.6}
\end{align}
We'd like to evaluate \(\langle v | U | w \rangle\) two ways. We know what \(U|w\rangle\) is, but not yet what \(\langle v|U\) is. So we first compute
\begin{equation}
\langle v| U = \langle v| e^{i\lambda}\text{.}\tag{5.5.9}
\end{equation}
We can now compute
\begin{equation}
\langle v | e^{i\lambda} | w \rangle
= \langle v | U | w \rangle
= \langle v | e^{i\mu} | w \rangle\tag{5.5.10}
\end{equation}
so that
\begin{equation}
(e^{i\lambda} - e^{i\mu}) \langle v | w \rangle = 0\text{.}\tag{5.5.11}
\end{equation}
Thus, if \(e^{i\lambda}\ne e^{i\mu}\text{,}\)\(|v\rangle\) must be orthogonal to \(|w\rangle\text{.}\)
As with Hermitian matrices, this argument can be extended to the case of repeated eigenvalues; it is always possible to find an orthonormal basis of eigenvectors for any unitary matrix.
