The eigenvalues and eigenvectors of unitary matrices have some special properties. If $U$ is unitary, then $UU^\dagger=I$. Thus, if \begin{equation} U |v\rangle = \lambda |v\rangle \label{eleft} \end{equation} then also \begin{equation} \langle v| U^\dagger = \langle v| \lambda^* . \label{eright} \end{equation} Combining (\ref{eleft}) and (\ref{eright}) leads to

\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 \end{equation} Assuming $\lambda\ne0$, we thus have \begin{equation} |\lambda|^2 = 1 . \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 $\alpha$.

Just as for Hermitian matrices, eigenvectors of unitary matrices corresponding to different eigenvalues must be orthogonal. The argument is essentially the same as for Hermitian matrices. Suppose that \begin{align} U |v\rangle &= e^{i\lambda} |v\rangle ,\\ U |w\rangle &= e^{i\mu} |w\rangle . \end{align} Then \begin{equation} \langle v | e^{i\lambda} | w \rangle = \langle v | U | w \rangle = \langle v | e^{i\mu} | w \rangle \end{equation} or equivalently \begin{equation} (e^{i\lambda} - e^{i\mu}) \langle v | w \rangle = 0 . \end{equation} Thus, if $e^{i\lambda}\ne e^{i\mu}$, $v$ must be orthogonal to $w$.

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.