## Section6.4Properties of Unitary Matrices

The eigenvalues and eigenvectors of unitary matrices have some special properties. If $U$ is unitary, then $UU^\dagger=I\text{.}$ Thus, if

\begin{equation} U |v\rangle = \lambda |v\rangle\label{eleft}\tag{6.4.1} \end{equation}

then also

\begin{equation} \langle v| U^\dagger = \langle v| \lambda^*\text{.}\label{eright}\tag{6.4.2} \end{equation}

Combining (6.4.1) and (6.4.2) 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\tag{6.4.3} \end{equation}

Assuming $\lambda\ne0\text{,}$ we thus have

\begin{equation} |\lambda|^2 = 1\text{.}\tag{6.4.4} \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 $\alpha\text{.}$

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 \amp = e^{i\lambda} |v\rangle ,\tag{6.4.5}\\ U |w\rangle \amp = e^{i\mu} |w\rangle\text{.}\tag{6.4.6} \end{align}

Then

\begin{equation} \langle v | e^{i\lambda} | w \rangle = \langle v | U | w \rangle = \langle v | e^{i\mu} | w \rangle\tag{6.4.7} \end{equation}

or equivalently

\begin{equation} (e^{i\lambda} - e^{i\mu}) \langle v | w \rangle = 0\text{.}\tag{6.4.8} \end{equation}

Thus, if $e^{i\lambda}\ne e^{i\mu}\text{,}$ $v$ must be orthogonal to $w\text{.}$

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.