Chapter 4: Special Matrices

### Properties of Unitary Matrices

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

$$\langle v | v \rangle = \langle v | U^\dagger U | v \rangle = \langle v | \lambda^* \lambda | v \rangle = |\lambda|^2 \langle v | v \rangle$$ Assuming $\lambda\ne0$, we thus have $$|\lambda|^2 = 1 .$$ 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 $$\langle v | e^{i\lambda} | w \rangle = \langle v | U | w \rangle = \langle v | e^{i\mu} | w \rangle$$ or equivalently $$(e^{i\lambda} - e^{i\mu}) \langle v | w \rangle = 0 .$$ 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.