Past exam of the mathematical methods I course of the University of Oxford / 2017 / 3 / ii / Solution

A unitary matrix preserves the Hermitian inner product and hence the norm. If $Uv=\lambda v$ with $v\ne 0$, thenDividing by the positive squared norm givesUnlike the Hermitian case, this constrains eigenvalues to the unit circle, not necessarily the real axis.