Solution
= Solution
If $Uv=\lambda v$ with $v\ne0$, unitarity gives
$$
\|v\|=\|Uv\|=\|\lambda v\|=|\lambda|\|v\|.
$$
\b[Hence $|\lambda|=1$. Every eigenvalue of a unitary matrix therefore lies on the unit circle.]
= Solution
If $Uv=\lambda v$ with $v\ne0$, unitarity gives
$$
\|v\|=\|Uv\|=\|\lambda v\|=|\lambda|\|v\|.
$$
\b[Hence $|\lambda|=1$. Every eigenvalue of a unitary matrix therefore lies on the unit circle.]