TITLE: Show that the eigenvalues of a unitary matrix have modulus $1$
QUESTION [28 upvotes]: Show that the eigenvalues of a unitary matrix have modulus $1$.

I know that a unitary matrix can be defined as a square complex matrix $A$, such that 
$$AA^*=A^*A=I$$
where $A^*$ is the conjugate transpose of $A$, and $I$ is the identity matrix. Furthermore, for a square matrix $A$, the eigenvalue equation is expressed by $$Av=\lambda v$$
If I use the relationship $u v=v^*u$ and take the conjugate transpose of this equation then
$$v^*A^*=\lambda^*v^*$$
But now I got stuck. Can someone help?

REPLY [9 votes]: A unitary matrix $U$ preserves the inner product:   $\langle Ux, Ux\rangle =\langle x,U^*Ux\rangle =\langle x,x\rangle $.
Thus if $\lambda $ is an eigenvalue,  $Ux=\lambda x$, we get $\vert\lambda \vert^2\langle x,x\rangle =\langle \lambda x,\lambda x\rangle =\langle Ux, Ux\rangle =\langle x,x\rangle $.
So $\vert \lambda\vert^2=1\implies \vert \lambda\vert=1$.