Show that the eigenvalues of a unitary matrix have modulus $1$ 
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?
 A: $\Delta$ as $\lambda$
$Av=\Delta v$
$(Av)^*=(\Delta v)^*$
$v^*A^*=\Delta^*v^*$
$v^*A^*Av=\Delta^*v^*\Delta v$
As $A^*A=I$
$v^*Iv=\Delta^*\Delta v^*v$
$v^*v=\Delta^*\Delta v^*v$
$(1-\Delta^*\Delta)v^*v=0$
Since $v$ is not equal to zero.
Hence
$1-\Delta\Delta^*=0\implies \Delta^*\Delta=1$
$|\Delta|^2=1\implies |\Delta|=1$.
A: You multiply your two relations to obtain 
\begin{align}
v^*A^*Av &=\lambda^* v^*\lambda v \\
v^*Iv &=\left(\lambda^*\lambda\right) v^*v \\
v^*v &=\left(\lambda^*\lambda\right) v^*v \\
||v||^2 &= |\lambda|^2 ||v||^2 \\
\sqrt{1} &=|\lambda| \\
1 &=|\lambda|
\end{align}

Recall that the modulus of a complex number $\lambda = a + bi$, also called the "complex norm", is denoted $|\lambda|$ and defined by $|\lambda| = |a + bi| = \sqrt{a^2 + b^2}$ and $\lambda^*\lambda = (a -bi)(a + bi) = a^2 + b^2$. Hence $\lambda^*\lambda = |\lambda|^2.$
A: The result that you seek follows from the following.
Lemma. If $A$ is unitary and $\vert \vert x \vert \vert_2 = 1$, then $\vert\vert Ax \vert\vert_2 =1$.
Proof. By definition, 
$$ \vert\vert Ax \vert\vert_2^2 =\langle Ax, Ax \rangle = (Ax)^*(Ax) = x^*A^*A x = x^*x = \vert\vert x \vert\vert_2^2 = 1.$$ 

If $(\lambda,v)$ is an eigenpair of $A$, then, without loss of generality, it may be assumed that $\vert\vert v \vert\vert_2 = 1$. Following the lemma above and the absolute homogeneity property of vector norms, notice that 
$$ |\lambda| 
= |\lambda|\vert\vert v \vert\vert_2 
= \vert\vert \lambda v \vert\vert_2 
= \vert\vert Av \vert\vert_2
= 1,$$
as desired. 
A: 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 $\vert \lambda\vert^2=1\implies \vert \lambda\vert=1$.
