Show that if $\lambda$ is a Real eigenvalue of U then $|\lambda| = 1$ Problem
Show that if $\lambda \in R$ and $\lambda$ is a eigenvalue of an orthogonal matrix U , then $| \lambda | = 1 \\$
Was given the hint 

use $U\cdot \hat{x} \ U\cdot \hat{x}$ and $|| \lambda \ \hat{x}|| = |\lambda| \ ||\hat{x}|| \\$


so I tried this
$U\cdot \hat{x} \ U\cdot \hat{x} =  \lambda \ \hat{x} \ \lambda \ \hat{x}
\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \   = {{\lambda}}^2 \ \hat{x} \ \hat{x}
\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \   = {{\lambda}}^2 {||\hat{x}||}^2$
I understand that we are using eigen value property   $A\hat{x} = \lambda \hat{x}$

but I'm not sure how to solve the rest
 I was told to use $ {||\hat{x}||}^2 \neq 0$ but I'm not sure why
 A: Our OP localplutonium was very close; indeed, right on top of a complete solution.  
$U$ is orthogonal if and only if $U^TU = UU^T = I$, by definiton.
If $\lambda \in \Bbb R$ is an eigenvalue of $U$, then there is some vector $\vec x \ne 0$ with
$U\vec x = \lambda \vec x; \tag 1$
since $U$ is orthogonal we have
$\langle U\vec x, U \vec x \rangle = \langle \vec x, U^TU \vec x \rangle = \langle \vec x, I \vec x \rangle = \langle \vec x, \vec x \rangle; \tag 2$
using (1) in (2) we thus find
$\langle \lambda \vec x, \lambda \vec x \rangle = \langle U \vec x, U \vec x \rangle = \langle \vec x, \vec x \rangle; \tag 3$
but
$\langle \lambda \vec x, \lambda \vec x \rangle = \lambda^2 \langle \vec x, \vec x \rangle = \vert \lambda \vert^2 \langle \vec x, \vec x \rangle; \tag 4$
combining (3) and (4),
$\vert \lambda \vert^2 \langle \vec x, \vec x \rangle = \langle\vec x, \vec x \rangle, \tag 5$
whence, since $\vec x \ne 0$, so that $\langle \vec x, \vec x \rangle \ne 0$ (and this is where we use $\Vert \vec x \Vert^2 \ne 0$, so we can cancel out $\langle \vec x,  \vec x \rangle = \Vert \vec x \Vert^2$)
$\vert \lambda \vert^2 = 1, \tag 6$
and so
$\vert \lambda \vert = 1 \tag 7$
as well.
