what are the eigenvalues in orthgonal matrix, How to explain? what are the possible eigenvalues of an orthogonal matrix?
I got the answer key which says its 1 and -1 but it doesn't explain well
 A: Another definition of an orthogonal matrix (from MathWorld) is that $A^T = A^{-1}$ (A is square) meaning $A^T A = I$
Recall that the transpose of a square matrix has the same eigenvalues as the original matrix (Meaning $A \vec v = \lambda \vec v = A^T \vec v$)
Taking our eigenvalue equation $A\vec v = \lambda \vec v$ and multiplying by $A^T$ on both sides then gives
$A^T A \vec v =\lambda A^T \vec v \implies  \vec v = \lambda^2 \vec v$
Which is true if and only if $\lambda = \pm 1$
A: Recall that if $U$ is an orthogonal matrix, we then have $\Vert Ux \Vert_2 = \Vert x \Vert_2$ for all $x \in \mathbb{R}^{N}$. 
Let $\lambda$ be an eigenvalue of $U$ and let $v$ be an eigenvector corresponding to $\lambda$. We then have
$$U v = \lambda v \implies \Vert Uv \Vert_2 = \vert \lambda \vert \Vert v \Vert_2$$
But since $U$ is orthogonal, we have $\Vert Uv \Vert_2 = \Vert v \Vert_2$. Hence, this gives us $\vert \lambda \vert  = 1$.
Hence, any real eigenvalue of an orthogonal matrix has to be either $+1$ or $-1$. It is important to note that the orthogonal matrix can have complex eigenvalues.
A: By your definition, $U^t=U\;$ for an orthogonal matrix $\;U\;$ , but since a matrix and its transpose have the same eigenvalues (why?), we get that if $\;v\;$ is an eigenvector of $\;U\;$ corresponding to eigenvalue $\;\lambda\;$ ,then
$$Uv=\lambda v\implies v=vI=U^tUv=U^t(\lambda v)=\lambda U^tv=\lambda^2v\implies \lambda^2=1$$
and we're done.
A: Here is an approach. Recalling the fact 

if $\lambda$ is the eigenvalue of $A$, then $\lambda^{-1}$ is the eigenvalue of $A^{-1}$,

we have
$$ \langle Ax,x \rangle =  \langle x, A^T x \rangle = \langle x, A^{-1} x \rangle $$
$$ \implies \langle \lambda x,x \rangle  = \langle x, \frac{1}{\lambda} x \rangle $$ 
$$ \implies \lambda \langle  x,x \rangle  =\frac{1}{\bar \lambda} \langle x,  x \rangle $$ 
$$ |\lambda|^2 = 1 .$$
