TITLE: Show that $0$ and $1$ are the only real eigenvalues of $A.$
QUESTION [6 upvotes]: Let $A\in M_n(\mathbb R)$ such that $A^2=A^T.$ Show that $0$ and $1$ are the only real eigenvalues of $A.$ 
All I can see is that $\det A=0$ or $1.$ I can't proceed any further.

REPLY [5 votes]: Try this:
suppose $\lambda$ is a real eigenvalue of $A$ with real eigenvector $x \ne 0$.  Then we have:
$Ax = \lambda x, \tag{1}$
whence
$\langle x, Ax \rangle = \langle x, \lambda x \rangle \tag{2}$
and
$\langle x, Ax \rangle = \langle A^Tx, x \rangle = \langle A^2x, x \rangle = \langle \lambda^2 x, x \rangle = \langle x, \lambda^2x \rangle; \tag{3}$
combining (2) and (3) yields
$\langle x, \lambda x \rangle =  \langle x, \lambda^2x \rangle \tag{4}$
or
$0 = \langle x, (\lambda - \lambda^2) x \rangle = (\lambda - \lambda^2)\langle x, x \rangle, \tag{5}$
and since $x \ne 0$, this implies
$\lambda - \lambda^2 = 0, \tag{6}$
forcing $\lambda = 0 \; \text{or} \; 1$.  QED.
Hope this helps.  Cheerio,
and as always, 
Fiat Lux!!!