Show that $0$ and $1$ are the only real eigenvalues of $A.$

Question

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.

Answer 1

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!!!