TITLE: Let $A$ be a $n× n$ real matrix with $A^2 = A^T$. Show that every real eigenvalue of $A$ is either $0$ or $1$.
QUESTION [7 upvotes]: Let $A$ be a $n×n$ real matrix with $A^2 = A^T$. Show that every real eigenvalue of $A$ is either $0$ or $1$.

My thoughts:
$A^2 = A^T$
$\implies$ $A.A=A^T$
$\implies$$(A.A)^T=A$
$\implies$ $A^TA^T=A$
$\implies$$A^2A^2=A$
$\implies$ $A^4-A=0$ .
so the real root of the equation $x^4-x=0$ are $0$ & $1$.
  Am I right?

REPLY [2 votes]: As pointed out by julien in his comment, you are going in the right direction. As $x^4-x$ is an annihilating polynomial of $A$, if $A$ has a real eigenvalue $\lambda$, then $\lambda$ must be a real root of $x^4-x$. Hence $\lambda=0$ or $1$.
Alternatively, you may prove the statement as follows. Let $(\lambda,v)$ be an eigenpair of $A$. By assumption, we have $v^TA^2v = v^TA^Tv$. However, since $A^2v=A(Av)=\lambda^2 v$ and $v^TA^T=(Av)^T=\lambda v^T$, the previous equation implies that $\lambda^2 v^Tv=\lambda v^Tv$. As $v$ is an eigenvector, $v^Tv=\|v\|^2\neq0$. Hence $\lambda^2=\lambda$, i.e. $\lambda=0$ or $1$.