Eigenvalues of a matrix whose square equals its transpose

Question

Let $A$ be a $n\times n$ matrix with $A^2=A^t$. Show that every real eigen value of $A$ is either $ 0$ or $1$

Answer 1

We have

$$A^2=A^t\implies (A^t)^2=(A^2)^2=A^4=A$$ hence the polynomial $P=x^4-x=x(x^3-1)=x(x-1)(x^2+x+1)$ annihilates $A$ and then the eigenvalues of $A$ belong to the set of roots $P$. The real roots are $0$ and $1$.

Answer 2

Hint: Note that the characteristic polynomial of $A$ is the same as the one of $A^t$. Therefore the eigenvalues of $A$ are the same as the ones of $A^t$.