possible eigenvalues of $A$ Let $A$ be an $n\times n$ matrix such that $A^2=A^t$. Then prove that possible real eigenvalues of $A$ are $0,1$.
Let $\lambda$ be an eigenvalue of $A$ then $\lambda^2$ is eigenvalue of $A^2$.
As $A^2=A^t$, $\lambda^2$ is eigenvalue of $A^t$..
Eigenvalue of $A$ are same as eigenvalue of $A^t$..
So, real eigenvalues of $A^t$  are $\{\lambda_1,\cdots,\lambda_r,\lambda_1^2,\cdots,\lambda_r^2\}$.
As number of real eigenvalues are fixed we must have $\lambda_i^2=\lambda_i$ or $\lambda_j$ or $\lambda_j^2$..
$\lambda_i^2=\lambda_j^2$ and $\lambda_i\neq \lambda_j$ implies $\lambda_i=-\lambda_j$ i do not see any contradiction here..
$\lambda_i^2=\lambda_j$ i do not know what to conclude...
$\lambda_i^2=\lambda_i$ then $\lambda_i=0$ or $1$.. which is what i want..
Help me to clear this...
 A: I think you should consider the fact that corresponding eigenvectors of $A$ and $A^2$ are the same. So these $\lambda$ and $\lambda^2$ are both eigenvalues of  the same eigenvectors. Since eigenvalues must be unique, then $\lambda = \lambda^2$, which leads $\lambda = 0$ or $\lambda = 1$.
EDIT:
Suppose $\lambda$ is a real eigenvalue of $A$, with eigenvector $v$; then $Av=\lambda v$ and, easily, also $A^2v=\lambda^2v$. Therefore
$$
A^tv=A^2v=\lambda^2v
$$
and so $v^tA=\lambda^2v^t$. Hence
$$
v^tAv=\lambda^2(v^tv)
$$
but, on the other hand,
$$
v^tAv=v^t(\lambda v)=\lambda(v^tv)
$$
Since $v^tv\ne0$, we get $\lambda^2=\lambda$.
A: If $\lambda$ is an eigenvalue, so is $\lambda^2$, and so is, for the same reason,  $\lambda^{4}$, $\lambda^{8}$... or any $\lambda^{2^k}$. 
Thus, in the case $\lambda$ is not in $\{-1,0,1\}$, it would generate an infinite spectrum, which cannot be.
It remains the case $\lambda=-1$ that has to be eliminated, because all eigenvalues of $A$ are eigenvalues of $A^2$, and these are $\geq 0$.
