Possible eigenvalues of a matrix satisfying certain conditions This problem appeared in my algebra textbook

Suppose $A$ is a real matrix such that $A^2 = A^t.$ What are the
  possible eigenvalues of $A?$

Here is what I tried in order to solve it.
Since the eigenvalues of $A$ and $A^t$ are the same and since $A^2 = A^t$ it follows that $$\det(A-\lambda I) = 0 \iff \det(A-\sqrt{\lambda}I)\det(A+\sqrt{\lambda}I) = 0$$
So if $\lambda$ is an eigenvalue of $A$ then so is $\sqrt{\lambda}$, with appropriate +/- sign.
As far as I can see this implies that all eigenvalues of $A$ are either 0 or 1 since if any other $\lambda$ is an eignevalue of $A$ then so is $\sqrt{\lambda},\sqrt{\sqrt{\lambda}},\ldots$ which cannot possibly hapen for a matrix of finite size?
Is my reasoning correct?
 A: You are correct that the eigenvalues of $A^t$ and of $A$ are equal. That means that if $\lambda$ is an eigenvalue of $A$, then $\lambda^2$ must be an eigenvalue of $A^2$, hence of $A^t$, hence of $A$. So if $\lambda$ is an eigenvalue of $A$, then so is $\lambda^2$, and hence $\lambda^4$, and hence $\lambda^8$, and hence $\lambda^{16}$ etc.
That means that the list $$\lambda,\lambda^2,\lambda^4,\lambda^8,\lambda^{16},\ldots,\lambda^{2^n},\ldots$$
must in fact be a finite list. This does not occur if $|\lambda|\gt 1$, because then the absolute value of these numbers is strictly increasing. It cannot happen if $0\lt|\lambda|\lt 1$ either, because then the absolute values of the terms of the sequence is strictly decreasing, so the list cannot be finite.
What does that leave?
P.S. This will reduce the possibilities to something tractable. You still need to figure out whether those possibilities can in fact occur.
A: You are given a relation between $A$ and $A^t$. 
Take the transpose on both sides of that relation to get a second relation between $A$ and $A^t$. 
Eliminate $A^t$ from the two relations to get an equation involving only $A$. 
From that equation you should be able to reduce the eigenvalue possibilities to four. 
Two of those four possibilities can certainly occur; I'm not sure about the other two.  
A: Let $A^2=A^t$, then by taking transpose on both sides $(A^t)^2=A$, so $(A^2)^2=A$.
Now $A^4=A$.
Let $Ax=\lambda x$. with x one of the eigenvectors of A.
Then $A^4x=\lambda x$.
So $A^3 \lambda x=\lambda x$.Expanding like this, $\lambda^4x = \lambda x$. So $(\lambda^4-\lambda)x=0$. x being one of the eigenvectors of A is not all zero, whereas the other factor is just a constant. So we get $\lambda^4 - \lambda = 0$. In other words, $\lambda$ must be either zero or one of the cube roots of unity (including 1 itself). 0 is obviously attained by the zero matrix and 1 by any identity matrix. For the rest, consider 
0   0   1
1   0   0
0   1   0

This satisfies the conditions in the question and its characteristic polynomial is $1-\lambda^3=0$. So its eigenvalues are the cube roots of unity.
