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?