If $A$ be a matrix with real entries and $A^2=A^t$, where $A^t$ denotes transpose of $A$, then the only real eigenvalues of $A$ are $0$ and $1$
I solved this problem using property that trace of a matrix is the sum of its eigenvalues. Then, I compared the trace of $A^2$ with $A^t$ which is equal to $A$. Then, I used Cauchy-Schwarz inequality on the eigenvalues and used the fact that square of a real number is always nonnegative to obtain the result.Am i right in that? I think we can also prove it using Jordan decomposition. Any ideas. Thanks beforehand.