If $A$ be a matrix with real entries and $A^2=A^t$, then the eigenvalues of $A$ are $0$ and $1$

Question

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.

Answer 1

You could also have done like this:

$ A={(A^t)}^{t}={(A^2)}^{t}={(A^t)}^{2}={(A^2)}^{2}=A^4 $

So the polynomial:

$P=X^4-X=X(X-1)(X^2+X+1)$ Cancels $A$

We hence conclude that:

$sp(A)$ ($sp(A)$ is the set of eigenvalues of $A$) is included in the roots of $P$, and the real roots of $P$ are $0$ and $1$. It follows that the only real roots that $A$ CAN have are $0$ and $1$