Proof that an involutory matrix has eigenvalues 1,-1 I'm trying to prove that an involutory matrix (a matrix where $A=A^{-1}$) has only eigenvalues $\pm 1$.

I've been able to prove that $det(A) = \pm 1$, but that only shows that the product of the eigenvalues is equal to $\pm 1$, not the eigenvalues themselves.

Does anybody have an idea for how the proof might go?

Thanks.
Let $\lambda$ a eigenvalue of A and $x \neq 0$ respective eigenvector, then

$Ax = \lambda x \Leftrightarrow A^{-1}A x= \lambda A^{-1} x \Leftrightarrow x = \lambda A x \Leftrightarrow x = \lambda^2 x \Leftrightarrow (1-\lambda^2)x = 0$

then $\lambda =\pm 1$
Suppose $\lambda$ is an eigenvalue of $A$ then we know that $1/\lambda$ is an eigenvalue of $A^{-1}$. But here, $A= A^{-1}$. So, for every $\lambda$ in $A$, $\lambda=1/\lambda$. And hence $\lambda^2 = 1$. And hence $\lambda= \pm 1$.