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.

You can easily prove the following statement:

Let $f: V\to V$ be an endomorphism. If $\lambda$ is an eigenvalue of $f$, then $\lambda^k$ is an engeinvalue of $\underbrace {f\ \circ\ ...\ \circ f}_{k \text{ times}}$

In this case, let $A$ be a matrix of an endomorphism $f$ such that $f\circ f = I$. This means that $A$ is an involutory matrix (Because $AA=I$). So if $\lambda$ is an eingenvalue of $f$, then $\lambda ^2$ is an eigenvalue for $f \circ f = I$. The only eigenbalue of the identity funcion is $1$, so $\lambda^2 = 1$ meaning that $\lambda = \pm1$.
