Prove that an involutory matrix has eigenvalues $\pm 1$

Question

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?

Answer 1

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 $\lambda =\pm 1$

Answer 2

Another approach is to note that, since $A^2 = I$, the minimal polynomial of an involutory matrix will divide $x^2 - 1 = (x-1)(x+1)$. The cases where the minimal polynomial is $(x-1)$ or $(x+1)$ correspond to the "degenerate" cases $A = I$ and $A = -I$. Here, the eigenvalues are all $1$ and all $-1$ respectively. All other cases result in $A$ having a mix of both $-1$ and $1$ eigenvalues, recognizing of course that there's no distinction between $-1$ and $1$ when $A$ is over a base field of characteristic two.

More generally, for a complex base field, this approach can be used to show that the set of eigenvalues of a matrix $m$-involution $A$ (for which $A^m=I$ for an integer $m>1$) belongs to the set of $m$-th roots of unity.

Answer 3

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 eigenvalue of the identity function is $1$, so $\lambda^2 = 1$ meaning that $\lambda = \pm1$.