Proof that an involutory matrix has eigenvalues 1,-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 $\lambda =\pm 1$

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.

Here's another approach with diagonalisation. Let $A=S\Lambda S^{-1}$, where $S$ has the eigenvectors of $A$ as its columns and $\Lambda$ is the matrix with eigenvalues on its main diagonal. Then $A^2=S\Lambda^2S^{-1}=I$, so $S\Lambda^2=S$ and $\Lambda^2=I$. Since the diagonal entries of $\Lambda^2$ are the eigenvalues squared, then $\lambda_i^2=1$ by comparing the entries of $\Lambda^2$ and $I$. So $\lambda_i=\pm1$.