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.
 A: 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.  
A: 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 $\lambda_i=\pm1$.
A: 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 $\lambda =\pm 1$
A: For a proof with determinants:
$$0 = \det(A- \lambda I) = 0$$
$$ = \det(A- \lambda (A^2))$$
$$= \det(A (I- \lambda A))$$
$$= \det(A) \det(I- \lambda A)$$
Hence
$$0 = \det(A) \det(I- \lambda A)$$
$$\iff 0 = (\pm1) \det(I- \lambda A)$$
$$\iff 0 = \det(I- \lambda A)$$
$$\iff 0 = \det((- \lambda)(\frac{1}{- \lambda}I+ A))$$
$$\iff 0 = (- \lambda)^n \det(\frac{1}{- \lambda}I+ A) \tag{1}$$
$$\iff 0 = \det(\frac{1}{- \lambda}I+ A)$$
$$\iff 0 = \det(-\frac{1}{ \lambda}I+ A)$$
$$\iff 0 = \det(A -\frac{1}{ \lambda}I)$$
Therefore, $\lambda$ is an eigenvalue of $A$ if and only if it equals its reciprocal assuming hopefully, that I didn't make a logical error and actually conclude only that $\lambda$ is an eigenvalue of $A$ if and only if its multiplicative inverse is too. If I did make an error, then I hope someone can tell me how to proceed.
QED

$(1)$ I don't know what $\lambda$ is, but I know what $\lambda$ isn't: $\lambda$ is nonzero because $\det(A)$ is nonzero because $\det(A) = \pm 1$
A: 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 $\lambda = \pm1$.
A: 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 $\lambda= \pm 1$.
