# 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 $\lambda =\pm 1$

• Thanks. I feel like such an idiot for trying to use determinants for hours. – spc38 Nov 25 '14 at 6:40
• @spc38 I tried determinants just now. See answer. – BCLC Oct 24 '18 at 5:57

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 $$\lambda_i=\pm1$$.

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 $$\det(A) = \pm 1$$

• Indeed, you proved that if $\lambda$ is an eigenvalue then $1-\lambda$ also is an eigenvalue, but not that they're equal. The key fallacy is in that $\det(A-\lambda I)=\det(A-(1/\lambda)I)$ does not imply $A-\lambda I=A- (1/\lambda)I$. – YiFan Oct 24 '18 at 6:04
• @user496634 (there are 2 sentences in the parenthetical remark) well ok so eigenvalues occur in pairs of reciprocals. Any suggestions on how to proceed? – BCLC Oct 24 '18 at 6:21
• Unfortunately I can't think of a way to proceed with the determinant approach. Good try though! – YiFan Oct 24 '18 at 9:32

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 $$\lambda = \pm1$$.

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 $$\lambda= \pm 1$$.

• This reasoning is incorrect. You have argued that if $\lambda$ is an eigenvalue of $A$ then so is $\lambda^{-1}$. But this does not mean that $\lambda = \lambda^{-1}$. – user279515 Dec 13 '18 at 19:15