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?



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 $\lambda =\pm 1$

  • 1
    $\begingroup$ Thanks. I feel like such an idiot for trying to use determinants for hours. $\endgroup$
    – spc38
    Nov 25 '14 at 6:40
  • $\begingroup$ @spc38 I tried determinants just now. See answer. $\endgroup$
    – 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:

$$0 = \det(A- \lambda I) = 0$$

$$ = \det(A- \lambda (A^2))$$

$$= \det(A (I- \lambda A))$$

$$= \det(A) \det(I- \lambda A)$$


$$0 = \det(A) \det(I- \lambda A)$$

$$\iff 0 = (\pm1) \det(I- \lambda A)$$

$$\iff 0 = \det(I- \lambda A)$$

$$\iff 0 = \det((- \lambda)(\frac{1}{- \lambda}I+ A))$$

$$\iff 0 = (- \lambda)^n \det(\frac{1}{- \lambda}I+ A) \tag{1}$$

$$\iff 0 = \det(\frac{1}{- \lambda}I+ A)$$

$$\iff 0 = \det(-\frac{1}{ \lambda}I+ A)$$

$$\iff 0 = \det(A -\frac{1}{ \lambda}I)$$

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.


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

  • 3
    $\begingroup$ 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 $A-\lambda I=A- (1/\lambda)I$. $\endgroup$
    – YiFan
    Oct 24 '18 at 6:04
  • 1
    $\begingroup$ @user496634 (there are 2 sentences in the parenthetical remark) well ok so eigenvalues occur in pairs of reciprocals. Any suggestions on how to proceed? $\endgroup$
    – BCLC
    Oct 24 '18 at 6:21
  • 2
    $\begingroup$ Unfortunately I can't think of a way to proceed with the determinant approach. Good try though! $\endgroup$
    – 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 $\lambda= \pm 1$.

  • 1
    $\begingroup$ 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 $\lambda = \lambda^{-1}$. $\endgroup$
    – user279515
    Dec 13 '18 at 19:15

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