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I'm currently reading a linear algebra book and there is this example without a solution: Let A be an invertible matrix. Is it possible that we have $A = A^{-1}$ with $A \ne I_n$?

If not, then what is an example/proof?

steven2005
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    Yes. These are called involutory matrices (matrices that are their own inverse -- so, $I_n$ would also be involutory). – twosigma May 22 '20 at 07:54
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    Consider a reflection of $\mathbb R^2$ about a line through the origin, such as the $2\times2$ matrix that swaps the two components of a vector. – amd May 22 '20 at 07:54
  • See also: https://math.stackexchange.com/questions/3115411/when-is-a-matrix-equal-to-its-own-inverse and https://en.wikipedia.org/wiki/Involutory_matrix. – Manuel Norman May 22 '20 at 07:54

4 Answers4

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Hint: Consider the often underappreciated $1\times 1$ case, and you will find a solution pretty quickly.

Arthur
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Yes. I'll even give you a recipe to generate such matrices:

Given $n$, you take a diagonal matrix with only $1$ or $-1$ on its diagonal. It's easy to see that in such a case:

$$A^2=I_n\Rightarrow A=A^{-1}$$

You don't have to stop there - there are lots more matrices with this property. You can actually take any invertible matrix $P$ and look at $A'=P^{-1}AP$. Notice that:

$$(A')^{-1}=(P^{-1}AP)^{-1}=P^{-1}A^{-1}(P^{-1})^{-1}=P^{-1}AP=A'$$

So this matrix also satisfies this property. Overall, you can use this method to create infinitely many such matrices (if your field is infinite).

GSofer
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Let $A = diag(a_1,a_2,...,a_n)$ with $a_j \in \{-1,1\}$ for $j=1,2,...,n.$

Then $A^2=I_n$. If there is an index $j$ such that $a_j=-1,$ then $A \ne I_n.$

Fred
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$$ \left( \begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix} \right)\left( \begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix} \right) = \left( \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right). $$

Gibbs
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