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?

$$

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).

$$

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

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:

$$A^2=I_nRightarrow A=A^{-1}$$

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).

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 $A ne I_n.$