Let A be an invertible matrix. Is it possible that we have $A = A^{-1}$ with $A \ne I_n$? 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?
 A: Hint: Consider the often underappreciated $1\times 1$ case, and you will find a solution pretty quickly.
A: 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).
A: 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 $A \ne I_n.$
A: $$
\left(
\begin{matrix}
0 & 1 \\
1 & 0
\end{matrix}
\right)\left(
\begin{matrix}
0 & 1 \\
1 & 0
\end{matrix}
\right)
=
\left(
\begin{matrix}
1 & 0 \\
0 & 1
\end{matrix}
\right).
$$
