# Math Genius: 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?

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

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 $$A ne I_n.$$

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