If $A^2$ is an orthogonal matrix, is $A$ also orthogonal?

Question

If $A^2$ is an orthogonal matrix, is $A$ also orthogonal?

I can only get $A$ is invertible, but from that, I can't figure out how to proceed. I thought this statement is false, but I tried and failed to come up with a counterexample.

Answer 1

You cannot prove it, since it is false. Let $a,b\in\mathbb R$ with $b\neq0$ and $a^2+b^2\neq1$. Then$$\begin{bmatrix} a & b \\ \frac{1-a^2}{b} & -a\end{bmatrix}^2=\begin{bmatrix}1&0\\0&1\end{bmatrix},$$which is orthogonal. However,$$\begin{bmatrix} a & b \\ \frac{1-a^2}{b} & -a\end{bmatrix}$$is not.

Answer 2

Doesn't have to be. Counterexample: $$ A = \begin{bmatrix} 1 & -2 \\ 0 & -1\end{bmatrix}. $$

Answer 3

If $A^2$ is orthogonal then it implies that $$A^2.(A^{T})^2=I \implies (AA^T)^2=I \implies AA^T=B,$$ where $B$ $(B^2=I)$ is an involutory matrix which may not be $I$. So $A$ may or may not be orthogonal. For instance, Pauli matrices are involutory like one in the comment of @egreg