If matrix $A$ orthogonal then $A^2$ is orthogonal?

Question

If matrices $A\in M_n(\mathbb R)$ orthogonal, then $A^2$ can be orthogonal but it does not have to be orthogonal?

My answer that $A^2$ is always orthogonal, I prove that $\|A^2x\|=\|x\|$, but is there better solution to prove that?

Answer 1

Yes, it has to be orthogonal. Suppose that $A$ is orthogonal. What this means is that $A^t.A=\operatorname{Id}$. But then\begin{align}(A^2)^t.A^2&=(A.A)^t.A.A\\&=A^t.A^t.A.A\\&=A^t.\operatorname{Id}.A\\&=A^t.A\\&=\operatorname{Id}.\end{align}Therefore, $A^2$ is orthogonal too.

Answer 2

$A$ is orthogonal $ \iff$ $A^T=A^{-1}$

$$ A^{-1}=A^T \implies A^{-2}=(A^T)^2=(A^2)^T$$

$$ (A^2)^{-1}=(A^2)^{T}$$

$A$ is orthogonal $\implies$ $A^2$ is orthogonal.

Answer 3

A square matrix is orthogonal if and only if its columns are orthonormal. For column indices $j,k \in \{1, \ldots, n\}$ we have

$$\sum_{i=1}^n (A^2)_{ij}(A^2)_{ik} = \sum_{i=1}^n \left(\sum_{r=1}^n A_{ir}A_{rj}\right)\left(\sum_{s=1}^n A_{is}A_{sk}\right) = \sum_{r=1}^n \sum_{s=1}^n \left(\sum_{i=1}^n A_{ir}A_{is}\right)A_{rj}A_{sk}\\ = \sum_{r=1}^n \sum_{s=1}^n \delta_{rs}A_{rj}A_{sk} = \sum_{r=1}^n A_{rj}A_{rk} = \delta_{jk}$$

so $A^2$ is orthogonal.