$\det(A)={\pm 1} \iff A\in\mathcal O_n(\Bbb R)$?

Question

I realise $A\in\mathcal O_n(\Bbb R) \iff AA^T= I_n$

Does this hold true? $\iff \det(A)={\pm 1}$

Or is it just $\implies \det(A)={\pm 1}$

Answer 1

$A\in\mathcal O_n(\Bbb R) \implies \det(A)={\pm 1}$

Indeed, $AA^T= I_n \implies \det(A)^2=\det(A)\det(A^T)=\det(AA^T)=1$.

The converse is not true, even for diagonal matrices: $$ A=\begin{bmatrix}a&0\\0&\pm \dfrac1a\end{bmatrix} $$ is not orthogonal but $\det(A)={\pm 1}$. The rows of $A$ are orthogonal but not orthonormal.

Answer 2

Not every matrix with $\det(A)=\pm 1$ is orthogonal. For instance, the matrix $$ A=\begin{bmatrix}1&1\\0&1\end{bmatrix} $$ is not orthogonal.