Simple question about special matrices

Question

Consider $A \in O(n)$, that's to say $A \in M_n(\mathbb R)$ st $AA_T=I$, and assume that $\det A=-1$. How can we demonstrate that there exists $A' \in SO(n)$, that's to say $A'\in O(n)$ and $\det A'=1$ and there exists $J$ which represents a reflection with respect to an hyperplane st $A'J=A$?

Answer 1

Since reflections satisfy $J^2=I$, we can restate the desired result as $A^\prime=AJ$. Then $A^\prime A^{\prime T}=AJJ^TA^T$. If $J$ is a self-transpose reflection, $A^\prime A^{\prime T}=I$. So we just need to choose a self-transpose reflection of determinant $-1$. The obvious choice is for $J$ to be diagonal, with one eigenvalue $-1$ while the others are $1$.