I was asked by my professor to show that if $A$ is an orthogonal matrix and $J$ is a matrix such that $J^2=-I$ ($I$ is the identity matrix) then $$AJ=JA.$$
I think my instructor made a mistake. As a (very) particular case, I computed that $$\begin{bmatrix} \cos(x) & \sin(x)\\ -\sin(x) & \cos(x) \end{bmatrix} \begin{bmatrix} 0 & -1\\ 1 & 0 \end{bmatrix} = \begin{bmatrix} \sin(x) & -\cos(x) \\ \cos(x) & \sin(x) \end{bmatrix} = \begin{bmatrix} 0 & -1\\ 1 & 0 \end{bmatrix} \begin{bmatrix} \cos(x) & \sin(x)\\ -\sin(x) & \cos(x) \end{bmatrix},$$ in this case $AJ=JA$. However, computing another example $$\begin{bmatrix} -1 & 0\\ 0 & 1 \end{bmatrix} \begin{bmatrix} 0 & -1\\ 1 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 1\\ 1 & 0 \end{bmatrix}$$ and $$\begin{bmatrix} 0 & -1\\ 1 & 0 \end{bmatrix} \begin{bmatrix} -1 & 0\\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 0 & -1\\ -1 & 0 \end{bmatrix}$$ therefore in this case $AJ=-JA$. So the instructor made a mistake. However, this me question something:
If $V$ is a vector space of finite dimension and $A \in \mathrm{O}(V)$, does it always holds that $AJ=\pm JA$? (It doesn’t matter if $\dim V$ is even or odd)
First of all, if you have a linear map $J: V\to V$ so that $J^2 = -I$, then taking dterminant gives
$$ (\det J)^2 = (-1)^n,$$ where $n$ is the dimension of $V$. This already implies that $n$ is even.
And it is not always true that $AJ = \pm JA$: for example, take $n=4$ and let
$$J = \begin{bmatrix} 0 & -I_2 \\ I_2 & 0 \end{bmatrix}, A = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0& -1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1\end{bmatrix}$$
then
$$JA = \begin{bmatrix} 0 & 0 & 1 & 0 \\ 0& 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0\end{bmatrix}, \ \ AJ = \begin{bmatrix} 0 & 0 & -1 & 0 \\ 0& 0 & 0 & 1 \\ -1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0\end{bmatrix}$$