suppose $A \in M_{2n}(\mathbb{R})$. and$J=\begin{pmatrix} 0 & E_n\\ -E_n&0 \end{pmatrix}$ where $E_n$ represents identity matrix.
if $A$ satisfies $AJA^T=J$
How to figure out $\det(A)=1$
My approach:
I have tried to separate $A$ into four submartix:$A=\begin{pmatrix}A_1&A_2 \\A_3&A_4 \end{pmatrix}$ and I must add a assumption that $A_1$ is invertible. by elementary transfromation:$\begin{pmatrix}A_1&A_2 \\ A_3&A_4\end{pmatrix}\rightarrow \begin{pmatrix}A_1&A_2 \\ 0&A_4-A_3A_1^{-1}A_2\end{pmatrix}$
we have: $\det(A)=\det(A_1)\det(A_4-A_3A_1^{-1}A_2)$ from$\begin{pmatrix}A_1&A_2 \\ A_3&A_4\end{pmatrix}\begin{pmatrix}0&E_n \\ -E_n&0\end{pmatrix}\begin{pmatrix}A_1&A_2 \\ A_3&A_4\end{pmatrix}^T=\begin{pmatrix}0&E_n \\ -E_n&0\end{pmatrix}$ we get two equalities:$A_1A_2^T=A_2A_1^T$ and $A_1A_4^T-A_2A_3^T=E_n$
then $\det(A)=\det(A_1(A_4-A_3A_1^{-1}A_2)^T)=\det(A_1A_4^T-A_1A_2^T(A_1^T)^{-1}A_3^T)=\det(A_1A_4^T-A_2A_1^T(A_1^T)^{-1}A_3^T)=\det(E_n)=1$
but I have no idea to deal with this problem when $A_1$ is not invertible...
Thanks