linear algebra - Why is the determinant of a symplectic matrix 1?

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

Answer

First, taking the determinant of the condition

$$\det AJA^T = \det J \implies \det A^TA = 1$$
using that $\det J \neq 0$. This immediately implies
$$\det A = \pm 1$$
if $A$ is real valued. The quickest way, if you know it, to show that the determinant is positive is via the Pfaffian of the expression $A J A^T = J$.