TITLE: Why is the determinant of a symplectic matrix 1?
QUESTION [41 upvotes]: 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.

REPLY [5 votes]: The most natural way to show [independent of the field $\mathbb{F}$ of characteristics $\neq 2$] that the symplectic group
$$Sp(2n,\mathbb{F})~:=~\{M\in {\rm Mat}_{2n\times 2n}(\mathbb{F})\mid MJM^T=J\} \tag{1}$$ consists of matrices $M$ with unit determinant is to use the following Pfaffian property
$$ {\rm Pf}(J)~\stackrel{(1)}{=}~{\rm Pf}(MJM^T)
=~{\rm Det}(M)~{\rm Pf}(J) \qquad \Longrightarrow\qquad  {\rm Det}(M)~=~+1,\tag{2}$$
as Willie Wong hints in his answer. 
An elementary proof [independent of the field $\mathbb{F}$ of characteristics $=0$] of the Pfaffian property (2) is e.g. given in my Math.SE answer here.