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.

There is an easy proof for real and complex case which does not require the use of  Pfaffians. This proof first appeared in a Chinese text. Please see http://arxiv.org/abs/1505.04240 for the reference.
I reproduce the proof for the real case here. The approach extends to complex symplectic matrices.
Taking the determinant on both sides of $A^T J A = J$,
$$\det(A^T J A) = \det(A^T) \det(J) \det(A) = \det(J).$$
So we immediately have that $\det(A) = \pm 1$.
Then let us consider the matrix $A^TA + I.$
Since $A^TA$ is symmetric positive definite,
its eigenvalues are real and greater than $1$.Therefore its determinant, being the product of its eigenvalues, has $\det(A^TA +I) > 1$. 
Now as $\det(A) \ne 0$, $A$ is invertible. Using this we may write
$$ A^TA + I = A^T( A + A^{-T}) = A^T(A + JAJ^{-1}).$$
Denote the four $N \times N$ subblocks of $A$ as follows,
$$
A = \begin{bmatrix} A_{11} & A_{12} \\ A_{21} & A_{22} \end{bmatrix},
    \text{ where } A_{11},A_{12},A_{21},A_{22} \in \mathbb{R}^{N \times N}. 
$$
Then we compute
$$ A + JAJ^{-1} = \begin{bmatrix} A_{11} & A_{12} \\ A_{21} & A_{22} \end{bmatrix}
+ \begin{bmatrix} O & I_N \\ -I_N & O \end{bmatrix}
   \begin{bmatrix} A_{11} & A_{12} \\ A_{21} & A_{22} \end{bmatrix}
\begin{bmatrix} O & - I_N \\  I_N & O \end{bmatrix} 
  = \begin{bmatrix} A_{11} & A_{12} \\ A_{21} & A_{22} \end{bmatrix}
+ \begin{bmatrix} A_{22} & -A_{21} \\ -A_{12} & A_{11} \end{bmatrix}
    = \begin{bmatrix} A_{11}+ A_{22} & A_{12} - A_{21} \\ - A_{12}+ A_{21}  & A_{11} + A_{22} \end{bmatrix}.$$
Writing the blocks as $C := A_{11} + A_{22}$ and $D:= A_{12} - A_{21}$, 
we make use of a unitary transform 
$$
A + JAJ^{-1} = \begin{bmatrix} C & D \\ -D & C \end{bmatrix} 
             =  
    \frac{1}{\sqrt{2}}\begin{bmatrix} I_N & I_N \\ iI_N & -iI_N \end{bmatrix}
             \begin{bmatrix} C + i D & O \\ O & C - i D \end{bmatrix}
    \frac{1}{\sqrt{2}} \begin{bmatrix} I_N & -iI_N \\ I_N & iI_N \end{bmatrix}.
$$
We plug this factorization into our identity.
Note that $C,D$ are both real. This allows the complex 
conjugation to commute with the determinant (as it is a polynomial of its 
entries)
$$0 < 1 < \det(A^TA + I)  = \det(A^T(A + JAJ^{-1})) \\
= \det(A) \det(C + i D) \det(C - iD) \\
= \det(A) \det(C + i D) \det\left(\overline{C + iD}\right)\\
= \det(A) \det(C + iD) \overline{\det(C + iD)} 
= \det(A) \left\lvert \det(C + iD)\right\rvert^2.
$$
Clearly, none of the two determinants on the RHS can be zero, 
so we may conclude
$\left\lvert \det(C + iD) \right\rvert^2 > 0$. Dividing this through on both sides,
we have $\det(A) > 0$, and thus $\det(A) = 1$.
