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.
 A: 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 $\det(A) = 1$.
A: The determinant is a continuous function, and the set of symplectic matrices with invertible $A_1$ is dense in the set of all symplectic matrices.  So if you've proven that it equals 1 for all invertible $A_1$, then it equals 1 for all $A_1$.
A: *

*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.
A: 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 $A J A^T = J$. 
A: Let me first restate your question in a somewhat more abstract way. Let $V$ be a finite dimensional real vector space. A sympletic form is a 2-form $\omega\in \Lambda^2(V^\vee)$ which is non-degenerate in the sense that $\omega(x,y)=0$ for all $y\in V$ implies that $x=0$. $V$ together with such a specified $\omega$ nondegerate 2-form is called a symplectic space. It can be shown that $V$ must be of even dimension, say, $2n$.
A linear operator $T:V\to V$ is said to be a symplectic transformations if $\omega(x,y)=\omega(Tx,Ty)$ for all $x,y\in V$. This is the same as saying $T^*\omega=\omega$. What you want to show is that $T$ is orientation preserving. Now I claim that $\omega^n\neq 0$. This can be shown by choosing a basis $\{a_i,b_j|i,j=1,\ldots,n\}$ such that $\omega(a_i,b_j)=\delta_{ij}$ and $\omega(a_i,a_j)=\omega(b_i,b_j)=0$, for all $i,j=1,\ldots,n $. Then $\omega=\sum_ia_i^\vee\wedge b_i^\vee$, where $\{a_i^\vee,b_j^\vee\}$ is the dual basis. We can compute $\omega^n=n!a_1^\vee\wedge b_1^\vee\wedge\dots\wedge a_n^\vee \wedge b_n^\vee$, which is clearly nonzero.
Now let me digress to say a word about determinants. Let $W$ be an n-dimensional vector space and $f:W\to W$ be linear. Then we have induced maps $f_*:\Lambda^n(W)\to \Lambda^n(W)$. Since $\Lambda^n(W)$ is 1-dimensional, $f_*$ is multiplication by a number. This is just the determinant of $f$. And the dual map $f^*:\Lambda^n(W^\vee)\to \Lambda^n(W^\vee)$ is also multiplication by the determinant of $f$. 
Since $T^*(\omega^n)=\omega^n$, we can see from the above argument that $\det(T)=1$. The key point here is that the sympletic form $\omega$ give a canonical orientation of the space, via the top from $\omega^n$.
A: Your approach can be remedied to work over any field $\mathbb F$. As you already know, if
$$
A=\pmatrix{X&Y\\ Z&W} \text{ where } X,Y,Z,W\in M_n(\mathbb F),
$$
the equation $A^TJA=J$ means precisely that $X^TZ,\,Y^TW$ are symmetric and $X^TW-Z^TY=I$. In particular, every matrix in the form of $M^T\oplus M^{-1}$ is a solution. Let $Z=PDQ$ where $P$ and $Q$ are some invertible matrices and $D$ is a diagonal matrix. Then $P^T\oplus P^{-1}$ and $Q^{-1}\oplus Q^T$ are members of $Sp(2n,\mathbb F)$. It follows that $A\in Sp(2n,\mathbb F)$ if and only if
$$
B=\pmatrix{P^T\\ &P^{-1}}A\pmatrix{Q^{-1}\\ &Q^T}=\pmatrix{\ast&\ast\\ D&\ast} \in Sp(2n,\mathbb F)
$$
and $\det(A)=1$ if and only if $\det(B)=1$.
In other words, we may assume without loss of generality that the sub-block $Z$ is a diagonal matrix. Let $t$ be an indeterminate. Then $A+tI$ is invertible over $\mathbb F(t)$. Therefore, using Schur complement and the conditions that $X^TZ$ is symmetric and $X^TW-Z^TY=I$, we obtain
\begin{aligned}
&\det\pmatrix{X+tI&Y\\ Z&W}\\
&=\det(X+tI)\det\left(W-Z(X+tI)^{-1}Y\right)\\
&=\det(X^T+tI)\det\left(W-Z(X+tI)^{-1}Y\right)\\
&=\det\left((X^T+tI)W-(X^T+tI)Z(X+tI)^{-1}Y\right)\\
&=\det\left(X^TW+tW-(X^TZ+tZ)(X+tI)^{-1}Y\right)\\
&=\det\left(X^TW+tW-(X^TZ+tZ^T)(X+tI)^{-1}Y\right)\ \text{(because $Z$ is diagonal)}\\
&=\det\left(X^TW+tW-(Z^TX+tZ^T)(X+tI)^{-1}Y\right)\ \text{(because $X^TZ$ is symmetric)}\\
&=\det\left(X^TW+tW-Z^TY\right)\\
&=\det(I+tW) \ \text{(because $X^TW-Z^TY=I$)}.
\end{aligned}
Put $t=0$, we get $\det(A)=\det(I)=1$.
Edit (2021-05-07). I have just learnt that a proof very similar in spirit to mine was given in Büngera and Rump, Yet more elementary proofs that the determinant of a symplectic matrix is 1, Linear Algebra and Its Applications, 515 (2017) 87-95. See proof II of their lemma 1 on p.92.
A: Every symplectic matrix is the product of two symplectic matrices with lower-left corner invertible. See: M. de Gosson, s Symplectic Geometry and Quantum Mechanics, Birkhäuser, Basel, series "Operator Theory: Advances and Applications" (subseries: "Advances in Partial Differential Equations"), Vol. 166 (2006)
