Maybe it's an elementary fact, but I couldn't find a point where it's explicitly stated.
Consider the $2n$-dimensional standard symplectic matrix, i.e.
$$\begin{bmatrix} 0 & I_n \\ -I_n & 0 \\ \end{bmatrix}$$
Where $I_n$ is the $n$-dimensional identity matrix.
- Could (in principle) any symplectic (real) matrix be brought in such form (by means of a linear trasformation represented by another real symplectic matrix)?
- Is this a conclusion derived from the Darboux theorem?
HINT: As I said earlier, you have to assume the matrix $A$ is $2n\times 2n$ with maximal rank, so the bilinear form $\langle x,y\rangle = x^\top Ay$ is nondegenerate. Then you will show that you can decompose $V=\Bbb R^{2n}$ as a sum of $n$ invariant $2$-dimensional subspaces, on each of which the bilinear form has the matrix $$\begin{bmatrix} 0&1\\-1&0\end{bmatrix}.$$
Choose $v_1,v_2\in V$ so that $\langle v_1,v_2\rangle\ne 0$. (It follows from nondegeneracy that you can do this.) By scaling $v_2$, we can assume $\langle v_1,v_2\rangle = 1$. Note that $v_1,v_2$ are linearly independent (why?). Then $v_1,v_2$ span the first subspace. Now write $V = \text{Span}(v_1,v_2) \oplus \left(\text{Span}(v_1,v_2)\right)^\perp$ and proceed by induction.
At the very end, you can reorder your basis vectors to get the matrix form you desire.