This is part of Exercise 2.14 in Kirillov's "An Introduction to Lie Groups and Lie algebras".
The reader is supposed to prove that $Sp(2n,\mathbb{R})$ acts transitively on the space $L_n$ of all Lagrangian linear subspaces ($L \subset \mathbb{R}^{2n}$ such that $\dim L = n, \omega|_L = 0$) to see the $L_n$ as a homogeneous space and to compute its dimension.
I have already proved, using a version of Gram--Schmidt procedure that the action is indeed transitive, being able to map the standard Lagrangian $\mathbb{R}^{n}\oplus 0$ to any other Lagrangian $L$ via a symplectic matrix. However, I am having trouble computing the isotropy subgroup of even the standard Lagrangian $\mathbb{R}^{n} \oplus 0$. We have a block matrix $$ g = \begin{pmatrix} A & B \\ C & D \end{pmatrix} $$ for which we immediately see that $C=0$ in order to preserve the first factor. This Imposing $g$ to be symplectic one gets the relations $D = (A^t)^{-1}$ and $B^t D - D^t B = 0$. The first condition gives $n^2$ free parameters (as $A\in GL(n,\mathbb{R})$), but I really can't handle the second.
Particularizing at $A = D = 1$ gives $B^t = B$ which is $\frac{n(n+1)}{2}$ extra parameters, yielding a total dimension of the homogeneous space $$ L_n = \frac{Sp(2n,\mathbb{R})}{G_*} $$ to be $\frac{n(3n+1)}{2}$.
I need help checking this, and if possible, getting a nicer description of the isotropy group and making all of this as rigorous as possible.