Let $(V, \omega)$ be a symplectic $2n$-dimensional vector space and $E\subset V$ any $n$-dimensional subspace. Prove there is a Lagrangian subspace $L\subset V$ with $V=E\oplus L$.
[here, Lagrangian means $L=L^\omega:=\{v\in V\mid \omega(v, w)=0\text{ for all }w\in L\}$]
I know the solution for this when $E$ is Lagrangian, which is simple if we use a compatible complex structure. I've tried to mimic the proof for a general $E$, but failed to do so.
Another thing I've tried was this: define $S:=\{I\subset V\mid I\text{ isotropic and }I\cap E=0\}$. $S$ is not empty, since $\{0\}\in S$. Now if $I_0\in S$ has maximal dimension, we must prove $\dim I_0=n$. I couldn't make this work either because $E$ has no special property, which makes it harder to come up with ideas.
Any suggestions?
The proposition is clear if $\dim V = 2$ so let's do an induction on $2m = \dim V$.
We look at the rank of $\omega$ restricted to $E$, say it's $2k$ : it means that there is a "orthonormal" basis $(e_1, \dots, e_m, z_1, \dots, z_m)$ so that $\omega = e_1 \wedge e_2 + \dots + e_{2k+1} \wedge z_1 + \dots + z_j \wedge z_l + \dots $. The proof is exactly the same as existence of orthonormal basis for a given scalar product.
If $k<m/2$, let $V'$ be the subspace generated by the $e_1, \dots, \widehat{e_{2k+1}}\dots ,e_m$ and the $z_2, \dots, z_m$, and $E' \subset V'$ generated by the $e_1, \dots, \widehat{e_{2k+1}}\dots ,e_m$. By construction, $(V', \omega')$ is symplectic where $\omega' = \omega - e_{2k+1} \wedge z_1$ and $2\dim E' = \dim V'$ so there is a lagrangian complement $E' \oplus L' = V'$ by the induction hypothesis. We have $(L')^{\omega} = L' \oplus \Bbb R\{e_{2k+1}\} \oplus \Bbb R\{z_1\}$, so we deduce that $L := L' \oplus \Bbb R\{z_1\}$ is a lagrangian complement of $E$.
If $k=m/2$, i.e $E$ is symplectic (in particular $k$ is even), then we can rephrase our problem : we have the canonical symplectic form $ \omega = e_1 \wedge f_1 + \dots + e_{4n} \wedge f_{4n}$ and we want to find a lagrangian complement to $(e_1, f_1, \dots, e_{2n}, f_{2n})$. This complement is given by $$L = \text{span}(e_{2n+1} + e_1, f_{2n+1} - f_1, e_{2n+2} +e_2, f_{2n+2} - f_2, \dots , f_{4n} - f_{2n})$$