# Symplectic basis $(A_i,B_i)$ such that $S=$ span$(A_1,…,A_k)$ for some $k$ when $S$ is isotropic

Let $(V,\omega)$ be a symplectic vector space of dimension $2n$. How can I show that for an isotropic subspace $S \subset V$, there exists a symplectic basis $(A_i,B_i)$ for $V$ such that $S=$ span$(A_1,...,A_k)$ for some $k$.

It is related to this thread. I understand the symplectic case, but I cannot apply it to isotropic case.

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Choose any basis $A_1,\dots, A_k$ of $S$. By non-degeneracy of $\omega$ there exist $C_i\in V$, $i=1\dots k$, s.t. $\omega(A_i,C_j)=\delta_{ij}$. You can the find $B_i$'s of the form $B_i=C_i+\sum_{j<i}a_{ij} A_k$ ($a_{ij}$'s are numbers) so that $\omega(B_i,B_j)=0$ ($\omega(A_i,B_j)=\delta_{ij}$ is automatic): find $a_{ij}$'s first for $i=1$, then $i=2$ etc. Now $\operatorname{span}(A_1,\dots,A_k,B_1,\dots,B_k)$ is a symplectic subspace, so use the answer you linked to (to complete $A_1,\dots,A_k,B_1,\dots,B_k$ to a symplectic basis of $V$).
I don't see where you used $S$ is isotropic. – user20353 Dec 30 '11 at 23:34
@user20353: to get $\omega(A_i,A_j)=0$, which is then also used to get $\omega(A_i,B_j)=\delta_{ij}$ (see the def. of $B_i$'s) – user8268 Dec 31 '11 at 7:37