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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

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    I don't see where you used $S$ is isotropic.2011-12-30
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    @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)2011-12-31