Let $(V,\omega)$ be a symplectic vector space of dimension $2n$. How can I show that for a symplectic subspace $S \subset V$, there exists a symplectic basis $(A_i,B_i)$ such that $S= $ span$(A_1,B_1,...,A_k,B_k)$ for some $k$.
I know that since $S$ is symplectic, $S \cap S^\perp= \lbrace0 \rbrace$ and this is true iff $\omega|_S$ is nondegenerate. Let $(\alpha^1, \beta^1,...,\alpha^n, \beta^n)$ be the corresponding dual basis for $V^\ast$. Then $\omega$ is defined by $\omega=\sum_{i=1}^n \alpha^i \wedge \beta^i$. I do not know how to continue.