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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.

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It's related to this thread: math.stackexchange.com/questions/95273/… –  user20353 Dec 30 '11 at 21:01

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up vote 2 down vote accepted

S is a symplectic space so it has a symplectic basis. Its orthogonal subspace is also a symplectic space, so it has a symplectic base. Now consider the union of these two bases.

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Could you give more details? –  user20353 Dec 30 '11 at 19:45
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I don't know what to add, really. You know that every symplectic space has a symplectic basis: Since S and $S^\perp$ are symplectic spaces, they have symplectic bases; now consider the union of these two bases: it is obviously a basis of the whole space, and you have to check that it is a symplectic basis. What part do you want details on? –  Mariano Suárez-Alvarez Dec 30 '11 at 19:51
    
Just to see the distinction between them, what happens if $S$ is not symplectic but isotropic? –  user20353 Dec 30 '11 at 20:12
1  
In that case S does not have a symplectic basis, nor is it linearly disjoint from its orthogonal complement, so we cannot even start the construction. –  Mariano Suárez-Alvarez Dec 30 '11 at 20:55

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