I have a question on Lagrangian subspaces of symplectic vector spaces:
Let $(V, \omega)$ be a symplectic vector space, $L,M \subset V$ lagrangian subspaces, s.t. $L \cap M = \{0\}$. I have the exercise to show that there exists a symplectic basis of $V$, $(e_1,...,e_n,f_1,..,f_n)$, s.t. $L= span(e_1,...,e_n)$ and $M= span(f_1,...,f_n)$.
My ansatz would be to use induction. For $n=1$, we can just choose some $v \in L$ and $w \in M$, s.t. $\omega(v,w) \neq 0$ and set $e_1=v, f_1= \dfrac{w}{\omega(v,w)}$. Then for some $n \geq 1$ I could look at a basis $v_1,...,v_n$ of $L$ and....
My problem with this ansatz is, how do I know that $L$ is $n$-dimensional?
The proof depends on your definition of Lagrangian: I will take it to mean maximal isotropic (another possible definition is that $L=\lbrace v\in V \mid \omega(v,w)=0 \ \forall w\in L\rbrace=:L^\omega$).
$\omega$ is non-degenerate, and so defines an injective (and hence bijective) mapping $\cdot^\flat:V\to V^*$ via $v^\flat := \omega(v,\cdot)$, with inverse $\cdot^\sharp:V^*\to V$. $L$ is an isotropic subspace of $V$, and so we know that $L^\flat\subset L^\circ$, the annihilator of $L$ in $V^*$.
In fact. $L^\flat = L^\circ$, since if not, we could choose $\alpha\in L^\circ - L^\flat$, and then $\alpha^\sharp\notin L$ would satisfy $\omega(\alpha^\sharp,v) = (\alpha^\sharp)^\flat(v) =\alpha(v)= 0$ for all $v\in L$. Since also $\omega(\alpha^\sharp, \alpha^\sharp) = 0$, the subspace $L\oplus \mathbb{R}\alpha^\sharp$ would be isotropic, contradicting maximality of $L$.
Now $\dim L^\flat = \dim L$ (since $\cdot^\flat$ is injective), while $\dim L^\circ = 2n - \dim L$. The only way these can be equal is if $\dim L = n$.