Passage in proving lagrangian subspaces have lagrangian complements The starting point is a $2n$-dimensional symplectic vector space $(E,\sigma)$ and a lagrangian subspace $L\leq E$. In order to show there exists a lagrangian $M\leq E$ satisfying $E=M\oplus L$, we first pick an isotropic $M\leq E:M\cap L=\{0\}$ and show there exists $e\in M^\sigma\smallsetminus(L+M)$, where $(\cdot)^\sigma$ denotes the orthocomplement w.r.t $\sigma$, i.e. $M^\sigma=\{v\in E:\sigma(v,w)=0\,\,\,\forall w\in M\}$. Then we proceed to show $M+\langle e\rangle$ is still isotropic and still trivially intersects $L$. The proof is then finished by induction on $\dim M$, since at every stem the dimension grows by 1 and eventually we reach an isotropic $n$-dimensional subspace which has to be lagrangian.
What I have trouble with is proving $(M+\langle e\rangle)\cap L=\{0\}$. If a vector is in that sum, then it is $v_M+ke$ for $v_M\in M$ and $k$ in whatever field the space $E$ is built upon. Suppose $v_M+ke\in L$. Then it is in $L^\sigma$, so if $v_L\in L$ we have $\sigma(v_M,v_L)=-k\sigma(e,v_L)$. So if either is 0 for all $v_L$, both are. But if either is 0 for all $v_L$, then $e\in L$, or $k=0$. $e\in L$ is false, so if either is 0 $k=0$, and if either is 0, $\sigma(v_M,v_L)=0$ for all $v_L$, so $v_M\in L$, but then $v_M\in M\cap L=\{0\}$, so if either is zero for all $v_L$, then the vector $v_M+ke$ is 0. If either is 0 for any $v_L$, one concludes $k=0$ and so $v_M\in L$ as above. But is it impossible for both to be always nonzero? So is it possible for $v_M$ and $ke$ to be $\sigma$-orthogonal to no vector in $L$ while $v_M+ke$ is $\sigma$-orthogonal to all of them?
 A: Actually this statement has a very nice proof using the compatible complex structure. Just for the sake of mathematical beauty I have to mention it here. I will only outline the necessary ingredients.
1) (Linear) complex structure on the vector space E is a linear endomorphism $J: E \rightarrow E$, such that $J^{2} = -1_{E}$. 
2) For every symplectic vector space $(E,\sigma)$ there exists so called symplectic basis $(e_{1}, \dots, e_{n}, f_{1},\dots,f_{n})$ of $E$ which satisfies $\sigma(e_{i},f_{j}) = \delta_{ij} = - \sigma(f_{j},e_{i})$ and it is zero for all other combinations. This is shown easily by induction on $n$. 
For $n = 1$, fix any non-zero vector $v \in E$. As $\sigma$ is non-degenerate, there must be another non-zero vector $w \in E$, such that $\sigma(v,w) \neq 0$. As $\sigma$ is skew-symmetric, $(v,w)$ have to linearly independent. Rescale $w$ so that $\sigma(v,w) = 1$ and the choice $(e_{1},f_{1}) = (v,w)$ gives the symplectic basis.
Let $n > 1$. Assume (induction hypothesis) that there exists a symplectic basis for any symplectic space with dimension strictly less then $n$. Again, fix a pair of non-zero vectors $(v,w)$, such that $\sigma(v,w) = 1$. Let $S =\mathbb{R}\{v,w\}$ be a two-dimensional space spanned by those two vectors. The restriction of $\sigma$ onto $S$ is obviously non-degenerate. It follows (this can be easily proved) that $E$ decomposes as 
$E = S \oplus S^{\perp}$ and the restriction $\sigma|_{S^{\perp}}$ is also non-degenerate. Then use the induction hypothesis on the symplectic space $(S^{\perp},\sigma|_{S^{\perp}})$ to find its symplectic basis $(e_{2},\dots,e_{n},f_{2},\dots,f_{n})$. Set $e_{1} = v$ and $f_{1} = w$ to finish the proof.
3) For a given symplectic space $(E,\sigma)$ a complex structure $J$ is called compatible with $\sigma$ if the formula $g(v,w) = \omega(v,J(w))$
defines a (positive definite) inner product on $V$.
There always exists a compatible complex structure. Simply find a symplectic basis $(e_{1},\dots,e_{n},f_{1},\dots,f_{n})$ and define $J(e_{i}) = f_{i}$ and $J(f_{i}) = -e_{i}$ for all $i \in \{1,\dots,n\}$. As a matter of fact, $(e_{1},\dots,e_{n},f_{1},\dots,f_{n})$ then forms an orthonormal basis for the inner product $g$.
4) Finally, for a given symplectic space $(E,\sigma)$ and a given Lagrangian subspace $L \subset E$, fix some compatible complex structure $J$ (it always exists) and set $M = J(L)$. Clearly $M$ has a dimension $n$ (the map $J$ is always invertible) and it is isotropic (a compatible complex structure $J$ must be a symplectomorphism). It remains to show that $L \cap M = \{0\}$. This follows from the observation that $J(L)$ is in fact an orthogonal complement to $L$ with respect to the positive inner product $g$, whence $L \cap J(L) = \{0\}$.
The concept of compatible complex structures is very useful e.g. to work with Lagrangian Grassmannians, or in Lie groups ($U(n)$ is maximal compact subgroup of $Sp(2n,\mathbb{R})$).
A: I have the impression I was making some confusion over here. Let us answer the real question:

Why is $(M+\langle e\rangle)\cap L=(0)$?

Suppose there exists $v\in(M+\langle e\rangle)\cap L$ and $v\neq0$. Then $v=v_M+ke$ for $v_M\in M$ and $k\in\mathbb{R}$ (or the field the spaces are built upon). If $k=0$, then $v\in M\cap L$, contradicting $v\neq0$. If $v_L=0$ then $v=ke$, but $L$ is lagrangian so if $v_L\in L$ we must have $\sigma(ke,v_L)=0$, yet $e\not\in L=L^\sigma$ so that is not possible. Suppose neither $v_M$ nor $k$ are 0. If that happens, we have $e=\frac{v-v_M}{k}$, but then $e\in M+L$, which contradicts the choice of $e\in M^\sigma\smallsetminus(M+L)$. So $(M+\langle e\rangle)\cap L=(0)$. $\square$
