# Without loss of generality in proof about subspaces in symplectic linear algebra

A linear symplectic space is a 2n-dimensional vector space $V$ with a symplectic two form $\omega.$ On this vector space $V$ is a canonical basis $(e_1,...,e_n,f_1,...f_n)$ with $\omega(e_i,f_j) = \delta_{i,j}$ and $\omega(e_i,e_j)= \omega(f_i,f_j)=0.$

Now, a Lagrangian subspace $L$ of $V$ is an n-dimensional subspace satisfying $L^{\perp} = L$ where the complement is taken with respect to the form $\omega.$

The theorem states now: Any Lagrangian subspace $L$ has an orthogonal complement $K$ that is also a Lagrangian subspace such that $L \oplus K = V.$

The proof starts like this: Let $L_0=(e_1,...,e_n)$ and set $M = L \cap L_0.$ Then $M$ is a $k \le n$ dimensional subspace of $L_0$. As any $k-$ dimensional subspace of $L_0$ has a $n-k$ dimensional orthogonal complement, we find such a $n-k$ dimensional space $T$ such that $M \oplus T = L_0.$

Now, the proof we had in the lecture says: Without loss of generality $T= \operatorname{span}\{e_1,...,e_{n-k}\}$ and I don't really understand why we are able to say this wlog here?

So, I think what the author is doing is picking a new symplectic basis for his/her proof at this step. Instead of using the standard symplectic basis, they would like to use a symplectic basis $\{\tilde e_i, \tilde f_i\}$ where the terms $\tilde e_1, \ldots, \tilde e_{n-k}$ span the space $T$. So, how do we construct the new basis $\tilde e_i$?
• Let's start by constructing the $\tilde e_i$. We can produce an orthonormal basis $\{\tilde e_i\}$ for $L$ with the first $n-k$ terms spanning $L_0$ by using Gram-Schmidt.
• Now, how do we construct the $\tilde f_i$? Write $\tilde e_i=\sum_k a^k_i e_k$. Then if $\tilde f_i=\sum_k a^k_i f_k$, $$\omega(\tilde e_i, \tilde f_j)=\omega\left( \sum_k a^k_i e_k, \sum_k a^k_j f_k\right)$$ which is $\sum_k a^k_i a^k_j = \tilde e_i \cdot \tilde e_j= \delta_{ij}$.
• How can $L_0$ be spanned by $n-k$ terms when $L_0$ is a n-dimensional space? – Surfaceint Aug 20 '15 at 12:22