Extension to Witt's bases Let $X$ be a real finite dimensional linear space. Let $B:X\times X \rightarrow \mathbf{R}$ be a bilinear symmetric non-degenerated form. Let $M$, $N$ be totally isotropic subspaces with the same dimension $r$ such that $M\cap N=\{0\}$. 
Assume that $e_1, ...,e_s$ are linearly indedendent in $M$, $f_1,...,f_s$ are linearly independent in $N$ (where $s< r$) and $B(e_i,f_j)=\delta_{i,j}$ for $i,j=1,...,s$.
Is it possible to extend $e_1,...,e_s$ to a basis $e_1,...,e_r$ in $M$, $f_1,...,f_s$ to a basis $f_1,...,f_r$ in $N$  in such a way that $B(e_i,f_j)=\delta_{i,j}$ for $i,j=1,...,r$?
I know only that for each basis $e_1,...,e_r$ in $M$ there exists a basis $f_1,...,f_r$ in $N$ such that $B(e_i,f_j)=\delta_{i,j}$ for $i,j=1,...,r$ (Bourbaki, Algebra, Chap.11, $\S$4,2, Prop.2)
Thanks.
 A: I think you need some additional assumption such as $X = M \oplus N$. 
For example, if $s=0$, you're just asking whether any 2 totally isotropic subspaces with intersection $\{0\}$ have bases satisfying $B(e_i, f_j) = \delta_{i,j}$, and this will be false if you take some totally isotropic subspace $U = \langle e_1, e_2 \rangle $ and split it into $M = \langle e_1 \rangle$ and $N = \langle e_2 \rangle$.  The whole space could still be big enough to be nondegenerate.
If you assume $X = M \oplus N$ (so $X$ has dimension $n=2r$), you can do it.  It suffices to show how to do a single step from $s$ to $s+1$.  Let $M_s$ be the span of $e_1, \ldots, e_s$ and let $N_s$ be the span of $f_1, \ldots, f_s$.  Pick some $e = e_{s+1}$ not in $M_s$.  By nondegeneracy (and the assumption $X = M \oplus N$), $N \cap e^\perp$ is a proper subspace of $N$.  Therefore there exists a vector in $N$ outside both $N \cap e^\perp$ and $N_s$ (the union of two proper subspaces of $N$ cannot be all of $N$). Choose such a vector and make it $f_{s+1}$; by construction, it lies in $N$, and $B(e_{s+1}, f_{s+1})$ is nonzero and therefore we can scale $f_{s+1}$ to make the inner product equal to  1.
