When there is an orthogonal complement of degenerate subspace Assume that $B$ is a bilinear symmetric form on a finite dimensional space $E$ Let $V$ be a subspace of $E$. 
If $V$ is nondegenerate (that is $V\cap V^\bot=0$) then there exist an orthogonal complement of $V$-a subspace $W\subset E$ such that $V \bigoplus W$ and $V\bot W$. In this case $W=V^\bot$.
What about the existence of an orthogonal complement of degenerate subspace $V$ in the case when the whole space $E$ is  degenerate?
 A: You have a bilinear form $B$ on the vector space $E$.  Let $V \leq E$.  From your notation you are assuming that $B$ is reflexive, that is, $B(u,v) = 0 \iff B(v,u) = 0$.
Here are the possible cases (note that we usually say that $B$ is degenerate instead of saying $E$ is degenerate): 


*

*$B$ is nondegenerate, $V$ is nondegenerate;

*$B$ is nondegenerate, $V$ is degenerate;

*$B$ is degenerate, $V$ is nondegenerate;

*$B$ is degenerate, $V$ is degenerate.


You comment that you know the result that if $B$ and $V$ are nondegenerate, then $V$ has an orthogonal complement; and that if $B$ is nondegenerate but $V$ is degenerate, there is no orthogonal complement.  And you also know the result that if $V$ is nondegenerate then it has an orthogonal complement.
The final case is that the form $B$ is degenerate, as is $V$.  Consider the possibilities for $\dim(V)$, $\dim(V^{\perp})$, and $\dim(V \oplus V^{\perp}) = \dim(V)+\dim(V^{\perp}) - \dim(V \cap V^{\perp})$ in comparison to $\dim(E)$.  This should answer your question once you work out the details. (it will depend on $\dim(E^{\perp})$ and $\dim(E^{\perp} \cap V^{\perp})$).
(I will happily fill in the details if you get stuck trying to work it out, just let me know)
