I have this proof for a homework exercise, which I can't figure out how to solve..
Let $V$ be a vector space over $F$ and let $B$ be a skew-symmetric or symmetric nondegenerate bilinear form on $V$. Assume that $W$ is a subspace of $V$ on which $B$ restricts to a nondegenerate form. Prove that the restriction of $B$ to the subspace
$W^{\perp}$ = {${v {\in}V : B(v,w) = 0 \;\text{for all}\; w{\in}W }$} is nondegenerate.
Thanks in advance!
I have found a proof for $\dim_F V<\infty$. We will show that in this case $$V=W\oplus W^\perp,$$ because then if $w\in W^\perp$ is orthogonal on all $W^\perp$, $w$ is orthogonal on $W+W^\perp=V$, so $w=0$ by the non-degeneratedness of $B$, i.e. $B$ is non-degenerate on $W^\perp$ as well.
So, why is $V=W\oplus W^\perp$? Well, $W\cap W^\perp=0$ follows directly from the fact that $B$ is non-degenerate on $W$. On the other hand, $W^\perp$ is the kernel of the linear map $$V\ni v\mapsto B(v,\cdot)|_W\in W^*,$$ which is also surjective, since $B$ is non-degenerate and $V$ is finite-dimensional. It follows $\dim W^\perp=\dim V-\dim W$ (the finite dimensionality is crucial for this!), completing the proof.
Unfortunately, I do not see, where the (anti-)symmetry of $B$ is needed. Maybe you do? :-)