I am trying to prove the following statement about isotropic subspaces. The definition I have is: Let $V$ be a finite-dimensional vector space on which there is defined a non-degenerate bilinear $B$. $W \subset V$ is said to be isotropic if $B(v,w) = 0$ for every $v,w \in W$.
I'm trying to prove that $$ \dim W \leq \frac{1}{2} \dim \dim V. $$ One option I considered is using the rank nullity theorem. The bilinear map $B$, restricted to $W \times W$, is a linear map $f$ from $W$ to $W^{*}$, sending $w \in W$ to $B(-, w): v \mapsto B(v,w)$. So: $$ \dim \text{W} = \dim \text{Im}(f) + \dim \text{Ker}(f). $$ Restricted to $W \times W$, $B$ is the zero map, so I should have $\dim W = \dim \text{ker}(f) = 0$ and $\dim \text{Im}(f) = 0$. I can't find a way to tie this back to $V$ or the dimension of the original map. Another idea I considered is writing down a basis $v_1, \ldots, v_n$ for $V$ and assessing the effect of $B$ on the basis.
You can do this using induction. $V$ can't be 1-dimensional since you are assuming the form is non-degenerate. If $V$ is 2-dimensional then let $W$ be a maximal isotropic subspace. $W\neq V$ by non-degeneracy, so $\dim(W)=0$ or $1\leq (1/2)2$.
Assume the result is true for vector spaces of dimension $\leq n$ and that $V$ has dimension $n+1$ and let $W$ be a maximal isotropic subspace. Pick your favorite nonzero $w\in W$. By non-degeneracy there is $v\in V$ so that $B(v,w)\neq 0$. By maximality, $W=W^\perp:=\{v\in V\mid B(v,w)=0\forall w\in W\}$, so we have that $v\notin W$. Let $U$ be the span of $v$ and $w$. By construction this is 2-dimensional.
Take the quotient $\tilde V=V/U$. $\tilde V$ has dimension $n-1$ and the image $\tilde W$ of $W$ in $\tilde V$ is isotropic and $\dim(W)=\dim(\tilde W)+1$. By induction $\dim(\tilde W)\leq 1/2(n-1)$ and hence $\dim(W)\leq 1/2(n+1)$.