Let $V$ be a real finite-dimensional vector space with a skew-symmetric bilinear form $B \colon V \times V \to \mathbb{R}$. In general, we assume that the form $B$ is degenerate. A subspace $S \subset V$ is called totally isotropic if $S \subset S^\perp$, where $S^\perp$ is the orthogonal complement of the subspace $S$ with respect to the form $B$. I need to prove the following statement: If $\dim S = (1/2) ( \dim V + \dim V^\perp )$, then $S$ is maximal totally isotropic.
As far as I understand, if $S \subset V$ is maximal totally isotropic, then it contains any totaly isotropic subspace. So, I begin my proof like this: Assume that there is a totally isotropic subspace $S_1 \subset V$ such that $S_1 \not\subset S$. This means that there is a vector $x_0 \in S_1$ such that $x_0 \notin S$. But I don’t understand what to do next? The only thing I was able to prove is that $S = S^\perp$.
I will be glad for any help.
Let
\begin{equation}\begin{array}{rcl}f&\colon &S \rightarrow {V}^{\ast }\\ &&x \mapsto B \left(x , \cdot \right) \end{array}\end{equation}
One has
\begin{equation}\text{Ker} f = S \cap {V}^{\perp } \quad \Longrightarrow \quad \text{dim Ker} f \leqslant \text{dim} \left({V}^{\perp }\right)\end{equation}
and
\begin{equation}\text{Image} \left(f\right) \subset {S'} \quad \Longrightarrow \quad \text{rank} \left(f\right) \leqslant \text{dim} \left(V\right)-\text{dim} \left(S\right)\end{equation}
where $ {S'}$ is the dual-orthogonal of $ S$. By the rank-nullity theorem it follows that
\begin{equation}\text{dim} \left(S\right) = \text{dim} \left(\text{Ker} f\right)+\text{rank} f \leqslant \text{dim} \left({V}^{\perp }\right)+\text{dim} \left(V\right)-\text{dim} \left(S\right)\end{equation}
Hence
\begin{equation}\text{dim} \left(S\right) \leqslant \frac{1}{2} \left(\text{dim} {V}^{\perp }+\text{dim} V\right)\end{equation}
Clearly the equality case implies that $ S$ is maximal.