I don't see why the following fact is true, although I thought it is purely a linear algebra problem.
Let $(M^{2n},\omega)$ be a symplectic manifold and $Y \subset M$ be a $(2n-1)$ hypersurface. We define the $\textit{characteristic foliation}$ of $Y$ as the subbundle $L_Y = \ker(\omega|_Y) \subset TY$, where $\ker(w|Y)_p = \{v \in T_pY \mid \omega(v,\cdot) \equiv 0 \text{ on } T_pY \}$ for each $p \in Y$.
The claim is that $L_Y$ is a rank 1 subbundle of $TY$. Of course, the linear algebra model is the following: Let $V$ be a $2n$-dimensional (real) vector space, $W \leqslant V$ be a $(2n-1)$-dimensional subspace, and $B: V \times V \to \mathbb{R}$ be a non-degenerate bilinear map, then $\dim_{\mathbb{R}}\{w \in W \mid B(w,\cdot) = 0 \text{ on } W\} = 1$. I tried to use the rank-nullity theorem for the mapping $W \to W^{*}$ given by $w \mapsto B(w,\cdot)$, but didn't know how to conclude the rank of this map is $(2n-2)$. I'd really appreciate any help on this.
I think the linear algebra argument goes as follows: you are considering the composition $$B: W \to V\to V^*\to W^*.$$ The part $V\to W^*$ is surjective and so for dimensional reasons, its kernel must be 1-dimensional. So $\dim\operatorname{ker}(B)\leq 1$. If this dimension were $0$, then $B$ would be a non-degenerate alternating form on $W$, i.e. a symplectic form. But $W$ has odd dimension and so we get a contradiction.