We have $V$ a $\mathbb{R}$-vector space of finite dimension $n$ with non-degenerate bilinear form $\phi:V\times V\rightarrow \mathbb{R}$ and subspace $U$ of $V$ with $U\subset U^{\perp}$. How do I show that the dimension of $U$ is at most $n/2$?
Edit $U^{\perp}=\{v\in V:\phi(v,u)=0\space \forall u\in U\}$
Non-degenerate means in this case that the linear map
$$V \rightarrow V^\ast; \quad v \mapsto \phi(v, \cdot)$$
is an isomorphism. Under this isomprphism, the subspace $U^\perp$ is isomprphic to what we call the anihilator of $U$ in $V^\ast$, i.e. the subspace $U' := \lbrace v^\ast \in V^\ast \, |\, v^\ast(u) = 0 \quad \forall u \in U \rbrace$ of $V^\ast$.
Now choose a basis $(u_1, \dots, u_k, w_{k+1}, \dots, w_n)$ of $V$ such that $(u_1, \dots, u_k)$ is a basis of $U$. Let $(u_1^\ast,\dots, u_k^\ast, w_{k+1}^\ast, \dots, w_n^\ast)$ be its dual basis in $V^\ast$. Then $(w_{k+1}^\ast, \dots, w_n^\ast)$ is a basis of $U'$. The assumption $U \subset U^\perp \cong U'$ then implies that $$k = \dim{U} \leq \dim{U^\perp} =\dim{U'} = n-k$$ which is equivalent to $k \leq \frac{n}{2}$.