Let $\phi$ be a bilinear form on a vector space $V$. Let $U$ be a vector subspace of $V$. Suppose that $\phi$ and $\phi_{|U\times U}$ are both non-degenerate. Prove that if $U$ is finite-dimensional, then $\phi_{|U^{\bot,L}\times U^{\bot,L}}$ is non-degenerate.
Here $U^{\bot,L} = \{ v\in V : \forall u\in U,\ \phi(v, u) = 0 \}$ and $U^{\bot,R} = \{ v\in V : \forall u\in U,\ \phi(u, v) = 0 \}$.
I know how to prove this if $V$ is finite dimensional but not sure when $V$ is infinite dimensional...
I can also prove that $ U^{\bot,L}\cap (U^{\bot,L})^{\bot,R} = \{0\}$, but for $U^{\bot,L}$ to be non degenerate we also need $U^{\bot,L}\cap (U^{\bot,L})^{\bot,L} = \{0\}$, and that seems far-fetched.
May I have some hint on this? Thank you.
$\boldsymbol{U^{\bot,L}\cap (U^{\bot,L})^{\bot,L} = \{0\}}$:
Since $\phi$ is non-degenerate, $V^{\bot,L}=\{0\}$.
$\Rightarrow$ $\forall u_k\in U^{\bot,L}-\{0\}$ $\exists v\in V$s.t. $\phi(u_k,v)\neq0$
Since $V = U \bigoplus U^{\bot,L}$, and $\forall u\in U,\ \phi(u_k, u)=0$, we must have $\exists v'\in U^{\bot,L},\ \phi(u_k,v')\neq0$.
$\boldsymbol{U^{\bot,L}\cap (U^{\bot,L})^{\bot,R} = \{0\}}$:
We also have $V = U \bigoplus U^{\bot,R}$.
Let $u_k\in U^{\bot,L}-\{0\}$, $\exists u \in U,\ u'\in U^{\bot,R}$, $u_k=u+u'$
$i.e. u_k - u = u'$
While $V^{\bot,R}=\{0\}\Rightarrow \exists v\in V,\ \phi(v,u_k-u)=\phi(v_1+v_2,u_k-u)\neq0$, for some $v_1\in U$, $v_2\in U^{\bot,L}$.
We have $\phi(v_1+v_2,u_k-u)=\phi(v_2,u_k-u)=\phi(v_2,u_k)\neq0$