Orthogonal complement of orthogonal complement of U equals U

Question

Let $V$ be of finite dimension over some field $F$. Let $ξ\in T_2(v)$ symetric bilinear form. Let $U\subset V$. suppose $ξ|_U$ is not degenerate, is it necessarily $(U^⊥)^⊥=U$ ?

my approach:

I managed to prove that $U\subset (U^⊥)^⊥$ and I also noticed that if $ξ|_{U^⊥}$ is also not degenerate we can get these:

1) $dim U +dim U^⊥=dim V$

2) $dim U^⊥ +dim (U^⊥)^⊥=dim V$

and thus, $dim U +dim U^⊥ -(dim U^⊥ +dim (U^⊥)^⊥)=dim V-dim V$

so $dim U=dim (U^⊥)^⊥$ and thus, $(U^⊥)^⊥=U$.

but we can't know for sure that $ξ|_{U^⊥}$ isn't degenerate.

please help me.

Answer 1

Try $F=\mathbb R$, $V=\mathbb R^2$, $\xi=\begin{bmatrix}1&0\\0&0\end{bmatrix}$, and $U=\text{span}\Biggl\{\begin{bmatrix}1\\0\end{bmatrix}\Biggr\}$.