How to show that $(W^\bot)^\bot=W$ (in a finite dimensional vector space) I need to prove that if $V$ is a finite dimensional vector space over a field K with a non-degenerate inner-product and $W\subset V$ is a subspace of V, then:
$$
(W^\bot)^\bot=W
$$
Here is my approach:
If $\langle\cdot,\cdot\rangle$ is the non-degenerate inner product of $V$ and $B={w_1, ... , w_n}$ is a base of $V$ where ${w_1, ... , w_r}$ is a base of $W$ then I showed that 
$$
\langle u,v\rangle=[u]^T_BA[v]_B
$$
for a symmetric, invertible matrix $A\in\mathbb{R}^{n\times n}$. Then $W^\bot$ is the solution space of $A_rx=0$ where $A_r\in\mathbb{R}^{r\times n}$ is the matrix of the first $r$ lines of $A$. Is all this true?
I tried to exploit this but wasn't able to do so. How to proceed further?

Hint It follows from the definition that $(W^\perp)^\perp \subset W$.
Hint 2: For every subspace $U$ of $V$ you have
$$\dim(U)+ \dim(U^\perp)=\dim(V)$$
What does this tells you about $\dim(W)$ and $\dim (W^\perp)^\perp$?
