Vector space is decomposed into direct sum of its subspace and its orthogonal complement

Question

Let $E$ be a finite-dimensional vector space over the field $k$ and let $g$ be a bilinear form which is symmetric, antisymmetric or hermitian. Let $V$ be a subspace of $E$. Let $V_1 = \{ x| x \in E, g(x,a)=0\ \forall a \in V\}$. Prove that $E$ is a direct sum of $V$ and $V_1$. This result is used in the proof as well-known but unfortunately I can't find the proof of it. The proof I know is in the case of $g$ being inner product.

Answer 1

Hint in the finite dimensional case. Let $\;n=\dim_kE=n\;$ and take a basis of $\;V_1\;,\;\{v_1,..,v_k\}$ . Complete this to a basis of the whole space: $\;\{v_1,..,v_k,v_{k+1},...,v_n\}\;$ . Using Gram-Schmidt we can assume this is an orthonormal basis, and then

$$\{v_{k+1},...,v_n\}\;\;\text{is a basis of}\;\;V$$

In fact, I can't see how this is really different from the usual proof for inner products ...