# Dimension of orthogonal subspace with possibily degenerate bilinear form

I'm having headaches because of this supposedly trivial problem:

• Given a bilinear form $$\phi$$ on the $$n-dimensional$$ vector space $$V$$ and a vector $$v$$, show that the subspace $$v^{\perp}$$ has either dimension $$n$$ or $$n-1$$.

Note that $$\phi$$ could be degenerate.

Maybe I should exploit Grassmann's relation, but in that case, I don't know how to argue that $$dim(L(v)+v^{\perp})=n$$.

As $$\phi: V \times V \rightarrow \mathbb{R}$$ is bilinear, we can form a map from $$V$$ to its dual space $$V^{\ast}$$ by just putting $$v$$ into its first slot, so
$$V \rightarrow V^{\ast},\ v \mapsto \phi(v,-),\quad \text{where }\phi(v,-): V \rightarrow \mathbb{R}.$$
(I'm guessing that the orthogonality refers to the form $$\phi$$?) The orthogonal subspace to $$v$$ is $$v^{T} = \ker(\phi(v,-)) = \{w \in V\, : \, \phi(v,w) = 0\}$$. Viewing $$\phi(v,-)$$ as a linear map from an $$n$$-dimensional vector space $$V$$ to a $$1$$-dimensional vector space $$\mathbb{R}$$, the Rank-Nullity Theorem gives
$$\text{rank}(\phi(v,-)) + \text{nullity}(\phi(v,-)) = \dim V \iff \text{nullity}(\phi(v,-)) = n - \text{rank}(\phi(v,-)).$$
Noting that the nullity here is just the dimension of the kernel, which moreover is the dimension of $$v^{T}$$, one gets that $$\dim(v^{T}) = n$$ if the rank is zero (i.e. $$\phi$$ is degenerate), and $$\dim(v^{T}) = n-1$$ if the rank is $$1$$ (i.e. $$\phi$$ is non-degenerate, so $$\phi(v,-)$$ has a trivial kernel).