Dimension of orthogonal subspace with possibily degenerate bilinear form I'm having headaches because of this supposedly trivial problem:

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*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 $dim(L(v)+v^{\perp})=n$.
 A: 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).
