Let $\mathbb{F}_2$ be the field with 2 elements, and consider $\mathbb{F}_2^n$, the space of all $n$-tuples over $\mathbb{F}_2$. This is an $n$-dimensional vector space over $\mathbb{F}_2$, and we may introduce the following 'dot product' for every pair of vectors $x,y$ in $\mathbb{F}_2^n$ ; if $x=(x_1,\ldots,x_n)$ and $y=(y_1,\ldots,y_n)$, we define
$$\langle x,y\rangle:=x_1y_1+\cdots+x_ny_n=\sum_{i=1}^{n}x_iy_i$$
Clearly this is a symmetric bilinear form, and my question is : when does this form become non-degenerate? More precisely,
For which subspaces $V$ (of $\mathbb{F}_2^n$) does the induced bilinear form $\langle \cdot,\cdot\rangle:V\times V\rightarrow\mathbb{F}_2$ become non-degenerate?
I've been struggling with this problem for quite a while, but still don't see how to approach in the right way. Any advice is welcome.
It will be easier to talk about subspaces on which it becomes degenerate. This means subspaces $V$ containing a nonzero vector $v$ such that $\langle v, w \rangle = 0$ for all $w \in V$. In particular, $v$ must satisfy $\langle v, v \rangle = 0$; such vectors are called null vectors or isotropic vectors, and for this particular bilinear form they are precisely the vectors $v \in \mathbb{F}_2^n$ with an even number of nonzero entries.
Given a nonzero vector $v$, the orthogonal complement
$$v^{\perp} = \{ w \in \mathbb{F}_2^n : \langle v, w \rangle = 0 \}$$
can be explicitly described as follows. If $v$ has nonzero entries in the indices $i_1, \dots i_k$, then $\langle v, w \rangle = 0$ iff among the $i_1, \dots i_k$th entries of $w$ an even number of them are nonzero.
Every degenerate subspace $V$ can be constructed by taking an isotropic vector $v$ and then letting $V$ be any subspace of $v^{\perp}$ containing $v$. This isn't quite as good as a classification, though. It would be nice to be able to count the degenerate subspaces, since there are finitely many of them. One relevant keyword here should be "isotropic Grassmannian."