Let $\mathbb{F}$ be a field such that $char(\mathbb{F}) \not= 2$ .
Let $q : \mathbb{F} ^n \rightarrow \mathbb{F}$ $(n \geq 2)$ be a quadratic form - defined such that there are $n-1$ linearly independent vectors $v _1 , \dots , v _{n-1}$ for which -
$\forall v \in Sp(\{v _1 , \dots , v _{n-1}\}) : q(v) = 0$ .
Can we say anything about the rank of $q$?
If not over a general field, does the answer change above $\mathbb{R}$ or $\mathbb{C}$?
NOTE:
The question has meaning, because we CAN always define such $q$ :
let $l _1 , l _2$ be non-zero linear transformations from $\mathbb{F} ^n$ to $\mathbb{F}$ .
Let us define $q: v \mapsto l _1(v) l _2(v)$ . $^{(***) }$ $^{see}$ $^{comment}$ $^{bellow.}$
We will demonstrate such a subspace of dimention $n-1$ that $q$ "sends" to $0$ .
$dim(Im(l _1)) = dim( \mathbb{F}) = 1$ .
we also know that: $dim(Ker(l _1)) + dim(Im(l _1)) = dim(\mathbb{F} ^n) = n$
So, we get $dim(Ker(l _1)) = n-1$ .
Indeed, $\forall v \in Ker(l _1) : q(v) = l_1(v)l_2(v) = 0l_2(v) = 0$ , and we are done.
$^{***}$ - $q$ indeed is a quadratic form over $\mathbb{F} ^n$ . it is fairly easy to show that if $f : \mathbb{F} ^n \times \mathbb{F} ^n \rightarrow \mathbb{F}$ is defined as $f(u,w) = l _1(u )l _2 ( w )$ for all $u,w \in \mathbb{F} ^n$, then $f$ is a bilinear form over $\mathbb{F} ^n$, and $f(v,v) = q(v)$ for all $v \in \mathbb{F} ^n$.
Because $\Bbb F$ is not of characteristic $2$, the polarization identity yields the associated symmetric bilinear form $$ b(v,u) = \frac 12 (q(u,v) - q(u) - q(v)). $$ Thus, $q$ can be associated with a symmetric matrix $A \in \Bbb F^{n \times n}$ (such that $q(v) = v^TAv$).
Complete the set $\{v_1,\dots,v_{n-1}\}$ to form a basis $\mathcal B = \{v_1,\dots,v_{n-1},v_n\}$. From the information given, we know that the matrix associated with this quadratic form has the form $$ A = \pmatrix{0_{(n-1) \times (n-1)} & x\\ x^T & \alpha}, $$ where $x \in \Bbb F^{n-1}$ and $\alpha \in \Bbb F$. It is easy to see that there are three possibilities for rank of this matrix (and therefore the rank of the quadratic form): it must either be $0$, $1$ (which holds iff $x = 0$), or $2$.
In fact, based on the form of the matrix, we can say that for some $v_1,v_2 \in \Bbb F^n$, the quadratic form must be $$ q(x) = x^T[v_1v_2^T + v_2^Tv_1] x = 2(v_1^Tx)(v_2^Tx). $$