Problem on a Bilinear pairing This is a problem from hatcher's that I am trying to solve.
Show that a nonsingular symmetric or skew-symmetric bilinear pairing over a field F, of the form $F^n\times F^n\rightarrow F$, cannot be identically zero when restricted to all pairs of vectors v,w in a k-dimensional subspace $V\subset F^n$ if k>n/2.
I am fairly rusty with linear algebra and don't know much about bilinear pairing barring the definitions. Could someone give me a hint on how to proceed. I would be most grateful.
 A: Let $\langle x, x'\rangle$ denote the symmetric or skew symmetric bilinear form for $x,x' \in \mathbb F^n$. which is your vector space.  The form is stated to be non-singular (non-degenerate) so there is no non-zero vector that is orthogonal to all other vectors under your bilinear form.  (I.e. the only null vector is the zero vector.)  Let $W$ denote a subspace where the form is identically zero, i.e. $\langle w, w'\rangle= 0$ for all $w,w'\in W$.
Suppose for contradiction that $\dim W = k \gt \frac{n}{2}$. Now build a bases for $W$ and extend to a basis for the vector space. I.e. we have
$\mathbf B =\bigg[\begin{array}{c|c|c|c|c|c|c|c}w_1 & \cdots &w_k& b_{k+1}&\cdots &b_n \end{array}\bigg]=\bigg[\begin{array}{c|c|c|c|c|c|c|c}b_1 & \cdots &b_k& b_{k+1}&\cdots &b_n \end{array}\bigg]$
Define $A$ such that $a_{i,j}:=\langle b_i, b_j\rangle$.  Note: if this were an inner product we would call $A$ a Gram Matrix.  Then
$A= \left[\begin{matrix}\mathbf 0_{k\times k} & *\\* & *\end{matrix}\right]=\left[\begin{matrix}\mathbf 0_{k\times k} & *\\\mathbf 0 & \mathbf 0\end{matrix}\right]+\left[\begin{matrix}\mathbf 0_{k\times k} & \mathbf 0\\* & *\end{matrix}\right]$
$\text{rank}\Big(A\Big)=\text{rank}\left(\left[\begin{matrix}\mathbf 0_{k\times k} & *\\\mathbf 0 & \mathbf 0\end{matrix}\right]+\left[\begin{matrix}\mathbf 0_{k\times k} & \mathbf 0\\* & *\end{matrix}\right]\right) $
$\leq \text{rank}\left(\left[\begin{matrix}\mathbf 0_{k\times k} & *\\\mathbf 0 & \mathbf 0\end{matrix}\right]\right)+ \text{rank}\left(\left[\begin{matrix}\mathbf 0_{k\times k} & \mathbf 0\\* & *\end{matrix}\right]\right) $
$\leq \big(n-k\big)+\big(n-k\big)$
$\lt \big(n-\frac{n}{2}\big)+\big(n-\frac{n}{2}\big)$
$=n$
justification:
sub-additivity of rank and the fact that $\left[\begin{matrix}\mathbf 0_{k\times k} & *\\\mathbf 0 & \mathbf 0\end{matrix}\right]$ has at most $n-k$ non-zero columns and  $\left[\begin{matrix}\mathbf 0_{k\times k} & \mathbf 0\\* & *\end{matrix}\right]$ has at most $n-k$ non-zero rows.
$\text{rank}\Big(A\Big)\lt n \implies \dim \ker A \geq 1$ by rank-nullity.  Thus there is some $\mathbf x \neq \mathbf 0$ such that
$\mathbf 0 = A\mathbf x$
but looking at the $i$th component shows
$0=\langle b_i, \sum_{j=1}^n x_j \cdot b_j\rangle$
and $\big(\sum_{j=1}^n x_j \cdot b_j\big) \neq 0$ by linear independence of basis elements, yet since this holds for all $i$, then $\big(\sum_{j=1}^n x_j \cdot b_j\big)$ is a null vector and we conclude the form is degenerate, a contradiction.
alternative proof:
the form $\langle , \rangle$ is totally isotropic on subspace $W$ where $\dim W=k$.  Then $W\subseteq W^\perp$ (orthogonal complement) so
$2\cdot k = 2\cdot\dim W \leq \dim W+\dim W^\perp = \dim \mathbb F^n=n\implies k\leq \frac{n}{2}$
(note: skew symmetry/symmetry isn't strictly needed here though without it we would need to run the argument separately for a right orthogonal complement and left orthogonal complement which is cumbersome)
The key result that $ \dim W+\dim W^\perp = n$ holds for any subspace $W$ of an $n$ dimensional vector space equipped with a non-degenerate (skew) symmetric bilinear form.  Proof: using $\mathbf B$ and $A$ as above, and $  P:=\left[\begin{matrix}I_{k} & \mathbf 0_{k\times n-k}\end{matrix}\right]$ (i.e first $k$ columns are the $k$ dimensional std basis vectors, followed by $n-k$ zero vectors)
$n = \text{rank}\big(P\big) + \dim \ker\big(P\big) = k + \dim \ker\big(P A\big)=\dim W+\dim W^\perp$
which holds by rank-nullity and the invertibility of $A$ (since the form is non-degenerate) and the fact that building a basis for $\ker PA$ gives all possible elements in $\mathbb F^n$ that annihilate $W$. i.e. mimicking the ending of the above
$\mathbf 0 = PA\mathbf x$
but looking at the $i$th component shows
$0=\langle b_i, \sum_{j=1}^n x_j \cdot b_j\rangle$
since this holds for all $i\in \big\{1,\dots,k\big\}$ conclude that $\big(\sum_{j=1}^n x_j \cdot b_j\big)\in W^\perp$.  Conversely any element $\in W^\perp$ may be written as $\big(\sum_{j=1}^n x_j \cdot b_j\big)\implies \mathbf x \in \ker PA$.
