What happens to symplectic basis if bilinearity condition is weak Let $B:V\times V \rightarrow K$ be a (weak!) bilinear form where $K$ is a finite field with base field $F$ and $V$ a vector space over $K$. Let $u,v \in V$ and $\lambda \in F (!)$
$B(u + v, w) = B(u, w) + B(v, w)$
$B(u, v + w) = B(u, v) + B(u, w)$
$B(λu, v) = B(u, λv) = λB(u, v)$ (actually this condition is not necessary because of above conditions.)
Alternating: $B(v, v) = 0$ holds for all $v ∈ V$, and
Nondegenerate: when $B(u, v) = 0$ holds regardless of $v, u$ must be zero.
My question is whether $V$ "can have" a symplictic basis or not.  (I think the answer is yes. But i am not sure.)+(i don't know there is a definition of "weak bilinear form", i just defined it for not misunderstanding.)
bilinear form
symplectic vector space
symplectic basis 
Edited part: has --> can have
 A: The answer is negative. Below is a slightly cheating counterexample. I say cheating, because you are probably looking for something more interesting.
Let $\overline{B}:V\times V\to F$ be a usual (non-degenerate) symplectic form over $F$ (among other things it thus has a symplectic basis over $F$), and let $e\in K\setminus F$ be a fixed element. Then the formula
$$
B(u,v)=e \overline{B}(u,v)
$$
gives another function $B:V\times V\to K$. This function inherits all the listed properties from $\overline{B}$.
But it cannot have a symplectic basis, because $B(u,v)\neq 1_K$ for all $u,v$.
===
A concrete toy example. Let $K=\Bbb{F}_4=\{0,1,a,a+1=a^2\}$. Let $V=K$. We view $V$ as a vector space over $F=\{0,1\}\subset K$ with basis $\mathcal{B}=\{1,a\}$. Let $\overline{B}$ be the bilinear mapping gotten by extending $\overline{B}(1,1)=0=\overline{B}(a,a)$, $\overline{B}(1,a)=1=\overline{B}(a,1)$ bilinearly over $F$. Then the function
$$
B(x,y)=a \overline{B}(x,y)
$$
is a weakly bilinear function. There is no symplectic basis w.r.t. $B$, because for all $x,y\in V$ we have $B(x,y)\in\{0,a\}$.
