This seems to be an elementary question, but I could not find an answer in google nor I am able to solve it.
Let $k$ be any field and let $V$ be a vector space over that field.
A bilinear map $B:V\times V\rightarrow k$ is called anti-symmetric if $b(u,v) = -b(v,u)$ for all $u,v\in V$.
Prove that any anti-symmetric bilinear form $B$ admits another bilinear form $L$ so that $B(u,v) = L(u,v)-L(v,u)$.
When $k$ is a field of characteristic $\mathrm{char} k\not = 2$, we can take $L:=\frac{B}{2}$ and use anti-symmetry.
When $k=\mathbb{F}_2$ (the field with $2$ elements) and $\dim V = 2$, I computed by hand that if $L=\begin{bmatrix} l_1 & l_2 \\ l_3 & l_4 \end{bmatrix}$, then $L(u,v)-L(v,u) = (l_2-l_3)u_2v_1+(l_3-l_2)u_1v_2$. Another computation shows that any anti-symmetric bilinear map takes this form (Edit: The last claim is apparently wrong unless one assumes in addition that $B(u,u)=0$ for all $u\in V$).
Even a general solution when $k=\mathbb{F}_2$ is appreciated.
Edit: With the additional assumption that $B(u,u)=0$ I can solve the claim for finite dimensional $V$.
Look at the homomorphism $\phi:\mathcal{M}_n(k)\rightarrow \mathcal{M}_n(k)$ which takes an $n\times n$ matrix $A$ and sends it to the matrix $\phi(A)$ defined by the form $(u,v)\mapsto u^T A v - v^T Au$.
The kernel of these maps consists of matrices $A$ satisfying $u^TAv = v^TAu$, hence $A=A^T$. This space has dimension $\frac{n^2+n}{2}$ (because $A$ is determined by its values on the diagonal and everything above it). Therefore the image has dimension $\frac{n^2-n}{2}$, which is the same as the dimension of anti-symmetric bilinear maps satisfying in addition $B(u,u)=0$ for all $u\in U$.
The result does not hold over fields with characteristic $2$.
It is not even true that every anti-symmetric bilinear map $\Bbb F_2^2$ takes the form you suggested. For instance, note that the bilinear map $$ b(v,u) = u_1v_1 + u_2v_2 $$ Is antisymmetric (in addition to being symmetric) over $\Bbb F_2$. Indeed, note that $x = -x$ for $x \in \Bbb F^2$, which means that $b(u,v) = b(v,u) \iff b(u,v) = -b(v,u)$.