Are there polarization-like identities for nonsymmetric bilinear forms?

Question

Let $B$ be a bilinear form on a vector space $V$. If $B$ is symmetric, there are polarization identities, e.g. $$2B(u,v)=B(u+v,u+v)-B(u,u)-B(v,v).$$ Are there any identities, similar in character, for $nonsymmetric$ ($B(u,v)\neq B(v,u)$ for some distinct $u,v\in V$) bilinear forms? By "similar in character", I mean they allow you to determine $B(u,v)$ with knowledge only of it's associated quadratic form $B(u,u).$

Answer 1

If the scalar field doesn't have characteristic 2, we can split any bilinear form $B$ uniquely into symmetric and antisymmetric parts $S$ and $A$: $$ B(u,v) = \underbrace{\tfrac12(B(u,v)+B(v,u))}_{=S(u,v)} + \underbrace{\tfrac12(B(u,v)-B(v,u))}_{=A(u,v)} $$

Or in other words: The vector space of all bilinear forms is a direct sum of the subspace of symmetric forms and the subspace of antisymmetric forms. (An "antisymmetric" form is by definition one that satisfies $A(u,v)+A(v,u)=0$).

Obviously $S(v,v) = B(v,v)$, so using the polarization identity you can recover $S(u,v)$ from knowing $B(v,v)$. (The subspace of symmetric forms is naturally isomorphic to the vector space of quadratic forms).

However, the antisymmetric summand is completely independent of how $B(v,v)$ behaves. In fact, in $B(u,v) = S(u,v)+A(u,v)$ you can replace the $A$ term with a completely arbitrary antisymmetric form, and that will make no difference at all to which quadratic form $B$ restricts to, because we have $A(u,v)=-A(v,u)$ and therefore $A(v,v)=0$.

So the polarization identity is the most that knowing the quadratic form will tell you about $B$, namely what its symmetric part is.