Prove that every bilinear map can be written as a sum of bilinear symmetric map and a bilinear anti-symmetric map.

Question

Show that every bilinear map $\phi: E \times E \longrightarrow F$ can be written uniquely as a sum of a bilinear symmetric map ($f(u,v) = f(v,u)$, for all $u,v \in E$) and a bilinear anti-symmetric map ($f(u,v) = -f(v,u)$, for all $u,v \in E$).

My attempt:

I defined $$L_2(E,F) := \{ \phi: E \times E \longrightarrow F \ ; \ \phi \ is \ bilinear \},$$ $$W_1 := \{f \in L_2(E,F) \ ; \ f \ is \ a \ bilinear \ symmetric \ map \},$$ $$W_2 := \{g \in L_2(E,F) \ ; \ g \ is \ a \ bilinear \ anti-symmetric \ map \}$$

and I thought to proof that $L_2(E,F) = W_1 \ \oplus \ W_2$. I got to prove that $W_1 \ \bigcap \ W_2 = \{ 0 \}$, but I dind't get prove that $L_2(E,F) = W_1 + W_2$. How can I show this? I tried to prove by contradiction unsuccessfully.

Thanks in advance!

Answer 1

I assume that $E,F$ are $k$-vector spaces, where the characteristic of the field $k$ is not $2$.

Let $\phi \in L_2(E,F)$. Consider

$$\phi(u,v) = \underbrace{\dfrac{\phi(u,v)+\phi(v,u)}{2}}_{=:\phi_1(u,v)}   \;\;+\;\; \underbrace{\dfrac{\phi(u,v)-\phi(v,u)}{2}}_{=:\phi_2(u,v)}$$

Can you take it from here?