Show non-degenerate form of subspaces sum

[ktatar156]

Let $$(E, d)$$ be nonzero bilinear space over $$K$$ and place conditions:
$$d(x,y) = d(y,x) \\ d(x,y) = - d(y,x)$$
for every $$x,y \in E$$. Show that:
if $$E_1$$ and $$E_2$$ are singular (degenerate?) bilinear subspaces relative with $$d$$ ( $$(E_1,d|(E_1 \times E_1)$$ and $$(E_2,d|(E_2 \times E_2)$$ are singular (degenerate?) bilinear spaces) with the same finite dimension and $$E_1 \cap E^d_2 = \{ \theta \}$$, then cutting the functional $$d$$ to the space $$(E_1 + E_2) \times (E_1 + E_2)$$ is non-degenerate form.

Can anyone can help me with that?
I know that space $$E$$ is symmetric and skew-symmetric, right? And I need to show that $$rad(E_1+E_2) = 0$$ or $$(E_1+E_2) \cap (E_1 + E_2)^d = \{ \theta \}$$, yes?

 

[jvicens]

Yes, you are correct. To prove that cutting the functional d to the space (E_1 + E_2) \times (E_1 + E_2) is non-degenerate, we need to show that the radical of (E_1 + E_2) is equal to zero or that the intersection of (E_1 + E_2) and (E_1 + E_2)^d is only the zero vector.

First, let's prove that the radical of (E_1 + E_2) is equal to zero. Assume that there exists a nonzero vector v \in (E_1 + E_2) such that d(v,w) = 0 for all w \in (E_1 + E_2). Since v \in (E_1 + E_2), we can write v = v_1 + v_2 where v_1 \in E_1 and v_2 \in E_2. Then, for any w \in (E_1 + E_2), we have d(v,w) = d(v_1 + v_2, w) = d(v_1,w) + d(v_2,w) = 0. This means that both d(v_1,w) and d(v_2,w) must be equal to zero. Since v_1 \in E_1 and v_2 \in E_2, this implies that v_1 = 0 and v_2 = 0. Therefore, v = v_1 + v_2 = 0, which contradicts our assumption that v is nonzero. Hence, the radical of (E_1 + E_2) is equal to zero.

Next, we need to show that the intersection of (E_1 + E_2) and (E_1 + E_2)^d is only the zero vector. Assume that there exists a nonzero vector v \in (E_1 + E_2) \cap (E_1 + E_2)^d. Then, for any w \in (E_1 + E_2), we have d(v,w) = 0 and d(w,v) = 0. Since v \in (E_1 + E_2), we can write v = v_1 + v_2 where v_1 \in E_1 and v_2 \in