# Prove that a linear map induced by a bilinear form is well-defined

Let $$\beta$$ be a symmetrical bilinear form on vector space $$V$$ over a field $$K$$ . Now let $$U= \{u\in V : \beta (u,v)=0,\forall v \in V\}$$ Now I have to show, that the linear map $$\bar\beta= V/U \times V/U \to K$$ $$(v_1+U,v_2+U) \mapsto \beta(v_1,v_2)$$ is well-defined. I have already showed that $$U$$ is a $$K$$-subspace of $$V$$ but have trouble showing that the linear map $$\bar\beta$$ is well-defined.

• You need to show that if $(v_1+U,v_2+U)=(w_1+U,w_2+U)$ then $\beta(v_1,v_2)=\beta(w_1,w_2)$. Jun 14 at 16:29

Expanding on my comment, if $$(v_1+U,v_2+U)=(w_1+U,w_2+U)$$ then $$v_1-w_1\in U$$ and $$v_2-w_2\in U$$. Thus $$\beta(v_1-w_1,v_2)=0$$ and $$\beta(v_2-w_2,w_1)=0$$ by definition of $$U$$. Now using the fact that $$\beta$$ is bilinear, we have $$\beta(v_1,v_2)-\beta(w_1,v_2)=0$$ and $$\beta(v_2,w_1)-\beta(w_2,w_1)=0$$. Thus $$\beta(v_1,v_2)=\beta(w_1,v_2)=\beta(v_2,w_1)=\beta(w_2,w_1)=\beta(w_1,w_2)$$ since $$\beta$$ is symmetric.