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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.

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

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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.

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