Let $V$ be a vector space over $K$, $\langle-.-\rangle$ a symmetric bilinear form on $V$ and $T \subseteq V$ with
$T :=$ {$v \in V$ | $\langle v, u\rangle = 0$ $\forall v \in V $}.
I have to show that there exists a bilinearform $\langle-.-\rangle'$ on the quotient space $V/T$ such that
$$\langle v,u \rangle = \langle [v], [u]\rangle'.$$
I already showed the linearity in both components and the symmetry, but I still haven't figured out how to show the well-definedness of the bilinear form. I know that I have to proof that
$[v] = [v']$ ^ $[u] = [u']$ $=>$ $\langle [v'],[u']\rangle'$ = $\langle [v],[u]\rangle',$
but I wasn't able to manage it yet. Does someone have a clear hint for me?
