Let $V$ be a vector space (over a field $K$) together with a symmetric bilinear form $\langle-,-\rangle$, and let $T\subseteq V$ be the orthogonal complement (since I'm not a native speaker, I hope this is the right term) with $$T :=\{v\in V|\ \langle u,v\rangle = 0 \text{ for every } u \in V\}.$$ Show that there exists a symmetric bilinear form $\langle-,-\rangle'$ on $V/T$ (this is the vector space that includes the equivalence classes of the equivalence relation defined by $T$ on $V$) such that $$\langle[u], [v]\rangle' =\langle u, v\rangle,$$ for every $u,v\in V$.
Yeah, that's the task, and I have to admit that I have certain difficulties with proving the existence of certain objects - never got into the right way of thinking for that I guess. Therefore, I would be glad to receive only a small hint so that I can try solving it on my own.
The problem is that I don't even know where to start from here. I know there exists a canonical projection
$$f:\ V\ \longrightarrow\ V/T,$$
with $f$ being linear. Shall I use this information and somehow proof the equation
$$\langle-,-\rangle=\langle-,-\rangle',$$
like I tried here:
The question already presents to you what the bilinear form should be; it is defined as $$\langle[u],[v]\rangle':=\langle u,v\rangle.$$ It remains to verify that this indeed defines a symmetric bilinear form.