# Showing well-definedness of addition and scalar multiplication for tensor product

Wikipedia defines the tensor product $$V\otimes W$$ of two vector spaces $$V$$, $$W$$ on a common field as partition $$F(V\times W)~/ \sim$$, not as a quotient space, so there is no a priori an obvious reason that the addition and scalar multiplication of the elements $$v\otimes w:=[(v,w)]$$ as defined there by $$(v_1\otimes w_1)+(v_2\otimes w_2):=[(v_1,w_1)+(v_2,w_2)]$$ and $$c(v\otimes w):=[c(v,w)]$$ are well-defined (in the sense that these don't depend on the choice of representative for any equivalence class $$v\otimes w$$.).

(If it were defined as a quotient space, then these definitions would have been well-defined.)

Question: How to show, using Wiki’s definition that these definitions are indeed well-defined?

• Well-defined means that $+: (V \otimes W)^{\times 2} \rightarrow V \otimes W$ is actually a map, i.e. each element in the domain is mapped to exactly one element in the target (same for the map $c$). The trick is to take an arbitrary equivalence class, take 2 representatives and show that the result does not depend on the choice of representative (as it should since the equivalence class is the same). This should follow directly from the definition of the equivalence relation $\sim$. – Mathphys meister May 26 at 12:02
• @Mathphysmeister Yes, I’m trying to show this for an hour now, but not making any progress... :( – Atom May 26 at 12:07
• Are you sure it is a different definition and not just an imprecise version of the usual one? I don't see how you could prove compatibility of the equivalence relation with the vector space structure without assuming it. – Stefan May 26 at 12:09
• Is this true: if $(v,w)\sim (v’,w’)$ then $v=cv’$ and $w’=cw$ for some scalar $c$? If this holds, then I think I’ve solved it. – Atom May 26 at 12:09
• @Stefan I also think the same! But I’m encountering this for the first time and that too for Quantum Physics, so don’t have much background... – Atom May 26 at 12:11