How to find the additive inverse and additive identity of an element of the tensor-product vector space?

The 8 axioms of holding for a vector space defined in Chapter 5 of the book Robertson's Basic Linear Algebra are easily checked for tensor product expect for existence of (unique?) additive inverse and identity:

(V3) there exists an element $$0 \in V$$ such that $$x + 0 = x$$ for every $$x \in V$$.

(V4) for every $$x \in V$$ there exists $$-x \in V$$ such that $$x + (-x) = 0$$;

For example, the problem is that sum is defined when at most one component differs. So in order $$(v,w) + 0 = (v,w)$$ to hold, then $$0=(0,w)$$ or $$0=(v,0)$$ not only not defining a unique $$0$$ but also it implies $$(0,w)=(v,0)$$. ...

How to check the mentioned axioms of a vector space, i.e. (V3) and (V4), for the tensor-product vector space?

PS I am learning tensor product from this link and I because it claimed that tensor product is a vector space "by force [=definition] I tired to check that by what I had learnt from Robertson's Basic Linear Algebra.

• Would you mind expanding on how the book constructs $V \otimes W$ ? – Guido A. Feb 19 at 22:37
• @GuidoA., I edited the post. thanks – 72D Feb 19 at 22:43

We have that

$$(-v,w) + (v,w) = (0,w) = (0,0),$$

since

$$(0,w) = (00,w) = (0,0w) = (0,0)$$

and likewise $$(v,0) = (0,0)$$. This is not a contradiction because on $$V \otimes W$$, all these pairs are 'declared' to be equal. Likewise

$$(0,0) + (v,w) = (v,0) + (v,w) = (v,0+w) = (v,w).$$

A more formal treatment of this could be done via equivalence relations, which I highly encourage you to read on. With this machinery, one takes the cartesian product $$V \times W$$ and then identifies some pairs as equivalent, which is what 'declaring' equality formally means in the article. Then, $$v \otimes w$$ is nothing more than the equivalence class of $$(v,w)$$.

• For example in tensor product $(-v,-w) + (v,w)$ is not allowed as at least one of 1st/2nd components must be equal. $(v_1,w) + (v_2,w)$ is allowed since second components are equal. – 72D Feb 19 at 23:17
• My bad, I meant to write something else. Fixed. As you can see, both $(-v,w)$ and $(v,-w)$ are inverses of $(v,w)$, but that's okay once again because $$(-v,w) = (-1v,w) = (v,-1w) = (v,-w).$$ – Guido A. Feb 19 at 23:25
• You could put in an extra step $(0, w) = (0 \cdot 0, w) = 0 (0, w)$ whereas $(0, 0) = (0, 0 \cdot w) = 0 (0, w)$. But really, this is beside the point, which is that elements of $V \otimes W$ are formal vector space expressions in terms of elements of $V \times W$ (modulo an equivalence relation as you said), so $0$ and $-(v, w)$ are perfectly valid expressions for elements of $V \otimes W$. – Daniel Schepler Feb 19 at 23:28