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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.

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

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  • $\begingroup$ 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. $\endgroup$ – 72D Feb 19 at 23:17
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    $\begingroup$ 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 $$(-v,w) = (-1v,w) = (v,-1w) = (v,-w).$$ $\endgroup$ – Guido A. Feb 19 at 23:25
  • $\begingroup$ 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 $V \otimes W$. $\endgroup$ – Daniel Schepler Feb 19 at 23:28

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