Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm trying to figure out exactly what the tensor product of vector spaces is. This is what I understand so far:

If $V, W$ are vector spaces over a field $R$ then the free vector space $C(V\times W)$ is a vector space which has an infinite basis (one element for each pair $(v,w)$ such that $v\in V, w\in W$. Then let the subgroup $Z$ of $C(V\times W)$ be generated by elements of the form:

1) $(v, w_1 + w_2)-(v,w_1)-(v,w_2)$

2) $(v_1+v_2,w)-(v_1,w)-(v_2,w)$

3) $(av,w)-a(v,w)$

4) $(v,aw)-a(v,w)$

Where $a\in R$, $v \in V$, $w \in W$. The tensor product $V\otimes W$ is the quotient group $C(V\times W)/Z$.

Apparently this group now obeys the rules $(v, w_1 + w_2)-(v,w_1)-(v,w_2)=0$, and the other corresponding rules from the above, and this follows from the definition of the quotient. I haven't seen this explained anywhere and it's not immediately apparent to me at any rate. Thanks for any replies!

share|cite|improve this question
It's literally immediately from the definition of quotient. If $V/W$ is a quotient space then for $x\in V$ we have $\bar{x}=0$ in $V/W$ iff $x\in W$. – Seth May 9 '14 at 23:34
What do you want to be explained exactly? You've explained how the tensor product is formed and why it obeys bilinearity. What does confuse you? – math.n00b May 9 '14 at 23:34
I saw that in an earlier post you mentioned that your background is mostly in linear algebra and you are doing this for a project. Quotient spaces are not really emphasized in a first course in linear algebra. They are more natural to study in the context of groups/rings/modules. But you may want to review the definition of quotient space. – Seth May 9 '14 at 23:40

It's literally immediately from the definition of quotient. If $V/W$ is a quotient space then for $x\in V$ we have $\bar{x}=0$ in $V/W$ iff $x\in W$.

So $(v,w_1+w_2)−(v,w_1)−(v,w_2)=0$ in $V\otimes W=C(V×W)/Z$ since $(v,w_1+w_2)−(v,w_1)−(v,w_2)\in Z$.

If you are still confused you might want to review the definition of quotient space.

share|cite|improve this answer
Thanks for the fast reply! I mean I guess my confusion is that as far as I know $(v,w_1+w_2)-(v,w_1)-(v,w_2)$ isn't an element in $V\otimes W$, it's an element in $Z$. I thought we're meant to get the linearity relation between the equivalence classes in the quotient group. – James Machin May 9 '14 at 23:57
Or is the class [(v,w_1 + w_2)-(v,w_1)-(v,w_2)] =0 and then you seperate it into the 3 different classes and get the result? – James Machin May 10 '14 at 0:26
Right, really what they are saying is [(v,w_1 + w_2)-(v,w_1)-(v,w_2)] =0. And from definition of quotient [(v,w_1 + w_2)-(v,w_1)-(v,w_2)] = [(v,w_1 + w_2)]-[(v,w_1)]-[(v,w_2)] – Seth May 10 '14 at 0:36
When I said "in $V\otimes W$" what I meant was to pretend that I had written the equivalence class. I was being a little sloppy with notation I guess. – Seth May 10 '14 at 0:37

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.