Possible Duplicate:
understanding of the “tensor product of vector spaces”

Take vector spaces $V, W$ over the field $\mathbb{K}$. I've come across two different definitions of tensor product $V \otimes W$ and was wondering whether they are the same thing.

Definition 1. $V \otimes W$ is the quotient space $F(V\times W) / \sim$ where $F(V\times W)$ is the free vector space over $V \times W$ and $\sim$ denotes the following relations:

$$(v_1+v_2, w)\sim (v_1, w)+(v_2,w), \qquad (v, w_1+w_2)\sim (v, w_1) + (v, w_2), $$ $$(\lambda v, w) \sim (v, \lambda w)\sim \lambda (v, w).$$

In other words $V \otimes W$ is the space of formal linear combinations

$$\sum_{j=1}^n \alpha_j v_j \otimes w_j, \qquad \alpha_j \in \mathbb{K}, v_j \in V, w_j \in W$$

where we explicitly require $\otimes$ to be bilinear.

Definition 2 (When $V, W$ are finite-dimensional.) $V \otimes W$ is the space of mappings from $V^\star\times W^\star$ into $\mathbb{K}$ that are linear in each variable.

Question. Under what circumstances do those two definitions coincide (up to canonical isomorphism)? I guess that this happens if and only if both $V$ and $W$ are finite-dimensional.


marked as duplicate by Rasmus, Zhen Lin, Asaf Karagila, J. M. is a poor mathematician, Zev Chonoles Oct 10 '11 at 6:10

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.


This answer has been moved to

understanding of the "tensor product of vector spaces".


They always coincide because they both fulfill the universal property of the tensor product.

  • 5
    $\begingroup$ The asker wants to go beyound the finite-dimensional case. $\endgroup$ – darij grinberg Oct 9 '11 at 13:26

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