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Given a field $\mathbb{F}$ and two $F$-vector spaces $V$ and $W$, it's true that if $\{v_i\}$ and $\{w_j\}$ are bases for $V$ and $W$, respectively, then the set $\{v_i \otimes w_j\}$ is a basis for $V\otimes_\mathbb{F}W$.

However, given an arbitrary basis $B$ for $V\otimes_\mathbb{F}W$, can we assume $B= \{v_i \otimes w_j \}$, for some $v_i \in V$ and $w_j \in W$? Off the top of my head, I can think of one case in which this is not true: If $\{v_i \otimes w_j \}$ is a basis, then so is $\{v_i \otimes w_j + a \otimes b \}$, for some fixed $a \in V$ and $b \in W$. But by subtracting $a \otimes b$ from each vector we obtain the original basis of pure tensors. So my question is: Are there any bases for $V \otimes_{\mathbb{F}} W$ that cannot somehow be "manipulated" into a basis of pure tensors?

I understand that "manipulate" here is not clearly defined, but hopefully the question is clear enough.

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  • $\begingroup$ This very much depends on what manipulations are allowed. $\endgroup$ – Jendrik Stelzner Jan 16 '16 at 22:17
  • $\begingroup$ I'm thinking of the usual addition and scalar multiplication. I guess maybe defining some kind of equivalence relation on the vectors could also make sense. $\endgroup$ – leibnewtz Jan 16 '16 at 22:19
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    $\begingroup$ If you allow each basis vector of a given basis $\mathcal{B}$ to be replaced by an arbitrary linear combination of all basis vectors in $\mathcal{B}$ (such that the new set is again a basis), then this is true, because every element of $V \otimes_\mathbb{F} W$ can be expressed as the linear combination of $\mathcal{B}$ — in particular the elements of a basis $\mathcal{C}$ consisting of simple tensors. That it is possible to go from $\mathcal{B}$ to $\mathcal{C}$ by carefully replacing one vector at a time is a basic statement from linear algebra. $\endgroup$ – Jendrik Stelzner Jan 16 '16 at 22:23
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So after some thought, I realized that in its clearest formulation my question is equivalent to the question of whether or not every element of $V \otimes_{\mathbb{F}} W$ can be expressed as a finite linear combination of pure tensors:

Letting $\{v_i \}$ and $\{w_j \}$ be bases for $V$ and $W$, the set $\{v_i \otimes w_j \}$ is a basis for $V \otimes_{\mathbb{F}} W$. Hence every element of the tensor product can be expressed as a finite linear combination of pure tensors. Further, as noted by Jendrik in the comments, by an elementary fact from linear algebra, any basis may be replaced by a basis of pure tensors.

Since my question was not clear to begin with, I think this is the best answer that can be given.

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  • $\begingroup$ To see that $V \otimes W$ is generated by pure tensors the statement that $\{v_i \otimes w_j\}$ is a basis of $V \otimes W$ is a bit overkill: The most common construction of $V \otimes W$ is as a quotient of the free vector space over the set $V \otimes W$. Because this free vector space has the set $V \times W$ as a basis it directly follows that $V \otimes W$ is generated by the residue classes of the tupels $(v, w)$, which are precisely the simple tensors. This argumentation also works for tensor product of $R$-modules, for a ring $R$, even bases not necessarily exist in this situation. $\endgroup$ – Jendrik Stelzner Jan 16 '16 at 22:59

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