Recognizing pure tensors in tensor product of vector spaces

Question

Let $V$ be a vector space and let $\{e_i\}$ be a basis for it. Then $\{e_I\equiv e_{i_1}\otimes...\otimes e_{i_r}\}$ is a basis for $V\otimes ... \otimes V$. Suppose I am given an element $w=\sum a_I e_I $ which is a finite sum of pure tensors.

What is a necessary condition on $a_I$ so that this element is a pure tensor i.e. $\exists w_i$ such that $w=w_1 \otimes ... \otimes w_r$ ?

I tried writing $w_i=\sum w_{il}e_l$ and equating coefficients but do not know how to proceed further. I guess that there must be some standard operating procedure for this which I am not aware of. Any reference will be of help.

We can assume that the characteristic of the underlying field is 0 and that it is algebraically closed, if these assumptions simplify matters.

Very sorry, I just realized that this is a duplicate question. There is a question already here. Somebody with high enough reputation, please close my question. — user90041, Nov 10 '13 at 19:45
I don't think it is a duplicate. Your question concerns tensors of any rank, whereas the other is only concern with rank 2. I could have answered the rank two question. But I have no idea how to answer yours. — Stephen Montgomery-Smith, Nov 10 '13 at 19:51
Oh, but I just saw a note in the other page that comments that your question is much harder and generally unsolved. Still, I think yours should remain open. — Stephen Montgomery-Smith, Nov 10 '13 at 19:53
It is more natural to take finitely many vector spaces $V_1,\dotsc,V_r$ and describe pure tensors in $V_1 \otimes \dotsc \otimes V_r$. — Martin Brandenburg, Nov 10 '13 at 20:21
@MartinBrandenburg Yes, we can assume that $V$ is finite dimensional. I agree with your other comment. — user90041, Nov 11 '13 at 01:53

Recognizing pure tensors in tensor product of vector spaces

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