# Does every element of tensor product look like this?

If $V\otimes W$ is the tensor product of vector spaces V and W, I know that for any basis $(v_i)_{i\in I}$ of V and $(w_j)_{j\in J}$ of W, $(v_i\otimes w_j)_{i\in I,j\in J}$ is a basis of $V\otimes W$, so any $a\in V\otimes W$ is a linear combination of some vectors $v_i\otimes w_j$.

But how can I prove that for every $a\in V\otimes W$ there exists $n\in \mathbb{N}$ and linear independent sets $\{v_1',...v_n'\}\in V$ and $\{w_1',...w_n'\}\in W$ such that $$a=v_1'\otimes w_1'+v_2'\otimes w_2'+...+v_n'\otimes w_n'?$$

This exercise is killing me: I have been trying to think of some way to construct these vectors starting with some fixed pair of bases and the upper fact, but I can't get anywhere! Help!

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Are you saying that the $v_i'$ and $w_i'$ depend on $a$? –  rschwieb Sep 26 '12 at 16:37
Yes, I have just edited the question, I formulated it unclearly the first time, sorry! –  sonjcy Sep 26 '12 at 16:41

Hint: Look at, say, $v_1$ and add up all the terms with $v_1$. Do that for each $v_i$. Argue that the resulting items in the right hand side are linearly independent.

I mean you should be using bilinearity. For example: $3(v_1\otimes w_2)+5(v_1\otimes w_3)=v_1\otimes(3w_2+5w_3)$.

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I think I get it (next answer). Thank you very much for your help! –  sonjcy Sep 26 '12 at 17:29
I think I got it: the upper fact implies that there exist linearly independent sets $\{v_1,...,v_n\}\in V$ and $\{w_1,...w_m\}\in W$ and $\alpha_{ij},\,i=1,...,m,\,j=1,...,n$, such that $$a=\sum_{i=1}^m\sum_{j=1}^n \alpha_{ij}v_i\otimes w_j=\sum_{i=1}^mv_i\otimes \sum_{j=1}^n \alpha_{ij}w_j.$$ If vectors $w_i'=\sum_{j=1}^n \alpha_{ij}w_j$ are linearly independent, we have what we need. If not, Let $(w_k'')_{k=1}^p$ be a basis for $span\{w_1',...w_n'\}$. Then $p<n$ and we have, for some $\beta_{ik}$, $$a=\sum_{i=1}^m v_i\otimes \sum_{k=1}^p \beta_{ik} w_k''=\sum_{k=1}^p((\sum_{i=1}^m\beta_{ik} v_i)\otimes w_k'').$$
Now, if $v_i'=\sum_{i=1}^m\beta_{ik} v_i$ are linearly independent, the end. If not, we take the basis for $span\{v_1',...,v_m'\}$ (whose dimension is $<m$) and continue like before.