# 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.

Since in every step of this algorithm the dimension of our spaces gets smaller, we have to get to linearly independent sets on both sides in finite number of steps. The end.

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Nice argument :-) – Mariano Suárez-Alvarez Sep 26 '12 at 17:53
the proof was awesome, writing 'the end' was possibly 'awesomer' :) – uncookedfalcon Feb 23 '13 at 20:02