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If $V,W$ are infinite-dimensional vector spaces with basis {${v_i}$} and {${w_j}$} respectively it holds that $V\otimes W$ has as basis {${v_i⊗w_j}$}.

What about the reciprocal? That is: if {${v_i}$} and {${w_j}$} are families of vectors in $V$ and $W$ respectively such that the family {${v_i⊗w_j}$} is a basis of $V\otimes W$, are {${v_i}$} and {${w_j}$} bases of $V$ and $W$ respectively?

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  • $\begingroup$ Doesn't this follow from the universal property of the tensor product? $\endgroup$ Commented Jun 2, 2020 at 22:42

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I am afraid this question might be downvoted for not being research-level, but let me quickly expand the comment by Padraig Ó Catháin. For a fixed $k$, consider $W_k=\operatorname{Span}(w_k)$ and the projection $W\to W_k$; by composing with the map $(v,\alpha w_k)\mapsto \alpha v$, you get a bilinear mapping $V\times W \to V \times W_k \to V$. Hence, by the universal property, a linear onto map $V\otimes W \to V$ showing that $\{v_i\}$ is a set of generators; linear independence follows from looking at $\{v_i\otimes w_k\}$.

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  • $\begingroup$ Thank you. I do not see the point in making use of the span of an element into W. Why is it not enough to use the projection? On the other hand, how do the uniqueness of the commutative diagram produces a set of generators? Thank you again. $\endgroup$
    – gibarian
    Commented Jun 3, 2020 at 7:08
  • $\begingroup$ I am not sure I follow your remark: the projection $(v,w)\mapsto v$ is not bilinear. The fact that $\{v_i\}$ is a set of generators follows from the surjectivity of the map $V\otimes W\to V$ described above; it is linearly independent because $\{v_i\otimes w_k\}$ (fixed $k$) is. $\endgroup$
    – Aurelio
    Commented Jun 3, 2020 at 8:51
  • $\begingroup$ Right. Understood. $\endgroup$
    – gibarian
    Commented Jun 3, 2020 at 9:37

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