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My problem lies in page 363 of Steven Roman's Advanced Linear Algebra (Here's a link). The author says that for each ordered pair $(e_i,f_j)$ where $\left\{e_i\right\}_{i\in I},\left\{f_j\right\}_{j\in J}$ are respective bases for vector spaces $U,V$, we "invent a new formal symbol" $e_i\otimes f_j$. He then defines $T$ (that will soon be taken as $U\otimes V$) to be the vector space with basis $$\mathcal{D}=\left\{ e_i\otimes f_j \mid i\in I,j\in J\right\}$$ Without further remarks. Why is $\mathcal{D}$ a basis? We don't know anything at all about $e_i\otimes f_j$ as they're just symbols..

Thanks in advance!

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He takes $D$ to be that set $D = \{e_i \otimes f_j\}$, where all elements are just formal symbols.

Then he is defining $U \otimes V$ to be the free vector space $\text{Free}(D)$, which is the set of all functions $f : D \to \mathbb{F}$ which are zero except at finitely many values of $D$. If you read about free vector spaces, you'll see that $D$ is indeed a basis of $\text{Free}(D)$ (basically by definition).

edit: Here are some references which I recommend on this subject: Dummit and Foote's Abstract Algebra and Werner Greub's Multilinear algebra.

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    $\begingroup$ you mean which are zero except at finitely many values of $D$ ? $\endgroup$ – mercio May 12 '14 at 13:54
  • $\begingroup$ Thanks. The author makes no mention of free vector spaces as a prerequisite for this chapter. Is it possible to avoid them and still follow the book's construction? $\endgroup$ – Sai May 12 '14 at 13:58
  • $\begingroup$ mercio: whoops, thanks, i'll fix that. Sai: I don't believe so, but if you're familiar with free groups then it's just like taking the free group on a set. You should also look into the universal property of the tensor product, i.e. bilinear maps $U \times V \to \mathbb{F}$ get turned into linear maps $U \otimes V \to \mathbb{F}$, but it should be noted that the universal property is not a construction. We actually need to do this construction to show that the tensor product exists. $\endgroup$ – nigel May 12 '14 at 13:59
  • $\begingroup$ Gotcha. The universal property for bilinearity preceded the construction in the book and I understand it. At any rate, from what I've just read, free vector spaces seem like nice things. I'll get on to them! Thanks again! $\endgroup$ – Sai May 12 '14 at 14:07

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