# Unclear Construction of Basis for Tensor Product

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..

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).
• you mean which are zero except at finitely many values of $D$ ? – mercio May 12 '14 at 13:54
• 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. – nigel May 12 '14 at 13:59