# Equivalence between these tensor product definitions

Let $$V$$ and $$W$$ be vector spaces. Then the tensor product $$V \otimes W$$ of $$V$$ and $$W$$ is the vector space $$V \otimes W$$ together with a bilinear map $$\phi: V \times W \rightarrow V \otimes W$$ such that for any vector space $$X$$ and any bilinear map $$f: V \times W \rightarrow X$$, there is a unique linear map $$\tilde{f}: V \otimes W \rightarrow X$$ such that $$f = \tilde{f} \circ \phi$$.

Now, when I look at finite dimensional vector spaces, I don't see how this definition relates to any of the material introduced in the finite dimensional case?

Let $$V, W$$ be finite dimensional vector spaces with dimensions $$n$$ and $$m$$ respectively. Let $${\{v_1,...,v_n}\}$$ be a basis for $$V$$ and $${\{w_1,...,w_m}\}$$ be a basis for $$W$$. Then, $$V \otimes W$$ has basis $${\{v_i \otimes w_j: 1 \leq i \leq n, 1 \leq j \leq m}\}$$ Where does this come from? How does this relate to the definition and the universal mapping property?

## 2 Answers

The correct conceptual definition is the first one, with the universal property.

The second one is basis dependent : how do you compare $$V\otimes W$$ with itself when you change the bases if you define it that way ?

It so happens (we're in luck !) that when $$V,W$$ have these bases, then in fact $$(v_i\otimes w_j)$$ is a basis for $$V\otimes W$$ but that shouldn't be its definition : very often we like to define $$V\otimes W$$ without having to specify bases for each space (for instance if we're interested in various bases).

Now why the two relate ? Well notice that if $$f: V\times W\to X$$ is bilinear, $$v\in V,w\in W$$ then $$f(v,w) = \displaystyle\sum_{i,j}\lambda_i\mu_j f(v_i,w_j)$$ for $$v= \sum_i \lambda_i v_i, w= \sum_j \mu_j w_j$$, therefore $$f$$ depends only on its values on the $$(v_i,w_j)$$, and conversely, if you specify values for these couples then you have a bilinear map.

Therefore if you have a vector space $$E$$ with basis $$(v_i\otimes w_j)$$ then $$f$$ factors (uniquely) through $$(v_i,w_j)\mapsto v_i\otimes w_j$$, which proves the universal property. Therefore, since the universal property characterizes objects up to unique isomorphism, it follows that $$V\otimes W$$ also has a basis consisting of the images of $$(v_i,w_j)$$ under $$\phi$$, which we denote $$v_i\otimes w_j$$.

(Note that this has nothing to do with finite dimensions or not)

The second definition is the same as saying that the tensor product is the set of all finite linear combinations of the elementary tensors $$v_i \otimes w_j$$.

$$\phi$$ is the obvious injection $$(v,w) \mapsto v \otimes w$$.

You define $$\tilde f(w \otimes v) = f(w,v)$$, and having defined $$\tilde f$$ on the basis vectors of $$V \otimes W$$ you extend it by linearity.

You can convince yourself that this does in fact have the universal mapping property required by the first definition. And by the uniqueness of universal objects, it must in fact be "the" tensor product up to unique isomorphism.