Editing Tensor product (section)

=== From bases ===
Let {{mvar|V}} and {{mvar|W}} be two [[vector space]]s over a [[field (mathematics)|field]] {{mvar|F}}, with respective [[basis (linear algebra)|bases]] <math>B_V</math> and {{tmath|1= B_W }}.

The ''tensor product'' <math>V \otimes W</math> of {{mvar|V}} and {{mvar|W}} is a vector space that has as a basis the set of all <math>v\otimes w</math> with <math>v\in B_V</math> and {{tmath|1= w \in B_W }}. This definition can be formalized in the following way (this formalization is rarely used in practice, as the preceding informal definition is generally sufficient): <math>V \otimes W</math> is the set of the [[function (mathematics)|functions]] from the [[Cartesian product]] <math>B_V \times B_W</math> to {{mvar|F}} that have a finite number of nonzero values. The [[pointwise operation]]s make <math>V \otimes W</math> a vector space. The function that maps <math>(v,w)</math> to {{math|1}} and the other elements of <math>B_V \times B_W</math> to {{math|0}} is denoted {{tmath|1= v\otimes w }}.

The set <math>\{v\otimes w\mid v\in B_V, w\in B_W\}</math> is then straightforwardly a basis of {{tmath|1= V \otimes W }}, which is called the ''tensor product'' of the bases <math>B_V</math> and {{tmath|1= B_W }}.

We can equivalently define <math>V \otimes W</math> to be the set of [[Bilinear form|bilinear forms]] on <math>V \times W</math> that are nonzero at only a finite number of elements of {{tmath|1= B_V \times B_W }}. To see this, given <math>(x,y)\in V \times W</math>  and a bilinear form {{tmath|1= B : V \times W \to F }},  we can decompose <math>x</math> and <math>y</math> in the bases <math>B_V</math> and <math>B_W</math> as:
<math display="block">x=\sum_{v\in B_V} x_v\, v \quad \text{and}\quad y=\sum_{w\in B_W} y_w\, w, </math>
where only a finite number of <math>x_v</math>'s and <math>y_w</math>'s are nonzero, and find by the bilinearity of <math>B</math> that:
<math display="block">B(x, y) =\sum_{v\in B_V}\sum_{w\in B_W} x_v y_w\, B(v, w)</math>

Hence, we see that the value of <math>B</math> for any <math>(x,y)\in V \times W</math> is uniquely and totally determined by the values that it takes on {{tmath|1= B_V \times B_W }}. This lets us extend the maps <math>v\otimes w</math> defined on <math>B_V \times B_W</math> as before into bilinear maps <math>v \otimes w : V\times W \to F</math> , by letting:
<math display="block">(v \otimes w)(x, y) : =\sum_{v'\in B_V}\sum_{w'\in B_W} x_{v'} y_{w'}\, (v \otimes w)(v', w') = x_v \, y_w .</math>

Then we can express any bilinear form <math>B</math> as a (potentially infinite) formal linear combination of the <math>v\otimes w</math> maps according to:
<math display="block">B = \sum_{v\in B_V}\sum_{w\in B_W} B(v, w)(v \otimes w)</math>
making these maps similar to a [[Schauder basis]] for the vector space <math>\text{Hom}(V, W; F)</math> of all bilinear forms on {{tmath|1= V \times W }}. To instead have it be a proper Hamel [[Basis (linear algebra)|basis]], it only remains to add the requirement that <math>B</math> is nonzero at an only a finite number of elements of {{tmath|1= B_V \times B_W }}, and consider the subspace of such maps instead. 

In either construction, the ''tensor product of two vectors'' is defined from their decomposition on the bases. More precisely, taking the basis decompositions of <math>x\in V </math> and <math>y \in W</math> as before:
<math display="block">\begin{align}
x\otimes y&=\biggl(\sum_{v\in B_V} x_v\, v\biggr) \otimes \biggl(\sum_{w\in B_W} y_w\, w\biggr)\\[5mu]
&=\sum_{v\in B_V}\sum_{w\in B_W} x_v y_w\, v\otimes w.
\end{align}</math>

This definition is quite clearly derived from the coefficients of <math>B(v, w)</math> in the expansion by bilinearity of <math>B(x, y)</math> using the bases <math>B_V</math> and {{tmath|1= B_W }}, as done above. It is then straightforward to verify that with this definition, the map <math>{\otimes} : (x,y)\mapsto x\otimes y</math> is a bilinear map from <math>V\times W</math> to <math>V\otimes W</math> satisfying the [[universal property]] that any construction of the tensor product satisfies (see below). 

If arranged into a rectangular array, the [[coordinate vector]] of <math>x\otimes y</math> is the [[outer product]] of the coordinate vectors of  <math>x</math> and {{tmath|1= y }}. Therefore, the tensor product is a generalization of the outer product, that is, an abstraction of it beyond coordinate vectors. 

A limitation of this definition of the tensor product is that, if one changes bases, a different tensor product is defined. However, the decomposition on one basis of the elements of the other basis defines a [[canonical isomorphism]] between the two tensor products of vector spaces, which allows identifying them. Also, contrarily to the two following alternative definitions, this definition cannot be extended into a definition of the [[tensor product of modules]] over a [[ring (mathematics)|ring]].