Editing Tensor product (section)

== Tensor product of linear maps ==
{{redirect|Tensor product of linear maps|the generalization for modules|Tensor product of modules#Tensor product of linear maps and a change of base ring}}
Given a linear map {{tmath|1= f : U\to V }}, and a vector space {{mvar|W}}, the ''tensor product:'' 
: <math>f\otimes W : U\otimes W\to V\otimes W</math>
is the unique linear map such that:
: <math>(f\otimes W)(u\otimes w)=f(u)\otimes w.</math>
The tensor product <math>W\otimes f</math> is defined similarly.

Given two linear maps <math>f : U\to V</math> and {{tmath|1= g : W\to Z }}, their tensor product: 
: <math>f\otimes g : U\otimes W\to V\otimes Z</math> 
is the unique linear map that satisfies: 
: <math>(f\otimes g)(u\otimes w)=f(u)\otimes g(w).</math>

One has: 
: <math>f\otimes g= (f\otimes Z)\circ (U\otimes g) = (V\otimes g)\circ (f\otimes W).</math>

In terms of [[category theory]], this means that the tensor product is a [[bifunctor]] from the [[category (mathematics)|category]] of vector spaces to itself.<ref>{{cite book| last1=Hazewinkel|first1=Michiel|last2=Gubareni|first2=Nadezhda Mikhaĭlovna| last3=Gubareni|first3=Nadiya|last4=Kirichenko|first4=Vladimir V.|title=Algebras, rings and modules|page=100|
publisher=Springer|year=2004|isbn=978-1-4020-2690-4}}</ref>

If {{mvar|f}} and {{mvar|g}} are both [[injective]] or [[surjective]], then the same is true for all above defined linear maps. In particular, the tensor product with a vector space is an [[exact functor]]; this means that every [[exact sequence]] is mapped to an exact sequence ([[tensor product of modules|tensor products of modules]] do not transform injections into injections, but they are [[right exact functor]]s).

By choosing bases of all vector spaces involved, the linear maps {{math|''f''}} and {{math|''g''}} can be represented by [[matrix (mathematics)|matrices]]. Then, depending on how the tensor <math>v \otimes w</math> is vectorized, the matrix describing the tensor product <math>f \otimes g</math> is the [[Kronecker product]] of the two matrices. For example, if {{math|''V'', ''X'', ''W''}}, and {{math|''U''}} above are all two-dimensional and bases have been fixed for all of them, and {{math|''f''}} and {{math|''g''}} are given by the matrices:
<math display="block">A=\begin{bmatrix}
 a_{1,1} & a_{1,2} \\
 a_{2,1} & a_{2,2} \\
 \end{bmatrix}, \qquad B=\begin{bmatrix}
 b_{1,1} & b_{1,2} \\
 b_{2,1} & b_{2,2} \\
 \end{bmatrix},</math>
respectively, then the tensor product of these two matrices is:
<math>
  \begin{align}
  \begin{bmatrix}
    a_{1,1} & a_{1,2} \\
    a_{2,1} & a_{2,2} \\
  \end{bmatrix}
  \otimes
  \begin{bmatrix}
    b_{1,1} & b_{1,2} \\
    b_{2,1} & b_{2,2} \\
  \end{bmatrix}
  &=
  \begin{bmatrix}
    a_{1,1} \begin{bmatrix}
      b_{1,1} & b_{1,2} \\
      b_{2,1} & b_{2,2} \\
    \end{bmatrix} & a_{1,2} \begin{bmatrix}
      b_{1,1} & b_{1,2} \\
      b_{2,1} & b_{2,2} \\
    \end{bmatrix} \\[3pt]
    a_{2,1} \begin{bmatrix}
      b_{1,1} & b_{1,2} \\
      b_{2,1} & b_{2,2} \\
    \end{bmatrix} & a_{2,2} \begin{bmatrix}
      b_{1,1} & b_{1,2} \\
      b_{2,1} & b_{2,2} \\
    \end{bmatrix} \\
  \end{bmatrix} \\
  &=
  \begin{bmatrix}
    a_{1,1} b_{1,1} & a_{1,1} b_{1,2} & a_{1,2} b_{1,1} & a_{1,2} b_{1,2} \\
    a_{1,1} b_{2,1} & a_{1,1} b_{2,2} & a_{1,2} b_{2,1} & a_{1,2} b_{2,2} \\
    a_{2,1} b_{1,1} & a_{2,1} b_{1,2} & a_{2,2} b_{1,1} & a_{2,2} b_{1,2} \\
    a_{2,1} b_{2,1} & a_{2,1} b_{2,2} & a_{2,2} b_{2,1} & a_{2,2} b_{2,2} \\
  \end{bmatrix}.
  \end{align}
</math>

The resultant rank is at most 4, and thus the resultant dimension is 4. {{em|rank}} here denotes the [[tensor rank]] i.e. the number of requisite indices (while the [[matrix rank]] counts the number of degrees of freedom in the resulting array). {{tmath|1= \operatorname{Tr} A \otimes B = \operatorname{Tr} A \times \operatorname{Tr} B }}.

A [[dyadic product]] is the special case of the tensor product between two vectors of the same dimension.