In mathematics, the **tensor product** of two vector spaces *V* and *W* (over the same field) is a vector space to which is associated a bilinear map that maps a pair to an element of denoted

An element of the form is called the **tensor product** of v and w. An element of is a tensor, and the tensor product of two vectors is sometimes called an *elementary tensor* or a *decomposable tensor*. The elementary tensors span in the sense that every element of is a sum of elementary tensors. If bases are given for *V* and *W*, a basis of is formed by all tensor products of a basis element of V and a basis element of W.

The tensor product of two vector spaces captures the properties of all bilinear maps in the sense that a bilinear map from into another vector space Z factors uniquely through a linear map (see Universal property).

Tensor products are used in many application areas, including physics and engineering. For example, in general relativity, the gravitational field is described through the metric tensor, which is a vector field of tensors, one at each point of the space-time manifold, and each belonging to the tensor product with itself of the cotangent space at the point.

The *tensor product* of two vector spaces is a vector space that is defined up to an isomorphism. There are several equivalent ways for defining it. Most consist of defining explicitly a vector space that is called a tensor product, and, generally, the equivalence proof results almost immediately from the basic properties of the vector spaces that are so defined.

The tensor product can also be defined through a universal property; see § Universal property, below. As for every universal property, all objects that satisfy the property are isomorphic through unique isomorphisms. When this definition is used, the other definitions may be viewed as constructions of objects satisfying the universal property and as proofs that there are objects satisfying the universal property, that is that tensor products exist.

Let V and W be two vector spaces over a field F, with respective bases and

The *tensor product* of V and W is a vector space which has as a basis the set of all with and This definition can be formalized in the following way (this formalization is rarely used in practice, as the preceding informal definition is generally sufficient): is the set of the functions from the Cartesian product to F that have a finite number of nonzero values. The pointwise operations make a vector space. The function that maps to 1 and the other elements of to 0 is denoted

The set is straightforwardly a basis of which is called the *tensor product* of the bases and

The *tensor product of two vectors* is defined from their decomposition on the bases. More precisely, if

are vectors decomposed on their respective bases, then the tensor product of x and y is

If arranged into a rectangular array, the coordinate vector of is the outer product of the coordinate vectors of x and y. Therefore, the tensor product is a generalization of the outer product.

It is straightforward to verify that the map is a bilinear map from to

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.

A construction of the tensor product that is basis independent can be obtained in the following way.

Let V and W be two vector spaces over a field F.

One considers first a vector space L that has the Cartesian product as a basis. That is, the basis elements of L are the pairs with and For getting such a vector space, one can define it as the vector space of the functions that have a finite number of nonzero values, and identifying with the function that takes the value 1 on and 0 otherwise.

Let R be the linear subspace of L that is spanned by the relations that the tensor product must satisfy. More precisely R is spanned by the elements of one of the forms

where and

Then, the tensor product is defined as the quotient space

and the image of in this quotient is denoted

It is straightforward to prove that the result of this construction satisfies the universal property considered below. (A very similar construction can be used for defining the tensor product of modules.)

In this section, the universal property satisfied by the tensor product is described. As for every universal property, two objects that satisfy the property are related by a unique isomorphism. It follows that this is a (non-constructive) way to define the tensor product of two vector spaces. In this context, the preceding constructions of tensor products may be viewed as proofs of existence of the tensor product so defined.

A consequence of this approach is that every property of the tensor product can be deduced from the universal property, and that, in practice, one may forget the method that has been used to prove its existence.

The "universal-property definition" of the tensor product of two vector spaces is the following (recall that a bilinear map is a function that is *separately* linear in each of its arguments):

- The
*tensor product*of two vector spaces V and W is a vector space denoted as together with a bilinear map from to such that, for every bilinear map there is a*unique*linear map such that (that is, for every and ).

Like the universal property above, the following characterization may also be used to determine whether or not a given vector space and given bilinear map form a tensor product.^{[1]}

**Theorem** — Let and be complex vector spaces and let be a bilinear map. Then is a tensor product of and if and only if^{[1]} the image of spans all of (that is, ), and also and are *-linearly disjoint*, which by definition means that for all positive integers and all elements and such that

- if all are linearly independent then all are and
- if all are linearly independent then all are

Equivalently, and are -linearly disjoint if and only if for all linearly independent sequences in and all linearly independent sequences in the vectors are linearly independent.

For example, it follows immediately that if and are positive integers then and the bilinear map defined by sending to form a tensor product of and ^{[2]} Often, this map will be denoted by so that denotes this bilinear map's value at

As another example, suppose that is the vector space of all complex-valued functions on a set with addition and scalar multiplication defined pointwise (meaning that is the map and is the map ). Let and be any sets and for any and let denote the function defined by If and are vector subspaces then the vector subspace of together with the bilinear map

form a tensor product of and

If *V* and *W* are vectors spaces of finite dimension, then is finite-dimensional, and its dimension is the product of the dimensions of *V* and *W*.

This results from the fact that a basis of is formed by taking all tensor products of a basis element of *V* and a basis element of *W*.

The tensor product is associative in the sense that, given three vector spaces there is a canonical isomorphism

that maps to

This allows omitting parentheses in the tensor product of more than two vector spaces or vectors.

The tensor product of two vector spaces and is commutative in the sense that there is a canonical isomorphism

that maps to

On the other hand, even when the tensor product of vectors is not commutative; that is in general.

The map from to itself induces a linear automorphism that is called a **braiding map. **
More generally and as usual (see tensor algebra), let denote the tensor product of n copies of the vector space V. For every permutation s of the first n positive integers, the map

induces a linear automorphism of which is called a braiding map.

Given a linear map and a vector space W, the *tensor product*

is the unique linear map such that

The tensor product is defined similarly.

Given two linear maps and their tensor product

is the unique linear map that satisfies

One has

In terms of category theory, this means that the tensor product is a bifunctor from the category of vector spaces to itself.^{[3]}

If f and 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 products of modules do not transform injections into injections, but they are right exact functors).

By choosing bases of all vector spaces involved, the linear maps *S* and *T* can be represented by matrices. Then, depending on how the tensor is vectorized, the matrix describing the tensor product is the Kronecker product of the two matrices. For example, if *V*, *X*, *W*, and *Y* above are all two-dimensional and bases have been fixed for all of them, and *S* and *T* are given by the matrices

respectively, then the tensor product of these two matrices is

The resultant rank is at most 4, and thus the resultant dimension is 4. Note that *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). Note

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

See also: Tensor |

For non-negative integers *r* and *s* a type tensor on a vector space *V* is an element of

Here is the dual vector space (which consists of all linear maps *f* from *V* to the ground field *K*).

There is a product map, called the *(tensor) product of tensors*^{[4]}

It is defined by grouping all occurring "factors" *V* together: writing for an element of *V* and for an element of the dual space,

Picking a basis of *V* and the corresponding dual basis of naturally induces a basis for (this basis is described in the article on Kronecker products). In terms of these bases, the components of a (tensor) product of two (or more) tensors can be computed. For example, if *F* and *G* are two covariant tensors of orders *m* and *n* respectively (i.e. and ), then the components of their tensor product are given by^{[5]}

Thus, the components of the tensor product of two tensors are the ordinary product of the components of each tensor. Another example: let **U** be a tensor of type (1, 1) with components and let **V** be a tensor of type with components Then

and

Tensors equipped with their product operation form an algebra, called the tensor algebra.

For tensors of type (1, 1) there is a canonical **evaluation map**

defined by its action on pure tensors:

More generally, for tensors of type with *r*, *s* > 0, there is a map, called tensor contraction,

(The copies of and on which this map is to be applied must be specified.)

On the other hand, if is *finite-dimensional*, there is a canonical map in the other direction (called the **coevaluation map**)

where is any basis of and is its dual basis. This map does not depend on the choice of basis.^{[6]}

The interplay of evaluation and coevaluation can be used to characterize finite-dimensional vector spaces without referring to bases.^{[7]}

The tensor product may be naturally viewed as a module for the Lie algebra by means of the diagonal action: for simplicity let us assume then, for each

where is the transpose of *u*, that is, in terms of the obvious pairing on

There is a canonical isomorphism given by

Under this isomorphism, every *u* in may be first viewed as an endomorphism of and then viewed as an endomorphism of In fact it is the adjoint representation ad(*u*) of

Given two finite dimensional vector spaces *U*, *V* over the same field *K*, denote the dual space of *U* as *U**, and the *K*-vector space of all linear maps from *U* to *V* as Hom(*U*,*V*). There is an isomorphism,

defined by an action of the pure tensor on an element of

Its "inverse" can be defined using a basis and its dual basis as in the section "Evaluation map and tensor contraction" above:

This result implies

which automatically gives the important fact that forms a basis for where are bases of *U* and *V*.

Furthermore, given three vector spaces *U*, *V*, *W* the tensor product is linked to the vector space of *all* linear maps, as follows:

This is an example of adjoint functors: the tensor product is "left adjoint" to Hom.

Main article: Tensor product of modules |

The tensor product of two modules *A* and *B* over a *commutative* ring *R* is defined in exactly the same way as the tensor product of vector spaces over a field:

where now is the free

More generally, the tensor product can be defined even if the ring is non-commutative. In this case *A* has to be a right-*R*-module and *B* is a left-*R*-module, and instead of the last two relations above, the relation

is imposed. If

The universal property also carries over, slightly modified: the map defined by is a middle linear map (referred to as "the canonical middle linear map".^{[8]}); that is, it satisfies:^{[9]}

The first two properties make *φ* a bilinear map of the abelian group For any middle linear map of a unique group homomorphism *f* of satisfies and this property determines within group isomorphism. See the main article for details.

Let *A* be a right *R*-module and *B* be a left *R*-module. Then the tensor product of *A* and *B* is an abelian group defined by

where is a free abelian group over and G is the subgroup of generated by relations

The universal property can be stated as follows. Let *G* be an abelian group with a map that is bilinear, in the sense that

Then there is a unique map such that for all and

Furthermore, we can give a module structure under some extra conditions:

- If
*A*is a (*S*,*R*)-bimodule, then is a left*S*-module where - If
*B*is a (*R*,*S*)-bimodule, then is a right*S*-module where - If
*A*is a (*S*,*R*)-bimodule and*B*is a (*R*,*T*)-bimodule, then is a (*S*,*T*)-bimodule, where the left and right actions are defined in the same way as the previous two examples. - If
*R*is a commutative ring, then*A*and*B*are (*R*,*R*)-bimodules where and By 3), we can conclude is a (*R*,*R*)-bimodule.

For vector spaces, the tensor product is quickly computed since bases of *V* of *W* immediately determine a basis of as was mentioned above. For modules over a general (commutative) ring, not every module is free. For example, **Z**/*n***Z** is not a free abelian group (**Z**-module). The tensor product with **Z**/*n***Z** is given by

More generally, given a presentation of some *R*-module *M*, that is, a number of generators together with relations

the tensor product can be computed as the following cokernel:

Here and the map is determined by sending some in the *j*th copy of to (in ). Colloquially, this may be rephrased by saying that a presentation of *M* gives rise to a presentation of This is referred to by saying that the tensor product is a right exact functor. It is not in general left exact, that is, given an injective map of *R*-modules the tensor product

is not usually injective. For example, tensoring the (injective) map given by multiplication with *n*, *n* : **Z** → **Z** with **Z**/*n***Z** yields the zero map 0 : **Z**/*n***Z** → **Z**/*n***Z**, which is not injective. Higher Tor functors measure the defect of the tensor product being not left exact. All higher Tor functors are assembled in the derived tensor product.

Main article: Tensor product of algebras |

Let *R* be a commutative ring. The tensor product of *R*-modules applies, in particular, if *A* and *B* are *R*-algebras. In this case, the tensor product is an *R*-algebra itself by putting

For example,

A particular example is when *A* and *B* are fields containing a common subfield *R*. The tensor product of fields is closely related to Galois theory: if, say, *A* = *R*[*x*] / *f*(*x*), where *f* is some irreducible polynomial with coefficients in *R*, the tensor product can be calculated as

where now

is isomorphic (as an

Square matrices with entries in a field represent linear maps of vector spaces, say and thus linear maps of projective spaces over If is nonsingular then is well-defined everywhere, and the eigenvectors of correspond to the fixed points of The *eigenconfiguration* of consists of points in provided is generic and is algebraically closed. The fixed points of nonlinear maps are the eigenvectors of tensors. Let be a -dimensional tensor of format with entries lying in an algebraically closed field of characteristic zero. Such a tensor defines polynomial maps and with coordinates

Thus each of the coordinates of is a homogeneous polynomial of degree in The eigenvectors of are the solutions of the constraint

and the eigenconfiguration is given by the variety of the minors of this matrix.^{[10]}

Main article: Tensor product of Hilbert spaces |

Hilbert spaces generalize finite-dimensional vector spaces to countably-infinite dimensions. The tensor product is still defined; it is the tensor product of Hilbert spaces.

Main article: Topological tensor product |

When the basis for a vector space is no longer countable, then the appropriate axiomatic formalization for the vector space is that of a topological vector space. The tensor product is still defined, it is the topological tensor product.

Main article: Graded vector space § Operations on graded vector spaces |

Some vector spaces can be decomposed into direct sums of subspaces. In such cases, the tensor product of two spaces can be decomposed into sums of products of the subspaces (in analogy to the way that multiplication distributes over addition).

Main article: Tensor product of representations |

Vector spaces endowed with an additional multiplicative structure are called algebras. The tensor product of such algebras is described by the Littlewood–Richardson rule.

Main article: Tensor product of quadratic forms |

Given two multilinear forms and on a vector space over the field their tensor product is the multilinear form

This is a special case of the product of tensors if they are seen as multilinear maps (see also tensors as multilinear maps). Thus the components of the tensor product of multilinear forms can be computed by the Kronecker product.

Main article: Sheaf of modules |

Main article: Vector bundle § Operations on vector bundles |

See also: tensor product bundle |

Main article: Tensor product of fields |

Main article: Tensor product of graphs |

It should be mentioned that, though called "tensor product", this is not a tensor product of graphs in the above sense; actually it is the category-theoretic product in the category of graphs and graph homomorphisms. However it is actually the Kronecker tensor product of the adjacency matrices of the graphs. Compare also the section Tensor product of linear maps above.

The most general setting for the tensor product is the monoidal category. It captures the algebraic essence of tensoring, without making any specific reference to what is being tensored. Thus, all tensor products can be expressed as an application of the monoidal category to some particular setting, acting on some particular objects.

A number of important subspaces of the tensor algebra can be constructed as quotients: these include the exterior algebra, the symmetric algebra, the Clifford algebra, the Weyl algebra, and the universal enveloping algebra in general.

The exterior algebra is constructed from the exterior product. Given a vector space *V*, the exterior product is defined as

Note that when the underlying field of

The image of in the exterior product is usually denoted and satisfies, by construction, Similar constructions are possible for (

The symmetric algebra is constructed in a similar manner, from the symmetric product

More generally

That is, in the symmetric algebra two adjacent vectors (and therefore all of them) can be interchanged. The resulting objects are called symmetric tensors.

Array programming languages may have this pattern built in. For example, in APL the tensor product is expressed as `○.×`

(for example `A ○.× B`

or `A ○.× B ○.× C`

). In J the tensor product is the dyadic form of `*/`

(for example `a */ b`

or `a */ b */ c`

).

Note that J's treatment also allows the representation of some tensor fields, as `a`

and `b`

may be functions instead of constants. This product of two functions is a derived function, and if `a`

and `b`

are differentiable, then `a */ b`

is differentiable.

However, these kinds of notation are not universally present in array languages. Other array languages may require explicit treatment of indices (for example, MATLAB), and/or may not support higher-order functions such as the Jacobian derivative (for example, Fortran/APL).