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).
The tensor product of two vector spaces is a vector space that is defined up to an isomorphism. There are several equivalent ways to define 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 a unique isomorphism that is compatible with the universal property. 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.
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.
As a quotient space
A construction of the tensor product that is basis independent can be obtained in the following way.
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 To get 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
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 separatelylinear 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.
Theorem — Let and be complex vector spaces and let be a bilinear map. Then is a tensor product of and if and only if 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
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  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
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.
Commutativity as vector space operation
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 permutations of the first n positive integers, the map
induces a linear automorphism of which is called a braiding map.
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.
There is a product map, called the (tensor) product of tensors
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
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
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 R is non-commutative, this is no longer an R-module, but just an abelian group.
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".); that is, it satisfies:
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.
Tensor product of modules over a non-commutative ring
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
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.
Computing the tensor product
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/nZ is not a free abelian group (Z-module). The tensor product with Z/nZ 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 jth 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/nZ yields the zero map 0 : Z/nZ → Z/nZ, 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.
where now f is interpreted as the same polynomial, but with its coefficients regarded as elements of B. In the larger field B, the polynomial may become reducible, which brings in Galois theory. For example, if A = B is a Galois extension of R, then
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).
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.
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 V does not have characteristic 2, then this definition is equivalent to
The image of in the exterior product is usually denoted and satisfies, by construction, Similar constructions are possible for (n factors), giving rise to the nth exterior power of V. The latter notion is the basis of differential n-forms.
That is, in the symmetric algebra two adjacent vectors (and therefore all of them) can be interchanged. The resulting objects are called symmetric tensors.
Tensor product in programming
Array programming languages
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).
Chen, Jungkai Alfred (Spring 2004), "Tensor product"(PDF), Advanced Algebra II (lecture notes), National Taiwan University, archived(PDF) from the original on 2016-03-04((citation)): CS1 maint: location missing publisher (link)