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Tensor product

In mathematics, the tensor product, which is denoted by <math>\otimes</math>, may be applied in different contexts to vectors, matrices, tensors and vector spaces. In each case the significance of the symbol is the same: the most general bilinear operation.

A representative case is the Kronecker multiplication of any two rectangular arrays, considered as matrices.


<math>\begin{bmatrix}a_1 & a_2 & a_3\end{bmatrix} \otimes \begin{bmatrix}b_1 & b_2 & b_3 & b_4\end{bmatrix} = \begin{bmatrix}a_1b_1 & a_2b_1 & a_3b_1 \\ a_1b_2 & a_2b_2 & a_3b_2 \\ a_1b_3 & a_2b_3 & a_3b_3 \\ a_1b_4 & a_2b_4 & a_3b_4\end{bmatrix}</math>

Resultant rank = 2, resultant dimension = 12.

Here rank denotes the number of requisite indices, while dimension counts the number of degrees of freedom in the resulting array.

Table of contents

Tensor product of two tensors

There is a general formula for the product of two (or more) tensors

<math>V \otimes U = V_{\left[ i_1,i_2,i_3,...i_n\right] }U_{\left[ j_1,j_2,j_3,...j_m\right] }</math>.

We are assuming here orthogonal tensors, with no distinction of covariant and contravariant indices, for simplicity.

The parameters introduced above work out like this:

<math>rank( U \otimes V )=rank(U)+rank(V)</math>

<math>dim( U \otimes V )=dim(U)dim(V)</math>

See also: Tensor-classical

Tensor product of multilinear maps

Given multilinear maps <math>f(x_1,...x_k)</math> and <math>g(x_1,... x_m)</math> their tensor product is the multilinear function (fg)<math>(x_1,...x_{k+m})=f(x_1,...x_k)g(x_{k+1},... x_{k+m})</math>

Tensor product of vector spaces

The tensor product VW of two vector spaces V and W has a formal definition by the method of generators and relations. The equivalence class of (v,w) is called a tensor and is denoted by vw. By construction, one can prove just as many identities between tensors, and sums of tensors, as follow from the relations used.

Take the vector space generated by V xW and apply (factor out the subspace generated by) the multilinear relations detiled just below. With this notation the relations take the form:

  • <math>(v_1+v_2)</math>⊗<math>w=v_1</math>⊗w<math>+v_2</math>⊗<math>w</math>

  • <math>v</math> ⊗<math>(w_1+w_2)=v</math> ⊗<math> w_1+v</math> ⊗<math>w_2</math>

  • <math>cv</math> ⊗<math>w=v</math> ⊗<math>cw=c(v</math> ⊗<math>w)</math>

Each element of the tensor product is a finite sum of tensors: more than one tensor is usually required to do that. It is simply shown how to construct a basis of VW.

Given bases for V and W, the set of tensors of basis vectors, one from V and one from W, forms a basis for VW. The dimension of the tensor product therefore is the product of dimensions.

Universal property of tensor product: The space of all multilinear maps from V xW to R is naturally isomorphic to the space of all linear maps from VW to R. That's because the multilinear maps are precisely those that respect the relations built into the construction.

Esoteric/Advanced/Pretentious Part -- (please make more accessible)

In abstract algebra, the subject of linear algebra is upgraded to multilinear algebra by introducing the tensor product of two vector spaces.

It is introduced to reduce the study of bilinear operators to that of linear operators. This is sufficient to do the same to all multilinear maps.

Formally, the tensor product of the two vector spaces V and W over the same base field F is defined by the following universal property: it is a vector space T over F, together with a bilinear operator ⊗: V x W -> T, such that for every bilinear operator B: V x W -> X there exists a unique linear operator L: T -> X with B = L o ⊗, i.e. B(x,y) = L(xy) for all x in V and y in W.

The tensor product is up to a unique isomorphism uniquely specified by this requirement, and we may therefore write VW instead of T. By direct construction, as suggested in the previous section, one can show that the tensor product for any two vector spaces exists.

The space VW is generated by the image of ⊗, and even more: if S is a basis of V and T is a basis of W, then { st : s in S and t in T} is a basis for VW. The dimension of the vector space VW is therefore given by the product of the dimensions of V and W.

It is possible to generalize the definition to a tensor product of any number of spaces. For example, the universal property of VWX is that every tri-linear operator on VxWxX corresponds to a unique linear operator on VWX. The binary tensor product is associative: (VW) ⊗ Z is naturally isomorphic to V ⊗ (WZ). The tensor product of all three may therefore be identified with either of those: the binary ⊗ will suffice. Tensor spaces allow us to use the theory of linear operators to study multi-linear operators, and this says the bilinear case is the main hurdle.

Relation with the dual space

Note that the space (VW)* (the dual space of VW containing all linear functionals on that space) corresponds naturally to the space of all bilinear functionals on VxW. In other words, every bilinear functional is a functional on the tensor product, and vice versa. There is a natural isomorphism between V*W* and (VW)*. So, the tensors of the linear functionals are bilinear functionals. This gives us a new way to look at the space of bilinear functionals, as a tensor product itself.

Types of tensors, e.g. alternating

Linear subspaces of the bilinear operators (or in general, multilinear operators) determine natural quotient spaces of the tensor space, which are frequently useful. See wedge product for the first major example. Another would be the treatment of algebraic forms as symmetric tensors.

Over more general rings

It is also possible to generalize the definition to tensor products of modules over the same ring. If the ring is non-commutative, we'll need to be careful about distinguishing right modules and left modules. We will write RM for a left module, and MR for a right module. If a module M has both a left module structure over a ring R and a right module structure over a ring S, and in addition for every m in M, r in R and s in S we have r(ms) = (rm)s, then we will say M is a bimodule[?], and will denote it by RMS. Note that every left-module is a bimodule module with Z as the right ring, and vice versa.

When defining the tensor product, we also need to be careful about the ring: most modules can be considered as modules over several different rings.

The most general form of the tensor product definition is as follows: let SMR and RNK be bimodules, then their tensor product over R is an abelian group P together with an R-bilinear operator T: M x N -> P such that for every R-bilinear operator B: M x N -> O there is a unique group homomorphism L: P -> O such that L o T = B. There is unique left-right module structure on P such that T is right-linear over S and left-linear over K, and if B is right-linear over S then so is L, and if B is left-linear over K then so is L. (P, T) are unique up to a unique isomorphism, and are called the "tensor product" of M and N.

Note that for a commutative ring R, and in particular for a field, a module is both a right module and a left module. Hence, the product of two modules over a commutative ring is again a module over that ring. Also note that this definition is also naturally associative, and we can use this to define the tensor product for any number of spaces.

Example: Consider the rational numbers Q and the integers modulo n Zn. Both can be considered as modules over the integers, Z. Let B: Q x Zn -> M be a Z-bilinear operator. Then B(q,i) = B(q/n, n*i) = B(q/p, 0) = 0, so every bilinear operator is identically zero. Therefore, if we define P to be the trivial module, and T to be the zero bilinear function, then we see that the properties for the tensor product are satisfied. Therefore, the tensor product of Q and Zn is {0}.

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