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

**Example:**

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.

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

- .

The parameters introduced above work out like this:

Given multilinear maps and
their tensor product
is the multilinear function (*f* ⊗ *g*)

Given bases for *V* and *W*, the set of tensors of basis vectors, one from *V* and one from *W*,
forms a basis for *V* ⊗*W*. 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* x*W* to **R** is naturally isomorphic to the space of all linear maps from *V* ⊗*W* to **R**. That's because the multilinear maps are precisely those that respect the relations built into the construction.

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*(*x*⊗*y*) 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 *V* ⊗ *W* 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 *V* ⊗ *W* is generated by the image of ⊗, and even more: if *S* is a basis of *V* and *T* is a basis of *W*, then { *s* ⊗ *t* : *s* in *S* and *t* in *T*} is a basis for *V* ⊗ *W*. The dimension of the vector space *V* ⊗ *W* 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 *V*⊗*W*⊗*X* is that every tri-linear operator on
*V*x*W*x*X* corresponds to a unique linear operator on
*V*⊗*W*⊗*X*. The binary tensor product is associative: (*V* ⊗ *W*) ⊗ *Z* is naturally isomorphic to *V* ⊗ (*W* ⊗ *Z*). 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.

Note that the space (*V*⊗*W*)^{*} (the dual space of *V*⊗*W* containing all linear functionals on that space)
corresponds naturally to the space of all
bilinear functionals on *V*x*W*. 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 (*V*⊗*W*)^{*}.
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.

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.

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 * _{R}M* for a left module,
and

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 * _{S}M_{R}* and

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* **Z**_{n}. Both can be considered as modules over the integers, **Z**.
Let *B*: **Q** x **Z**_{n} -> *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 **Z**_{n} is {0}.