Tensor Map Product Basis Formula

The Tensor Map Product Basis Formula explains how tensor products of bases form a basis for tensor map spaces, key in multilinear algebra.

Tensor Map Product Basis Formula is the explicit rule expressing the tensor product map $f \otimes g$ on the basis tensors of $V \otimes W$ in terms of the images of the individual basis vectors of $V$ under $f$ and of $W$ under $g$ , and it is the starting point from which every componentwise description of $f \otimes g$ is derived.

Basis Data

Bases of the Four Spaces

Let $f : V \to V^{'}$ and $g : W \to W^{'}$ be linear maps between finite-dimensional vector spaces over a field $K$ , and fix bases

{e_{1}, \dots, e_{n}} of V, {e_{1}', \dots, e_{p}'} of V^{'}

{f_{1}, \dots, f_{m}} of W, {f_{1}', \dots, f_{q}'} of W^{'}

so that ${e_{i} \otimes f_{j}}$ , ranging over all $1 \leq i \leq n$ and $1 \leq j \leq m$ , is a basis of $V \otimes W$ of dimension $n m$ , and likewise ${e_{a}' \otimes f_{b}'}$ is a basis of $V^{'} \otimes W^{'}$ of dimension $p q$ .

Expansion of the Images Under f and g

Each image vector $f (e_{i})$ expands in the target basis of $V^{'}$ as

f (e_{i}) = \sum_{a = 1}^{p} F_{i}^{a} e_{a}'

and each image vector $g (f_{j})$ expands as

g (f_{j}) = \sum_{b = 1}^{q} G_{j}^{b} f_{b}'

where $F_{i}^{a}$ and $G_{j}^{b}$ are the entries of the matrices representing $f$ and $g$ relative to the chosen bases.

The Basis Formula for the Tensor Product Map

Statement of the Formula

By the defining property of the tensor product map on simple tensors, the action on a basis tensor of $V \otimes W$ is

(f \otimes g) (e_{i} \otimes f_{j}) = f (e_{i}) \otimes g (f_{j})

and substituting the two expansions above and using bilinearity of the tensor product to distribute the sums gives the basis formula

(f \otimes g) (e_{i} \otimes f_{j}) = \sum_{a = 1}^{p} \sum_{b = 1}^{q} F_{i}^{a} G_{j}^{b} e_{a}' \otimes f_{b}'

which expresses the image of every basis tensor of the domain as an explicit linear combination of the basis tensors of the codomain, with coefficients given by products of entries of $F$ and $G$ .

Uniqueness by Linearity

Because ${e_{i} \otimes f_{j}}$ is a basis of $V \otimes W$ , the values $(f \otimes g) (e_{i} \otimes f_{j})$ prescribed by the basis formula determine $f \otimes g$ completely and uniquely as a linear map on all of $V \otimes W$ , since any element of $V \otimes W$ is a linear combination of these basis tensors and the value of a linear map on a linear combination is forced by linearity.

Matrix Reading of the Basis Formula

Entry of the Combined Matrix

Reading the basis formula as specifying one entry of a matrix at a time, the entry of the matrix representing $f \otimes g$ in row $(a, b)$ and column $(i, j)$ , when the combined bases are ordered lexicographically, is

{[f \otimes g]}_{(a b), (i j)} = F_{i}^{a} G_{j}^{b}

which is exactly the defining rule of the Kronecker product of the matrix of $f$ with the matrix of $g$ , so that the basis formula and the Kronecker product construction agree entry by entry once an ordering of the combined basis is fixed.

Independence From the Ordering Convention

Choosing a different ordering of the combined basis, for instance ordering pairs by $j$ first and $i$ second rather than the reverse, permutes the rows and columns of the resulting matrix but does not change which coefficient the basis formula assigns to which basis tensor, since the basis formula itself refers to basis tensors directly rather than to positions in an ordered list.

Worked Illustration

A Small Example

Take $V = K^{2}$ with basis $e_{1}, e_{2}$ , and $W = K^{2}$ with basis $f_{1}, f_{2}$ , with $f (e_{1}) = e_{1}'$ , $f (e_{2}) = e_{2}'$ , $g (f_{1}) = f_{1}' + f_{2}'$ , and $g (f_{2}) = f_{2}'$ . The basis formula then gives directly

(f \otimes g) (e_{1} \otimes f_{1}) = e_{1}' \otimes (f_{1}' + f_{2}') = e_{1}' \otimes f_{1}' + e_{1}' \otimes f_{2}'

with the remaining three basis tensors $e_{1} \otimes f_{2}$ , $e_{2} \otimes f_{1}$ , and $e_{2} \otimes f_{2}$ handled in exactly the same way, term by term, without ever needing to invoke a general element of $V \otimes W$ .