2.5.2 Tensor Basis Spanning Property

The Tensor Basis Spanning Property ensures that any tensor can be expressed as a linear combination of basis tensors in a tensor algebra.

Tensor Basis Spanning Property is the guarantee that every element of a tensor space can be written as a linear combination of the tensor products formed from a chosen basis of the underlying vector space and its dual, with no tensor lying outside the reach of such combinations. It is one of the two properties, alongside linear independence, required to establish that the induced tensor products actually form a basis, and it is the property responsible for the completeness of the coordinate description of tensors.

Statement of the Property

Setting

Let $V$ be a vector space of dimension $n$ over a field $F$ , with basis $e_{1}, \dots, e_{n}$ and dual basis $e^{1}, \dots, e^{n}$ . For a tensor space $T_{s}^{r} (V)$ , the spanning property asserts that every element $T$ of this space is a linear combination of the basis tensor products

e_{i_{1}} \otimes \dots \otimes e_{i_{r}} \otimes e^{j_{1}} \otimes \dots \otimes e^{j_{s}}

taken over every admissible index tuple.

Proof of the Spanning Property

Step One: Defining Candidate Coefficients

Given an arbitrary $T \in T_{s}^{r} (V)$ , define scalars by evaluating $T$ on the basis and dual basis:

T_{j_{1} \dots j_{s}}^{i_{1} \dots i_{r}} = T (e^{i_{1}}, \dots, e^{i_{r}}, e_{j_{1}}, \dots, e_{j_{s}})

Step Two: Constructing the Candidate Combination

Form the linear combination

T' = T_{j_{1} \dots j_{s}}^{i_{1} \dots i_{r}} e_{i_{1}} \otimes \dots \otimes e^{j_{s}}

with summation implied over every repeated index, and let $T^{'}$ denote this candidate sum, which must be shown to equal $T$ itself.

Step Three: Verifying Agreement on Basis Inputs

Evaluate $T^{'}$ on an arbitrary basis input tuple $(e^{k_{1}}, \dots, e_{l_{1}}, \dots)$ . Each basis tensor product term in $T^{'}$ evaluates, by the duality relation $e^{i} (e_{j}) = δ_{j}^{i}$ , to $1$ if its indices match the input exactly, and to $0$ otherwise. Only a single term survives, leaving

T' (e^{k_{1}}, \dots, e_{l_{1}}, \dots) = T_{l_{1} \dots}^{k_{1} \dots} = T (e^{k_{1}}, \dots, e_{l_{1}}, \dots)

showing that $T^{'}$ and $T$ agree on every basis input.

Step Four: Extending Agreement to All Inputs

Since both $T$ and $T^{'}$ are multilinear, and multilinear maps agreeing on all basis input tuples must agree on every input tuple by expanding arbitrary vectors and covectors in the basis and dual basis and applying linearity in each argument, it follows that $T = T^{'}$ as multilinear maps. This confirms that the candidate combination reproduces $T$ exactly.

Why Multilinearity Is Essential

The Role of Linearity in Each Slot

The extension from agreement on basis inputs to agreement on all inputs depends critically on multilinearity: a general function of several vector and covector arguments need not be determined by its values on basis inputs, but a multilinear one is, since linearity in each slot allows an arbitrary input to be decomposed into a sum of basis contributions, one slot at a time.

Failure Without Multilinearity

If the objects under consideration were merely functions rather than multilinear maps, the same spanning argument would fail, since knowledge of a function's values on basis vectors places no general constraint on its values elsewhere. The spanning property of the tensor basis is therefore a direct consequence of restricting attention to multilinear maps, not a generic feature of arbitrary function spaces.

Consequences of the Spanning Property

Completeness of the Coordinate Description

Because every tensor is spanned by the basis tensor products, no tensor requires additional descriptive data beyond its finite array of components relative to a chosen basis; the spanning property is precisely what guarantees this array is sufficient.

Finite Generation

The spanning property shows that $T_{s}^{r} (V)$ is finitely generated, by the $n^{r + s}$ basis tensor products, even though the space is defined abstractly as a set of multilinear maps on a potentially infinite domain of vector and covector tuples.

Prerequisite for Dimension Counting

Spanning alone establishes only that the dimension of $T_{s}^{r} (V)$ is at most $n^{r + s}$ ; combined separately with the linear independence of the same set, it yields the exact dimension formula, since a spanning set that is also independent is a minimal spanning set and therefore a basis.