2.7.5 Tensor Linear Combination Span Relation

Tensor Linear Combination Span Relation explores how tensors span vector spaces through linear combinations, establishing foundational connections in algebraic structures.

Tensor Linear Combination Span Relation is the precise correspondence between the set of all linear combinations formed from a given collection of tensors and the span of that collection, establishing the span as exactly, and only, the reachable set of such combinations, and thereby fixing the smallest subspace that contains the original collection. It converts the informal idea of "everything buildable from a set of tensors" into a rigorously defined subspace with predictable algebraic behavior.

The Defining Relation

Setting

Let $V$ be a vector space over a field $F$ , and let $S \subset T_{s}^{r} (V)$ be any collection of tensors of a fixed type $(r, s)$ . The span relation identifies

span (S) = \{\sum_{k = 1}^{N} α_{k} T_{k} : T_{k} \in S, α_{k} \in F, N \in ℕ\}

exactly with the set of all finite linear combinations of elements of $S$ , with no tensor belonging to $span (S)$ unless it is expressible in this form, and every such combination belonging to $span (S)$ .

Span as the Smallest Enclosing Subspace

Span Contains the Original Set

Every element $T \in S$ is itself a trivial linear combination, using a single term with coefficient $1$ , so $S \subset span (S)$ .

Span Is Closed Under Linear Combination

A linear combination of elements already in $span (S)$ is again in $span (S)$ , since substituting the defining combination of each such element back into the outer combination produces a single, larger linear combination of elements of $S$ itself. This shows $span (S)$ is a subspace.

Minimality

If $W$ is any subspace of $T_{s}^{r} (V)$ containing $S$ , then $W$ must contain every linear combination of elements of $S$ , by closure of a subspace under addition and scalar action, so $span (S) \subset W$ . This establishes $span (S)$ as the smallest subspace containing $S$ .

Basic Properties of the Span Relation

Monotonicity

If $S_{1} \subset S_{2}$ , then $span (S_{1}) \subset span (S_{2})$ , since any linear combination available using terms from the smaller set is also available using terms from the larger set.

Idempotence

Taking the span of a span produces nothing new:

span (span, (S)) = span (S)

since, as noted above, a linear combination of elements of $span (S)$ collapses back into a linear combination of elements of $S$ directly.

Span of a Union

The span of a union of two sets equals the sum of their individual spans:

span (S_{1} \cup S_{2}) = span (S_{1}) + span (S_{2})

where the right-hand sum denotes the set of all tensors expressible as the sum of an element of the first span and an element of the second, since any combination drawing on both sets separates cleanly into a combination of terms from $S_{1}$ and a combination of terms from $S_{2}$ .

Span of the Empty Set

By convention, the empty linear combination, with no terms, equals the zero tensor, so $span (\emptyset) = \{0\}$ , the smallest possible subspace of $T_{s}^{r} (V)$ .

Span Relation and Spanning Sets

Definition of a Spanning Set

A set $S$ is called a spanning set for $T_{s}^{r} (V)$ precisely when the span relation yields the entire space:

span (S) = T_{s}^{r} (V)

The tensor basis spanning property is the specific instance of this relation in which $S$ consists of the basis tensor products.

Dimension Bound from a Finite Spanning Set

If $S$ is a finite spanning set of size $m$ , the span relation implies $dim T_{s}^{r} (V) \leq m$ , since no linearly independent set can exceed the size of a spanning set; equality holds exactly when $S$ is also linearly independent.

Span Relation Under Removal and Addition of a Generator

Removing a Dependent Generator

If some $T \in S$ already lies in $span (S ∖ \{T\})$ , then $span (S) = span (S ∖ \{T\})$ , since any occurrence of $T$ in a combination can be replaced by its own expression in terms of the remaining elements of $S$ , so removing a dependent generator never shrinks the span.

Adding a Generator Outside the Span

Conversely, if $T \notin span (S)$ , then $span (S \cup \{T\})$ strictly contains $span (S)$ , and its dimension increases by exactly one, since $T$ is then linearly independent from every element already spanned by $S$ .

Span Relation and Subspace Membership Testing

Reformulating Membership as a Linear Combination Problem

Determining whether a specific tensor $T$ belongs to $span (S)$ is, by the span relation, exactly the question of whether coefficients $α_{k}$ exist such that $T = \sum_{k = 1}^{N} α_{k} T_{k}$ , which, using the coordinate form of a linear combination, reduces to solving a system of linear equations over $F$ in the unknown coefficients.