linear span

Given a vector space {{mvar|V}} over a field {{mvar|K}}, the span of a set {{mvar|S}} of vectors (not necessarily finite) is defined to be the intersection {{mvar|W}} of all subspaces of {{mvar|V}} that contain {{mvar|S}}. It is thus the smallest (for set inclusion) subspace containing {{mvar|S}}. It is referred to as the subspace spanned by {{mvar|S}}, or by the vectors in {{mvar|S}}. Conversely, {{mvar|S}} is called a spanning set of {{mvar|W}}, and we say that {{mvar|S}} spans {{mvar|W}}.

It follows from this definition that the span of {{mvar|S}} is the set of all finite linear combinations of elements (vectors) of {{mvar|S}}, and can be defined as such.{{Harvard citation text|Hefferon|2020}} p. 100, ch. 2, Definition 2.13{{Harvard citation text|Axler|2015}} pp. 29-30, §§ 2.5, 2.8{{Harvard citation text|Roman|2005}} pp. 41-42 That is, $$\backslash operatorname\{span\}(S)\; =\; \backslash biggl\; \backslash \{\; \backslash lambda\_1\; \backslash mathbf\; v\_1\; +\; \backslash lambda\_2\; \backslash mathbf\; v\_2\; +\; \backslash cdots\; +\; \backslash lambda\_n\; \backslash mathbf\; v\_n\; \backslash mid\; n\; \backslash in\; \backslash N,\backslash ;\; \backslash mathbf\; v\_1,...\backslash mathbf\; v\_n\; \backslash in\; S,\; \backslash ;\; \backslash lambda\_1,...\backslash lambda\_n\; \backslash in\; K\; \backslash biggr\; \backslash \}$$

When {{mvar|S}} is empty, the only possibility is {{math|1=n = 0}}, and the previous expression for $\backslash operatorname\{span\}(S)$ reduces to the empty sum.{{efn| This is logically valid as when {{math|1= n = 0}}, the conditions for the vectors and constants are empty, and therefore vacuously satisfied.}} The standard convention for the empty sum implies thus $\backslash text\{span\}(\backslash empty)\; =\; \backslash \{\backslash mathbf\; 0\backslash \},$ a property that is immediate with the other definitions. However, many introductory textbooks simply include this fact as part of the definition.

When $S=\backslash \{\backslash mathbf\; v\_1,\backslash ldots,\; \backslash mathbf\; v\_n\backslash \}$ is finite, one has

$$\backslash operatorname\{span\}(S)\; =\; \backslash \{\; \backslash lambda\_1\; \backslash mathbf\; v\_1\; +\; \backslash lambda\_2\; \backslash mathbf\; v\_2\; +\; \backslash cdots\; +\; \backslash lambda\_n\; \backslash mathbf\; v\_n\; \backslash mid\; \backslash lambda\_1,...\backslash lambda\_n\; \backslash in\; K\; \backslash \}$$

The real vector space $\backslash mathbb\; R^3$ has {(−1, 0, 0), (0, 1, 0), (0, 0, 1)} as a spanning set. This particular spanning set is also a basis. If (−1, 0, 0) were replaced by (1, 0, 0), it would also form the canonical basis of $\backslash mathbb\; R^3$.

Another spanning set for the same space is given by {(1, 2, 3), (0, 1, 2), (−1, {{frac|1|2}}, 3), (1, 1, 1)}, but this set is not a basis, because it is linearly dependent.

The set {{math|{(1, 0, 0), (0, 1, 0), (1, 1, 0)}}} is not a spanning set of $\backslash mathbb\; R^3$, since its span is the space of all vectors in $\backslash mathbb\; R^3$ whose last component is zero. That space is also spanned by the set {(1, 0, 0), (0, 1, 0)}, as (1, 1, 0) is a linear combination of (1, 0, 0) and (0, 1, 0). Thus, the spanned space is not $\backslash mathbb\; R^3.$ It can be identified with $\backslash mathbb\; R^2$ by removing the third components equal to zero.

The empty set is a spanning set of {(0, 0, 0)}, since the empty set is a subset of all possible vector spaces in $\backslash mathbb\; R^3$, and {(0, 0, 0)} is the intersection of all of these vector spaces.

The set of monomials {{mvar|xⁿ}}, where {{mvar|n}} is a non-negative integer, spans the space of polynomials.

The set of all linear combinations of a subset {{mvar|S}} of {{mvar|V}}, a vector space over {{mvar|K}}, is the smallest linear subspace of {{mvar|V}} containing {{mvar|S}}.

:Proof. We first prove that {{math|span S}} is a subspace of {{mvar|V}}. Since {{mvar|S}} is a subset of {{mvar|V}}, we only need to prove the existence of a zero vector {{math|0}} in {{math|span S}}, that {{math|span S}} is closed under addition, and that {{math|span S}} is closed under scalar multiplication. Letting $S\; =\; \backslash \{\; \backslash mathbf\; v\_1,\; \backslash mathbf\; v\_2,\; \backslash ldots,\; \backslash mathbf\; v\_n\; \backslash \}$, it is trivial that the zero vector of {{mvar|V}} exists in {{math|span S}}, since $\backslash mathbf\; 0\; =\; 0\; \backslash mathbf\; v\_1\; +\; 0\; \backslash mathbf\; v\_2\; +\; \backslash cdots\; +\; 0\; \backslash mathbf\; v\_n$. Adding together two linear combinations of {{mvar|S}} also produces a linear combination of {{mvar|S}}: $(\backslash lambda\_1\; \backslash mathbf\; v\_1\; +\; \backslash cdots\; +\; \backslash lambda\_n\; \backslash mathbf\; v\_n)\; +\; (\backslash mu\_1\; \backslash mathbf\; v\_1\; +\; \backslash cdots\; +\; \backslash mu\_n\; \backslash mathbf\; v\_n)\; =\; (\backslash lambda\_1\; +\; \backslash mu\_1)\; \backslash mathbf\; v\_1\; +\; \backslash cdots\; +\; (\backslash lambda\_n\; +\; \backslash mu\_n)\; \backslash mathbf\; v\_n$, where all $\backslash lambda\_i,\; \backslash mu\_i\; \backslash in\; K$, and multiplying a linear combination of {{mvar|S}} by a scalar $c\; \backslash in\; K$ will produce another linear combination of {{mvar|S}}: $c(\backslash lambda\_1\; \backslash mathbf\; v\_1\; +\; \backslash cdots\; +\; \backslash lambda\_n\; \backslash mathbf\; v\_n)\; =\; c\backslash lambda\_1\; \backslash mathbf\; v\_1\; +\; \backslash cdots\; +\; c\backslash lambda\_n\; \backslash mathbf\; v\_n$. Thus {{math|span S}} is a subspace of {{mvar|V}}.

:It follows that $S\; \backslash subseteq\; \backslash operatorname\{span\}\; S$, since every {{math|v_i}} is a linear combination of {{mvar|S}} (trivially). Suppose that {{mvar|W}} is a linear subspace of {{mvar|V}} containing {{mvar|S}}. Since {{mvar|W}} is closed under addition and scalar multiplication, then every linear combination $\backslash lambda\_1\; \backslash mathbf\; v\_1\; +\; \backslash cdots\; +\; \backslash lambda\_n\; \backslash mathbf\; v\_n$ must be contained in {{mvar|W}}. Thus, {{math|span S}} is contained in every subspace of {{mvar|V}} containing {{mvar|S}}, and the intersection of all such subspaces, or the smallest such subspace, is equal to the set of all linear combinations of {{mvar|S}}.

Every spanning set {{mvar|S}} of a vector space {{mvar|V}} must contain at least as many elements as any linearly independent set of vectors from {{mvar|V}}.

:Proof. Let $S\; =\; \backslash \{\; \backslash mathbf\; v\_1,\; \backslash ldots,\; \backslash mathbf\; v\_m\; \backslash \}$ be a spanning set and $W\; =\; \backslash \{\; \backslash mathbf\; w\_1,\; \backslash ldots,\; \backslash mathbf\; w\_n\; \backslash \}$ be a linearly independent set of vectors from {{mvar|V}}. We want to show that $m\; \backslash geq\; n$.

:Since {{mvar|S}} spans {{mvar|V}}, then $S\; \backslash cup\; \backslash \{\; \backslash mathbf\; w\_1\; \backslash \}$ must also span {{mvar|V}}, and $\backslash mathbf\; w\_1$ must be a linear combination of {{mvar|S}}. Thus $S\; \backslash cup\; \backslash \{\; \backslash mathbf\; w\_1\; \backslash \}$ is linearly dependent, and we can remove one vector from {{mvar|S}} that is a linear combination of the other elements. This vector cannot be any of the {{math|w_i}}, since {{mvar|W}} is linearly independent. The resulting set is $\backslash \{\; \backslash mathbf\; w\_1,\; \backslash mathbf\; v\_1,\; \backslash ldots,\; \backslash mathbf\; v\_\{i-1\},\; \backslash mathbf\; v\_\{i+1\},\; \backslash ldots,\; \backslash mathbf\; v\_m\; \backslash \}$, which is a spanning set of {{mvar|V}}. We repeat this step {{mvar|n}} times, where the resulting set after the {{mvar|p}}th step is the union of $\backslash \{\; \backslash mathbf\; w\_1,\; \backslash ldots,\; \backslash mathbf\; w\_p\; \backslash \}$ and {{mvar|m - p}} vectors of {{mvar|S}}.

:It is ensured until the {{mvar|n}}th step that there will always be some {{math|v_i}} to remove out of {{mvar|S}} for every adjoint of {{math|v}}, and thus there are at least as many {{math|v_i}}'s as there are {{math|w_i}}'s—i.e. $m\; \backslash geq\; n$. To verify this, we assume by way of contradiction that . Then, at the {{mvar|m}}th step, we have the set $\backslash \{\; \backslash mathbf\; w\_1,\; \backslash ldots,\; \backslash mathbf\; w\_m\; \backslash \}$ and we can adjoin another vector $\backslash mathbf\; w\_\{m+1\}$. But, since $\backslash \{\; \backslash mathbf\; w\_1,\; \backslash ldots,\; \backslash mathbf\; w\_m\; \backslash \}$ is a spanning set of {{mvar|V}}, $\backslash mathbf\; w\_\{m+1\}$ is a linear combination of $\backslash \{\; \backslash mathbf\; w\_1,\; \backslash ldots,\; \backslash mathbf\; w\_m\; \backslash \}$. This is a contradiction, since {{mvar|W}} is linearly independent.

Let {{mvar|V}} be a finite-dimensional vector space. Any set of vectors that spans {{mvar|V}} can be reduced to a basis for {{mvar|V}}, by discarding vectors if necessary (i.e. if there are linearly dependent vectors in the set). If the axiom of choice holds, this is true without the assumption that {{mvar|V}} has finite dimension. This also indicates that a basis is a minimal spanning set when {{mvar|V}} is finite-dimensional.

Generalizing the definition of the span of points in space, a subset {{mvar|X}} of the ground set of a matroid is called a spanning set if the rank of {{mvar|X}} equals the rank of the entire ground set{{sfnp|Oxley|2011|p=28}}

The vector space definition can also be generalized to modules.{{Harvard citation text|Roman|2005}} p. 96, ch. 4{{Harvard citation text|Mac Lane|Birkhoff|1999}} p. 193, ch. 6 Given an {{mvar|R}}-module {{mvar|A}} and a collection of elements {{math|a₁}}, ..., {{math|a_n}} of {{mvar|A}}, the submodule of {{mvar|A}} spanned by {{math|a₁}}, ..., {{math|a_n}} is the sum of cyclic modules

$$Ra\_1\; +\; \backslash cdots\; +\; Ra\_n\; =\; \backslash left\backslash \{\; \backslash sum\_\{k=1\}^n\; r\_k\; a\_k\; \backslash bigg|\; r\_k\; \backslash in\; R\; \backslash right\backslash \}$$

consisting of all R-linear combinations of the elements {{math|a_i}}. As with the case of vector spaces, the submodule of A spanned by any subset of A is the intersection of all submodules containing that subset.

In functional analysis, a closed linear span of a set of vectors is the minimal closed set which contains the linear span of that set.

Suppose that {{mvar|X}} is a normed vector space and let {{mvar|E}} be any non-empty subset of {{mvar|X}}. The closed linear span of {{mvar|E}}, denoted by $\backslash overline\{\backslash operatorname\{Sp\}\}(E)$ or $\backslash overline\{\backslash operatorname\{Span\}\}(E)$, is the intersection of all the closed linear subspaces of {{mvar|X}} which contain {{mvar|E}}.

One mathematical formulation of this is

:

The closed linear span of the set of functions xⁿ on the interval [0, 1], where n is a non-negative integer, depends on the norm used. If the L² norm is used, then the closed linear span is the Hilbert space of square-integrable functions on the interval. But if the maximum norm is used, the closed linear span will be the space of continuous functions on the interval. In either case, the closed linear span contains functions that are not polynomials, and so are not in the linear span itself. However, the cardinality of the set of functions in the closed linear span is the cardinality of the continuum, which is the same cardinality as for the set of polynomials.

The linear span of a set is dense in the closed linear span. Moreover, as stated in the lemma below, the closed linear span is indeed the closure of the linear span.

Closed linear spans are important when dealing with closed linear subspaces (which are themselves highly important, see Riesz's lemma).

Let {{mvar|X}} be a normed space and let {{mvar|E}} be any non-empty subset of {{mvar|X}}. Then

{{ordered list

|list-style-type=lower-alpha

| $\backslash overline\{\backslash operatorname\{Sp\}\}(E)$ is a closed linear subspace of X which contains E,

| $\backslash overline\{\backslash operatorname\{Sp\}\}(E)\; =\; \backslash overline\{\backslash operatorname\{Sp\}(E)\}$, viz. $\backslash overline\{\backslash operatorname\{Sp\}\}(E)$ is the closure of $\backslash operatorname\{Sp\}(E)$,

| $E^\backslash perp\; =\; (\backslash operatorname\{Sp\}(E))^\backslash perp\; =\; \backslash left(\backslash overline\{\backslash operatorname\{Sp\}(E)\}\backslash right)^\backslash perp.$

| $(E^\backslash perp)^\backslash perp\; =\; ((\backslash operatorname\{Sp\}(E))^\backslash perp)^\backslash perp\; =\; \backslash overline\{\backslash operatorname\{Sp\}(E)\}.$

}}

(So the usual way to find the closed linear span is to find the linear span first, and then the closure of that linear span.)

• Affine hull
• Conical combination
• Convex hull

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{{Linear algebra}}

Category:Abstract algebra

Category:Linear algebra