Linear span

File:Basis for a plane.svg

The cross-hatched plane is the linear span of u and v in both R² and R³, here shown in perspective.

In mathematics, the linear span (also called the linear hull<ref>Template:Harvard citation text. Linear Hull.</ref> or just span) of a set <math>S</math> of elements of a vector space <math>V</math> is the smallest linear subspace of <math>V</math> that contains <math>S.</math> It is the set of all finite linear combinations of the elements of Template:Mvar,<ref>Template:Harvard citation text p. 29, § 2.7</ref> and the intersection of all linear subspaces that contain <math>S.</math> It is often denoted Template:Math<ref name=":0">Template:Harvard citation text pp. 29-30, §§ 2.5, 2.8</ref> or <math>\langle S \rangle.</math>

For example, in geometry, two linearly independent vectors span a plane.

To express that a vector space Template:Mvar is a linear span of a subset Template:Mvar, one commonly uses one of the following phrases: Template:Mvar spans Template:Mvar; Template:Mvar is a spanning set of Template:Mvar; Template:Mvar is spanned or generated by Template:Mvar; Template:Mvar is a generator set or a generating set of Template:Mvar.

Spans can be generalized to many mathematical structures, in which case, the smallest substructure containing <math>S</math> is generally called the substructure generated by <math>S.</math>

Definition

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

It follows from this definition that the span of Template:Mvar is the set of all finite linear combinations of elements (vectors) of Template:Mvar, and can be defined as such.<ref>Template:Harvard citation text p. 100, ch. 2, Definition 2.13</ref><ref name=":02">Template:Harvard citation text pp. 29-30, §§ 2.5, 2.8</ref><ref>Template:Harvard citation text pp. 41-42</ref> That is, <math display="block"> \operatorname{span}(S) = \biggl \{ \lambda_1 \mathbf v_1 + \lambda_2 \mathbf v_2 + \cdots + \lambda_n \mathbf v_n \mid n \in \N,\; \mathbf v_1,...\mathbf v_n \in S, \; \lambda_1,...\lambda_n \in K \biggr \}</math>

When Template:Mvar is empty, the only possibility is Template:Math, and the previous expression for <math>\operatorname{span}(S)</math> reduces to the empty sum.Template:Efn The standard convention for the empty sum implies thus <math>\text{span}(\empty) = \{\mathbf 0\}, </math> a property that is immediate with the other definitions. However, many introductory textbooks simply include this fact as part of the definition.

When <math>S=\{\mathbf v_1,\ldots, \mathbf v_n\}</math> is finite, one has <math display="block"> \operatorname{span}(S) = \{ \lambda_1 \mathbf v_1 + \lambda_2 \mathbf v_2 + \cdots + \lambda_n \mathbf v_n \mid \lambda_1,...\lambda_n \in K \}</math>

Examples

The real vector space <math>\mathbb R^3</math> 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 <math>\mathbb R^3</math>.

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

The set Template:Math} is not a spanning set of <math>\mathbb R^3</math>, since its span is the space of all vectors in <math>\mathbb R^3</math> 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 <math>\mathbb R^3.</math> It can be identified with <math>\mathbb R^2</math> 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 <math>\mathbb R^3</math>, and {(0, 0, 0)} is the intersection of all of these vector spaces.

The set of monomials Template:Mvar, where Template:Mvar is a non-negative integer, spans the space of polynomials.

Theorems

Equivalence of definitions

The set of all linear combinations of a subset Template:Mvar of Template:Mvar, a vector space over Template:Mvar, is the smallest linear subspace of Template:Mvar containing Template:Mvar.

Proof. We first prove that Template:Math is a subspace of Template:Mvar. Since Template:Mvar is a subset of Template:Mvar, we only need to prove the existence of a zero vector Template:Math in Template:Math, that Template:Math is closed under addition, and that Template:Math is closed under scalar multiplication. Letting <math>S = \{ \mathbf v_1, \mathbf v_2, \ldots, \mathbf v_n \}</math>, it is trivial that the zero vector of Template:Mvar exists in Template:Math, since <math>\mathbf 0 = 0 \mathbf v_1 + 0 \mathbf v_2 + \cdots + 0 \mathbf v_n</math>. Adding together two linear combinations of Template:Mvar also produces a linear combination of Template:Mvar: <math>(\lambda_1 \mathbf v_1 + \cdots + \lambda_n \mathbf v_n) + (\mu_1 \mathbf v_1 + \cdots + \mu_n \mathbf v_n) = (\lambda_1 + \mu_1) \mathbf v_1 + \cdots + (\lambda_n + \mu_n) \mathbf v_n</math>, where all <math>\lambda_i, \mu_i \in K</math>, and multiplying a linear combination of Template:Mvar by a scalar <math>c \in K</math> will produce another linear combination of Template:Mvar: <math>c(\lambda_1 \mathbf v_1 + \cdots + \lambda_n \mathbf v_n) = c\lambda_1 \mathbf v_1 + \cdots + c\lambda_n \mathbf v_n</math>. Thus Template:Math is a subspace of Template:Mvar.

It follows that <math>S \subseteq \operatorname{span} S</math>, since every Template:Math is a linear combination of Template:Mvar (trivially). Suppose that Template:Mvar is a linear subspace of Template:Mvar containing Template:Mvar. Since Template:Mvar is closed under addition and scalar multiplication, then every linear combination <math>\lambda_1 \mathbf v_1 + \cdots + \lambda_n \mathbf v_n</math> must be contained in Template:Mvar. Thus, Template:Math is contained in every subspace of Template:Mvar containing Template:Mvar, and the intersection of all such subspaces, or the smallest such subspace, is equal to the set of all linear combinations of Template:Mvar.

Size of spanning set is at least size of linearly independent set

Every spanning set Template:Mvar of a vector space Template:Mvar must contain at least as many elements as any linearly independent set of vectors from Template:Mvar.

Proof. Let <math>S = \{ \mathbf v_1, \ldots, \mathbf v_m \}</math> be a spanning set and <math>W = \{ \mathbf w_1, \ldots, \mathbf w_n \}</math> be a linearly independent set of vectors from Template:Mvar. We want to show that <math>m \geq n</math>.

Since Template:Mvar spans Template:Mvar, then <math>S \cup \{ \mathbf w_1 \}</math> must also span Template:Mvar, and <math>\mathbf w_1</math> must be a linear combination of Template:Mvar. Thus <math>S \cup \{ \mathbf w_1 \}</math> is linearly dependent, and we can remove one vector from Template:Mvar that is a linear combination of the other elements. This vector cannot be any of the Template:Math, since Template:Mvar is linearly independent. The resulting set is <math>\{ \mathbf w_1, \mathbf v_1, \ldots, \mathbf v_{i-1}, \mathbf v_{i+1}, \ldots, \mathbf v_m \}</math>, which is a spanning set of Template:Mvar. We repeat this step Template:Mvar times, where the resulting set after the Template:Mvarth step is the union of <math>\{ \mathbf w_1, \ldots, \mathbf w_p \}</math> and Template:Mvar vectors of Template:Mvar.

It is ensured until the Template:Mvarth step that there will always be some Template:Math to remove out of Template:Mvar for every adjoint of Template:Math, and thus there are at least as many Template:Math's as there are Template:Math's—i.e. <math>m \geq n</math>. To verify this, we assume by way of contradiction that <math>m < n</math>. Then, at the Template:Mvarth step, we have the set <math>\{ \mathbf w_1, \ldots, \mathbf w_m \}</math> and we can adjoin another vector <math>\mathbf w_{m+1}</math>. But, since <math>\{ \mathbf w_1, \ldots, \mathbf w_m \}</math> is a spanning set of Template:Mvar, <math>\mathbf w_{m+1}</math> is a linear combination of <math>\{ \mathbf w_1, \ldots, \mathbf w_m \}</math>. This is a contradiction, since Template:Mvar is linearly independent.

Spanning set can be reduced to a basis

Let Template:Mvar be a finite-dimensional vector space. Any set of vectors that spans Template:Mvar can be reduced to a basis for Template:Mvar, 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 Template:Mvar has finite dimension. This also indicates that a basis is a minimal spanning set when Template:Mvar is finite-dimensional.

Generalizations

Generalizing the definition of the span of points in space, a subset Template:Mvar of the ground set of a matroid is called a spanning set if the rank of Template:Mvar equals the rank of the entire ground setTemplate:Sfnp

The vector space definition can also be generalized to modules.<ref>Template:Harvard citation text p. 96, ch. 4</ref><ref>Template:Harvard citation text p. 193, ch. 6</ref> Given an Template:Mvar-module Template:Mvar and a collection of elements Template:Math, ..., Template:Math of Template:Mvar, the submodule of Template:Mvar spanned by Template:Math, ..., Template:Math is the sum of cyclic modules <math display="block">Ra_1 + \cdots + Ra_n = \left\{ \sum_{k=1}^n r_k a_k \bigg| r_k \in R \right\}</math> consisting of all R-linear combinations of the elements Template:Math. 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.

Closed linear span (functional analysis)

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 Template:Mvar is a normed vector space and let Template:Mvar be any non-empty subset of Template:Mvar. The closed linear span of Template:Mvar, denoted by <math>\overline{\operatorname{Sp}}(E)</math> or <math>\overline{\operatorname{Span}}(E)</math>, is the intersection of all the closed linear subspaces of Template:Mvar which contain Template:Mvar.

One mathematical formulation of this is

<math>\overline{\operatorname{Sp}}(E) = \{u\in X | \forall\varepsilon > 0\,\exists x\in\operatorname{Sp}(E) : \|x - u\|<\varepsilon\}.</math>

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.

Notes

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).

A useful lemma

Let Template:Mvar be a normed space and let Template:Mvar be any non-empty subset of Template:Mvar. Then Template:Ordered list(E)</math> is a closed linear subspace of X which contains E,

| <math>\overline{\operatorname{Sp}}(E) = \overline{\operatorname{Sp}(E)}</math>, viz. <math>\overline{\operatorname{Sp}}(E)</math> is the closure of <math>\operatorname{Sp}(E)</math>,
| <math>E^\perp = (\operatorname{Sp}(E))^\perp = \left(\overline{\operatorname{Sp}(E)}\right)^\perp.</math>
| <math>(E^\perp)^\perp = ((\operatorname{Sp}(E))^\perp)^\perp = \overline{\operatorname{Sp}(E)}.</math>

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(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.)

See also

• Affine hull
• Conical combination
• Convex hull

Citations

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Sources

Textbooks

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• Lay, David C. (2021) Linear Algebra and Its Applications (6th Edition). Pearson.

Web

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External links

• Linear Combinations and Span: Understanding linear combinations and spans of vectors, khanacademy.org.
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