Linear span

Given a vector space V over a field K, the span of a set S of vectors (not necessarily finite) is defined to be the intersection W of all subspacesofV that contain S. W is referred to as the subspace spanned by S, or by the vectors in S. Conversely, S is called a spanning setofW, and we say that S spans W.

Alternatively, the span of S may be defined as the set of all finite linear combinations of elements (vectors) of S, which follows from the above definition.^[4][5][6][7]

$${\displaystyle \operatorname {span} ($$S)=\left\{{\left.\sum _{i=1}^{k}\lambda _{i}\mathbf {v} _{i}\;\right|\;k\in \mathbb {N} ,\mathbf {v} _{i}\in S,\lambda _{i}\in K}\right\}.}  

In the case of infinite S, infinite linear combinations (i.e. where a combination may involve an infinite sum, assuming that such sums are defined somehow as in, say, a Banach space) are excluded by the definition; a generalization that allows these is not equivalent.

The real vector space ${\displaystyle \mathbb {R} ^{3}}$  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 basisof${\displaystyle \mathbb {R} ^{3}}$ .

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

The set {(1, 0, 0), (0, 1, 0), (1, 1, 0)} is not a spanning set of ${\displaystyle \mathbb {R} ^{3}}$ , since its span is the space of all vectors in ${\displaystyle \mathbb {R} ^{3}}$  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 ${\displaystyle \mathbb {R} ^{3}.}$  It can be identified with ${\displaystyle \mathbb {R} ^{2}}$  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 ${\displaystyle \mathbb {R} ^{3}}$ , and {(0, 0, 0)} is the intersection of all of these vector spaces.

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

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

Proof. We first prove that span S is a subspace of V. Since S is a subset of V, we only need to prove the existence of a zero vector 0inspan S, that span S is closed under addition, and that span S is closed under scalar multiplication. Letting ${\displaystyle S=\{\mathbf {v} _{1},\mathbf {v} _{2},\ldots ,\mathbf {v} _{n}\}}$ , it is trivial that the zero vector of V exists in span S, since ${\displaystyle \mathbf {0} =0\mathbf {v} _{1}+0\mathbf {v} _{2}+\cdots +0\mathbf {v} _{n}}$ . Adding together two linear combinations of S also produces a linear combination of S: ${\displaystyle (\lambda _{1}\mathbf {v} _{1}+\cdots +\lambda _{n}\mathbf {v} _{n})+(\mu _{1}\mathbf {v} _{1}+\cdots +\mu _{n}\mathbf {v} _{n})=(\lambda _{1}+\mu _{1})\mathbf {v} _{1}+\cdots +(\lambda _{n}+\mu _{n})\mathbf {v} _{n}}$ , where all ${\displaystyle \lambda _{i},\mu _{i}\in K}$ , and multiplying a linear combination of S by a scalar ${\displaystyle c\in K}$  will produce another linear combination of S: ${\displaystyle c(\lambda _{1}\mathbf {v} _{1}+\cdots +\lambda _{n}\mathbf {v} _{n})=c\lambda _{1}\mathbf {v} _{1}+\cdots +c\lambda _{n}\mathbf {v} _{n}}$ . Thus span S is a subspace of V.

Suppose that W is a linear subspace of V containing S. It follows that ${\displaystyle S\subseteq \operatorname {span} S}$ , since every v_i is a linear combination of S (trivially). Since W is closed under addition and scalar multiplication, then every linear combination ${\displaystyle \lambda _{1}\mathbf {v} _{1}+\cdots +\lambda _{n}\mathbf {v} _{n}}$  must be contained in W. Thus, span S is contained in every subspace of V containing S, and the intersection of all such subspaces, or the smallest such subspace, is equal to the set of all linear combinations of S.

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

Proof. Let ${\displaystyle S=\{\mathbf {v} _{1},\ldots ,\mathbf {v} _{m}\}}$  be a spanning set and ${\displaystyle W=\{\mathbf {w} _{1},\ldots ,\mathbf {w} _{n}\}}$  be a linearly independent set of vectors from V. We want to show that ${\displaystyle m\geq n}$ .

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

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

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

Generalizing the definition of the span of points in space, a subset X of the ground set of a matroid is called a spanning set if the rank of X equals the rank of the entire ground set^[8]

The vector space definition can also be generalized to modules.^[9][10] Given an R-module A and a collection of elements a₁, ..., a_nofA, the submoduleofA spanned by a₁, ..., a_n is the sum of cyclic modules $${\displaystyle Ra_{1}+\cdots +Ra_{n}=\left\{\sum _{k=1}^{n}r_{k}a_{k}{\bigg |}r_{k}\in R\right\}}$$  consisting of all R-linear combinations of the elements 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.

Infunctional analysis, a closed linear span of a setofvectors is the minimal closed set which contains the linear span of that set.

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

One mathematical formulation of this is

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

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 spaceofsquare-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 X be a normed space and let E be any non-empty subset of X. Then

1. ${\displaystyle {\overline {\operatorname {Sp} }}($E)}   is a closed linear subspace of X which contains E,
2. ${\displaystyle {\overline {\operatorname {Sp} }}($E)={\overline {\operatorname {Sp} (E)}}}  , viz. ${\displaystyle {\overline {\operatorname {Sp} }}($E)}   is the closure of ${\displaystyle \operatorname {Sp} ($E)}  ,
3. ${\displaystyle E^{\perp }=(\operatorname {Sp} ($E))^{\perp }=\left({\overline {\operatorname {Sp} (E)}}\right)^{\perp }.}  

(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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