Pengguna:Kekavigi/bak pasir

Bidang yang direntang oleh vektor u dan v di R³.

Dalam aljabar linear, rentang linear atau span dari sebarang himpunan $S$ berisi vektor-vektor (yang berasal dari suatu ruang vektor) adalah himpunan semua kombinasi linear dari vektor-vektor di $S$. Rentang linear dari $S$ umum disimbolkan dengan ${\text{span}}(S).$ Sebagai contoh, dua vektor yang saling bebas linear akan merentang suatu bidang. Rentang dapat dikarakterisasikan sebagai irisan dari semua subruang (vektor) yang mengandung $S$, maupun sebagai subruang yang mengandung $S.$ Alhasil, rentang dari himpunan vektor menghasilkan suatu ruang vektor. Rentang dapat diperumum untuk matroid dan modul.

Untuk menyatakan bahwa suatu ruang vektor $V$ adalah rentang linear dari subset $S$, beberapa pernyataan berikut umum digunakan: $S$ merentang $V$, $S$ adalah himpunan merentang dari $V$, $V$ direntang/dibangkitkan oleh $S$, atau $S$ adalah pembangkit atau himpunan pembangkit dari $V.$

Definisi

Untuk sebarang ruang vektor $V$ atas lapangan $K$, rentang dari suatu himpunan $S$ yang beranggotakan vektor-vektor (tidak harus berhingga) didefinisikan sebagai irisan $W$ dari semua subruang dari $V$ yang mengandung $S$. Irisan $W$ disebut sebagai subruang yang direntang oleh $S$, atau oleh vektor-vektor di $S$. Kebalikannya, $S$ disebut himpunan merentang dari $W$, dan kita katakan $S$ merentang $W$.

Rentang dari $S$ juga dapat didefinisikan sebagai himpunan dari semua kombinasi linear terhingga dari vektor-vektor di $S$. Secara matematis, ini dituliskan sebagai

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

Pada kasus $S$ berukuran tak-hingga, syarat kombinasi linear yang tak-terhingga (yakni, keadaan ketika kombinasi menggunakan konsep penjumlahan tak-hingga, dengan mengasumsikan penjumlahan seperti itu dapat didefinisikan) tidak disertakan dalam definisi.

Contoh

Ruang vektor riil $\mathbb {R} ^{3}$ dapat direntang oleh himpunan $\{(-1,0,0),\,(0,1,0),\,(0,0,1)\}$. Himpunan ini juga merupakan suatu basis dari $\mathbb {R} ^{3}$. Jika $(-1,0,0)$ digantikan dengan $(1,0,0)$, himpunan tersebut merupakan basis standar dari $\mathbb {R} ^{3}$. Contoh himpunan pembangkit lain dari $\mathbb {R} ^{3}$ adalah $\{(1,2,3),\,(0,1,2),\,(-1,{\tfrac {1}{2}},3),\,(1,1,1)\}$, namun himpunan ini bukan basis karena bersifat bergantung linear.

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

Teorema

Equivalence of definitions

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

Suppose that W is a linear subspace of V containing S. It follows that $S\subseteq \operatorname {span} S$, since every v_i is a linear combination of S (trivially). Since W is closed under addition and scalar multiplication, then every linear combination $\lambda _{1}\mathbf {v} _{1}+\cdots +\lambda _{n}\mathbf {v} _{n}$ 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.

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

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

Since S spans V, then $S\cup \{\mathbf {w} _{1}\}$ must also span V, and $\mathbf {w} _{1}$ must be a linear combination of S. Thus $S\cup \{\mathbf {w} _{1}\}$ 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 $\{\mathbf {w} _{1},\mathbf {v} _{1},\ldots ,\mathbf {v} _{i-1},\mathbf {v} _{i+1},\ldots ,\mathbf {v} _{m}\}$, 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 $\{\mathbf {w} _{1},\ldots ,\mathbf {w} _{p}\}$ 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. $m\geq n$. To verify this, we assume by way of contradiction that $m<n$. Then, at the mth step, we have the set $\{\mathbf {w} _{1},\ldots ,\mathbf {w} _{m}\}$ and we can adjoin another vector $\mathbf {w} _{m+1}$. But, since $\{\mathbf {w} _{1},\ldots ,\mathbf {w} _{m}\}$ is a spanning set of V, $\mathbf {w} _{m+1}$ is a linear combination of $\{\mathbf {w} _{1},\ldots ,\mathbf {w} _{m}\}$. This is a contradiction, since W is linearly independent.

Spanning set can be reduced to a basis

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.

Generalizations

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.

The vector space definition can also be generalized to modules. Given an R-module A and a collection of elements a₁, ..., a_n of A, the submodule of A spanned by a₁, ..., a_n is the sum of cyclic modules

$$Ra_{1}+\cdots +Ra_{n}=\left\{\sum _{k=1}^{n}r_{k}a_{k}{\bigg |}r_{k}\in R\right\}$$

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.

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

One mathematical formulation of this is

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

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

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

(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.)

See also

• Affine hull
• Conical combination
• Convex hull

Sources

Textbooks

• Axler, Sheldon Jay (2015). Linear Algebra Done Right (edisi ke-3rd). Springer. ISBN 978-3-319-11079-0. 
• Hefferon, Jim (2020). Linear Algebra (edisi ke-4th). Orthogonal Publishing. ISBN 978-1-944325-11-4. 
• Lane, Saunders Mac; Birkhoff, Garrett (1999) [1988]. Algebra (edisi ke-3rd). AMS Chelsea Publishing. ISBN 978-0821816462. 
• Roman, Steven (2005). Advanced Linear Algebra (edisi ke-2nd). Springer. ISBN 0-387-24766-1. 
• Rynne, Brian P.; Youngson, Martin A. (2008). Linear Functional Analysis. Springer. ISBN 978-1848000049. 
• Lay, David C. (2021) Linear Algebra and Its Applications (6th Edition). Pearson.

Web

• Lankham, Isaiah; Nachtergaele, Bruno; Schilling, Anne (13 February 2010). "Linear Algebra - As an Introduction to Abstract Mathematics" (PDF). University of California, Davis. Diakses tanggal 27 September 2011. 
• Weisstein, Eric Wolfgang. "Vector Space Span". MathWorld. Diakses tanggal 16 Feb 2021. 
• "Linear hull". Encyclopedia of Mathematics. 5 April 2020. Diakses tanggal 16 Feb 2021. 

External links

• Linear Combinations and Span: Understanding linear combinations and spans of vectors, khanacademy.org.
• Sanderson, Grant (August 6, 2016). "Linear combinations, span, and basis vectors". Essence of Linear Algebra. Diarsipkan dari versi asli tanggal 2021-12-11 – via YouTube.