Pengguna:Kekavigi/bak pasir
Dalam aljabar linear, rentang linear atau span dari sebarang himpunan berisi vektor-vektor (yang berasal dari suatu ruang vektor) adalah himpunan semua kombinasi linear dari vektor-vektor di .[1] Rentang linear dari umum disimbolkan dengan [2] Sebagai contoh, dua vektor yang saling bebas linear akan merentang suatu bidang. Rentang dapat dikarakterisasikan sebagai irisan dari semua subruang (vektor) yang mengandung , maupun sebagai subruang yang mengandung Alhasil, rentang dari himpunan vektor menghasilkan suatu ruang vektor. Rentang dapat diperumum untuk matroid dan modul.
Untuk menyatakan bahwa suatu ruang vektor adalah rentang linear dari subset , beberapa pernyataan berikut umum digunakan: merentang , adalah himpunan merentang dari , direntang/dibangkitkan oleh , atau adalah pembangkit atau himpunan pembangkit dari
Definisi
Untuk sebarang ruang vektor atas lapangan , rentang dari suatu himpunan yang beranggotakan vektor-vektor (tidak harus berhingga) didefinisikan sebagai irisan dari semua subruang dari yang mengandung . Irisan disebut sebagai subruang yang direntang oleh , atau oleh vektor-vektor di . Kebalikannya, disebut himpunan merentang dari , dan kita katakan merentang .
Rentang dari juga dapat didefinisikan sebagai himpunan dari semua kombinasi linear terhingga dari vektor-vektor di .[3][4][5][6] Secara matematis, ini dituliskan sebagai
Contoh
Ruang vektor riil dapat direntang oleh himpunan . Himpunan ini juga merupakan suatu basis dari . Jika digantikan dengan , himpunan tersebut merupakan basis standar dari . Contoh himpunan pembangkit lain dari adalah , 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 , since its span is the space of all vectors in 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 It can be identified with 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 , 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 , it is trivial that the zero vector of V exists in span S, since . Adding together two linear combinations of S also produces a linear combination of S: , where all , and multiplying a linear combination of S by a scalar will produce another linear combination of S: . Thus span S is a subspace of V.
- Suppose that W is a linear subspace of V containing S. It follows that , since every vi is a linear combination of S (trivially). Since W is closed under addition and scalar multiplication, then every linear combination 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 be a spanning set and be a linearly independent set of vectors from V. We want to show that .
- Since S spans V, then must also span V, and must be a linear combination of S. Thus 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 wi, since W is linearly independent. The resulting set is , 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 and m - p vectors of S.
- It is ensured until the nth step that there will always be some vi to remove out of S for every adjoint of v, and thus there are at least as many vi's as there are wi's—i.e. . To verify this, we assume by way of contradiction that . Then, at the mth step, we have the set and we can adjoin another vector . But, since is a spanning set of V, is a linear combination of . 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[butuh rujukan].
The vector space definition can also be generalized to modules.[7][8] Given an R-module A and a collection of elements a1, ..., an of A, the submodule of A spanned by a1, ..., an is the sum of cyclic modules
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 or , is the intersection of all the closed linear subspaces of X which contain 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.
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
- is a closed linear subspace of X which contains E,
- , viz. is the closure of ,
(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
Citations
- ^ (Axler 2015) p. 29, § 2.7
- ^ (Axler 2015) pp. 29-30, §§ 2.5, 2.8
- ^ (Hefferon 2020) p. 100, ch. 2, Definition 2.13
- ^ (Axler 2015) pp. 29-30, §§ 2.5, 2.8
- ^ (Roman 2005) pp. 41-42
- ^ (MathWorld 2021) Vector Space Span.
- ^ (Roman 2005) p. 96, ch. 4
- ^ (Lane & Birkhoff 1999) p. 193, ch. 6
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.