DOCSLIB.ORG

Subspaces, Basis, Dimension and Rank Subspaces

Total Page：16

File Type：pdf, Size：1020Kb

Download full-text PDF Read full-text
Subspaces, Basis, Dimension and Rank Subspaces We recall Theorem 4, Sec 1.3.LetA be an m × n matrix. Then the following is equivalent: a. For each b ∈ Rm,theequationAx = b has a solution. b. Each b ∈ Rm is a linear combination of the columns of A. c. The columns of A span Rm. Subspaces, Basis, Dimension and Rank d. A has a pivot position in every row. n Theorem 8, Sec 1.7 Any {v1,...vp} in R is linearly dependent if p > n. 1 A consequence of Theorem 4 above is that if A has less than m column vectors, the set of columns of A cannot span Rm 2 A consequence of Theorem 8 above is that The maximum number of independent vectors in Rn is n. Any other vector is linear combination of the vectors in the maximum independent set of vectors. Subspaces Definition. A subspace of Rn is any set H in Rn that has three properties: a. 0 ∈ H. b. u + v ∈ H for all u, v ∈ H. c. cu ∈ H for all c ∈ Rn and u ∈ H. A subspace is closed under addition and scalar multiplication. Example 1. - Planes and lines through the origin in R3 are R3 subspaces of . Plane: H = Span{u, v} is a subspace of R3. Rn is a subspace of itself. Basis for a Subspace Basis for a Subspace n Definition. A basis for a subspace H of R is a linearly It is consequence that the vectors independent set in H that spans H. ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ n 1 0 0 Example 2. Let V = {a1,...,an} be a set of vectors in R . ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ 0 ⎥ ⎢ 1 ⎥ ⎢ . ⎥ Suppose that they are the columns of an invertible n × n matrix A. e1 = ⎢ ⎥ , e2 = ⎢ ⎥ , ··· , en = ⎢ . ⎥ ⎣ . ⎦ ⎣ . ⎦ ⎣ ⎦ Then V is a basis for Rn. 0 Solution. By the big Invertible Matrix Theorem - IMT we have 0 0 1 that columns of A form a basis for Rn, because the identity matrix is invertible. form a linearly independent set Thm 8 (e) and We also have the converse. If set V is a basis for Rn, i.e., V is span Rn Thm8 (h). linearly independent set that spans Rn, then matrix A is invertible. V Rn So is a basis for . That also follows from IMT. Basis for a Subspace Vector Coordinates relative to a Basis B {b , ··· , b } H Given a basis = 1 p for a subspace suppose Definition. Let B be a basis for H. For each x in H,the x that vector can be generated in two ways: coordinates of x relative to the basis B are the weights p c1,...,cp such that x = c1b1 + ···cpbp, and the vector in R x = c1b1 + ···cpbp and x = d1b1 + ···+ dpbp ⎡ ⎤ c1 . ⎢ . ⎥ [x]B = ⎣ . ⎦ Then subtracting gives . cp 0=x − x =(c1 − d1)b1 + ···(cp − dp)bp (1) is called the coordinate vector of x (relative to B)orthe Since B is linearly independent all weights in (1) must be B-coordinate vector of x. zero. So cj = dj for j =1, ··· , p. Coordinate systems Coordinate systems T T T Example 3. Let v1 =[3, 6, 2] , v2 =[−1, 0, 1] , x =[3, 12, 7] , Rowreductionshowsthat and B = {v1, v2}.ThenB is a basis for H = Span{v1, v2} because ⎡ ⎤ ⎡ ⎤ 3 −13 102 v1 and v2 are linearly independent. ⎣ 6012⎦ ⎣ 013⎦ (a) Is x ∈ H? (b) If so, find the B-coordinate vector of x. 21 7 000 (a) x ∈ H if and only if the system Thus c1 =2, c2 = 3 is the solution and x is in H. T T T T c1[3, 6, 2] + c2[−1, 0, 1 ]=[3, 12, 7] (b) The B-coordinates of x is [x]B =[2, 3] . The basis B determines a coordinate system on H. See figure is consistent. below. Coordinate systems Coordinate systems The correspondence x → [x]B is a one-to-one correspondence between H and R2 that preserves linear combinations. Such a correspondence is called an isomorphism.WesaythatH is isomorphic to R2. In general, if B = {b1, ··· , bk } is a basis of H, then the mapping k x → [x]B is a one-to-one correspondence between H and R ,even though the vectors in H mayhavemorethankentries.SoH is isomorphic to Rk . Basis and the number of basic vectors Dimension and Rank Proposition 1. IfasubspaceH has a basis with k elements so every other basis has exactly k vectors. Sketch of the proof. Remark. Any set of k linearly independent vectors form a basis for Rk . H k Rk A subspace with a basis with vectors is isomorphic to . Definition. The dimension of a nonzero subspace H, denoted by k So we need to show that all basis for R has exactly k vectors. dim H, is the number of vectors in any basis for H. The dimension By Theorem 8, Section 1.7, the maximum number of vectors of the zero space is zero. Rk p Rk abasisfor can have is .Soabasisfor must have at Definition. Given an m × n matrix A,therank of A is the k most (independent) vectors. maximum number of linearly independent column vectors in A. k Suppose a basis for R has l vectors with l < k.ByTheorem This number is denoted by rankA = r. 4, placing the vectors of the basis as the column vectors of a It can be shown that the number of independent columns in a matrix A, the matrix must have one pivot at each row to span matrix A is also equal to the number os independent row of A. Rk , i.e., k pivots. But A can have at most l < k pivots, because it has l columns. A contradiction. Therefore, a basis for Rk must have exactly k vectors. Dimension and Rank Theorem 1. - The Basis Theorem Let H be a k dimensional subspace of Rn. Any linearly independent set of exactly k vectors in H is automatically a basis for H.Alsoasetofk elements that spans H form a basis for H. Theorem 2. - Invertible Matrix Theorem (continued)Let A be an n × n square matrix. Then the following are equivalent. a. A is invertible m. Columns of A form a basis for Rn. n. rank A = n .
Load more

Recommended publications
  • ADDITIVE MAPS on RANK K BIVECTORS 1. Introduction. Let N 2 Be an Integer and Let F Be a Field. We Denote by M N(F) the Algeb
    Electronic Journal of Linear Algebra, ISSN 1081-3810 A publication of the International Linear Algebra Society Volume 36, pp. 847-856, December 2020. ADDITIVE MAPS ON RANK K BIVECTORS∗ WAI LEONG CHOOIy AND KIAM HEONG KWAy V2 Abstract. Let U and V be linear spaces over fields F and K, respectively, such that dim U = n > 2 and jFj > 3. Let U n−1 V2 V2 be the second exterior power of U. Fixing an even integer k satisfying 2 6 k 6 n, it is shown that a map : U! V satisfies (u + v) = (u) + (v) for all rank k bivectors u; v 2 V2 U if and only if is an additive map. Examples showing the indispensability of the assumption on k are given. Key words. Additive maps, Second exterior powers, Bivectors, Ranks, Alternate matrices. AMS subject classifications. 15A03, 15A04, 15A75, 15A86. 1. Introduction. Let n > 2 be an integer and let F be a field. We denote by Mn(F) the algebra of n×n matrices over F. Given a nonempty subset S of Mn(F), a map : Mn(F) ! Mn(F) is called commuting on S (respectively, additive on S) if (A)A = A (A) for all A 2 S (respectively, (A + B) = (A) + (B) for all A; B 2 S). In 2012, using Breˇsar'sresult [1, Theorem A], Franca [3] characterized commuting additive maps on invertible (respectively, singular) matrices of Mn(F). He continued to characterize commuting additive maps on rank k matrices of Mn(F) in [4], where 2 6 k 6 n is a fixed integer, and commuting additive maps on rank one matrices of Mn(F) in [5].
    [Show full text]
  • Do Killingâ•Fiyano Tensors Form a Lie Algebra?
    University of Massachusetts Amherst ScholarWorks@UMass Amherst Physics Department Faculty Publication Series Physics 2007 Do Killing–Yano tensors form a Lie algebra? David Kastor University of Massachusetts - Amherst, [email protected] Sourya Ray University of Massachusetts - Amherst Jennie Traschen University of Massachusetts - Amherst, [email protected] Follow this and additional works at: https://scholarworks.umass.edu/physics_faculty_pubs Part of the Physics Commons Recommended Citation Kastor, David; Ray, Sourya; and Traschen, Jennie, "Do Killing–Yano tensors form a Lie algebra?" (2007). Classical and Quantum Gravity. 1235. Retrieved from https://scholarworks.umass.edu/physics_faculty_pubs/1235 This Article is brought to you for free and open access by the Physics at ScholarWorks@UMass Amherst. It has been accepted for inclusion in Physics Department Faculty Publication Series by an authorized administrator of ScholarWorks@UMass Amherst. For more information, please contact [email protected]. Do Killing-Yano tensors form a Lie algebra? David Kastor, Sourya Ray and Jennie Traschen Department of Physics University of Massachusetts Amherst, MA 01003 ABSTRACT Killing-Yano tensors are natural generalizations of Killing vec- tors. We investigate whether Killing-Yano tensors form a graded Lie algebra with respect to the Schouten-Nijenhuis bracket. We find that this proposition does not hold in general, but that it arXiv:0705.0535v1 [hep-th] 3 May 2007 does hold for constant curvature spacetimes. We also show that Minkowski and (anti)-deSitter spacetimes have the maximal num- ber of Killing-Yano tensors of each rank and that the algebras of these tensors under the SN bracket are relatively simple exten- sions of the Poincare and (A)dS symmetry algebras.
    [Show full text]
  • 21. Orthonormal Bases
    21. Orthonormal Bases The canonical/standard basis 011 001 001 B C B C B C B0C B1C B0C e1 = B.C ; e2 = B.C ; : : : ; en = B.C B.C B.C B.C @.A @.A @.A 0 0 1 has many useful properties. • Each of the standard basis vectors has unit length: q p T jjeijj = ei ei = ei ei = 1: • The standard basis vectors are orthogonal (in other words, at right angles or perpendicular). T ei ej = ei ej = 0 when i 6= j This is summarized by ( 1 i = j eT e = δ = ; i j ij 0 i 6= j where δij is the Kronecker delta. Notice that the Kronecker delta gives the entries of the identity matrix. Given column vectors v and w, we have seen that the dot product v w is the same as the matrix multiplication vT w. This is the inner product on n T R . We can also form the outer product vw , which gives a square matrix. 1 The outer product on the standard basis vectors is interesting. Set T Π1 = e1e1 011 B C B0C = B.C 1 0 ::: 0 B.C @.A 0 01 0 ::: 01 B C B0 0 ::: 0C = B. .C B. .C @. .A 0 0 ::: 0 . T Πn = enen 001 B C B0C = B.C 0 0 ::: 1 B.C @.A 1 00 0 ::: 01 B C B0 0 ::: 0C = B. .C B. .C @. .A 0 0 ::: 1 In short, Πi is the diagonal square matrix with a 1 in the ith diagonal position and zeros everywhere else.
    [Show full text]
  • Multivector Differentiation and Linear Algebra 0.5Cm 17Th Santaló
    Multivector differentiation and Linear Algebra 17th Santalo´ Summer School 2016, Santander Joan Lasenby Signal Processing Group, Engineering Department, Cambridge, UK and Trinity College Cambridge [email protected], www-sigproc.eng.cam.ac.uk/ s jl 23 August 2016 1 / 78 Examples of differentiation wrt multivectors. Linear Algebra: matrices and tensors as linear functions mapping between elements of the algebra. Functional Differentiation: very briefly... Summary Overview The Multivector Derivative. 2 / 78 Linear Algebra: matrices and tensors as linear functions mapping between elements of the algebra. Functional Differentiation: very briefly... Summary Overview The Multivector Derivative. Examples of differentiation wrt multivectors. 3 / 78 Functional Differentiation: very briefly... Summary Overview The Multivector Derivative. Examples of differentiation wrt multivectors. Linear Algebra: matrices and tensors as linear functions mapping between elements of the algebra. 4 / 78 Summary Overview The Multivector Derivative. Examples of differentiation wrt multivectors. Linear Algebra: matrices and tensors as linear functions mapping between elements of the algebra. Functional Differentiation: very briefly... 5 / 78 Overview The Multivector Derivative. Examples of differentiation wrt multivectors. Linear Algebra: matrices and tensors as linear functions mapping between elements of the algebra. Functional Differentiation: very briefly... Summary 6 / 78 We now want to generalise this idea to enable us to find the derivative of F(X), in the A ‘direction’ – where X is a general mixed grade multivector (so F(X) is a general multivector valued function of X). Let us use ∗ to denote taking the scalar part, ie P ∗ Q ≡ hPQi. Then, provided A has same grades as X, it makes sense to define: F(X + tA) − F(X) A ∗ ¶XF(X) = lim t!0 t The Multivector Derivative Recall our definition of the directional derivative in the a direction F(x + ea) − F(x) a·r F(x) = lim e!0 e 7 / 78 Let us use ∗ to denote taking the scalar part, ie P ∗ Q ≡ hPQi.
    [Show full text]
  • 28. Exterior Powers
    28. Exterior powers 28.1 Desiderata 28.2 Definitions, uniqueness, existence 28.3 Some elementary facts 28.4 Exterior powers Vif of maps 28.5 Exterior powers of free modules 28.6 Determinants revisited 28.7 Minors of matrices 28.8 Uniqueness in the structure theorem 28.9 Cartan's lemma 28.10 Cayley-Hamilton Theorem 28.11 Worked examples While many of the arguments here have analogues for tensor products, it is worthwhile to repeat these arguments with the relevant variations, both for practice, and to be sensitive to the differences. 1. Desiderata Again, we review missing items in our development of linear algebra. We are missing a development of determinants of matrices whose entries may be in commutative rings, rather than fields. We would like an intrinsic definition of determinants of endomorphisms, rather than one that depends upon a choice of coordinates, even if we eventually prove that the determinant is independent of the coordinates. We anticipate that Artin's axiomatization of determinants of matrices should be mirrored in much of what we do here. We want a direct and natural proof of the Cayley-Hamilton theorem. Linear algebra over fields is insufficient, since the introduction of the indeterminate x in the definition of the characteristic polynomial takes us outside the class of vector spaces over fields. We want to give a conceptual proof for the uniqueness part of the structure theorem for finitely-generated modules over principal ideal domains. Multi-linear algebra over fields is surely insufficient for this. 417 418 Exterior powers 2. Definitions, uniqueness, existence Let R be a commutative ring with 1.
    [Show full text]
  • Preserivng Pieces of Information in a Given Order in HRR and GA$ C$
    Proceedings of the Federated Conference on ISBN 978-83-60810-22-4 Computer Science and Information Systems pp. 213–220 Preserivng pieces of information in a given order in HRR and GAc Agnieszka Patyk-Ło´nska Abstract—Geometric Analogues of Holographic Reduced Rep- Secondly, this method of encoding will not detect similarity resentations (GA HRR or GAc—the continuous version of between eye and yeye. discrete GA described in [16]) employ role-filler binding based A quantum-like attempt to tackle the problem of information on geometric products. Atomic objects are real-valued vectors in n-dimensional Euclidean space and complex statements belong to ordering was made in [1]—a version of semantic analysis, a hierarchy of multivectors. A property of GAc and HRR studied reformulated in terms of a Hilbert-space problem, is compared here is the ability to store pieces of information in a given order with structures known from quantum mechanics. In particular, by means of trajectory association. We describe results of an an LSA matrix representation [1], [10] is rewritten by the experiment: finding the alignment of items in a sequence without means of quantum notation. Geometric algebra has also been the precise knowledge of trajectory vectors. Index Terms—distributed representations, geometric algebra, used extensively in quantum mechanics ([2], [4], [3]) and so HRR, BSC, word order, trajectory associations, bag of words. there seems to be a natural connection between LSA and GAc, which is the ground for fututre work on the problem I. INTRODUCTION of preserving pieces of information in a given order. VER the years several attempts have been made to As far as convolutions are concerned, the most interesting O preserve the order in which the objects are to be remem- approach to remembering information in a given order has bered with the help of binding and superposition.
    [Show full text]
  • 8 Rank of a Matrix
    8 Rank of a matrix We already know how to figure out that the collection (v1;:::; vk) is linearly dependent or not if each n vj 2 R . Recall that we need to form the matrix [v1 j ::: j vk] with the given vectors as columns and see whether the row echelon form of this matrix has any free variables. If these vectors linearly independent (this is an abuse of language, the correct phrase would be \if the collection composed of these vectors is linearly independent"), then, due to the theorems we proved, their span is a subspace of Rn of dimension k, and this collection is a basis of this subspace. Now, what if they are linearly dependent? Still, their span will be a subspace of Rn, but what is the dimension and what is a basis? Naively, I can answer this question by looking at these vectors one by one. In particular, if v1 =6 0 then I form B = (v1) (I know that one nonzero vector is linearly independent). Next, I add v2 to this collection. If the collection (v1; v2) is linearly dependent (which I know how to check), I drop v2 and take v3. If independent then I form B = (v1; v2) and add now third vector v3. This procedure will lead to the vectors that form a basis of the span of my original collection, and their number will be the dimension. Can I do it in a different, more efficient way? The answer if \yes." As a side result we'll get one of the most important facts of the basic linear algebra.
    [Show full text]
  • Math 395. Tensor Products and Bases Let V and V Be Finite-Dimensional
    Math 395. Tensor products and bases Let V and V 0 be finite-dimensional vector spaces over a field F . Recall that a tensor product of V and V 0 is a pait (T, t) consisting of a vector space T over F and a bilinear pairing t : V × V 0 → T with the following universal property: for any bilinear pairing B : V × V 0 → W to any vector space W over F , there exists a unique linear map L : T → W such that B = L ◦ t. Roughly speaking, t “uniquely linearizes” all bilinear pairings of V and V 0 into arbitrary F -vector spaces. In class it was proved that if (T, t) and (T 0, t0) are two tensor products of V and V 0, then there exists a unique linear isomorphism T ' T 0 carrying t and t0 (and vice-versa). In this sense, the tensor product of V and V 0 (equipped with its “universal” bilinear pairing from V × V 0!) is unique up to unique isomorphism, and so we may speak of “the” tensor product of V and V 0. You must never forget to think about the data of t when you contemplate the tensor product of V and V 0: it is the pair (T, t) and not merely the underlying vector space T that is the focus of interest. In this handout, we review a method of construction of tensor products (there is another method that involved no choices, but is horribly “big”-looking and is needed when considering modules over commutative rings) and we work out some examples related to the construction.
    [Show full text]
  • Concept of a Dyad and Dyadic: Consider Two Vectors a and B Dyad: It Consists of a Pair of Vectors a B for Two Vectors a a N D B
    1/11/2010 CHAPTER 1 Introductory Concepts • Elements of Vector Analysis • Newton’s Laws • Units • The basis of Newtonian Mechanics • D’Alembert’s Principle 1 Science of Mechanics: It is concerned with the motion of material bodies. • Bodies have different scales: Microscropic, macroscopic and astronomic scales. In mechanics - mostly macroscopic bodies are considered. • Speed of motion - serves as another important variable - small and high (approaching speed of light). 2 1 1/11/2010 • In Newtonian mechanics - study motion of bodies much bigger than particles at atomic scale, and moving at relative motions (speeds) much smaller than the speed of light. • Two general approaches: – Vectorial dynamics: uses Newton’s laws to write the equations of motion of a system, motion is described in physical coordinates and their derivatives; – Analytical dynamics: uses energy like quantities to define the equations of motion, uses the generalized coordinates to describe motion. 3 1.1 Vector Analysis: • Scalars, vectors, tensors: – Scalar: It is a quantity expressible by a single real number. Examples include: mass, time, temperature, energy, etc. – Vector: It is a quantity which needs both direction and magnitude for complete specification. – Actually (mathematically), it must also have certain transformation properties. 4 2 1/11/2010 These properties are: vector magnitude remains unchanged under rotation of axes. ex: force, moment of a force, velocity, acceleration, etc. – geometrically, vectors are shown or depicted as directed line segments of proper magnitude and direction. 5 e (unit vector) A A = A e – if we use a coordinate system, we define a basis set (iˆ , ˆj , k ˆ ): we can write A = Axi + Ay j + Azk Z or, we can also use the A three components and Y define X T {A}={Ax,Ay,Az} 6 3 1/11/2010 – The three components Ax , Ay , Az can be used as 3-dimensional vector elements to specify the vector.
    [Show full text]
  • 2.2 Kernel and Range of a Linear Transformation
    2.2 Kernel and Range of a Linear Transformation Performance Criteria: 2. (c) Determine whether a given vector is in the kernel or range of a linear trans- formation. Describe the kernel and range of a linear transformation. (d) Determine whether a transformation is one-to-one; determine whether a transformation is onto. When working with transformations T : Rm → Rn in Math 341, you found that any linear transformation can be represented by multiplication by a matrix. At some point after that you were introduced to the concepts of the null space and column space of a matrix. In this section we present the analogous ideas for general vector spaces. Definition 2.4: Let V and W be vector spaces, and let T : V → W be a transformation. We will call V the domain of T , and W is the codomain of T . Definition 2.5: Let V and W be vector spaces, and let T : V → W be a linear transformation. • The set of all vectors v ∈ V for which T v = 0 is a subspace of V . It is called the kernel of T , And we will denote it by ker(T ). • The set of all vectors w ∈ W such that w = T v for some v ∈ V is called the range of T . It is a subspace of W , and is denoted ran(T ). It is worth making a few comments about the above: • The kernel and range “belong to” the transformation, not the vector spaces V and W . If we had another linear transformation S : V → W , it would most likely have a different kernel and range.
    [Show full text]
  • 23. Kernel, Rank, Range
    23. Kernel, Rank, Range We now study linear transformations in more detail. First, we establish some important vocabulary. The range of a linear transformation f : V ! W is the set of vectors the linear transformation maps to. This set is also often called the image of f, written ran(f) = Im(f) = L(V ) = fL(v)jv 2 V g ⊂ W: The domain of a linear transformation is often called the pre-image of f. We can also talk about the pre-image of any subset of vectors U 2 W : L−1(U) = fv 2 V jL(v) 2 Ug ⊂ V: A linear transformation f is one-to-one if for any x 6= y 2 V , f(x) 6= f(y). In other words, different vector in V always map to different vectors in W . One-to-one transformations are also known as injective transformations. Notice that injectivity is a condition on the pre-image of f. A linear transformation f is onto if for every w 2 W , there exists an x 2 V such that f(x) = w. In other words, every vector in W is the image of some vector in V . An onto transformation is also known as an surjective transformation. Notice that surjectivity is a condition on the image of f. 1 Suppose L : V ! W is not injective. Then we can find v1 6= v2 such that Lv1 = Lv2. Then v1 − v2 6= 0, but L(v1 − v2) = 0: Definition Let L : V ! W be a linear transformation. The set of all vectors v such that Lv = 0W is called the kernel of L: ker L = fv 2 V jLv = 0g: 1 The notions of one-to-one and onto can be generalized to arbitrary functions on sets.
    [Show full text]
  • Review a Basis of a Vector Space 1
    Review • Vectors v1 , , v p are linearly dependent if x1 v1 + x2 v2 + + x pv p = 0, and not all the coefficients are zero. • The columns of A are linearly independent each column of A contains a pivot. 1 1 − 1 • Are the vectors 1 , 2 , 1 independent? 1 3 3 1 1 − 1 1 1 − 1 1 1 − 1 1 2 1 0 1 2 0 1 2 1 3 3 0 2 4 0 0 0 So: no, they are dependent! (Coeff’s x3 = 1 , x2 = − 2, x1 = 3) • Any set of 11 vectors in R10 is linearly dependent. A basis of a vector space Definition 1. A set of vectors { v1 , , v p } in V is a basis of V if • V = span{ v1 , , v p} , and • the vectors v1 , , v p are linearly independent. In other words, { v1 , , vp } in V is a basis of V if and only if every vector w in V can be uniquely expressed as w = c1 v1 + + cpvp. 1 0 0 Example 2. Let e = 0 , e = 1 , e = 0 . 1 2 3 0 0 1 3 Show that { e 1 , e 2 , e 3} is a basis of R . It is called the standard basis. Solution. 3 • Clearly, span{ e 1 , e 2 , e 3} = R . • { e 1 , e 2 , e 3} are independent, because 1 0 0 0 1 0 0 0 1 has a pivot in each column. Definition 3. V is said to have dimension p if it has a basis consisting of p vectors. Armin Straub 1 [email protected] This definition makes sense because if V has a basis of p vectors, then every basis of V has p vectors.
    [Show full text]