DOCSLIB.ORG

Row Rank Equals Column Rank

Total Page：16

File Type：pdf, Size：1020Kb

Download full-text PDF Read full-text
Row Rank Equals Column Rank 316 MATHEMATICSMAGAZINE Proof Referto FIGURE3. We findthat there are twelve pairsof triangles,each pair of which has a commonside or a commonangle (or two congruentangles one from each triangleof the pair).Applying P1 andP2 on these triangleswe have AM Thus, if we let IA = a, IC = c, IM = m, IN = n, we get a-m which simplifiesto 1 Hencea butterflyinscribed in a quadrilateralsatisfies the samerelation (1) as a butterfly inscribedin a circle.Equivalently, the conclusionof the theoremindicates that the ratio of the ratios, (AM/IM)/(CN/IN),is the same as the ratioIA/IC, or thatthe harmonic meanof IC andIM equalsthe harmonicmean of IA andIN. In eithercase, if IC = IA, we haveIM = IN therebythe analogof the usualbutterfly theorem for quadrilaterals. Acknowledgment. The authorwould like to thank the referees for their helpful suggestions, and Joseph Kung for preparationof the article. REFERENCE 1. Leon Bankoff, The metamorphosisof the butterflyproblem, this MAGAZINE60 (1987), 195-210. Row RankEquals Column Rank WILLIAM P. WARDLAW U. S. Naval Academy Annapolis, MD 21402-5000 [email protected] Dedicatedto GeorgeMackiw, a goodfriend and an excellentmathematical expositor The purpose of this note is to present a short (perhaps shortest?) proof that the row rank of a matrix is equal to its column rank. The proof is elementary and accessi- ble to students in a beginning linear algebra course. It requires only the definition of Mathematical Association of America is collaborating with JSTOR to digitize, preserve, and extend access to Mathematics Magazine ® www.jstor.org VOL. 78, NO. 4, OCTOBER 2005 317 matrixmultiplication and the fact that a minimalspanning set is a basis. It differsin approachfrom proofs given in textbooksas well as from some interestingproofs in MAAjournals [1, 2]. And, unlikethe latter,this proofis validover any field of scalars. But first,recall thatif the m x n matrixA = BC is a productof the m x r matrix B andthe r x n matrixC, thenit follows fromthe definitionof matrixmultiplication thatthe ith row of A is a linearcombination of the r rows of C with coefficientsfrom the ith row of B, andthe jth columnof A is a linearcombination of the r columnsof B with coefficientsfrom the jth columnof C. (If you have troubleunderstanding this or the nextparagraph, you shouldconstruct several examples of smallmatrix products, say, a 3 x 2 times a 2 x 3 matrix,etc., with smallinteger as well as symbolicentries.) On the otherhand, if any collectionof r row vectorsc1, spansthe row space of A, an r x n matrixC can be formedby takingthese vectorsas its rows.Then the ith row of A is a linearcombination of the rows of C, say bil1- birCr. This means A = BC, where B = (bij) is the m x r matrix whose ith row, bir), is formed from the coefficients giving the ith row of A as a linearcombination of the r rows of C. Similarly,if any r columnvectors span the columnspace of A, and B is the m x r matrixformed by these columns,then the r x n matrixC formedfrom the appropriate coefficientssatisfies A = BC. Now the four sentenceproof. THEOREM.If A is an m x n matrix, then the row rank of A is equal to the column rank of A. Proof If A = 0, thenthe row andcolumn rank of A areboth 0; otherwise,let r be the smallestpositive integer such thatthere is an m x r matrixB andan r x n matrix C satisfyingA = BC. Thus the r rows of C form a minimalspanning set of the row space of A and the r columnsof B form a minimalspanning set of the columnspace of A. Hence,row andcolumn ranks are both r. E Severalother properties of the rankof a matrixover a field arealso very easy to ob- tain.The factorizationsA = ImA= AInshow that r < m andr < n, whichproves that the rankof A is less thanor equalto the numberof rows andthe numberof columns of A. Since A = BC implies AT = CT BT, the transpose clearly has the same rank as the original matrix. Since A = BC and D - EF implies AD = B(CD) = (AE)F, the rankof AD mustbe less thanor equalto the rankof A andto the rankof D. These conceptssuggest the followingdefinition [5, p. 123]: DEFINITION.Let R be a commutativering with identityand let A be an m x n matrix over R. Then the spanning rank of A is 0 if A = 0 and otherwise is the smallest positive integer r such that there is an m x r matrix B and an r x n matrix C satisfying A = BC. This definitionis one way of extendingthe notionof rankto matricesover commu- tativerings. Even if the ringhas no identity,it can be embeddedin a ringwith identity so thatthe definitioncan be used.Care must be takenin consideringrank over commu- tative rings, because several different extensions of the definitions over a field can give different results over rings, even though they all give the standardconcept of rank over a field. Nonetheless, if the above definition is used, matrices over rings automatically have row rank equal to column rank, have rank less than or equal to the number of rows and the number of columns, the rank of the transpose is equal to the rank of the matrix, and the rank of a product is less than or equal to the rank of either factor. Another application of the spanning rank, first used by the author in a problem [3] and later a Note [5] in the MAGAZINE, is the proof that a matrix over a commutative 318 MATHEMATICSMAGAZINE ring with spanningrank r satisfiesa polynomialequation of degreeat most r + 1. For if A = BC is an n x n matrix of spanning rank r, then D = CB is an r x r matrix with characteristic polynomial fD (x) = det(xI - D) of degree r and fD(D) = 0 fol- lows fromthe Cayley-HamiltonTheorem. (The authorhas shown [4] thatthe Cayley- HamiltonTheorem holds for matricesover commutativerings.) Thus thereis a poly- nomial m(x) of smallest positive degree such that m(D) = 0. Then p(x) = xm(x) is a polynomial of degree 1 such that p(A) = Am(A) = BCm(BC) = Bm(CB)C = Bm(D)C = 0. REFERENCES 1. H. Liebeck, A proof of the equalityof column and row rankof a matrix,Amer Math.Monthly 73 (1966), 1114. 2. G. Mackiw, A note on the equality of column and row rankof a matrix,this MAGAZINE68 (1995), 285-286. 3. W. Wardlaw,problem 1179, this MAGAZINE56 (1983), 326, and solution 1179, this MAGAZINE57 (1984), 303. 4. - A transferdevice for matrixtheorems, this MAGAZINE59 (1986), 30-33. 5. - Minimumand characteristicpolynomials of low-rankmatrices, this MAGAZINE68 (1995), 122-127. A Modern Approachto a Medieval Problem AWANI KUMA R, Director Lucknow Zoological Garden Lucknow 226001 INDIA [email protected] The following problem from Lilavati [1], a mathematicaltreatise written by Bhaskaracharya,a 12th-centuryIndian mathematicianand astronomer,deserves a modemapproach: A snake'shole is at the foot of pillar,nine cubitshigh, anda peacockis perched on its summit.Seeing a snakeat the distanceof thricethe pillargliding towards his hole, he pouncesobliquely upon him. Say quicklyat how manycubits from the snake'shole they meet,both proceeding an equaldistance. Since both proceedan equal distance,it is reasonableto assumethat their speeds are equal.Readers are invitedto solve this problembefore proceeding. Assumingthat the peacockflies alongthe hypotenuseof a right-angledtriangle and knowsthe PythagoreanTheorem, it will grabthe snakeat a distanceof 12 cubitsfrom the pillar.Practically, however, such a thing does not happen.Why shoulda peacock know-a priori-to fly along the hypotenuseof a right-angledtriangle having a base of 12 cubits?A morepeacock-like behavior would be to keepan eagle eye on the snake andchange its directionat everyinstant, always aiming toward the snake. This type of pursuit problem has a history of over five hundred years. However, this particularproblem is a bit different from most. Instead of the prey running away from the predator, here prey and the predator are moving toward each other. Even so, the results are startling. Although the reader may have seen similar problems, I offer a general analysis. We assume that the snake and the peacock move at different, but constant, speeds: the snake in a straight line toward its hole and the peacock along a curve, changing its direction at every instant so as to be flying directly toward the snake. .
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]
  • 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]
  • 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]
  • CME292: Advanced MATLAB for Scientific Computing
    CME292: Advanced MATLAB for Scientific Computing Homework #2 NLA & Opt Due: Tuesday, April 28, 2015 Instructions This problem set will be combined with Homework #3. For this combined problem set, 2 out of the 6 problems are required. You are free to choose the problems you complete. Before completing problem set, please see HomeworkInstructions on Coursework for general homework instructions, grading policy, and number of required problems per problem set. Problem 1 A matrix A P Rmˆn can be compressed using the low-rank approximation property of the SVD. The approximation algorithm is given in Algorithm1. Algorithm 1 Low-Rank Approximation using SVD Input: A P Rmˆn of rank r and approximation rank k (k ¤ r) mˆn Output: Ak P R , optimal rank k approximation to A 1: Compute (thin) SVD of A “ UΣVT , where U P Rmˆr, Σ P Rrˆr, V P Rnˆr T 2: Ak “ Up:; 1 : kqΣp1 : k; 1 : kqVp:; 1 : kq Notice the low-rank approximation, Ak, and the original matrix, A, are the same size. To actually achieve compression, the truncated singular factors, Up:; 1 : kq P Rmˆk, Σp1 : k; 1 : kq P Rkˆk, Vp:; 1 : kq P Rnˆk, should be stored. As Σ is diagonal, the required storage is pm`n`1qk doubles. The original matrix requires mn storing mn doubles. Therefore, the compression is useful only if k ă m`n`1 . Recall from lecture that the SVD is among the most expensive matrix factorizations in numerical linear algebra. Many SVD approximations have been developed; a particularly interesting one from [1] is given in Algorithm2.
    [Show full text]
  • Some Key Facts About Transpose
    Some key facts about transpose Let A be an m × n matrix. Then AT is the matrix which switches the rows and columns of A. For example 0 1 T 1 2 1 01 5 3 41 5 7 3 2 7 0 9 = B C @ A B3 0 2C 1 3 2 6 @ A 4 9 6 We have the following useful identities: (AT )T = A (A + B)T = AT + BT (kA)T = kAT Transpose Facts 1 (AB)T = BT AT (AT )−1 = (A−1)T ~v · ~w = ~vT ~w A deeper fact is that Rank(A) = Rank(AT ): Transpose Fact 2 Remember that Rank(B) is dim(Im(B)), and we compute Rank as the number of leading ones in the row reduced form. Recall that ~u and ~v are perpendicular if and only if ~u · ~v = 0. The word orthogonal is a synonym for perpendicular. n ? n If V is a subspace of R , then V is the set of those vectors in R which are perpendicular to every vector in V . V ? is called the orthogonal complement to V ; I'll often pronounce it \Vee perp" for short. ? n You can (and should!) check that V is a subspace of R . It is geometrically intuitive that dim V ? = n − dim V Transpose Fact 3 and that (V ?)? = V: Transpose Fact 4 We will prove both of these facts later in this note. In this note we will also show that Ker(A) = Im(AT )? Im(A) = Ker(AT )? Transpose Fact 5 As is often the case in math, the best order to state results is not the best order to prove them.
    [Show full text]
  • Electronic Reprint Wedge Reversion Antisymmetry and 41 Types of Physical Quantities in Arbitrary Dimensions Iucr Journals
    electronic reprint ISSN: 2053-2733 journals.iucr.org/a Wedge reversion antisymmetry and 41 types of physical quantities in arbitrary dimensions Venkatraman Gopalan Acta Cryst. (2020). A76, 318–327 IUCr Journals CRYSTALLOGRAPHY JOURNALS ONLINE Copyright c International Union of Crystallography Author(s) of this article may load this reprint on their own web site or institutional repository provided that this cover page is retained. Republication of this article or its storage in electronic databases other than as specified above is not permitted without prior permission in writing from the IUCr. For further information see https://journals.iucr.org/services/authorrights.html Acta Cryst. (2020). A76, 318–327 Venkatraman Gopalan · Wedge reversion antisymmetry research papers Wedge reversion antisymmetry and 41 types of physical quantities in arbitrary dimensions ISSN 2053-2733 Venkatraman Gopalan* Department of Materials Science and Engineering, Department of Physics, and Department of Engineering Science and Mechanics, Pennsylvania State University, University Park, PA 16802, USA. *Correspondence e-mail: [email protected] Received 18 October 2019 Accepted 16 February 2020 It is shown that there are 41 types of multivectors representing physical quantities in non-relativistic physics in arbitrary dimensions within the formalism of Clifford algebra. The classification is based on the action of three Edited by S. J. L. Billinge, Columbia University, symmetry operations on a general multivector: spatial inversion, 1, time- USA reversal, 10, and a third that is introduced here, namely wedge reversion, 1†.Itis shown that the traits of ‘axiality’ and ‘chirality’ are not good bases for extending Keywords: multivectors; wedge reversion antisymmetry; Clifford algebra. the classification of multivectors into arbitrary dimensions, and that introducing 1† would allow for such a classification.
    [Show full text]
  • Spacetime Algebra As a Powerful Tool for Electromagnetism
    Spacetime algebra as a powerful tool for electromagnetism Justin Dressela,b, Konstantin Y. Bliokhb,c, Franco Norib,d aDepartment of Electrical and Computer Engineering, University of California, Riverside, CA 92521, USA bCenter for Emergent Matter Science (CEMS), RIKEN, Wako-shi, Saitama, 351-0198, Japan cInterdisciplinary Theoretical Science Research Group (iTHES), RIKEN, Wako-shi, Saitama, 351-0198, Japan dPhysics Department, University of Michigan, Ann Arbor, MI 48109-1040, USA Abstract We present a comprehensive introduction to spacetime algebra that emphasizes its prac- ticality and power as a tool for the study of electromagnetism. We carefully develop this natural (Clifford) algebra of the Minkowski spacetime geometry, with a particular focus on its intrinsic (and often overlooked) complex structure. Notably, the scalar imaginary that appears throughout the electromagnetic theory properly corresponds to the unit 4-volume of spacetime itself, and thus has physical meaning. The electric and magnetic fields are combined into a single complex and frame-independent bivector field, which generalizes the Riemann-Silberstein complex vector that has recently resurfaced in stud- ies of the single photon wavefunction. The complex structure of spacetime also underpins the emergence of electromagnetic waves, circular polarizations, the normal variables for canonical quantization, the distinction between electric and magnetic charge, complex spinor representations of Lorentz transformations, and the dual (electric-magnetic field exchange) symmetry that produces helicity conservation in vacuum fields. This latter symmetry manifests as an arbitrary global phase of the complex field, motivating the use of a complex vector potential, along with an associated transverse and gauge-invariant bivector potential, as well as complex (bivector and scalar) Hertz potentials.
    [Show full text]
  • Methods for Finding Bases
    Methods for Finding Bases 1 Bases for the subspaces of a matrix Row-reduction methods can be used to find bases. Let us now look at an example illustrating how to obtain bases for the row space, null space, and column space of a matrix A. To begin, we look at an example, the matrix A on the left below. If we row reduce A, the result is U on the right. 1 1 2 0 1 0 3 −2 A = 2 4 2 4 ⇐⇒ U = 0 1 −1 2 . (1) 2 1 5 −2 00 0 0 Let the rows of A be denoted by rj, j = 1, 2, 3, and the columns of A by ak, k = 1, 2, 3, 4. Similarly, ρj denotes the rows of U. (We will not need the columns of U.) 1.1 Row space The row spaces of A and U are identical. This is because elementary row operations preserve the span of the rows and are themselves reversible op- erations. Let’s see in detail how this works for A. The row operations we used to row reduce A are these. step 1: R2 = R2 − 2R1 = r2 − 2r1 R3 = R3 − 2R1 = r3 − 2r1 1 step 2: R3 = R3 + 2 R2 = 0 1 1 step 3: R2 = 2 R2 = 2 r2 − r1 1 step 4: R1 = R1 − R2 = 2r1 − 2 r2 Inspecting these row operations shows that the rows of U satisfy 1 1 ρ = 2r − r ρ = r − r ρ = 0. 1 1 2 2 2 2 2 1 3 It’s not hard to run the row operations backwards to get the rows of A in terms of those of U.
    [Show full text]
  • Math 4326 Fall 2018 Linear Transformations and the Rank
    Math 4326 Linear Transformations Fall 2018 and the Rank-Nullity Theorem In these notes, I will present everything we know so far about linear transformations. This material comes from sections 1.7, 1.8, 4.2, 4.5 in the book, and supplemental stuff that I talk about in class. The order of this material is slightly different from the order I used in class. We will start with the initial definition. Definition 1 A function T : V ! W from a vector space V to a vector space W is called a linear transformation if T (u + v) = T (u) + T (v); T (cu) = cT (u); for all u; v 2 V and all real c. The following facts will be used so frequently, they should be memorized as if they too were part of the definition. Lemma 1 If T : V ! W is a linear transformation, then T (0) = 0 and T (c1v1 + c2v2 + ··· + cnvn) = c1T (v1) + ··· + cnT (vn): Associated with each linear transformation are two vector spaces: Ker(T ) = fv 2 V j T (v) = 0g and Range(T ) = fT (v) j v 2 V g = fw 2 W j w = T (v) for some vg: You should be able to verify that these two sets are, indeed, subspaces of the appropriate spaces. Make sure you know where each of these subspaces \lives." Ker(T) is a subspace of V and Range(T) is a subspace of W . If I asked you for a basis for the Range of T and you listed vectors from V instead of W , there would be a considerable penalty on an exam.
    [Show full text]