linear algebra and its applications pearson

This title-specific access card provides access to the Lay/Lay/McDonald, Linear Algebra and Its Applications 6/e accompanying MyLab course ONLY. He has also served as a visiting professor at the University of Amsterdam, the Free University in Amsterdam, and the University of Kaiserslautern, Germany. (175.8MB), Download Accessible PowerPoint Lecture Slides (application/zip) This is a great introduction to linear algebra for students and those with practical uses for it. She has been an active member of the International Linear Algebra Society and the Association for Women in Mathematics throughout her career and has also been a member of the Canadian Mathematical Society, the American Mathematical Society, the Mathematical Association of America, and the Society for Industrial and Applied Mathematics. Unable to add item to List. Linear Algebra and Its Applications (3rd Edition), Student Study Guide for Linear Algebra and Its Applications, Introduction to Linear Algebra, Fifth Edition (Gilbert Strang), Linear Algebra Done Right (Undergraduate Texts in Mathematics), A First Course in Differential Equations with Modeling Applications, Vector Mechanics for Engineers: Statics, 11th Edition, Materials Science and Engineering: An Introduction, Computer Systems: A Programmer's Perspective. 13 Hint: Take functions f and g in C Œ0; , and fix an integer m Write the Fourier coefficient of f C g that involves cos mt , and write the Fourier coefficient that involves sin mt m > 0/ 15 [M] The cubic curve is the graph of g.t / D :2685 C 3:6095t C 5:8576t :0477t The velocity at t D 4:5 seconds is g 4:5/ D 53:4 ft=sec This is about 7% faster than the estimate obtained in Exercise 13 in Section 6.6 Chapter Supplementary Exercises, page 392 a g m s F T T F b h n T T F c i o T F F d j p F T T e k q F T T f l r T F F Hint: If fv1 ; v2 g is an orthonormal set and x D c1 v1 C c2 v2 , then the vectors c1 v1 and c2 v2 are orthogonal, and kxk2 D kc1 v1 C c2 v2 k2 D kc1 v1 k2 C kc2 v2 k2 D jc1 jkv1 k/2 C jc2 jkv2 k/2 D jc1 j2 C jc2 j2 (Explain why.) and Ph.D. from the University of California at Los Angeles. I figured that, they are trying to keep the printing costs down by offering optional material online, but I assumed for free or at least covered in the outrageous $143 cost of this thin textbook. Linear Algebra and Its Applications, 5th Edition, ©2016 He has been elected by the university students to membership in Alpha Lambda Delta National Scholastic Honor Society and Golden Key National Honor Society. Upon his return to the States in 1998, he joined the mathematics faculty at Lee University (Tennessee) and has been there ever since. MyLab Math is the world’s leading online tutorial, and assessment program designed to help you learn and succeed in your mathematics course. Check with the seller prior to purchase. In order to navigate out of this carousel please use your heading shortcut key to navigate to the next or previous heading. Instructors seem to agree that certain concepts (such as linear independence, spanning, subspace, vector space, and linear transformations) are not easily understood and require time to assimilate. Linear Algebra and Its Applications (Subscription) David C. Lay holds a B.A. She has been an active member of the International Linear Algebra Society and the Association for Women in Mathematics throughout her career and has also been a member of the Canadian Mathematical Society, the American Mathematical Society, the Mathematical Association of America, and the Society for Industrial and Applied Mathematics. If you're a seller, Fulfillment by Amazon can help you grow your business. Buyer beware: e-book is nearly impossible to read in the Kindle Reader. Lay, Lay & McDonald Judi J. McDonald joins the authorship team after working closely with David on the fourth edition. Steven J. Leon is a Chancellor Professor Emeritus at the University of Massachusetts Dartmouth. Pearson published this exclusive edition for the benefit of students outside the United States and Canada. Instead, our system considers things like how recent a review is and if the reviewer bought the item on Amazon. Learn more about the program. in Mathematics from the University of Alberta, and an M.A. We work hard to protect your security and privacy. This change in format supports the authors’ hallmark — using modern practical application to make key concepts tangible and demonstrating how mathematics is used in the real world. Well written book which you can learn from yourself, Reviewed in the United States on June 15, 2017. Please try again. 21 The right singular vector v1 is an eigenvector for the largest eigenvalue of AT A By Theorem in Section 7.3, the largest eigenvalue, , is the maximum of xT AT A/x over all unit vectors orthogonal to v1 Since xT AT A/x D jjAxjj2 , the square root of , which is the second largest eigenvalue, is the maximum of jjAxjj over all unit vector orthogonal to v1 23 Hint: Use a column–row expansion of U †/V T 25 Hint: Consider the SVD for the standard matrix of T —say, A D U †V T D U †V Let B D fv1 ; : : : ; g and C D fu1 ; : : : ; um g be bases constructed from the columns of V and U , respectively Compute the matrix for T relative to B and C , as in Section 5.4 To this, you must show that V vj D ej , the j th column of In :57 :65 :42 :27 :63 :24 :68 :29 7 27 [M] :07 :63 :53 :56 :34 :29 :73 :51 16:46 0 0 12:16 0 07 0 4:87 05 0 4:31 :10 :61 :21 :52 :55 :39 :29 :84 :14 :19 7 :74 :27 :07 :38 :49 7 :41 :50 :45 :23 :58 :36 :48 :19 :72 :29 29 [M] 25.9343, 16.7554, 11.2917, 1.0785, 00037793; = D 68;622 C XN D P T XN w X1 T By definition XN w D P D T That is, Y1 C C YN D 0, so the Yk are in mean-deviation form b Hint: Because the Xj are in mean-deviation form, the covariance matrix of the Xj is 1/ X1 1=.N XN X1 XN T Compute the covariance matrix of the Yj , using part (a) O1 13 If B D X SD D N 1 N O N , then X BB T D N X 1 N O kX O Tk D X O1 X N On X N X Xk OT X :1 : : O TN X 7 M/.Xk M /T Chapter Supplementary Exercises, page 434 a T g F m T b h n F T F c i o T F T d j p F F T e k q F F F f l F F If rank A D r , then dim Nul A D n r , by the Rank Theorem So is an eigenvalue of multiplicity n r Hence, of the n terms in the spectral decomposition of A, exactly n r are zero The remaining r terms (corresponding to the nonzero eigenvalues) are all rank matrices, as mentioned in the discussion of the spectral decomposition If Av D v for some nonzero , then v D Av D A v/, which shows that v is a linear combination of the columns of A SECOND REVISED PAGES Section 8.2 Hint: If A D RTR, where R is invertible, then A is positive definite, by Exercise 25 in Section 7.2 Conversely, suppose that A is positive definite Then by Exercise 26 in Section 7.2, A D B TB for some positive definite matrix B Explain why B admits a QR factorization, and use it to create the Cholesky factorization of A If A is m n and x is in Rn , then xTATAx D Ax/T Ax/ D kAxk2 Thus ATA is positive semidefinite By Exercise 22 in Section 6.5, rank ATA D rank A 11 Hint: Write an SVD of A in the form A D U †V T D PQ, where P D U †U T and Q D UV T Show that P is symmetric and has the same eigenvalues as † Explain why Q is an orthogonal matrix 13 a If b D Ax, then xC D AC b D AC Ax By Exercise 12(a), xC is the orthogonal projection of x onto Row A b From (a) and then Exercise 12(c), AxC D A.AC Ax/ D AAC A/x D Ax D b c Since xC is the orthogonal projection onto Row A, the Pythagorean Theorem shows that kuk2 D kxC k2 C ku xC k2 Part (c) follows immediately 3 14 13 13 :7 6 :7 14 13 13 7 C 7 7 7, xO D 15 [M] A D :8 40 4 :8 75 12 6 :6 Ä A The reduced echelon form of is the same as the xT reduced echelon form of A, except for an extra row of zeros So adding scalar multiples of the rows of A to xT can produce the zero vector, which shows that xT is in Row A 3 17 607 7 7 Basis for Nul A: 6 7, 05 415 0 a p1 Span S , but p1 … aff S b p2 Span S , and p2 aff S c p3 … Span S , so p3 … aff S Ä Ä v1 D and v2 D Other answers are possible 11 See the Study Guide 13 Span fv2 v1 ; v3 v1 g is a plane if and only if fv2 v1 ; v3 v1 g is linearly independent Suppose c2 and c3 satisfy c2 v2 v1 / C c3 v3 v1 / D Show that this implies c2 D c3 D 15 Let S D fx W Ax D bg To show that S is affine, it suffices to show that S is a flat, by Theorem Let W D fx W Ax D 0g Then W is a subspace of Rn , by Theorem in Section 4.2 (or Theorem 12 in Section 2.8) Since S D W C p, where p satisfies Ap D b, by Theorem in Section 1.5, S is a translate of W , and hence S is a flat 17 A suitable set consists of any three vectors that are not collinear and have as their third entry If is their third entry, they lie in the plane ´ D If the vectors are not collinear, their affine hull cannot be a line, so it must be the plane 19 If p; q f S/, then there exist r; s S such that f r/ D p and f s/ D q Given any t R, we must show that z D t/p C t q is in f S/ Now use definitions of p and q, and the fact that f is linear The complete proof is presented in the Study Guide 21 Since B is affine, Theorem implies that B contains all affine combinations of points of B Hence B contains all affine combinations of points of A That is, aff A B 23 Since A A [ B/, it follows from Exercise 22 that aff A aff A [ B/ Similarly, aff B aff A [ B/, so Œaff A [ aff B aff A [ B/ 25 To show that D E \ F , show that D E and D The complete proof is presented in the Study Guide Section 8.1, page 444 Some possible answers: y D 2v1 1:5v2 C :5v3 , y D 2v1 2v3 C v4 , y D 2v1 C 3v2 7v3 C 3v4 5 a p1 D 3b1 to See the Study Guide b3 aff S since the coefficients sum b p2 D 2b1 C 0b2 C b3 … aff S since the coefficients not sum to c p3 D b1 C 2b2 C 0b3 aff S since the coefficients sum to 3v3 D The set is affinely independent If the points are called v1 , v2 , v3 , and v4 , then fv1 ; v2 ; v3 g is a basis for R3 and v4 D 16v1 C 5v2 3v3 , but the weights in the linear combination not sum to y D 3v1 C 2v2 C 2v3 The weights sum to 1, so this is an affine sum b2 F Section 8.2, page 454 Affinely dependent and 2v1 C v2 Chapter A51 4v1 C 5v2 4v3 C 3v4 D The barycentric coordinates are 2; 4; 1/ 11 When a set of five points is translated by subtracting, say, the first point, the new set of four points must be linearly dependent, by Theorem in Section 1.7, because the four points are in R3 By Theorem 5, the original set of five points is affinely dependent SECOND REVISED PAGES A52 Answers to Odd-Numbered Exercises 13 If fv1 ; v2 g is affinely dependent, then there exist c1 and c2 , not both zero, such that c1 C c2 D and c1 v1 C c2 v2 D Show that this implies v1 D v2 For the converse, suppose v1 D v2 and select specific c1 and c2 that show their affine dependence The details are in the Study Guide Ä Ä 15 a The vectors v2 v1 D and v3 v1 D are 2 not multiples and hence are linearly independent By Theorem 5, S is affinely independent b p1 $ ; ; , p2 $ 0; 12 ; 12 , p3 $ 148 ; 58 ; 18 , 8 p4 $ 68 ; 58 ; 78 , p5 $ 14 ; 18 ; 58 c p6 is ; ; C/, p7 is 0; C; /, and p8 is C; C; / 17 Suppose S D fb1 ; : : : ; bk g is an affinely independent set Then equation (7) has a solution, because p is in aff S Hence equation (8) has a solution By Theorem 5, the homogeneous forms of the points in S are linearly independent Thus (8) has a unique solution Then (7) also has a unique solution, because (8) encodes both equations that appear in (7) The following argument mimics the proof of Theorem in Section 4.4 If S D fb1 ; : : : ; bk g is an affinely independent set, then scalars c1 ; : : : ; ck exist that satisfy (7), by definition of aff S Suppose x also has the representation x D d1 b1 C C dk bk and x D c1 d1 /b1 C 25 The intersection point is3x.4/ D 3 5:6 :1 C :6 C :5 D 6:0 : 3:4 It is not inside the triangle Section 8.3, page 461 See the Study Guide None are in conv S p1 D 16 v1 C 13 v2 C 23 v3 C 16 v4 , so p1 … conv S p2 D 13 v1 C 13 v2 C 16 v3 C 16 v4 , so p2 conv S a The barycentric coordinates of p1 , p2 , p3 , and p4 are, respectively, 13 ; 16 ; 12 , 0; 12 ; 12 , 12 ; 14 ; 34 , and ; ; 14 b p3 and p4 are outside conv T p1 is inside conv T p2 is on the edge v2 v3 of conv T p1 and p3 are outside the tetrahedron conv S p2 is on the face containing the vertices v2 , v3 , and v4 p4 is inside conv S p5 is on the edge between v1 and v3 d1 C C dk D (7a) 11 See the Study Guide C ck dk /bk (7b) 13 If p, q f S/, then there exist r, s S such that f r/ D p and f s/ D q The goal is to show that the line segment y D t/p C t q, for Ä t Ä 1, is in f S/ Use the linearity of f and the convexity of S to show that y D f w/ for some w in S This will show that y is in f S/ and that f S/ is convex for scalars d1 ; : : : ; dk Then subtraction produces the equation 0Dx the denominator is twice the area of 4abc This proves the formula for r The other formulas are proved using Cramer’s rule for s and t The weights in (7b) sum to because the c ’s and the d ’s separately sum to This is impossible, unless each weight in (8) is 0, because S is an affinely independent set This proves that ci D di for i D 1; : : : ; k 19 If fp1 ; p2 ; p3 g is an affinely dependent set, then there exist scalars c1 , c2 , and c3 , not all zero, such that c1 p1 C c2 p2 C c3 p3 D and c1 C c2 C c3 D Now use the linearity of f Ä Ä Ä a1 b1 c 21 Let a D ,bD , and c D Then a2 b2 c a1 b1 c1 det Œ aQ bQ cQ  D det a2 b2 c2 D 1 a1 a2 det b1 b2 5, by the transpose property of the c1 c2 determinant (Theorem in Section 3.2) By Exercise 30 in Section 3.3, this determinant equals times the area of the triangle with vertices at a, b, and c r Q then Cramer’s rule gives 23 If Œ aQ bQ cQ 4 s D p, t r D det Œ pQ bQ cQ = det Œ aQ bQ cQ  By Exercise 21, the numerator of this quotient is twice the area of 4pbc, and 15 p D 16 v1 C 12 v2 C 13 v4 and p D 12 v1 C 16 v2 C 13 v3 17 Suppose A B , where B is convex Then, since B is convex, Theorem implies that B contains all convex combinations of points of B Hence B contains all convex combinations of points of A That is, conv A B 19 a Use Exercise 18 to show that conv A and conv B are both subsets of conv A [ B/ This will imply that their union is also a subset of conv A [ B/ b One possibility is to let A be two adjacent corners of a square and let B be the other two corners Then what is conv A/ [ conv B/, and what is conv A [ B/?