(Solution manual) contemporary linear algebra by howard anton, robert c busby

www.elsolucionario.org CHAPTER Systems of Linear Equations EXERCISE SET 2.1 (a) and {c) are linear (b) is not linear due to the x1x3 term {d) is not' linear due to the x} term (a) a.nd (d) are linear (b) is not linear because of the xyz term (c) is not linear because of the x 3/5 term (a) is linear (b) is linear if k # (c) is linear only if k = (a) is linear (b) is linear jf m =f: (c) is linear only if m = (a), (d), and (c) are solutions; these sets of values satisfy all three equations (b) and (c) are not solutions (b) , (d), and (e) are solutions; these sets of values satisfy all three equations (a) and (c) are not solutions The tluee lines intersect at the point {1, 0) (see figure) The values x equations and this is the unique solution of the system = 1, y = satisfy all three 3.x-3y"' :c The augmented matrix of the system is l ~ ~ ~]· r231 -3 !3 Add -2 times row to row and add -3 times row to row 3: [~ =! !~] Multiply row by - j and add times the new row t.o row 3: [~ ~ ~] From the last row we s~ that the system is redundant (reduces to only two equations) From the second row we see that y =0 and, from back substitution, it follows that x = - 2y = 22 23 Exercise Set 2.1 The three lines not intersect in a common point (see figure) This system has no solution )' The augmented matrix of the system is and the reduced row echelon form of this matrix (details omitted) is: The last row corresponds to the equation = 1, so the system is jnconsistent (a) The solution set of the equation 7x - 5y = can be described parametrically by (for example) solving the equation for x in terms of y and then making y into a parameter This leads to x == 3'*:,5t, y = t , where -oo < t < oo (b ) The solution set of 3x - 5xz + 4x3 = can be described by solving the equation for x1 in terms of xz and x , then making Xz and x into parameters This leads to x = I t 5;- - 5); thus the eigenvalues = v'5 is a singular value of A = [~ ~] are A1 = liJ and A2 = 9; thus u = v'i6 = and Exercise Set 8.6 307 ~ The eigenvalues of A7'A [_ ~ ~] [~ -~] [~ ~]are > = = of mulliplicaty 2); thus the 's ingular values of A are crt = = ('7, ~}[v; ~] = [Jz v'22] are >.1 = and > The eigemalucs of ATA values of A are cr = J4 - [~] = [~] and v2 of A: A• The = /2 ar:d u2 = J2 is > =2 ~ [-J ] [- -·1 = [~] and v ] -4 = [~] = [90 ~1 [l0 OJI _UEVT [v2 v2 16 ] are )q resperLively The = 16 and >.2 = 9, with corresponding smgular values of A are u = and =3 A The t:i~~:envnhti'S eigen\'t.!Liors " o2 = l] We hnvc u ,~ = ,1,A v = ~[-~ -~l [~] = [-~]· r-~ -~J = [_~ -~1 [~ ~] [~ ~] = UEVT c 16 ] = [ of A r A = [' ] ( =[ I ~] 2.J L1 ct = [- ~] and v::: arf' \ = 6-t und r~sp~>cth·ely \~• = -1, with correspomling umt The :;mgular values of A arc cr = and ![~ ~] [~] and u + = [-jg] -!., • L;\v = c ~ (; -~] [~] = [~],and This res ults in t.he following singular value decomposition \Vp IHl,·e Ut : = is an eigeJJ.value J5 and cr2 :: JS and > l tl v$ - v5 L.4v (1, = ![ 6) [•-};] = _JA' vr> [-~] l .;:; This results in the fo llowing singular value decomposition: The ('tgcnvn.lucs of AT A eigenwcLors we have u Vt - =; [~] Av1 I [ anrl = ~ [33 3 3 ] [ ] 3 3 "2 J ] = (18 ~] 18 I,., = r-~] [~] = 7i are > = 36 and >.2 = 0, with corresponding unit rPspccth·ely The only smgular value of A is [+1 72J Thf' vector is an or thonormal bas1s for R , e.g., u = (- ~] mu~t =6, and u2 must be chosen so thaL {u1 , u 2} This results in the following singular value decomposition A = Ut [33 33] = [~ -~] [6 OJ [ ~ ~] =UEVT 72 72 0 - 72 72 www.elsolucionario.org 308 Chapter Th~ eigenvalues of AT A = [-~ -~ -~] [=~ ~1 = [_: -:] are > = 18 and A2 = 0, with corre- -2 [-~) spondtl'g uniL etgenvectors v1 = A ~u the and v2 = [~] = ,/18 = 3,/2, and we have u = ;, Av, = veo~ts u respectively The only singular value of ~ =! _:] [- ~] = [ Ul Wemus',choose and u so that { u 1, u 2, u } is an o•thono•mal has;s fo• R , e.g., u, = [i]•nd u , = [ -:] This results io the foUowU.g singuiOI" value deoomposWon' A= t -1] [ ~ ~] [-2 2] [ ~ - = -2 ~ _.£ ! 3 Note The singular value decomposition is not unique It depends on the choice of the (extended) orthonormal basis for R This is just one possibility = [=: 10 Theeigenvalu"',"f ATA ve1 is a unit eigenvedo3 = The vecto A2 = 2, with corresponding I •tlspectively The singular values or A are u = J3 and Exercise Set 8.6 309 , singular value decomposition: I [ l [' fi I A= l = -1 ? unit eigenvectors v = [ o2 '- u, = -73 '+s] [J3~ ~l[~ ~] 76 I I 72 -"76 72 =UEVT ·, I [~ ~ :1 [: i] = r;: ::1 are ~ = 64 and ~, = 4, with conespondlng 12 The eigenvalues or AT A = and ~ = We have ~] UJ [!l and v = [_ ~~] resp,.ctively The sinf?:nlar valtlf'.s fA are~=~ [~ ~][f] ~s~J~]J u~ = ~ [~ ~1 [_ ~] = ~-~]-:e k and vf> .! .~ ~~ :iS choose oJS so that {u, , u,, u,} Is an octhonocma\-basis roc R3 This cesults m the rollowmg singular value decornposttton A =[~ ~] =[~ ·I ] -::-sI v5 Using the smgular vrdtH' decomposition A = 0][80]['2 1] = l/EVT ll /, _"J 0 ll $ J.J~2s [s0 [0~ 6)composition nf A: 14 Using the singular value decowpo.q = 25 and ~ = 0, with corresponding unit eigen- respectively The only singular value of AT is [i] = [1] ,1 T = [3 •J = 0'! =5, and This results in the singular value decomposition III [5 OJ [-! ll = UEVT www.elsolucionario.org 314 Chapter ThP corresponding reduced singular value decomposition is AT = (3 4] = fll [5] [l ~) = UlEl vr and from this we obtam (AT>+ = (f) "•Ei"•ur = [i] UJ 111 = [:~] = (A+>r ~ [ 1~ ~!] are At= 215 and A2 = 0, with corresponcling unit and v2 = [-~] respectively The only singular value of A+ is a = t• The eigenvalues of (A+)T A+ = eigenvectors v and we have ~ u - [i] ct , A+v• = 5[,\ [!] = fs) [1) This results in the singular value decompo- sition The corresponding reduced singular value decomposition is 2~] A+= [ = u.E vt [1) [t) (~ ~) = and from thts we obtam (a) [l ~] [~ -~il [l ;] -30 : 11- = = (b) A:\ = [~ ]_ 30 -1] [i ~][~ r~ ~j [~ ~] ~ [I~ ~I] 11 - 30 ~1] [~ - Jo s =A ~1] [! I '2!1 3o - ygI l -.t 13 15 l =[5 (c) AA' = [; i] [~ -1 ~1] [~: ~ -~]E z 1] [; i] ~ [~ ~~ (d ) (e) The eigenvalues of AAT = unol eog•nvocto.s is symmetric, thus (AA +)T -= AA+ -1$ is 'ymme.,ic; thus [~2 1~4 3:] are A1 v, ~ },;; [·~]· v, ~ W = 15, >.2 = *' Hl· A)T ~A+ A '2, and AJ and , = f,o = 0, nl with corresponding The singular values of Exercise Set 8.7 315 *' ~2 = Ji Setting u , = :, AT v, = ~-jg, [: ~ ~] [':] = u, = :/rv, = ~;+ [: ~ :J [-!] = f,; [_:].we have AT a" u, = Jj5 and kJ-u - 't"' 1~1 726 *' 1:1 and vr1 and from this it foUows t.hat (C) = - The eignn-.;tlues of (A")rAi- [ i& - 1~\l 1; :15] ~~ ~ - 2~5 46 -m 40 43 arc ns conespondmg uniteigenw>CIO