9th - CHAPTER 7 Dimension and Structure EXERCISE SET T21 2....

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Unformatted text preview: CHAPTER 7 Dimension and Structure EXERCISE SET T21 2. {a} v1 = in {h} V3 = 5v; — 5v; [1:] v4 = {1191+ 41m + 8?; 6. {a} Any one of the vectors {1,3}, {2.6L (—1,—3} forms a basis for the line I = :1 y = at. {b} Any two of the vectors [1,1,3], {1,—1‘2L {2,0,5} form a. basis for the plate .7; =r1+r«2, y =[1—tg1 Z: 3t1+2t3. 1 2—2 1 3H] “‘3'1‘1'u dhedoed hlt‘ ‘ ‘ nfl ter u rower: eon mm It]. The augmented rnntrnt ofthe eye-temle 2 3 4 3 Lien U l 'D—l III} 1 o o e—gio _ I _ I of this matrix is n 1 fl 1 1:0 . Thus the general solutionis u o 1 o—§.n o o n n nio x= 35+ gen-r 43mm} = s[—31l,[},l.fl:| +n[%1 e1,§,n,1) where —oo -=: 3,: a: oo. The solution spnee is 2-dimensionsl 1with canonical basis {vhvn} where v1=E—3rllfl1llfl}m'd 1'? = {%r_11 grflllj' D2. Yen1 theyr are linearly independent. If we write the 1reetorn in the order Iv" = {utfl'flifllllilva = {Drfltfltlttlll11v2 = [fl‘fll li‘!‘l‘]lvl = t t i t * iirii thEfl,lhECfl-1L‘ifi of the positions of the lending le+ it is clear that none of these vector-e can be written no n lJnesr oomhination of the preoeding ones in the list. P31 1“: 55 '1 thefl {hair}: = kin-x} :I] if and only iin-x = fl:[,h|1fi|:ka}J- 2 ae EXERCISE SET 12 S. [a] An arbitrary vector x = (Ly, 2] can be written an F an + [y -z}n+ [r 'I‘I'iV3+UV-i thus the vectors in. v2, Val and in. open R3. On the other hand, since in; = 2v: +1»;T the vectors in, V2. v1: and v4 are. linearly dependent and do not iorm a home for H3. {h} The vector equation it = [311?], + oat-'2 + cava + e41“ is equivalent to the linear ayetem l 1 l l l D 1 U D EMILE 1 = 2 3 The augmented matrix of this system is GMT-p: “Ml—l and the reduced row eeheloo form of this matrix 3 -1 is U 2 1 l 1 {II I] {1' 1 I] I] I] l TlaL-uoT setting :24 = L a genfll'fll solution of the exertion is given by C1=3,Cg=—1"211{:3=—1—tlc4=1 where—oofltéoo.Thu5 V: 3V1—{1+2E}V2 —{1+E}V3+1Vq for 3113' “ill-IE flf I. For example. mrresponding 15.3. t = Dr :2 _1r and t: _%1 “"3 have v: 3V1-V2 -V3.V=3v1 +192 —v., and v=3VI_ %v3_ %v4_ 16. {a} Sinoe u o n = [1H2] + {1H1} + [3][-—l} = I], the vector u is orthogonal to n. Thus V iswntained in W. {h} The line V is parallel to u = [1,21 —5}. and the plane W has nonnal vector 11 = {312,1}. Since 1.1- n = [13(3}+[2]{2]+{-5}{1}= 2 a! t}. the vector u is not orthogonal to n. Thus if is not contained in W. 18. (a) The vector equation e1{l_. l. l] + m[1.1,fl} + e3[110.[l] = (4,3,1?!) is equivalent to the linear sve- t-E‘III 1 l 1 {:1 a]. l I l] e; = 3 l D [I (:3 fl whieh hes solution c1= n. a. z 3. c3 = 1. Thus {4. in}: = on, 1, 1) + 3(1.1,e}+1{1,e.e} and, by linearity. we have Tl‘iaifll = Ul312.0.1i+ 3(2.1.3.—1i+1i5,*2.l.fl}=(11,1,1u.—3) {b} The vector equation e1{1,1.l}+ C:[l,11fl} + ea{1,fl,tl} = {mt-,e} is equivalent to the linear sys- tem HHI—n l 1 u: 1 [I 1132 = b 0 I] 133 C which has solution :e; = e, I: = h — c. 1:. = n — h. Thus {01:11 a} = £111 1| + _ Elli-1r + {a '— fl! and so. by linearity, we have T{a.h,e} = e[312,[}_.l}l + {h —e]{2,l,3.-—1}+[e—hj{5,—2.l,fl] ={ea—35H.~2u+3t+c.a+2t—3c.-t+2c} 5 —3 l (e) From the formula ehove,urehave[T]= "'3 g l n —1 2 D1. [:1] True. Any set of more than n vectors in R" is linearly.r dependent. [b] True. Any set of less than n. vectors helmet be spanning set for R“. {1:} True. II" every.r vector in R“ can he expressed in exact];r one way as a linear combination of the vectors in S. then S is a basis for R“ and thus must contain exactlyr as vectors. {:1} True. If Ax = I) has infiniter Imam},r attentions1 then detEA} = D and so the roar vectors of A are linearly dependent. {e} True. If V E W {or W E V} and if dimll’} = diln[W], then 1’ = W. Dd. Let A be the matrix having the vectors v1 and v; as its eelumns. Then '11'313'l4‘1i|='l5‘i|12 fl — $1123} - {0063 a - 6052.3] = floosfio + eos2fi‘ and detlifl} 5% I] if and only if easier 5! (1052131 i.e.. if and only if or 75 :hfl + law where it = I]. i1, i=2. For that! 1values of a and :3. the vectors v1 and v; form a heels for R2. P5. Since 13 = {V1,V2,....Vn} is a basis for R“, siren.r 1hector x in R" can be expressed as s linear combination x = c1vl+ t:ng + r - - + ens" for exactly one choice of sealers c1,e3,...,cn. Thus it makes sense to define s transformation T: R“ —> R” by setting Tix} = Ticlvl + ‘2‘“? + ' " + Cnth = cit-'1 + new; + - -- + there 19. “I- It is easy to check that T is linear. For eitsnnpleI it‘x : 2 131.3 and I = div? than i=1 :1 I 11 Ttx +3} = 112.19 + djivJ-j = Z {cj +d3}w, =29“ + Zsjwj- = Tic +rse 1:1 j=l j=1 3i=1 and so T is sdditite. Finally, the transformation T has the |::I1't:I[:.~cI't3:r that TVJ: = wj for each 3' = 1.....n. and it is clear [rem the defining formula that T is Luu'quely determined by this property. EXERCISE SET 13 8. Here W is the line ufintersecticn of the planes at + y + z = El and 2: — y + e = t}. Parametric equations for this line are x = -t, y = I]1 z = t} thus W corresponds to vectors of the form Ll = tit—1,12], 1}. It follows that E sector w = {:1:,y.z} is in WJ- if and only it's: 2 2, Le. if and oral}r ifw is of the fonn w = (135.1"): 111, 0,1) + slit], 11 [It] Parametric equations for this plane are given by s: = r, y = s, z z r. 23. The reduced row echelon form of the matrix A is 1 e I] i —E c 1 e t g R: 1 3 t] ID I] 0 it Thus the vectors r1 = {1,13101 }1 —§), r3 = [D.l.fl,%,§}, and r3 = {t}.fl,l,§. g} form n. basis for the row space of A. We also conclude from an inspection of R that the null space of A {solutions of A1: = (1} consists of vectors of the form 3‘: Sir—i1 _%r_%illul +tt§r"§r—%Ti}+ll Thus the motors n1 2 —fi,—%,e§,l,fl}, :13 = [E,—%, -§1(},1] formshssis for the null space of A. It is teams}.? to check that r; - n}; = I] for all 1', 3'. Thus row{A} end nulI{A]I are orthogonal suhspscte of R“. n l I} I: 32. {a} IfA :10 I} and x= y , then Ax=fl ifanrlonlp ifxzyzfl. Thus thenull spacecfA I] U U z corresponds to points on the z—axis. 0n the other hand, the column space of A consists of all sectors that are of the form H II tr: Chi—n: + H {SDI- + N as: H GflH which corresponds to points on the mgr—plane. {h} The matrix B = U 'l} U E! l 1]] has the specified null space and column spare. I] I] l D1. DIE+ Do. {a} {a} Pulse. This statement is true if and only it tho nonzero town of A are linsaa'lgl-r innhspendnnt1 to.r if there are no additional zero rows in an echelon Form. {1:} True. If E is an slants-titan”r matrix, then E is invertible and an EA; = I] if and mil},r if A): = II. {n} Thus. If A has rank n, then there can he no zero rows in an echelon form for A; thus the reduced row echelon form is the identity matrix. [d] False. For instant-ls1 if m = n and A is invertible, than row{A) = R” and nuiltA] = [e] True. This follows from the fast [Theorem 13.3} that Si is a. subspace of R". {a} Trno. If A is invertible1 then the rows of A and the columns of A each form a basis for R”; thus l'l'IWL‘lJ = oolfA] = H“. [h] False. In Each the opposite is true: If H" is s. snhspsoe of V1 than t”- is a subspace of H”- sinoe ever}! vector that is orthogonal to V 1arill also he orthogonal to W. {-3} False. The specifiod condition implies that rowfsl} I; rowiB] from which it [011m that nul1{A] = msEAJi ; mw[B:JJ' = nulHB) hut in general the inclusion Can he prop-er. is) False. For example A = i“ 1 0 fl and B = [E ‘11] have the same row space but different column spaces. {e} True This is in first true for antr invertible matrix E. The rows of EA are linear combinations of the rows of A; thus it is always true that row{EJl.J E tourist). If E is intros-tibia1 than A = E‘1I[EA] and so we also have roar{A] Q row[EA]. if nuilfifl} is a line through the originr then row{A}I = nul]{A}|J- is the plane through the origin that is perpendicular to that line. If oohffl] is a line through the origin1 than nullEAT} = oolEA'lllJ' is the plane through the origin that is perpendicular to that line. (11} P3. If F is invertible1 then (PM.th = PIAJL} = fl ifand only it Ax. = I]: thus the matrices PA and A have P4. the same null space, and so nuilityljFA} = nullitflAt From Theorem I13. it follows that PA and A have also have the same rot:r spans. Thus rank[PA} = dimfnowiPA}}I = dimtrowmj} = rankL‘l]. From Theorem "(3.4 we have SJ- = spanl[S}J-, and [spanESJIJ'J'L =span{S] since spsn{S] is a subspace. Thus [SJ'F' =[spanlISl1JJ‘ = spanES] ...
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9th - CHAPTER 7 Dimension and Structure EXERCISE SET T21 2....

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