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Lecture11and12filled

Lecture11and12filled - University of Toronto at Scarborough...

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Unformatted text preview: University of Toronto at Scarborough Department of Computer 8.: Mathematical Sciences MAT A23 Winter 2008 Lecture 11 and 12 Section 1.6 Sophie Chrysostomou THEOREM 0.1. Every consistent system with fewer equation than unknowns has inﬁnitely many solutions. Let M’s-”F be a. msistem} sjgﬁem (it hose: {was one, PROOF. SOlUﬁ/ioni wH:_h less quadrant Man unknownsm This means that A is mm with ““4“- li: £Alb]~£ch] and H is in PREP or REEF :7H has art west-m piwts that one wimmvxs in H withom. PM?” ﬁ-the. soithh 17v Pvt-=1? is in Terms OF Fammetm “AT-1'5 has iniihttdﬂ W‘Wj so [Wong DEFENI’I‘ION 0.2. A homogeneous linear system is of the form anal + 0:12:62 + —1— 0,1an = 0 (121531 + (12293:; + ~i~ agnstn : 0 a1172,1333 "i" am2w2 "l‘ + amnmn m 0 or it is the system Ax 2: 0 where A is the coefficient matrix. Note: i) every homogeneous system is consistent. CH in KEFfi) If AN H and H has a pivot in every column, then Ax— — 0 has the trivial soiution only 0 is. ﬁne 0N3 solution to AY“O AYZ'D’ has Sole in pan-ti color, is? is o Sqluﬁbn called the trivial solution. DEFENITION 0.3. Let A be an m >< n matrix. i)The nullspace of A is the set of all solutions to AX = 0. in a. set nota— tion, the nulispace of A is N, the subset of R” given by: N:{XE}R”AX=G} ii) The row sgace of A is the span of the row vetors of A. The row space of A is a subset of ER”. iii) The column sgace of A is the span of the column vectors of A. The column space of A is a subset of W”. 2 m6 -2 4 EXAMPLE 0.4. Find the solstice to Ax = 0, if A := [ -1 3 3 2 ] (Find {ht W1 3 7 10 1.4, .1 q. ORE); l “77 “I 7.0 “1%) «Hi5 «~17, o [M6314 3 3‘2, 0%50 Ol40m--~7 no no “17>7WOWZ‘WOOGW’OW’4VW 00000 I J; 0 LP ‘7 ._ WWI-7 o o i 1 0 784—5 0 o o 0 0 X151 X3=’7.5 W: n: XFVrt’ﬂ-S PiwL (Jim 74; 7ft’45 1‘7 ”‘0"? X1 .- ‘t— .4 3 —t 5 X1, .. ~15 at O “T ,7, J 6% m 5 0 ' 77 «s a 5 .- X5 H \ Mfg-6‘ :{ 3ft; 1,55%}— SFW([5]’[&]) vaious ’7? o 1 1n we; L} }L x g 1, 4r 2 s .4 '5 'I I9 "g 0 i nullspoce. mtA % New, T523; ‘15 a Fawn/aim SO‘MHOn‘t/o Mfr-1?. 0 THEOREM 0.5. Let if AX m b be a linear system with a particular solution 13. Then: i) If h is in the nulispsce of A, then 1:) + h is also a solution to Ax = b. ii) If q is any solution to Ax = b, then q = p + h for some h in the ____, nullspece of A. i .V 19 0‘ smut/onto M‘z‘E’Je A =3 Why’ﬁ'a nunspovz. 0’? Pi 4- ATE; Mﬂﬁk-AW Pit-amen? 9-"??? is o. solwr/wn to “:1; . WW; is a. Sam/ion to A‘X‘z‘ﬁ mist kab e x“ __, :7 EHFAFWJFFM“ [\$235350 3-1}: E huHsPoce. 0*? A and tiff—t? DEFINITION 0.6. Let V1,Vg, - -- ,vk 6 R”. Then ine vow bin W D n . i) If mvl + rgvg +l- -- +01;ka = 0 has exactly one solution then we say that the vectors V1,V2, - - ‘ ,vk are linearly indegendent, ii) If rlvl + Y'ng + + new 2 0 has more than one solution, then we say that the vectors V],V2, - - - ,vk are Einearlx degeneient. Note: Two vectors in R” are linearly independent if they are nonzero and nonparallel. In Noemi—3 indqzemdm sets, no?5 EXAMPLE 0.7. Determine ifthe vectors v1 = [2, -1, —1],V2 = [M2,3, 7'],v3 = [2, 1, 5} are linearly independent or linearly dependent. So‘u'tiow L61? vii/”I Jr “,7; 'i‘ YLV: 2T; Owd 4?in VI, Y1K;- Y:[1,—-\,-G + Y»E‘1;%1]+Y1E2,LS]SE\7) o) o] [ZYVZYH’Lm-«hﬁvwm“YW‘lYﬁSYslqjo, o, o] . " 2h ~1Y1+7xh 7517 “Y! +3“ “PYE :3. 0 “"Yi «PTV‘L‘Q'SﬁT—ZO 1 J}, 'L O I 0 2- 0 “4 77 i 0 ”9'9""? 0 i 1 0 .4 '1 S - 0 O 0- O 0 “2:5 Y1:“Y3:“\$ ﬁz—erx—ZS .....-u w.» w“ (’15)V;«-S~Vz+ 3V3“? , SEW’L Wit 0. unique solution ~‘v“\'/'T,V‘£,Vi owe, mt weary independent, EXeroiée: Vetevmine {JP V1:[1,*|,"£1,K=[1,1231,V5=[7’"’}1 0““ [mega/135 immendQ/MZ. HOW TO FIND IF THE VECTORS v1, v2, - ‘ — ,vk ES LINEARLY DEPENDENT OR LINEARLY INDEPENDENT. @Form P‘ wiﬂn W‘uwms W,V;,~.,W in ﬁhis order @Pmd H Sc A~H,H is. in away MEF. (91%: H how a piwt In wag wluwn \$7V7,Vi,...,'\7§ owe 1080ka Independent_ wheywigg, 171/031, M, Vi ave linemb dapandem,‘ DEFINITION 0.8. If W c: R“ and i) W is nonempty, (doswe under ii)ifu7V6W,thenu«+~v Em, and mddiﬂon) (dowwe under iii)ifuewandTER,thenruEI/V, scoJ Ox W‘UW) _ then we say that W 18 a subsgace of R“. EXAMPLE 0.9. If A is an m X n matrix and N is the nuHSpace of A, then N is a subspace of R". N: {74 Eﬁn§‘ P6516 {42(1)} 3§§ since. A636 N is honaemm (%)1'F 1on E N527 Aﬁﬁﬁ, ATP—<6“ A(ﬂ4~V‘)= Mﬁ+ Maﬁa—"0’: Tia-We N. W) Hag, raw—x :7 max-7‘ and. AW)=HP\W=W‘V’ YMEN («LLWLCWBhow Nis 0L subsgace 0F WK. EXAMPLE 0.10. If W m 3P(V1,V2, - -- ,vm) where V1,V2, - - >vm e R”, then W is a subspace of R”. 66w EXAMPLE 0.11. If W = { [m1,m2,m3,x4] E R4lx1= 2:3 _ .734, mg m :33 + 3:4 }, ciatermine if W is a subspace of R4. W:{[.S”‘t, 5+1, 5, 11‘ smeﬂzﬁ So\u't\0n=(‘v'1[0, ”2., I, I] EUU , 521:: l W is. homemw (WI-F Til-57 E Wszﬁf/sIZSwt, 3-H; Sgt], V;[a-b, (M 250,, b] m7ﬁ+7= [SAtJr a;— b, s+1+arr b) 5+0L, “(3+ b1 :[(s+a.)»(m+w, (mm—raw), sthvrbl “1i+‘\7 EW (Mil? ﬁ=ES-—t, 5+1, Sgt/1, VERA, w?» = I new, rs-r th, Y3; H91 6 W WEDAW) W is 0» subspace. 0’? W". EXAMPLE 0.12. If W m {[mz,w2,\$3] 6 R31 221 + x3 : mg + 3}, determine if W is a subspace of R3. C Fa) SQ) DEFINITION 0.13. Let W be a subspace of 1R“. If B : {b1,b2,‘ - - ,bk} is a subset of W, then we say that 15’ is a basis for W if every vector in W can be written uniguelx as a linear combination of the vectors in B 17 The plural for the word basis is ” bases . wu- £13 EXAMPLE 0.14. Leth 3;] m,yER Wisasubspace OfR4. —y 1 1 0 Is 13’: i , (1) , ”é abesisforW'? ~1 e 1 '2- I I o I I o ‘11::[3’ EW “72:0 ::\+7- 0 ”'3 '3 :5 i "’3 37+? ‘3 I ”"7 ‘ 0 l ”I o I 1 I 67 o I I o B...‘ "‘sx+)'«-°+%"‘ 7‘ w-O I +7‘ I £3 0 ( 5 I 3 t o "3 “l U I “I o I Bismtabasbﬁww. THEOREM 0.15. 13 = {131,132, - ~ - ,bk} be a basis for the subspace W of IR“ {22? if and only if B m {131,132, - -~ ,bk} is a linearly independent set of vectors and W = 8p(b:,b2r ~ ﬁat-SP8! V—F—dﬂéis abasis JFww “A W is a subs‘mca 0? 9R“ O EUUCFYOWL) Lw'é’ﬁraﬁvrnﬁi-hwf “(5: is uniowa. .‘,Y‘:Y),:-»:’YI¢=O \$259“; owe “\$60ng indﬁff/nO‘W.® H: ﬂew,t\nen‘(ﬁ is» 01. uniowe. linear wwbihmorw o¥‘5,..,.,r,’ﬁ_ .-, UU=5VH5J (2—D ' ~ VFW)? E is Hmong" 1'? Raw , since: 362(9):: W,%m “TL‘JYtﬁ-i' YLW’V‘"“{*WB; SwﬁNyL W‘sf'ﬁ’l’ 5151+m'4' Suﬁ: ﬁzﬁvﬁ:("faw\$1)ﬁ+UI-‘S’IOW+---~+(IN—"SHE: a4 T515 \meowb ﬁndePe/v‘dm *6 has only} one so‘wu‘on Mn Yi"51==0, Y'u'Swf-O, m, Yk’Sk-JO W Y1\$SH Y1\$S1,~~> YIN-15k 3-17» is. a mum/«e, lin. WWW; 0‘? ‘51-, ‘53,”) 5": EXAMPLE 0.16. Find a basis for 1 4 3 m1 0 l 1 9 _ Wzsp 1 4 mwg, 3 WW3, W1 “W4, 0 1 2 1 __ So \u‘tion SFIW’ WM”; 00:; w;)=~ W Th” does NOT ““80"“ W|2m2WB W9 W5 aver)» box/51\$ WW W6 need £0 dam/k \m wade? ii\\ :Lryuso ... [WWW]: 0 ~ : 0: Wm 5:35, 0*2"‘ 00900.»- 2W}: HOP-MC) "9“: 61%;“RWrr'Z/LW3’MZW: W2:W([email protected]”1WE Wse\$ptﬁ,Wi,W&) W§::%:W§ WWCSFLW)W§UW¢) ((9 W: SP[W1”"W3)¢5?{W,W1,W}) @419 Wimﬁvﬁ‘meabmsiskaw 10 TO FIND A BASIS FOR W = 3p(W1,W2, - -- ,wk) W1 W2 Wk: . t l i ii) Row reduce A to get H in a. REF or in a REEF. So A N H, A and H are row equivalent. '1) Construct the matrix A = iii) A basis for W is the set 13 which consist only of vectors wi, such that wi E B if and only if the ookumn i of H has a pivot. OR iii) A basis for W consists oniy of all the columns of A, corresponding to the pivotal columns of H. TO FIND A BASIS FOR THE COLUMN SPACE OF AN m x n MATRIX A 11 THEOREM 0.17. Any two bases for a subspace W of ER”, contain the same number of vectors ' PROOF. (page 139 of text book). DEFINITION 0.18 Let W be a subspace of R”. The number of elements in a basis for W is called the dimension of W, denoted by dim(W). THEOREM 0.19. Existence and Determination of Base i) Every subspace W of R” has a basis and dim(W) g n. ii) Every linearly independent set of vectors in R” can be enlarged , if nec— essary, to become e basis for 1%.“. iii) If W is a subspace of ER” and dim(W) = k, then a) every independent set of k vectors in W is a basis for W, and ’0) every set of k vectors that spans W, is a basis for W. 12 THEOREM 0.20 If V is a subspace (3ka with dimﬂf) :2 n. va1,v2, . _ . ,vm E V are tinearty independent 23> m g n. PROOF. (By contradiction) Let us assume that m > n. If dim(V) = n then 3 15’ C V st. 13 is a basis for V and B has n vectors in it. ' Let B = {331, . . . , ion}. It follows that we can write each of the v1- ’s as a unique linear combination of the b1- ’3: V1. = T11b1 “I“ ' ' ' "5“ Tinbn V2 = Tglbl + ' ' ' +72an (*) Tmibl + ‘ ' ' + Tmnbn <5 3 II The numbers Tjj ’s above are unique fixed real numbers. Now let us form the homogeneous system of n equations in the in un- knowns 2:1, .932, \$3, ..., am where the coefﬁcients are the just—discovered scalars Tij ’3 , T111131 +T21\$2 + ‘ ‘ ‘ ‘1' Tmlﬂfm Z 0 712562 + T223212 't— ' ' ' “i” ngﬂlm = 0 (**) Sr'lnml "3" 7.21232 "I“ ' ‘ ’ ”é” Tmnmm = 0 There are are n homogeneous equations in m unknowns and n < m. Thus, there are nontrivial solutions to this system. Let a nontrivial solution of this system he: 31,32, , -- ,sm. So some of these si ’s are nonzero. Now consider the tinear combination 33v; ~l- - - - + smvm. Using (2:) above we get: 81V1 + ‘ ‘ ' + Sme = (33?"11 + 32T21+ 83T31.+' " + Sme1)b1 + (31?"12 + .5ng2 “3*" 83T32 + ' ' ' “i“ Sme2)b2 + (Slrin "i” 327‘27; + S3'r3n ' ' ' + Smrmn)bn 13 Since the coeﬁcient of each ba- in (a: as at) is given by (ink) to be 0 then we have: 81v1+---+smvm=0 Of course this is contradictions since some of the 33- ’s are nonzero and the vectors v1,v2, - -- ,vm are linearly independent. This shows that our assumption is wrong and therefore m S n. 14 ...
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