{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Lecture11and12filled - University of Toronto at Scarborough...

Info icon This preview shows pages 1–14. Sign up to view the full content.

View Full Document Right Arrow Icon
Image of page 1

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 2
Image of page 3

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 4
Image of page 5

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 6
Image of page 7

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 8
Image of page 9

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 10
Image of page 11

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 12
Image of page 13

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 14
This is the end of the preview. Sign up to access the rest of the document.

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 ...
View Full Document

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern