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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 westm piwts that one wimmvxs in H withom. PM?”
ﬁthe. soithh 17v Pvt=1? is in Terms OF Fammetm “AT1'5 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 panti 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 ._
WWI7 o o i 1 0 784—5
0 o o 0 0 X151
X3=’7.5
W: n: XFVrt’ﬂS
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 \ Mfg6‘ :{ 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 Mfr1?.
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; MﬂﬁkAW Pitamen?
9"??? is o. solwr/wn to “:1; . WW; is a. Sam/ion to A‘X‘z‘ﬁ mist
kab e x“ __,
:7 EHFAFWJFFM“ [$235350
31}: E huHsPoce. 0*? A and tiff—t? DEFINITION 0.6. Let V1,Vg,   ,vk 6 R”. Then ine vow bin W D n .
i) If mvl + rgvg +l  +01;ka = 0 has exactly one solution then we say that the vectors V1,V2,   ‘ ,vk are linearly indegendent, ii) If rlvl + Y'ng + + new 2 0 has more than one solution, then we
say that the vectors V],V2,    ,vk are Einearlx degeneient. Note: Two vectors in R” are linearly independent if they are
nonzero and nonparallel. In Noemi—3 indqzemdm sets, no?5 EXAMPLE 0.7. Determine ifthe vectors v1 = [2, 1, —1],V2 = [M2,3, 7'],v3 =
[2, 1, 5} are linearly independent or linearly dependent. So‘u'tiow L61? vii/”I Jr “,7; 'i‘ YLV: 2T; Owd 4?in VI, Y1K;
Y:[1,—\,G + Y»E‘1;%1]+Y1E2,LS]SE\7) o) o]
[ZYVZYH’Lm«hﬁvwm“YW‘lYﬁSYslqjo, o, o] . " 2h ~1Y1+7xh 7517
“Y! +3“ “PYE :3. 0
“"Yi «PTV‘L‘Q'SﬁT—ZO 1 J}, 'L O I 0 2 0
“4 77 i 0 ”9'9""? 0 i 1 0
.4 '1 S  0 O 0 O 0
“2:5
Y1:“Y3:“$ 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’: TiaWe N.
W) Hag, raw—x :7 max7‘ 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
(WIF Til57 E Wszﬁf/sIZSwt, 3H; Sgt], V;[ab, (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, rsr 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‘ wO 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 ~ ﬁatSP8! V—F—dﬂéis abasis JFww “A
W is a subs‘mca 0? 9R“ O EUUCFYOWL) Lw'é’ﬁraﬁvrnﬁihwf “(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‘JYtﬁi' YLW’V‘"“{*WB;
SwﬁNyL W‘sf'ﬁ’l’ 5151+m'4' Suﬁ:
ﬁzﬁvﬁ:("faw$1)ﬁ+UI‘S’IOW+~+(IN—"SHE: a4
T515 \meowb ﬁndePe/v‘dm
*6 has only} one so‘wu‘on Mn Yi"51==0, Y'u'SwfO, m, Yk’SkJO
W Y1$SH Y1$S1,~~> YIN15k 317» is. a mum/«e, lin. WWW; 0‘? ‘51, ‘53,”) 5": EXAMPLE 0.16. Find a basis for 1 4 3 m1
0 l 1 9 _
Wzsp 1 4 mwg, 3 WW3, W1 “W4,
0 1 2 1 __ So \u‘tion SFIW’ WM”; 00:; w;)=~ W
Th” does NOT ““80"“ W2m2WB W9 W5 aver)» box/51$ WW
W6 need £0 dam/k \m wade? ii\\ :Lryuso ...
[WWW]: 0 ~ : 0: Wm 5:35, 0*2"‘ 00900.» 2W}: HOPMC) "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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 Spring '10
 Sophie
 Linear Algebra, Algebra

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