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section3.3

# section3.3 - '1 2 3 2 1 1 2 0 5[i 3 iΒ"I h 3 7[I(l 1...

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Unformatted text preview: '1 2 3 2 1 1 2 0 5 [i 3 iΒ» "I h 3 7 [I (l 1 βl U , . 1 2 4 1 2 O [l {l 0 1 We Will enmlate Exercise 23 to find the 2 4 β3 l 2 U [l [l 1') (1 2 1 U U (l 5 (l ,, l l (l l 2 (l 5 I I 1 o s n Kernel: (l 0 1 "1 (l 1 then {J J. *1 I] l) 0 (l [J 1 D l [l U l [J U G U (i U 0 0 (l 2 1' βl β6 l 3 1 . g β So a basis of ker{A) is 0 j 71. and a basis of iIn[A) is 3 1 J 7 d D 7.1 l 4 2 9 D 0 2 9 u 11 X 4 matrix A with the given vectors as its columns. β7e find that 1β1βef(A) 2 L1, 27. Form a. _ _ ,1 7β . so that the vectors do indeed form a bass of R , by bmnmary 3.3,9. 1'1 1 I 1131 U . T . a . 1; (1:2 (1 an; 1 32. We need to find all vectors :1: m 111 such that - _ :: 0 and β :- G. 173 *- 1 33,3 2 '14 1 I j 3 ,. . . :1: .1"-Β», is :L β 0 _ . 1.1113 amounts to solving the system % 1 9 " 1 1 , which 111 turn 1 #7173 ,, 14' *2 , . t 1 (l 771 1 amounts to finding the kernel oi 10 1 2 3]. 1 *1 . , . . β2 β3 Usmg kyle Numbers: we ο¬nd the bas1s 1 1 U 0 1 β such that 17-51β = U, 01'1β1131't β232 + ' β ' β1 β"5ββ = 0β 35. we need to ο¬nd all vectors :3 in R7 A f the vector 17. These vectors form a hyperplane m Rβ where the o; are the components 0 β (see Exercise 33), so that the dimension of the space is n e 1. 38. at The rank of a 3 X 5 matrix A is 0,1,2, 01β 3, so that (lim(l~ier(A)) = 5 iiβanMA) is 273,4, or 5. b The rank of a T X 4 matrix A is at most 4, so that dim(im(A)) -: rank(A) is 0,133} or 4β 1 U #1 r2 . 0 1 2 3 52,. Melt/l) = 0 0 0 0 0 U 0 0 By Exercises 50 and 51a, {1 O 7- 1 - β2], [U 1 2 31 is a basis of the row space of A. 61. a, Note that rank(B) g 2, so that dim(ker(B)) = 5 e'iranldB) 2 3 and d1m(ker(AB)) 2 3 since ker(B) g lqer(AB). Since ker(AB) is an subspace of R5, dim(l<er(AB)) could be 1 0 3,4β or 5. It is easy to give an example for each case; for example, if A z 3 (l) and 0 0 1 0 U 0 0 410000 ._01000β.. J B _ [O 1 O U 0], then AB β 0 0 O U 0 and chm(ke1(AB)) i 3. 0 0 U 0 U 13, Since dim(im(AB)) : 5 β dim(ker(AB)), the possible values of dim(im(AB)) are 0,1, and 2; by part a. ...
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