HW03

# HW03 - 1 4 W a ‘ 1 R2 ~ 18 By Fact 3.1.3 im(A span ink...

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Unformatted text preview: 1 4 W a ‘ 1 . . R2 ' ~ 18. By Fact 3.1.3, im(A) : span , ink) «— span X3} (a lme m ) 24. '1m(T) is the plane as + 2y + 32 == 0, and ker(T) is the line perpendicular to this plane, 1 spanned by the vector 2 (compare with Examples 5 and 9). 3 7 ‘ 32‘ By Fact 31-3, A = 6 does the job. There are many other correct answers: any nonzero 5 7 3 x 71 matrix A whose column vectors are scalar multiples of 6 O 5 49. Ifffandwareinker(T),thenT(q7+qﬁ :T v +T -. “ ~ ~__ w. w a m. ker(T) as well. ) (d) (w) " 0 + 0 " 0v *0 that ’U + 'w IS 111 If 17 is in ker(T) and k is an arbitrar m .. ~ - y scalar, then T A. z a z a = ‘4 . in ker(T) as well. ( U) “101) ho 0’ 8,0 that [W ls 3-2 ’4.san". "‘ -—"‘ “’ p (m, ..,vm) - 11n[vl ’Um] is a subspace of R", by Fact 3 2 2 t8» 6. a. Yes! F0] 20. O l is redundant, simply because it is the zero vector. :01 1 0 is our ﬁrst non-zero vector, and thus, is not redundant. 0 3 1 0 2 3 0 and is redundant. 0 0 0 1 1 is not a multiple of 0 and is not redundant. O 0 4 1 O 5 = 4 0 + 5 1 and is redundant. 0 0 O 6 1 0 Similarly, 7 z 6 0 + 7 1 and is also redundant. 0 0 0 O 1 0 However, by inspection, 0 is not a linear combination of 0 and 1 , meaning 1 0 O that this last vector is not redundant. Thus, the seven vectors are linearly dependent. 6 6 1 3 1 3 6 _’ 24. 5 is redundant, because 5 = 3 1 + 1 2 . Thus, 3 1 + 1 2 — 1 5 z 0 4 4 1 1 d 1 1 4 3 1 3 b and 1 is in the kernel of 1 2 5 ——1 1 1 4 1 l 1 28. The three column vectors are linearly independent, since rref 1 2 5 = 13. 1 3 7 Therefore, the three columns form a basis of iin(A)(= 1R3): 1 1 1 1 , 2 , 5 1 3 7 Another sensible choice for a basis of ini(A) is 61,52, 63. {’21:} t" Yes; the vectors are linearly independent. The vectors in the list 271, . . . , '17," are linearly independent (and therefore non-redundant), and r? is non-redundant since it fails to be . in the span of ~17] , . . . , 17m. ‘ 42} Vie can use the hint and form the dot product of Hi and both sides of the relation 01771 + ' ‘ ' ‘l‘ 6177i + ‘ ' ' + Cmﬁm : 63 (C1171 +' ‘ '+CiUi+' ' ‘+Cm’l:m)"l7i SO that C} -’l7i)+~ - -+Cii(’l—Ili~'l7i)+‘ - "i-Cm (an‘ﬁg) 2 0. Since '0;- is perpendicular to all the other '53, we will have 17¢ - Q = 0 whenever j # 2'; since 17.1 is a unit vector, we will have 271 ~01- : 1. Therefore, the equation above simpliﬁes to Ci = 0. Since this reasoning applies to all i = 1, . . . ,m, we have only the trivial relation among the vectors 171,172, . . . , 17m, so that these vectors are linearly independent, as claimed. 50. The veriﬁcation of the three properties listed in Deﬁnition 3.2.1 is straightforward. Alternatively, we can choose a basis 17], . . . ,177, of V and a basis 2171, . . . ,u7q of W (see Exercise 38a) and show that V + W : spanwh...,'r7p,zU],...,rﬁq) (compare with Exercise 4). . Indeed, if17+u7 is in V+W, then '5' is a linear combination of 171 , . . . ,EP and ~15 is a linear combination of 131, . . . , rﬁq, so that 'U—HD is a linear combination 01.171, . . . (171,, 215] , . . . ,diq. Conversely, iii: is in span(171, . . . 47],, 1171 , . . . , u’iq), then 1" = (c1171 +' - -+cp'Up)+ ((11231 + ---+ dquiq), so that it is in V + W. If V and W are distinct lines in R3 (spanned by 1U and 217, respectively), then V + W is the plane spanned by 17 and “LU. 2 1 0 O 0 ~l 0 5 - 1 O 1 ~2 O «1 O 1 .— 0 ~1 0 O 0 1 5 0 ’ and 0 0 1 5 0 O O 0 0 1 0 0 O 0 1 2 ~l 1 O 1 O 0 Then 0 , 5 is a basis of the kernel and 0 , l , O is a basis of the O ~1 O 0 1 O 0 4/!» “(I [4 8 1 1 6 1 2 0 0 0 _ 3 6 1 2 5 _ 0 0 1 0 0 ‘ _ .w I f 24.11efi2 4 1 9 10 w 0 O 0 1 O . Heie on: kernel is the span of onl) one 11 2 3 2 0 0 0 0 0 1 vector: 31 4 1 1 6 0 1'1 b'“ ftl ‘ ‘ fA‘ 3 1 2 5 0 ,wnea dblSO 1e1n1ageo IS 2 , 1 , 9 , 10 2 0 0 1 3 28. Form a 4 x 4 matrix A with the given vectors as its columns. The matrix A reduces to O 2 0 3 1 4 0 COP-‘0 1 0 0 0 16—29 This matrix can be reduced further to I4 if (and only if) k '— 29 ;£ 0, that is, if k # ‘29. By Summary 3.3.9, the four given vectors form a basis of R4 unless A: z 29. 131 1 II?) 0 32. We need to ﬁnd all vectors :3 in R4 such that \$2 - O :2 0 and \$2 1 = O. 133 -1 173 2 x4 1 1:4 3 This amounts to solving the system m1 - \$3 + \$4 = 0 which in turn 1:2 + 21.3 + 33:4 = 0 ’ amounts to ﬁnding the kernel of 1 0 #1 1 O 1 2 3 ' 1 —1 . _ . —2 —3 Usnig Kyle Numbers, we ﬁnd the basxs 1 , 0 0 1 1 1 46. Use Exercise 45 with 171 2 g , 172 = , and LUZ z e. 101 L: 1,2,3,4 4 8 1 1 1 1 0 0 0 1 0 2 0 0 “Z 2 4 0 1 O O 0 1 —1 O 0 l N = 4 owrref360010 00010_% 1 1 0 O . . b . , . 2 4 1 O Picking the non-redundant columns gives the basw 3 , 6 , 0 , 1 L - 8 0 O 60. 16 O to W1“ 11 W p 1 7 *— 111 ~ 2 1111(14) lb the p18,] 11 C e 10 ect SO Iallk(11) d111](1 . {31‘ £5??? ...
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