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~ 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 313, 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 32 ’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 nonzero 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 nonredundant), and r? is nonredundant 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)+‘  "iCm (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 nonredundant 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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