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**Unformatted text preview: **Wk 37%
E“ 20%, Sa{u+;ov1$ 23: 1(12962H411c‘é‘1ﬂ0
at: tenwizzﬁ
3'2; Hihloﬂitiﬂﬂlﬁl (3.3 #3010 1“
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“MM J 20. Not necessarin true; (A ~ B)(A + B) = A2 + AB - BA —— 32 % A2 —- B2 if AB 7é BA.
24. True; (In + A)(In + AW) 2 13+ A + A4 + A1414: 21,1 + A + A4. [2 EMS 3H: 2
4
[2 M3 M: Z hMCKDU‘ ‘ 324. We car; write ABiAB)“1 = A(B(AB)'1) = In and (.43)“1AB : ((AB}—1A)B 2 LE, By 2.4.9, A and B are invertible. 25 35 2-1 ——1—1 [35 15...; 1
10. Solving the system A3? : 5 we ﬁnd that ker(A} 2 span -3 .
0 18. By Fact 3.1.3, im(A) : span , [142]) : span {a line in R2). 24. im(T) is the plane :5 + 23; + 32 = 0, and ker(T) is the line perpendicular to this plaixe,
1
spanned by the vector [2] (compare with Examples 5 and 9).
3 4i).A=(AB)B—1:({AB>-1)_;B_1:[I 3]”[1 2]:[_5 ﬁn 2}:[ 4 5]. [Mini 2‘“ ' °P$ g w Soweto“ 7 .
32. By Fact 3.1.3. A = 6 does the job. There are many other correct answers: any nonzero
5
7
3 x n matrix A whose column vectors are scalar multiples of 6
5 34. To describe a subset of R3 as a kernel means to describe it as an intersection of planes
‘ (think about it). By inspection, the given line is the intersection of the planes z + y z 0 and
_ 25:: + z 2 0.
L ‘ a: 3+
This means that the line is the kernel of the linear transformation T y z: [ 22: +32]
- - z
.ﬁ‘om R3 to R2. . 27ml is a subspace of R”, by Fact 3.2.2. o The zero vector is in V H W, since 5 is in both V and W. o If f and ﬂare in Vﬂ W, then both 55 and g] are in V, so that 5+ y] is in V as well, since V
is a subspace of lit". Likewise, a? + 37 is in W, so that a? + 3} is in V n W. e If 55 is in V 0 W and it is an arbitrary scalar, then its? is in both V and IV, since they are
subspaces of R“. Therefore, k5 is in V ﬂ W. b. No; asa counterexample consider V = span(é§) and W : spade?) in R2. 26;. l
l
l
l S: , ,
G] is redundant, simply because it is the zero vector.
0 GM is our ﬁrst nonzero vector, and thus, is not redundant. CD 03 and is redundant. 6:)
H
0:)
moi—a O 1
is not a multiple of 0 and is not redundant. 0
1l “l
r = 4 OJ + 5 1 i and is redundant.
0 0i Qéw‘i‘, a a y... l
l
l in CDC?! g.1 Lou‘lta—‘ 42. 28. NM GAR ’ 37$ 3 ’ €ol¢9m$
6 1 0
Similarly, 7 2 6 0 + 7 1 and is also redundant.
O 0 O
0 1 0
However, by inspection, 0 is not a linear combination of 0 and 1 , meaning
1 O 0 that this last vector is not redundant. Thus, the seven vectors are linearly dependent. 5 6 1 3 1 3 5
5 is redundant, because 5 = 3 1 +1 2 . Thus, 3 1 + 1 2 — 1 5 r 6
4 4 1 l l 1 4
1 l 1
The three column vectors are linearly independent, since rref 1 2 5 2: 13
1 3 7 Therefore, the three columns form a basis of i1n(A)(= 1R3): 1
1, 2, 5
1 :3 7 Another sensiblechoice for a basis of im(A) is 51, 632, 53. We can use the hint and form the dot product of if, and both sides of the relation
c1271 +---+c,-s;~+-~+cma7m =5; (6151+ - --%~C{£3}+- ' ~+cmﬂm)-5i xii-23¢, so that 61(61-6i)+< v ~-§-c,{2§‘,-5,-}+- — -+Cm{27msqji}:
0. Since 271' is perpendicular to all the other 27}, we will have 272‘ 473‘ z: 0 Whenever j % 2'; since 17',» is a unit vector, we will have 17, ~17,- = 1. Therefore, the equation above simpliﬁes Since this reasoning appiies to all 2‘ = i, . , . ,m, we have only the triviai relation among
the vectors 211,122, . . . , am, so that these vectors are linearly independent, as claimed.
a b d i‘ 1 0 0
52. If a, c and f are nonzero, then rref G C e 2 Q 1 0 - and the three vectors are
0 0 f [0 e 1 , 7
0 G 0 O 0 ...

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- Fall '05
- HUI
- Math, Linear Algebra, Algebra