Algebra lineal david c lay solucionario y

Solucionario Algebra Lineal David C. Libro de Algebra Lineal y sus Aplicaciones David C. Libro de Calculo Varias Variables James Stewart 6t. Solucionario Raymond Chang 10ma edicion. Al-Zaytoonah University of Jordan P.O.Box 130 Amman 11733 Jordan Telephone: 00962-6-422-6-4291511 Fax: 00962-6-4291432. Email: president@zuj.edu.jo. Student Inquiries استفسارات الطلاب: registration@zuj.edu.jo. Shed the societal and cultural narratives holding you back and let step-by-step Linear Algebra and Its Applications textbook solutions reorient your old paradigms. NOW is the time to make today the first day of the rest of your life.

Solucionario Algebra Lineal David C Lay 3ra Edicion

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Algebra Lineal David C Lay Solucionario En

335
6.1 SOLUTIONS
Notes: The first half of this section is computational and is easily learned. The second half concerns the concepts of orthogonality and orthogonal complements, which are essential for later work. Theorem 3 is an important general fact, but is needed only for Supplementary Exercise 13 at the end of the chapter and in Section 7.4. The optional material on angles is not used later. Exercises 2731 concern facts used later.
1. Since 12
u and
4,
6
v 2 2( 1) 2 5 u u , v u = 4(1) + 6(2) = 8, and 8 .5
v u
u u
2. Since 315
w and 62 ,3
x 2 2 23 ( 1) ( 5) 35 w w , x w = 6(3) + (2)(1) + 3(5) = 5, and
5 1.
35 7
x w
w w
3. Since 31 ,5
w 2 2 23 ( 1) ( 5) 35 w w , and 3/ 35
1 1/ 35 .1/ 7
ww w
4. Since 1
,
2
u 2 2( 1) 2 5 u u and 1/ 51 .2 /5
uu u
5. Since 12
u and
4,
6
v u v = (1)(4) + 2(6) = 8, 2 24 6 52, v v and
4 8 /132.
6 12 /1313
u vv
v v
6. Since 623
x and 31 ,5
w x w = 6(3) + (2)(1) + 3(5) = 5, 2 2 26 ( 2) 3 49, x x and
6 30 / 495 2 10 / 49 .49
3 15/ 49
x wx
x x
336 CHAPTER 6 Orthogonality and Least Squares
7. Since 31 ,5
w 2 2 2|| || 3 ( 1) ( 5) 35. w w w
8. Since 62 ,3
x 2 2 2|| || 6 ( 2) 3 49 7. x x x
9. A unit vector in the direction of the given vector is
2 2
30 30 3/ 51 140 40 4 / 550( 30) 40
10. A unit vector in the direction of the given vector is
2 2 2
6 / 616 61 14 4 4 / 61
61( 6) 4 ( 3) 3 3 3 61
11. A unit vector in the direction of the given vector is
2 2 2
7 / 697 / 4 7 / 41 11/ 2 1/ 2 2 / 69
69 /16(7 / 4) (1/ 2) 1 1 1 4 / 69
12. A unit vector in the direction of the given vector is
2 2
8/ 3 8/ 3 4 / 51 12 2 3/ 5100 / 9(8 / 3) 2
13. Since 10
3
x and
1,
5
y 2 2 2|| || [10 ( 1)] [ 3 ( 5)] 125 x y and dist ( , ) 125 5 5. x y
14. Since 052
u and 41 ,8
z 2 2 2 2|| || [0 ( 4)] [ 5 ( 1)] [2 8] 68 u z and
dist ( , ) 68 2 17. u z
15. Since a b = 8(2) + (5)( 3) = 1 0, a and b are not orthogonal. 16. Driver monitor samsung u28e590. Since u v= 12(2) + (3)( 3) + (5)(3) = 0, u and v are orthogonal. 17. Since u v = 3(4) + 2(1) + (5)( 2) + 0(6) = 0, u and v are orthogonal. 18. Since y z= (3)(1) + 7(8) + 4(15) + 0(7) = 1 0, y and z are not orthogonal. 19. a. True. See the definition of || v ||.
b. True. See Theorem 1(c). c. True. See the discussion of Figure 5.
6.1 Solutions 337
d. False. Counterexample: 1 1
.
0 0
e. True. See the box following Example 6.
20. a. True. See Example 1 and Theorem 1(a). b. False. The absolute value sign is missing. See the box before Example 2. c. True. See the defintion of orthogonal complement. d. True. See the Pythagorean Theorem. e. True. See Theorem 3.
21. Theorem 1(b): ( ) ( ) ( )T T T T T u v w u v w u v w u w v w u w v w The second and third equalities used Theorems 3(b) and 2(c), respectively, from Section 2.1. Theorem 1(c): ( ) ( ) ( ) ( )T Tc c c c u v u v u v u v The second and third equalities used Theorems 3(c) and 2(d), respectively, from Section 2.1. 22. Since u u is the sum of the squares of the entries in u, u u0. The sum of squares of numbers is zero
if and only if all the numbers are themselves zero.
23. One computes that u v = 2(7) + (5)( 4) + (1)6 = 0, 2 2 2 2|| || 2 ( 5) ( 1) 30, u u u 2 2 2 2|| || ( 7) ( 4) 6 101, v v v and 2|| || ( ) ( ) u v u v u v
2 2 2(2 ( 7)) ( 5 ( 4)) ( 1 6) 131.
24. One computes that
2 2 2|| || ( ) ( ) 2 || || 2 || || u v u v u v u u u v v v u u v v and
2 2 2|| || ( ) ( ) 2 || || 2 || || u v u v u v u u u v v v u u v v so
2 2 2 2 2 2 2 2|| || || || || || 2 || || || || 2 || || 2 || || 2 || || u v u v u u v v u u v v u v
25. When ,a
b
v the set H of all vectors x
y
that are orthogonal to is the subspace of vectors whose
entries satisfy ax + by = 0. If a 0, then x = (b/a)y with y a free variable, and H is a line through the
origin. A natural choice for a basis for H in this case is .ba
If a = 0 and b 0, then by = 0. Since
b 0, y = 0 and x is a free variable. The subspace H is again a line through the origin. A natural choice
for a basis for H in this case is 1
,
0
but ba
is still a basis for H since a = 0 and b 0. If a = 0
and b = 0, then H = 2 since the equation 0x + 0y = 0 places no restrictions on x or y.
26. Theorem 2 in Chapter 4 may be used to show that W is a subspace of 3, because W is the null space of the 1 3 matrix .Tu Geometrically, W is a plane through the origin.
338 CHAPTER 6 Orthogonality and Least Squares
27. If y is orthogonal to u and v, then y u = y v = 0, and hence by a property of the inner product, y (u + v) = y u + y v = 0 + 0 = 0. Thus y is orthogonal to u+ v.
28. An arbitrary w in Span{u, v} has the form 1 2c c w u v . If y is orthogonal to u and v, then u y = v y = 0. By Theorem 1(b) and 1(c),
1 2 1 2( ) ( ) ( ) 0 0 0c c c c w y u v y u y v y
29. A typical vector in W has the form 1 1 .p pc c w v v If x is orthogonal to each ,jv then by Theorems 1(b) and 1(c),
1 1 1 1( ) ( ) ( ) 0p p p pc c c c w x v v y v x v x So x is orthogonal to each w in W.
30. a. If z is in ,W u is in W, and c is any scalar, then (cz) u= c(zu) c 0 = 0. Since u is any element of W, c z is in .W
b. Let 1z and 2z be in .W
Then for any u in W, 1 2 1 2( ) 0 0 0. z z u z u z u Thus 1 2z z is in .W
c. Since 0 is orthogonal to every vector, 0 is in .W Thus W is a subspace.
31. Suppose that x is in W and .W Since x is in ,W x is orthogonal to every vector in W, including x itself. So x x = 0, which happens only when x = 0.
32. [M] a. One computes that 1 2 3 4|| || || || || || || || 1 a a a a and that 0i j a a for i j. b. Answers will vary, but it should be that || Au|| = || u|| and || Av|| = || v||. c. Answers will again vary, but the cosines should be equal. d. A conjecture is that multiplying by A does not change the lengths of vectors or the angles between
vectors.
33. [M] Answers to the calculations will vary, but will demonstrate that the mapping ( )T
x v
x x vv v
(for v0) is a linear transformation. To confirm this, let x and y be in n, and let c be any scalar. Then
( ) ( ) ( )( )T
x y v x v y v
x y v vv v v v
( ) ( )T T
x v y v
v v x yv v v v
and
( ) ( )( ) ( )c cT c c cT
x v x v x v
x v v v xv v v v v v
34. [M] One finds that
5 11 0 5 0 1/ 31 4
, 0 1 1 0 4 / 31 00 0 0 1 1/ 30 1
0 3
N R
6.2 Solutions 339
The row-column rule for computing RN produces the 3 2 zero matrix, which shows that the rows of R are orthogonal to the columns of N. This is expected by Theorem 3 since each row of R is in Row A and each column of N is in Nul A.
6.2 SOLUTIONS
Notes: The nonsquare matrices in Theorems 6 and 7 are needed for the QR factorizarion in Section 6.4. It is important to emphasize that the term orthogonal matrix applies only to certain square matrices. The subsection on orthogo