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8.4 • Solutions 467
8.4 SOLUTIONS
Notes:
In this section lines and planes are generalized to higher dimensions using the notion of
hyperplanes. Important topological ideas such as open, closed, and compact sets are introduced.
1. Let
12
13
and .
41
−
⎡⎤ ⎡⎤
==
⎢⎥ ⎢⎥
⎣⎦ ⎣⎦
vv
Then
21
314
.
14 3
−
⎡
⎤⎡⎤⎡⎤
−= − =
⎢
⎥⎢⎥⎢⎥
−
⎣
⎦⎣⎦⎣⎦
vv
Choose n to be a vector orthogonal to
2. Let
12
12
and .
41
−
⎡⎤ ⎡ ⎤
==
⎢⎥ ⎢ ⎥
−
⎣⎦ ⎣ ⎦
vv
Then
21
21 3
.
14 5
−−
⎡
⎤⎡⎤⎡ ⎤
−= − =
⎢
⎥⎢⎥⎢ ⎥
−−
⎣
⎦⎣⎦⎣ ⎦
vv
Choose n to be a vector orthogonal
3. a. The set is open since it does not contain any of its boundary points.
b. The set is closed since it contains all of its boundary points.
4. a. The set is closed since it contains all of its boundary points.
b. The set is open since it does not contain any of its boundary points.
5. a. The set is not compact since it is not closed, however it is convex.
b. The set is compact since it is closed and bounded. It is also convex.
6. a. The set is compact since it is closed and bounded. It is not convex.
b. The set is not compact since it is not closed. It is not convex.
468 CHAPTER 8 • The Geometry of Vector Spaces
7. a. Let
12 3
12 1
1, 4, 2,
31 5
a
b
c
−
⎡⎤ ⎡⎤ ⎡ ⎤ ⎡⎤
⎢⎥ ⎢⎥ ⎢ ⎥ ⎢⎥
== =−=
⎢⎥ ⎢⎥ ⎢ ⎥ ⎢⎥
⎢⎥ ⎢⎥ ⎢ ⎥ ⎢⎥
⎣⎦ ⎣⎦ ⎣ ⎦ ⎣⎦
vv v n
and compute the translated points
[1 3 2] 0,
a
b
c
⎡⎤
⎢⎥
−=
⎢⎥
⎢⎥
⎣⎦
[2 3 2] 0
a
b
c
⎡⎤
⎢⎥
−− =
⎢⎥
⎢⎥
⎣⎦
.
b. The linear functional is
123 2 3
(, , ) 2 3fxx x x x=+
, so d = f
(1, 1, 3) = 2 + 9 = 11. As a check,
evaluate f at the other two points on the hyperplane: f
(2, 4, 1) = 8 + 3 = 11 and
8. a. Find a vector in the null space of the transpose of [v
2
-v
1
v
3
-v
1
]. For example, take
4
3.
6
⎡⎤
⎢⎥
=⎢⎥
⎢⎥
−
⎣⎦
n
9. a. Find a vector in the null space of the transpose of [v
2
-v
1
v
3
-v
1
v
4
-v
1
]. For example, take
n =
3
1.
⎡⎤
⎢⎥
−
⎢⎥
8.4 • Solutions 469
10. a. Find a vector in the null space of the transpose of [v
2
-v
1
v
3
-v
1
v
4
-v
1
]. For example, take
n =
2
3.
−
⎡⎤
⎢⎥
⎢⎥
11.
12 3
2; 02; 52; 22; 2.==< => =−< =np n0 nv nv nv
ii i i i
Hence v
2
is on the same side of H
as 0, v
1
is on the other side, and v
3
is in H.
12. Let H = [ f : d ], where f (x
1
, x
2
, x
3
) = 3x
1
+ x
2
− 2x
3
. f (a
1
) = −5, f (a
2
) = 4. f (a
3
) =3, f (b
1
) =7,
13. H
1
= {x : n
1
⋅ x = d
1
} and H
2
= {x : n
2
⋅ x = d
2
}. Since p
1
∈ H
1
, d
1
= n
1
⋅ p
1
= 4. Similarly,
d
2
= n
2
⋅ p
2
= 22. Solve the simultaneous system [1 2 4 2]x = 4 and [2 3 1 5]x = 22:
12424 1 010 432
−
⎡⎤⎡ ⎤
14. Since each of F
1
and F
2
can be described as the solution sets of A
1
x=b
1
and A
2
x=b
2
respectively,
15. f (x
1
, x
2
, x
3
) =Ax= x
1
– 3x
2
+ 4x
3
– 2x
4
and d =b= 5
17. Since by Theorem 3 in Section 6.1, Row B=(Nul B)
⊥
, choose a nonzero vector n∈ Nul B . For
1
⎡⎤
18. Since by Theorem 3, Section 6.1, Row B=(Nul B)
⊥
, choose a nonzero vector n∈ Nul B . For
11
−
⎡⎤
19. Theorem 3 in Section 6.1 says that (Col B)
⊥
= Nul B
T
. Since the two columns of B are clearly linear
independent, the rank of B is 2, as is the rank of B
T
. So dim Nul B
T
= 1, by the Rank Theorem, since
there are three columns in B
T
. This means that Nul B
T
is one-dimensional and any nonzero vector n
20. Since by Theorem 3, Section 6.1, Col B=(Nul B
T
)
⊥
, choose a nonzero vector n∈ Nul B
T
. For
6
−
⎡⎤
21. a. False. A linear functional goes from R
n
to R. See the definition at the beginning of this section.
b. False. See the discussion of (1) and (4). There is a 1×n matrix A such that f (x) = Ax for all x in
22. a. True. See the statement after (3).
b. False. The vector n must be nonzero. If n = 0, then the given set is empty if d ≠ 0 and the set
23. Notice that the side of the triangle closest to p is
23
vv
. A vector orthogonal to
23
vv
.is n=
3
2
⎡⎤
⎢⎥
−
⎣⎦
.
8.4 • Solutions 471
24. Notice that the side of the triangle closest to p is
13
vv
A vector orthogonal to
13
vv
is n=
2
3
−
⎡⎤
⎢⎥
.
25. Let L be the line segment from the center of B(0, 3) to the center of B(p, 1). This is on the line
through the origin in the direction of p. The length of L is (4
2
+ 1
2
)
1/2
4.1231. This exceeds the
sum of the radii of the two disks, so the disks do not touch. If the disks did touch, there would be no
hyperplane (line) strictly separating them, but the line orthogonal to L through the point of tangency
26. The normal to the separating hyperplane has the direction of the line segment between p and q. So,
let n = p − q =
4
2
⎡⎤
⎢⎥
−
⎣⎦
. The distance between p and q is
20
, which is more than the sum of the radii
27. Exercise 2(a) in Section 8.3 gives one possibility. Or let S = {(x, y) : x
2
y
2
= 1 and y > 0}. Then
28. One possibility is B = {(x, y) : x
2
y
2
= 1 and y > 0} and A = {(x, y) : |
x
| ≤ 1 and y = 0}.
29. Let x, y ∈ B(
p,
δ
) and suppose z = (1 − t)
x + t
y, where 0 ≤ t ≤ 1. Then
30. Let S be a bounded set. Then there exists a
δ
> 0 such that S
⊂
B(0,
δ
). But B(0,
δ
) is
8.5 SOLUTIONS
Notes:
A polytope is the convex hull of a finite number of points. Polytopes and simplices are important
in linear programming, which has numerous applications in engineering design and business
management. The behavior of functions on polytopes is studied in this section.
1. Evaluate each linear functional at each of the three extreme points of S. Then select the extreme
point(s) that give the maximum value of the functional.
2. Evaluate each linear functional at each of the three extreme points of S. Then select the point(s) that
give the maximum value of the functional.
3. Evaluate each linear functional at each of the three extreme points of S. Then select the point(s) that
give the minimum value of the functional.
4. Evaluate each linear functional at each of the three extreme points of S. Then select the point(s) that
give the maximum value of the functional.
a. f
(p
1
) = −1, f
(p
2
) = 3, and f
(p
3
) = 3, so m = –1 at the point p
1
.
5. The two inequalities are (a) x
1
+ 2x
2
≤ 10 and (b) 3x
1
+ x
2
≤ 15. Line (a) goes from (0,5) to (10,0).
Line (b) goes from (0,15) to (5,0). One vertex is (0,0). The x
1
-intercepts (when x
2
= 0) are 10 and 5,
8.5 • Solutions 473
6. The two inequalities are (a) 2x
1
+ 3x
2
≤ 18 and (b) 4x
1
+ x
2
≤ 16. Line (a) goes from (0,6) to (9,0).
Line (b) goes from (0,16) to (4,0). One vertex is (0,0). The x
1
-intercepts (when x
2
= 0) are 9 and 4, so
7. The three inequalities are (a) x
1
+ 3x
2
≤ 18, (b) x
1
+ x
2
≤ 10, and (c) 4x
1
+ x
2
≤ 28. Line (a) goes from
(0,6) to (18,0). Line (b) goes from (0,10) to (10,0). And line (c) goes from (0,28) to (7,0). One
8. The three inequalities are (a) 2x
1
+ x
2
≤ 8, (b) x
1
+ x
2
≤ 6, and (c) x
1
+ 2x
2
≤ 7. Line (a) goes from
(0,8) to (4,0). Line (b) goes from (0,6) to (6,0). And line (c) goes from (0,3.5) to (7,0). One vertex is
9. The origin is an extreme point, but it is not a vertex. It is an
extreme point since it is not in the interior of any line segment
10. One possibility is a ray. It has an extreme point at one end.
11. One possibility is to let S be a square that includes part of the boundary but not all of it. For example,
include just two adjacent edges. The convex hull of the profile P is a triangular region.
12. a. f
0
(S
5
) = 6, f
1
(S
5
) = 15, f
2
(S
5
) = 20, f
3
(S
5
) = 15, f
4
(S
5
) = 6, and 6 − 15 + 20 − 15 + 6 = 2.
b.
f
0
f
1
f
2
f
3
f
4
S
1
2
S
2
3 3
S
conv P =
474 CHAPTER 8 • The Geometry of Vector Spaces
13. a. To determine the number of k-faces of the 5-dimensional hypercube C
5
, look at the pattern that is
followed in building C
4
from C
3
. For example, the 2-faces in C
4
include the 2-faces of C
3
and
the 2-faces in the translated image of C
3
. In addition, there are the 1-faces of C
3
that are
b. The general formula is
!
()2 ,where !( )!
nnk
k
na
a
fC kb
ba b
−
⎛⎞ ⎛⎞
==
⎜⎟ ⎜⎟ −
⎝⎠ ⎝⎠
is the binomial coefficient.
14. a. X
1
is a line segment X
2
is a parallelogram
b. f
0
(X
3
) = 6, f
1
(X
3
) = 12, f
2
(X
3
) = 8. X
3
is an octahedron.
15. a. f
0
(P
n
) = f
0
(Q) +
1
16. a. True. See the definition at the beginning of this section.
17. a. False. It has six facets (faces).
b. True. See Theorem 14.
18. Let v be an extreme point of the convex set S and let T = {y ∈ S : y ≠ v}. If y and z are in T, then
0 v
1
v
1
v
2
8.6 • Solutions 475
19. Let S be convex and let x ∈ cS + dS, where c > 0 and d > 0. Then there exist s
1
and s
2
in S such that
x = cs
1
+ ds
2
. But then
12 1 2
()
cd
cd cd
cd cd
⎛⎞
=+ =+ +
⎜⎟
++
⎝⎠
xss s s
.
20. For example, let S = {1, 2} in
1
R
. Then 2S = {2, 4}, 3S = {3, 6} and (2 + 3)S = {5, 10}.
21. Suppose A and B are convex. Let x, y ∈ A
+
B. Then there exist a, c ∈ A and b, d ∈ B such that
x = a + b and y = c + d. For any t such that 0 ≤ t ≤ 1, we have
22. a. Since each edge belongs to two facets, kr is twice the number of edges: k
r = 2e. Since each edge
has two vertices, s
v = 2e.
c. A polygon must have at least three sides, so k ≥ 3. At least three edges meet at each vertex,
so s ≥ 3. But both k and s cannot both be greater than 3, for then the left side of the equation
8.6 SOLUTIONS
Notes:
This section moves beyond lines and planes to the study of some of the curves that are used to
model surfaces in engineering and computer aided design. Notice that these curves have a matrix
representation.
1. The original curve is x(t) = (1 – t)
3
p
0
+ 3t(1 – t)
2
p
1
+ 3t
2
(1 – t)p
2
+ t
3
p
3
(0 < t < 1). Since the
curve is determined by its control points, it seems reasonable that to translate the curve, one
should translate the control points. In this case, the new Bézier curve y(t) would have the
equation
2. a. Equation (15) reveals that each polynomial weight is nonnegative for 0 < t < 1, since 4 − 3t >
0. For the sum of the coefficients, use (15) with the first term expanded: 1 – 3t + 6t
2
− t
3
.
The 1 here plus the 4 and 1 in the coefficients of p
1
and p
2
, respectively, sum to 6, while the
other terms sum to 0. This explains the 1/6 in the formula for x(t), which makes the
3. a. x'
(t) = (–3 + 6t – 3t
2
)p
0
+ (3 –12t + 9t
2
)p
1
+ (6t – 9t
2
)p
2
+ 3t
2
p
3
, so x'
(0) = –3p
0
+ 3p
1
=3(p
1
– p
0
),
and x'
(1) = –3p
2
+ 3p
3
= 3(p
3
– p
2
). This shows that the tangent vector x'
(0) points in the
b. x''
(t) = (6 – 6t)p
0
+ (–12 + 18t)p
1
+ (6 – 18t)p
2
+ 6tp
3
, so that
x''
(0) = 6p
0
– 12p
1
+ 6p
2
= 6(p
0
– p
1
) + 6(p
2
– p
1
)
4. a. x'
(t) =
()()()
2222
01 23
1
6
363 912 963 3tt t t tt t
⎡⎤
−+− + − +−++ +
⎣⎦
pp pp
x'
(0) =
()
20
1
2
−pp
and x'
(1) =
()
31
1
2
−pp
(Verify that, in the first part of Fig. 10, a line drawn through p
0
and p
2
is parallel to the
8.6 • Solutions 477
= 0, the curve always moves away from p
3
toward p
0
for 0 < t < 1.
b. x''
(t) = (1 – t)p
0
+ (–2 + 3t)p
1
+ (1 – 3t)p
2
+ tp
3
x''
(0) = p
0
– 2p
1
+ p
2
= (p
0
− p
1
) + (p
2
− p
1
)
and x''
(1) = p
1
– 2p
2
+ p
3
= (p
1
− p
2
) + (p
3
− p
2
)
For a picture of x''
(0), construct a coordinate system with the origin at p
1
, temporarily, label p
0
as
p
0
− p
1
, and label p
2
as p
2
− p
1
. Finally, construct a line from this new origin to the sum of p
0
− p
1
5. a. From Exercise 3(a) or equation (9) in the text,
x'
(1) = 3(p
3
– p
2
)
b. If x'
(1) = y'
(0) = 0, then p
2
= p
3
and p
3
= p
4
. Thus, the “line segment” from p
2
to p
4
is just
the point p
3
. [Note: In this case, the combined curve is still C
1
continuous, by definition.
6. a. With x(t) as in Exercise 2,
x(0) = (p
0
+ 4p
1
+ p
2
)/6 and x(1) = (p
1
+ 4p
2
+ p
3
)/6
b. From Exercise 4(a),
7. From Exercise 3(b),
x''
(0) = 6(p
0
– p
1
) + 6(p
2
– p
1
) and x''
(1) = 6(p
1
– p
2
) + 6(p
3
– p
2
)
Use x''
(0) with the control points for y(t), to get
478 CHAPTER 8 • The Geometry of Vector Spaces
8. From Exercise 4(b), x''
(0) = p
0
– 2p
1
+ p
2
and x''
(1) = p
1
– 2p
2
+ p
3
. Use the formula for x''
(0), with
the shifted control points for y(t), to get
9. Write a vector of the polynomial weights for x(t), expand the polynomial weights and factor the
vector as M
B
u(t):
234
14 6 4 1
14641
tt tt
⎡⎤
−+ − +
⎡
⎤
−−
⎡⎤
14 6 4 1
−−
⎡
⎤
10. Write a vector of the polynomial weights for x(t), expand the polynomial weights, taking care to
write the terms in ascending powers of t, and factor the vector as M
S
u(t):
23
13 3 1
1331
ttt
⎡⎤
−+ −
⎡
⎤
−−
⎡⎤
1331
−−
⎡
⎤
11. a. True. See equation (2).
12. a. False. The essential properties are preserved under translations as well as linear transformations.
See the comment after Figure 1.
13. a. From (12),
111
10 10 1 0
222
()−= − = −qq pp p p
. Since
1
001 10
2
, ( )==+qpq pp
.
8.6 • Solutions 479
14. a. 3(r
3
− r
2
) = z'(1), by (9) with z'(1) and r
j
in place of x'(1) and p
j
.
b. From part (a), 6(r
3
− r
2
) = 3(p
3
− p
2
),
11 11
32 3 2 3 3 2 2
22 22
, and .−= − − + =rr p p r p p r
c. 3(r
− r
) = z'
(0), by (9) with z'(0) and r
in place of x'(0) and p
.
d. Part (c) and (10) show that 3(r
1
− r
0
) =
3
8
(−p
0
− p
1
+ p
2
+ p
3
). Multiply by
8
3
and rearrange to
e. From (8), 8r
0
= p
0
+ 3p
1
+ 3p
2
+ p
3
. Substitute into the equation from part (d), and obtain
8r
1
= 2p
1
+ 4p
2
+ 2p
3
. Divide by 8 and use part (b) to obtain
15. a. From (11), y'(1) = .5x'(.5) = z'(0).
b. Observe that y'(1) = 3(q
3
– q
2
). This follows from (9), with y(t) and its control points in
16. A Bézier curve is completely determined by its four control points. Two are given directly: p
0
=
x(0) and p
3
= x(1). From equation (9), x'
(0) = 3(p
1
– p
0
) and x'
(1) = 3(p
3
– p
2
). Solving gives
17. a. The quadratic curve is w(t) = (1 – t)
2
p
0
+ 2t(1 − t)p
1
+ t
2
p
2
. From Example 1, the tangent
vectors at the endpoints are w'
(0) = 2p
1
− 2p
0
and w'
(1) = 2p
2
− 2p
1
. Denote the control
480 CHAPTER 8 • The Geometry of Vector Spaces
b. Write the standard formula (7) in this section, with r
i
in place of p
i
for i = 1, …, 4, and then
replace r
0
and r
3
by p
0
and p
2
, respectively:
x(t) = (1 – 3t + 3t
2
– t
3
)p
0
+ (3t – 6t
2
+ 3t
3
)r
1
+ (3t
2
– 3t
3
)r
2
+ t
3
p
2
(iii)
Use the formulas (i) and (ii) for r
1
and r
2
to examine the second and third terms in (iii):
23 23 23
12
101
33
(3 6 3 ) (3 6 3 ) (3 6 3 )
tt t tt t tt t
−+ = −+ + −+
rp p
0
01
33
⎡⎤
⎢⎥
−+
p
pp
