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The Geometry of
Vector Spaces
8.1 SOLUTIONS
Notes
. This section introduces a special kinds of linear combination used to describe the sets created
when a subspace is shifted away from the origin. An affine combination is a linear combination in which
1.
12 3 4
12035
,,4,,
22473
−
⎡⎤ ⎡ ⎤ ⎡⎤ ⎡⎤ ⎡⎤
== = ==
⎢⎥ ⎢ ⎥ ⎢⎥ ⎢⎥ ⎢⎥
⎣⎦ ⎣ ⎦ ⎣⎦ ⎣⎦ ⎣⎦
vv v vy
⎣⎦⎣ ⎦⎣ ⎦⎣ ⎦
The general solution is c
2
= 1.5c
4
−1.5, c
3
= −2.5c
4
+ .5, with c
4
free. When c
4
= 0,
y − v
1
= −1.5(v
2
− v
1
) + .5(v
3
− v
1
) and y = 2v
1
− 1.5v
2
+ .5v
3
2.
12 3
1135
,,,
1227
−
⎡⎤ ⎡ ⎤ ⎡⎤ ⎡⎤
== ==
⎢⎥ ⎢ ⎥ ⎢⎥ ⎢⎥
⎣⎦ ⎣ ⎦ ⎣⎦ ⎣⎦
vv v y
, so
21 31 1
22 4
,,and
11 6
−
⎡
⎤⎡⎤ ⎡⎤
−= −= −=
⎢
⎥⎢⎥ ⎢⎥
⎣
⎦⎣⎦ ⎣⎦
vv vv yv
3. Row reduce the augmented matrix [v
2
-v
1
v
3
-v
1
y-v
1
] to find a solution for writing y-v
1
in terms of
4. Row reduce the augmented matrix [v
2
-v
1
v
3
-v
1
y-v
1
] to find a solution for writing y-v
1
in terms of
5. Since {b
1
, b
2
, b
3
} is an orthogonal basis, use Theorem 5 from Section 6.2 to write
12 3
12 3
11 2 2 3 3
= +
jj j
j
+
pb pb pb
pbbb
bb bb bb
ii i
ii i
6. Since {b
1
, b
2
, b
3
} is an orthogonal basis, use Theorem 5 from Section 6.2 to write
12 3
12 3
11 2 2 3 3
= +
jj j
j
+
pb pb pb
pbbb
bb bb bb
ii i
ii i
7. The matrix [v
1
v
2
v
3
p
1
p
2
p
3
] row reduces to
100 2 2 2
010 1 4 2
001132
000 0 0 5
⎡
⎤
⎢
⎥
−
⎢
⎥
⎢
⎥
−
⎢
⎥
−
⎢
⎥
⎣
⎦
.
8.1 • Solutions 455
8. The matrix [v
1
v
2
v
3
p
1
p
2
p
3
] row reduces to
100 3 0 2
010106
001 1 0 3
000010
−
⎡
⎤
⎢
⎥
−
⎢
⎥
⎢
⎥
−
⎢
⎥
⎢
⎥
⎣
⎦
.
Parts a., b., and c. use columns 4, 5, and 6, respectively, as the “augmented” column.
9. Choose v
1
and v
2
to be any two point on the line x=x
3
u+p. For example, take x
3
=0 and x
3
=1 to get
1
3
0
−
⎡⎤
=⎢⎥
⎣⎦
v
and
2
1
2
⎡
⎤
=
⎢
⎥
−
⎣
⎦
v
respectively. Other answers are possible.
10. Choose v
1
and v
2
to be any two point on the line x=x
3
u+p. For example, take x
3
=0 and x
3
=1 to get
11. a. True. See the definition at the beginning of this section.
b. False. The weights in the linear combination must sum to one. See the definition.
12. a. False. If S = {x}, then aff S = {x}. See the definition at the beginning of this section.
b. True. Theorem 2.
13. Span {v
2
− v
1
, v
3
− v
1
} is a plane if and only if {v
2
− v
1
, v
3
− v
1
} is linearly independent. Suppose c
2
2
1
3
1
14. Since {v
1
, v
2
, v
3
} is a basis for R
3
, the set W = Span {v
2
− v
1
, v
3
− v
1
} is a plane in R
3
, by
15. Let S = {x : Ax = b}. To show that S is affine, it suffices to show that S is a flat, by Theorem 3.
Let W = {x : Ax = 0}. Then W is a subspace of R
n
, by Theorem 2 in Section 4.2 (or Theorem 12
16. Suppose p, q ∈ S and t ∈ R. Then, by properties of the dot product (Theorem 1 in Section 6.1),
17. A suitable set consists of any three vectors that are not collinear and have 5 as their third entry. If
5 is their third entry, they lie in the plane x
3
= 5. If the vectors are not collinear, their affine hull
18. A suitable set consists of any four vectors that lie in the plane 2x
1
+ x
2
− 3x
3
= 12 and are not col-
linear. If the vectors are not collinear, their affine hull cannot be a line, so it must be the plane.
19. If p, q ∈ f (S), then there exist r, s ∈ S such that f (r) = p and f (s) = q. Given any t ∈ R, we must
show that z = (1 − t)p + t q is in f (S). Since f is linear,
20. Given an affine set T, let S = {x ∈ R
n
: f (x) ∈ T}. Consider x, y ∈ S and t ∈ R. Then
21. Since B is affine, Theorem 1 implies that B contains all affine combinations of points of B. Hence
B contains all affine combinations of points of A. That is, aff A ⊂ B.
23. Since A ⊂ (A ∪ B), it follows from Exercise 22 that aff A ⊂ aff (A ∪ B).
Similarly, aff B ⊂ aff (A ∪ B), so [aff A ∪ aff B] ⊂ aff (A ∪ B).
26. One possibility is to let A = {(0, 1)} and B = {(0, 2)}. Then both aff A and aff B are equal to the
x-axis. But A ∩ B = ∅, so aff (A ∩ B) = ∅.
8.2 SOLUTIONS
Notes:
Affine dependence and independence are developed in this section. Theorem 5 links affine
independence to linear independence. This material has important applications to computer graphics.
1. Let
123
302
,,.
360
⎡⎤ ⎡⎤ ⎡⎤
===
⎢⎥ ⎢⎥ ⎢⎥
−
⎣⎦ ⎣⎦ ⎣⎦
vvv
Then
21 31
31
,.
93
−−
⎡
⎤⎡⎤
−= −=
⎢
⎥⎢⎥
⎣
⎦⎣⎦
vv vv
Since v
3
− v
1
is a multiple
25 3 3 5
,, . ,
−−
⎡⎤ ⎡⎤ ⎡ ⎤ ⎡⎤ ⎡ ⎤
=== −=−=
vv v vvvv
3. The set is affinely independent. If the points are called v
1
, v
2
, v
3
, and v
4
, then row reduction of
[v
1
v
2
v
3
v
4
] shows that {v
1
, v
2
, v
3
} is a basis for R
3
and v
4
= 16v
1
+ 5v
2
− 3v
3
. Since there is
sum to one, v
4
is not an affine combination of the first three vectors.
4. Name the points v
1
, v
2
, v
3
, and v
4
. Then
21 31 41
230
8, 7, 2
496
⎡
⎤⎡⎤⎡⎤
⎢
⎥⎢⎥⎢⎥
−=− −=− −=
⎢
⎥⎢⎥⎢⎥
⎢
⎥⎢⎥⎢⎥
−−
⎣
⎦⎣⎦⎣⎦
vv vv vv
. To study
the linear independence of these points, row reduce the augmented matrix for Ax = 0:
2 3 00 2 3 00 2300 10 .60
8720~0 520~0520~01.40
4960 01560 0000 00 00
−
⎡⎤⎡⎤⎡⎤⎡⎤
⎢⎥⎢⎥⎢⎥⎢⎥
−−
⎢⎥⎢⎥⎢⎥⎢⎥
⎢⎥⎢⎥⎢⎥⎢⎥
−− − −
⎣⎦⎣⎦⎣⎦⎣⎦
. The first three columns
Alternative solution: Name the points v
1
, v
2
, v
3
, and v
4
. Use Theorem 5(d) and study the
homogeneous forms of the points. The first step is to move the bottom row of ones (in the
augmented matrix) to the top to simplify the arithmetic:
1111 1001.2
5327 001.4
376 3 000 0
⎡
⎤⎡ ⎤
⎢
⎥⎢ ⎥
−−
⎢
⎥⎢ ⎥
−−
⎢
⎥⎢ ⎥
⎣
⎦⎣ ⎦
5. −
4v
1
+ 5v
2
– 4v
3
+ 3v
4
= 0 is an affine dependence relation. It can be found by row reducing the
matrix
[]
1234
,vv vv
and proceeding as in the solution to Exercise 4.
6. The set is affinely independent, as the following calculation with homogeneous forms shows:
[]
1234
1 1 11 10 00
1023 0100
~~
3155 0010
1220 0001
⎡
⎤⎡ ⎤
⎢
⎥⎢ ⎥
⎢
⎥⎢ ⎥
⎢
⎥⎢ ⎥
−
⎢
⎥⎢ ⎥
−
⎢
⎥⎢ ⎥
⎣
⎦⎣ ⎦
vv vv
7. Denote the given points as v
1
, v
2
, v
3
, and p. Row reduce the augmented matrix for the equation
x
1
ṽ
1
+ x
2
ṽ
2
+ x
3
ṽ
3
= p.̃ Remember to move the bottom row of ones to the top as the first step to
simplify the arithmetic by hand.
Alternative solution: Another way that this problem can be solved is by “translating” it to the
origin. That is, compute v
2
– v
1
, v
3
– v
1
, and p – v
1
, find weights c
2
and c
3
such that
c
2
(v
2
– v
1
) + c
3
(v
3
– v
1
) = p – v
1
and then write p = (1 – c
2
– c
3
)v
1
+ c
2
v
2
+ c
3
v
3
. Here are the calculations for Exercise 7:
8.2 • Solutions 459
8. Denote the given points as v
1
, v
2
, v
3
, and p. Row reduce the augmented matrix for the equation
x
1
ṽ
1
+ x
2
ṽ
2
+ x
3
ṽ
3
= p̃.
1111 1002
⎡⎤⎡⎤
9. a. True. Theorem 5 uses the point v
1
for the translation, but the paragraph after the theorem
points out that any one of the points in the set can be used for the translation.
b. False, by (d) of Theorem 5.
10. a. False. By Theorem 5, the set of homogeneous forms must be linearly dependent, too.
b. True. If one statement in Theorem 5 is false, the other statements are false, too.
11. When a set of five points is translated by subtracting, say, the first point, the new set of four
By Theorem 5, the original set of five points is affinely dependent.
12. Suppose v
1
, …, v
p
are in R
n
and p ≥ n + 2. Since p −
1 ≥ n + 1, the points v
2
− v
1
, v
3
− v
1
, … , v
p
− v
1
are linearly dependent, by Theorem 8 in Section 1.7. By Theorem 5, {v
1
, v
2
, …, v
p
} is affinely
dependent.
13. If {v
1
, v
2
} is affinely dependent, then there exist c
1
and c
2
, not both zero, such that c
1
+ c
2
= 0, and
c
1
v
1
+ c
2
v
2
= 0. Then c
1
= – c
2
0 and c
1
v
1
= – c
2
v
2
= c
1
v
2
, which implies that v
1
= v
2
.
14. Let S
1
consist of three (distinct) points on a line through the origin. The set is affinely dependent
460 CHAPTER 8 • The Geometry of Vector Spaces
15. a. The vectors v
2
− v
1
=
1
2
⎡⎤
⎢⎥
⎣⎦
and v
3
− v
1
=
3
2
⎡
⎤
⎢
⎥
−
⎣
⎦
are not multiples and hence are linearly
independent. By Theorem 5, S is affinely independent.
16. a. The vectors v
2
− v
1
=
1
4
⎡⎤
⎢⎥
⎣⎦
and v
3
− v
1
=
4
2
⎡
⎤
⎢
⎥
⎣
⎦
are not
multiples and hence are linearly independent. By
Theorem 5, S is affinely independent.
b.
55103
24 2 22
12 3
777 7 7 7 777
( ,,), (, , ), (,,)↔− ↔ − ↔pp p
c.
45
(,,), (,,),↔+−− ↔++−pp
17. Suppose S = {b
1
, …, b
k
} is an affinely independent set. Then (7) has a solution, because p is in
aff S. Hence (8) has a solution. By Theorem 5, the homogeneous forms of the points in S are
linearly independent. Thus (8) has a unique solution. Then (7) also has a unique solution,
because (8) encodes both equations that appear in (7).
The following argument mimics the proof of Theorem 7 in Section 4.4. If S = {b
1
, …, b
k
} is
an affinely independent set, then scalars c
1
, …, c
k
exist that satisfy (7), by definition of aff S.
Suppose x also has the representation
18. Let
.
x
y
z
⎡⎤
⎢⎥
=⎢⎥
⎢⎥
⎣⎦
p
Then
00 0
0010
00 0
xa
xyz xyz
yb
abc abc
zc
⎡⎤ ⎡⎤ ⎡⎤ ⎡⎤ ⎡⎤
⎛⎞
⎢⎥ ⎢⎥ ⎢⎥ ⎢⎥ ⎢⎥
=+++−−−
⎜⎟
⎢⎥ ⎢⎥ ⎢⎥ ⎢⎥ ⎢⎥
⎝⎠
⎢⎥ ⎢⎥ ⎢⎥ ⎢⎥ ⎢⎥
⎣⎦ ⎣⎦ ⎣⎦ ⎣⎦ ⎣⎦
So the barycentric coordi-
19. If {p
1
, p
2
, p
3
} is an affinely dependent set, then there exist scalars c
1,
c
2
, and c
3
, not all zero, such
that c
1
p
1
+ c
2
p
2
+ c
3
p
3
= 0 and c
1
+ c
2
+ c
3
= 0. But then, applying the transformation f,
v
3
v
2
p
1
p
3
p
5
p
7
8.2 • Solutions 461
123
20. If the translated set {p
1
+ q, p
2
+ q, p
3
+ q} were affinely dependent, then there would exist real
numbers c
1,
c
2
, and c
3
, not all zero and with c
1
+ c
2
+ c
3
= 0, such that
c
1
(p
1
+ q) + c
2
(p
2
+ q) + c
3
(p
3
+ q) = 0.
But then,
21. Let
11 1
22 2
,, and .
ab c
ab c
⎡⎤ ⎡⎤ ⎡⎤
== =
⎢⎥ ⎢⎥ ⎢⎥
⎣⎦ ⎣⎦ ⎣⎦
ab c
Then det
[ã b̃ c̃] =
111 12
222 12
12
1
det det 1
111 1
abc aa
abc bb
cc
⎡
⎤⎡ ⎤
⎢
⎥⎢ ⎥
=
⎢
⎥⎢ ⎥
⎢
⎥⎢ ⎥
⎣
⎦⎣ ⎦
,
22. If p is on the line through a and b, then p is an affine combination of a and b, so p̃ is a linear
combination of ã and b̃. Thus the columns of [ã b̃ p̃] are linearly dependent. So the determinant
23. If [ã b̃ c̃]
r
s
t
⎡⎤
⎢⎥
⎢⎥
⎢⎥
⎣⎦
= p̃, then Cramer’s rule gives r = det
[p̃ b̃ c̃] /
det
[ã b̃ c̃]. By Exercise 21, the
24. Let p = (1 − x)q + x
a, where q is on the line segment from b to c. Then, because the determinant
is a linear function of the first column when the other columns are fixed (Section 3.2),
8.3 SOLUTIONS
Notes:
The notion of convexity is introduced in this section and has important applications in computer
graphics. Bézier curves are introduced in Exercises 21-24 explored in greater detail in Section 8.6.
1. The set 0:0 1Vy
y
⎧⎫
⎡⎤
⎪⎪
=≤<
⎨⎬
⎢⎥
⎪⎪
⎣⎦
⎩⎭
is the vertical line segment from (0,0) to
2. a. Conv S includes all points p of the form
1/ 2 1/ 2 ( 1/ 2)
(1 ) 21/ 2(21/)
xtx
tt
xtx
+−
⎡⎤⎡⎤⎡ ⎤
=− + =
⎢⎥⎢⎥⎢ ⎥
−−
⎣⎦⎣⎦⎣ ⎦
p
, where
b. Recall that for any integer n,
()
sin( 2 ) sinxn x
π
+=
. Then
22
(1 ) conv S.
sin( ) sin( 2 ) sin( )
xxnxnt
tt
xxn x
ππ
π
++
⎡⎤⎡ ⎤⎡ ⎤
=− + = ∈
⎢⎥⎢ ⎥⎢ ⎥
+
⎣⎦⎣ ⎦⎣ ⎦
p
c. Conv S includes all points p of the form
0
xx
⎡
⎤⎡⎤
3. From Exercise 5, Section 8.1,
2
2
8.3 • Solutions 463
4. From Exercise 5, Section 8.1,
5. Row reduce the matrix
[]
123412
vv vvpp
to obtain the barycentric coordinates
6. Let W be the subspace spanned by the orthogonal set S = {v
1
, v
2
, v
3
}. As in Example 1, the
barycentric coordinates of the points p
1
, …, p
4
with respect to S are easy to compute, and they
determine whether or not a point is in Span S, aff S, or conv S.
a.
13
11 1 2
1123
11 2 2 3 3
proj
W
⋅
⋅⋅
=+ +
⋅⋅ ⋅
pv
pv pv
pvvv
vv v v vv
b. Similarly,
99 9
44 2
21231232
11 1
proj 99 9 442
W
=++ =++=pvvvvvvp
. This shows that p
2
c.
31231233
9918
proj 2
999
W=+−=+−=p v v vvvvp
. Thus p
3
is in Span S. However, since
the coefficients do not sum to one, p
3
is not in aff S and certainly not in conv S.
7.
1231234
1242320
,,,,,,
0311202
−
⎡ ⎤ ⎡⎤ ⎡⎤ ⎡⎤ ⎡⎤ ⎡⎤ ⎡⎤
=======
⎢ ⎥ ⎢⎥ ⎢⎥ ⎢⎥ ⎢⎥ ⎢⎥ ⎢⎥
⎣ ⎦ ⎣⎦ ⎣⎦ ⎣⎦ ⎣⎦ ⎣⎦ ⎣⎦
vvvpppp
, T = {v
1
, v
2
, v
3
}
a. Use an augmented matrix (with four augmented columns) to write the homogeneous forms of
464 CHAPTER 8 • The Geometry of Vector Spaces
1231234
111
322
3
11 1
62 4 4
3
11 1
22 4 4
100 0
1111111
~ 1242320~010
0311202 001
⎡
⎤
⎡⎤
⎢
⎥
⎢⎥
⎡⎤
⎢
⎥
−−
⎣⎦
⎢⎥
⎢
⎥
⎢⎥
⎣⎦
⎢
⎥
−
⎣
⎦
vv vpppp
b. p
3
and p
4
are outside conv T, because in each case at least one of the barycentric coordinates is
negative. p
is inside conv T, because all of its barycentric coordinates are positive. p
is on the
8. a. The barycentric coordinates of p
1
, p
2
, p
3
, and p
4
are, respectively,
()
3
12 2
13 13 13
,, ,−
()
83
2
13 13 13
,, ,
9. The points p
1
and p
3
are outside the tetrahedron conv S since their barycentric coordinates contain
10. The point q
1
is inside conv S because the barycentric coordinates are all positive. The point q
2
is
11. a. False. In order for y to be a convex combination, the c’s must also all be nonnegative. See the
12. a. True. See the definition prior to Theorem 7.
b. True. Theorem 9.
13. If p, q ∈ f
(S), then there exist r, s ∈ S such that f
(r) = p and f
(s) = q. The goal is to show that the
line segment y = (1 − t)p + t
q, for 0 ≤ t ≤ 1, is in f
(S). Since
f
is linear,
14. Suppose r, s ∈ S and 0 ≤ t ≤ 1. Then, since f is a linear transformation,
f
[(1 − t)r + t
s ] = (1 − t)
f
(r) + t
f
(s)
15. It is straightforward to confirm the equations in the problem: (1)
11 1 1
1234
33 6 6
+++ =vv v vp
and
(2) v
1
– v
2
+ v
3
– v
4
= 0. Notice that the coefficients of v
1
and v
3
in equation (2) are positive. With
the notation of the proof of Caratheodory’s Theorem, d
1
= 1 and d
3
= 1. The corresponding
coefficients in equation (1) are
1
13
c=
and
1
36
c=
. The ratios of these coefficients are
1
11 3
/cd=
16.
1234
103 11
,,, ,
031 12
−
⎡⎤ ⎡⎤ ⎡⎤ ⎡⎤ ⎡⎤
=====
⎢⎥ ⎢⎥ ⎢⎥ ⎢⎥ ⎢⎥
−
⎣⎦ ⎣⎦ ⎣⎦ ⎣⎦ ⎣⎦
vvvvp
It is straightforward to confirm the
Notice that the coefficients of v
1
and v
3
in equation (2) are positive. With the notation of the proof of
Caratheodory’s Theorem, d
1
= 10 and d
3
= 7. The corresponding coefficients in equation (1) are
1
1121
c=
and
37
3121
c=
. The ratios of these coefficients are
11
11 121 1210
/10cd=÷=
and
17. Suppose A ⊂ B, where B is convex. Then, since B is convex, Theorem 7 implies that B contains
18. Suppose A ⊂ B. Then A ⊂ B ⊂ conv B. Since conv B is convex, Exercise 17 shows that
conv A ⊂ conv B.
20. a. Since (A ∩ B) ⊂ A, Exercise 18 shows that conv (A ∩ B) ⊂ conv A. Similarly,
conv (A ∩ B) ⊂ conv B. Thus, conv (A ∩ B) ⊂ [(conv A) ∩ (conv B)].
b. One possibility is to let A be a pair of opposite vertices of a square and let B be the other pair
21. 22.
23. g(t) = (1 − t)f
0
(t) + t
f
1
(t)
= (1 − t)[(1 − t)p
0
+ t
p
1
] + t[(1 − t)p
1
+ t
p
2
] = (1 − t)
2
p
0
+ 2t(1 − t)p
1
+ t
2
p
2
.
24. h(t) = (1 − t)g
1
(t) + t
g
2
(t). Use the representation for g
1
(t) from Exercise 23, and the analogous
representation for g
2
(t), based on the control points p
1
, p
2
, and p
3
, and obtain
h(t) = (1 − t)[(1 − t)
2
p
0
+ 2t(1 − t)p
1
+ t
2
p
2
] + t
[(1 − t)
2
p
1
+ 2t(1 − t)p
2
+ t
2
p
3
]
= (1 − t)
3
p
0
+ 2t(1 − 2t + t
2
)p
1
+ (t
2
− t
3
)p
2
+ t
(1 − 2t + t
2
)
p
1
+ 2t
2
(1 − t)p
2
+ t
3
p
3
p
1
0
p
()
1
12
f
p
1
0
p
2
()
3
14
f
