r
4.1 SOLUTIONS
Notes:
This section is designed to avoi
h
o
1. a. If u and v are in V, then their en
t
nonnegative, the vector u + v h
a
2. a. If
x
y
⎡⎤
=⎢⎥
⎣⎦
u
is in W, then the vec
t
since xy ≥ 0.
3. Example: If
.5
.5
⎡⎤
=⎢⎥
⎣⎦
u
and c = 4, the
n
multiplication, H is not a subspace
o
r
d the standard exercises in which a student is asked
t
ries are nonnegative. Since a sum of nonnegative nu
m
a
s nonnegative entries. Thus u + v is in V.
t
or
xcx
cc
ycy
⎡⎤ ⎡ ⎤
==
⎢⎥ ⎢ ⎥
⎣⎦ ⎣ ⎦
u
is in W because
2
()() (cx cy c x
y
=
n
u is in H but cu is not in H. Since H is not closed u
n
o
f
2
.
to check ten
m
bers is
)0
y
≥
n
der scalar
198 CHAPTER 4 • Vector Spaces
5. Yes. Since the set is
2
Span{ }t
, the set is a subspace by Theorem 1.
7. No. The set is not closed under multiplication by scalars which are not integers.
8. Yes. The zero vector is in the set H. If p and q are in H, then (p + q)(0) = p(0) + q(0) = 0 + 0 = 0,
9. The set H = Span
{v}, where
2
5
−
⎡⎤
⎢⎥
=⎢⎥
v
. Thus H is a subspace of
3
by Theorem 1.
10. The set H = Span
{v}, where
3
0
⎡⎤
⎢⎥
=⎢⎥
v
. Thus H is a subspace of
3
by Theorem 1.
11. The set W = Span
{u, v}, where
2
1
⎢
⎢
⎣
⎡⎤
⎢⎥
=−
u
and
3
0
⎡
⎤
⎢
⎥
=
v
. Thus W is a subspace of
3
by Theorem 1.
⎢
⎢
⎣
2
2
⎡⎤
⎢⎥
⎢⎥
4
0
⎡
⎤
⎢
⎥
⎢
⎥
4
13. a. The vector w is not in the set
123
{, , }vv v
. There are 3 vectors in the set
123
{, , }.vv v
b. The set
123
Span{ , , }vvv
contains infinitely many vectors.
14. The augmented matrix is found as in Exercise 13c. Since
1241 100 5
−
⎡⎤⎡⎤
4.1 • Solutions 199
15. Since the zero vector is not in W, W is not a vector space.
17. Since a vector w in W may be written as
210
031
abc
−
⎡⎤ ⎡⎤ ⎡⎤
⎢⎥ ⎢⎥ ⎢⎥
−
⎢⎥ ⎢⎥ ⎢⎥
=++
w
18. Since a vector w in W may be written as
43 0
00 0
abc
⎡⎤ ⎡⎤ ⎡ ⎤
⎢⎥ ⎢⎥ ⎢ ⎥
⎢⎥ ⎢⎥ ⎢ ⎥
=++
w
19. Let H be the set of all functions described by
12
() cos sin .yt c t c t
ωω
=+
Then H is a subset of the
20. a. The following facts about continuous functions must be shown.
2. The sum of two continuous functions is continuous.
b. Let H = {f in C[a, b]: f(a) = f(b)}.
1. Let g(t) = 0 for all t in [a, b]. Then g(a) = g(b) = 0, so g is in H.
21. The set H is a subspace of
22
.M×
The zero matrix is in H, the sum of two upper triangular matrices is
22. The set H is a subspace of
24
.M×
The 2 × 4 zero matrix 0 is in H because F0 = 0. If A and B are
23. a. False. The zero vector in V is the function f whose values f(t) are zero for all t in .
b . False. An arrow in three-dimensional space is an example of a vector, but not every arrow is a
vector.
24. a. True. See the definition of a vector space.
b . True. See statement (3) in the box before Example 1.
25. 2, 4
26. a. 3
27. a. 8
28. a. 4
b . 7
29. Consider u + (–1)u. By Axiom 10, u + (–1)u = 1u + (–1)u. By Axiom 8, 1u + (–1)u = (1 + (–1))u =
30. By Axiom 10 u = 1u. Since c is nonzero,
11cc
−=
, and
1
()cc
−
=uu
. By Axiom 9,
31. Any subspace H that contains u and v must also contain all scalar multiples of u and v, and hence
must also contain all sums of scalar multiples of u and v. Thus H must contain all linear
32. Both H and K contain the zero vector of V because they are subspaces of V. Thus the zero vector of V
is in H ∩ K. Let u and v be in H ∩ K. Then u and v are in H. Since H is a subspace u + v is in H.
Likewise u and v are in K. Since K is a subspace u + v is in K. Thus u + v is in H ∩ K. Let u be in H
∩ K. Then u is in H. Since H is a subspace cu is in H. Likewise u is in K. Since K is a subspace cu is
33. a. Given subspaces H and K of a vector space V, the zero vector of V belongs to H + K, because 0 is
in both H and K (since they are subspaces) and 0 = 0 + 0. Next, take two vectors in H + K, say
111
=+wuv
and
222
=+wuv
where
1
u
and
2
u
are in H, and
1
v
and
2
v
are in K. Then
121122 12 12
()()+ =++ + = + + +ww uvu v uu vv
because vector addition in V is commutative and associative. Now
12
+uu
is in H and
12
+vv
is
in K because H and K are subspaces. This shows that
12
+ww
is in H + K. Thus H + K is closed
under addition of vectors. Finally, for any scalar c,
11111
()cc cc=+=+wuvuv
34. A proof that
11
Span{ , , , , , }
pq
HK+= … …uuvv
has two parts. First, one must show that H + K is a
subset of
11
Span{ , , , , , }.
pq
……uuvv
Second, one must show that
11
Span{ , , , , , }
pq
……uuvv
is a
subset of H + K.
(1) A typical vector in H has the form
11 pp
cc+…+uu
and a typical vector in K has the form
11
.
qq
dd+…+vv
The sum of these two vectors is a linear combination of
11
,, ,,,
pq
……uuvv
(2) Each of the vectors
11
,, ,,,
pq
……uuvv
belongs to H + K, by Exercise 33(b), and so any linear
combination of these vectors belongs to H + K, since H + K is a subspace, by Exercise 33(a).
35. [M] Since
84 79 1001
43 64 0102
−−
⎡⎤⎡⎤
⎢⎥⎢⎥
−− −
⎢⎥⎢⎥
36. [M] Since
3594 1001/5
−−− −
⎡⎤⎡⎤
37. [M] The graph of f(t) is given below. A conjecture is that f(t) = cos 4t.
–1
–
0.5
1
–
0.5
1
4.2 • Solutions 203
38. [M] The graph of f(t) is given below. A conjecture is that f(t) = sin 3t.
The graph of g(t) is given below. A conjecture is that g(t) = cos 4t.
The graph of h(t) is given below. A conjecture is that h(t) = sin 5t.
4.2 SOLUTIONS
Notes:
This section provides a review of Chapter 1 using the new terminology. Linear tranformations are
introduced quickly since students are already comfortable with the idea from
n
. The key exercises are
17–26, which are straightforward but help to solidify the notions of null spaces and column spaces.
1. One calculates that
3531 0
−−
⎡⎤⎡⎤⎡⎤
2. One calculates that
26410
⎡⎤⎡⎤⎡⎤
1 2 3 4 5 6
–1
0.5
1 2 3 4 56
–1
–
1 2 3 4 56
–1
–
1
204 CHAPTER 4 • Vector Spaces
3. First find the general solution of Ax = 0 in terms of the free variables. Since
10 2 40
⎡⎤
the general solution is
134
24xxx=−
,
234
32xxx=− +
, with
3
x
and
4
x
free. So
1
2
24
32
x
xxx
−
⎡⎤ ⎡⎤ ⎡⎤
⎢⎥ ⎢⎥ ⎢⎥
−
⎢⎥ ⎢⎥ ⎢⎥
and a spanning set for Nul A is
24
32
⎧⎫−
⎡⎤⎡⎤
⎪⎪
⎢⎥⎢⎥
−
⎪⎪
⎢⎥⎢⎥
4. First find the general solution of Ax = 0 in terms of the free variables. Since
13000
⎡⎤
the general solution is
12
3xx=
,
30x=
, with
2
x
and
4
x
free. So
1
2
30
10
x
xxx
⎡⎤ ⎡⎤ ⎡⎤
⎢⎥ ⎢⎥ ⎢⎥
and a spanning set for Nul A is
30
10
⎧⎫
⎡⎤⎡⎤
⎪⎪
⎢⎥⎢⎥
5. First find the general solution of Ax = 0 in terms of the free variables. Since
140200
−
⎡⎤
4.2 • Solutions 205
1
2
42
10
x
x
−
⎡⎤ ⎡⎤ ⎡ ⎤
⎢⎥ ⎢⎥ ⎢ ⎥
⎢⎥ ⎢⎥ ⎢ ⎥
and a spanning set for Nul A is
42
10
⎧⎫
−
⎡⎤⎡ ⎤
⎪⎪
⎢⎥⎢ ⎥
6. First find the general solution of Ax = 0 in terms of the free variables. Since
10 5 6 10
−
⎡⎤
the general solution is
1345
56xxxx=− + −
,
234
3xxx=−
, with
3
x
,
4
x
, and
5
x
free. So
1
561
310
x
x
−−
⎡⎤ ⎡⎤ ⎡⎤ ⎡⎤
⎢⎥ ⎢⎥ ⎢⎥ ⎢⎥
−
and a spanning set for Nul A is
561
310
⎧⎫
−−
⎡⎤⎡⎤⎡⎤
⎪⎪
⎢⎥⎢⎥⎢⎥
−
⎪⎪
⎢⎥⎢⎥⎢⎥
7. The set W is a subset of
3
. If W were a vector space (under the standard operations in
3
), then it
8. The set W is a subset of
3
. If W were a vector space (under the standard operations in
3
), then it
9. The set W is the set of all solutions to the homogeneous system of equations p – 3q – 4s = 0,
⎣
1340
⎡
⎤
4
206 CHAPTER 4 • Vector Spaces
10. The set W is the set of all solutions to the homogeneous system of equations 3a + b – c = 0,
11. The set W is a subset of
4
. If W were a vector space (under the standard operations in
4
), then it
12. The set W is a subset of
4
. If W were a vector space (under the standard operations in
4
), then it
13. An element w on W may be written as
1616
0101
⎡
⎢
⎢
⎢
⎣
c
cd d
−−
⎡⎤ ⎡ ⎤ ⎡ ⎤
⎡⎤
⎢⎥ ⎢ ⎥ ⎢ ⎥
=+ = ⎢⎥
w
14. An element w on W may be written as
1313
1212
5151
s
st t
−−
⎡⎤ ⎡⎤⎡ ⎤
⎡⎤
⎢⎥ ⎢⎥⎢ ⎥
=+−=−
⎢⎥
⎢⎥ ⎢⎥⎢ ⎥
⎣⎦
⎢⎥ ⎢⎥⎢ ⎥
−−
⎣⎦ ⎣⎦⎣ ⎦
w
⎡
⎢
⎢
⎢
⎣
15. An element in this set may be written as
⎢
⎢
⎣
⎡
⎢
⎢
⎢
⎢
⎣
021021
112112
r
⎡⎤ ⎡⎤ ⎡⎤⎡ ⎤
⎡
⎤
⎢⎥ ⎢⎥ ⎢⎥⎢ ⎥
−−
⎢
⎥
16. An element in this set may be written as
4.2 • Solutions 207
⎢
⎢
⎢
⎣
⎡
⎢
⎢
⎢
⎢
⎣
110110
20 3203
b
−−
⎡⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤
⎡
⎤
⎢⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥
17. The matrix A is a 4 × 2 matrix. Thus
.
18. The matrix A is a 4 × 3 matrix. Thus
.
19. The matrix A is a 2 × 5 matrix. Thus
20. The matrix A is a 1 × 5 matrix. Thus
21. Either column of A is a nonzero vector in Col A. To find a nonzero vector in Nul A, find the general
solution of Ax = 0 in terms of the free variables. Since
12/30
000
−
⎡⎤
⎢⎥
22. Any column of A is a nonzero vector in Col A. To find a nonzero vector in Nul A, find the general
solution of Ax = 0 in terms of the free variables. Since
10 10
⎡⎤
⎢⎥
208 CHAPTER 4 • Vector Spaces
23. Consider the system with augmented matrix
[]
Aw
. Since
[]
121
,
000
A−−
⎡⎤
∼
⎢⎥
⎣⎦
w
24. Consider the system with augmented matrix
[]
Aw
. Since
100 11
010 11
−
⎡⎤
⎢⎥
−
25. a. True. See the definition before Example 1.
b . False. See Theorem 2.
26. a. True. See Theorem 2.
b . True. See Theorem 3.
27. Let A be the coefficient matrix of the given homogeneous system of equations. Since Ax = 0 for
3
2
1
⎡⎤
⎢⎥
=⎢⎥
⎢⎥
−
x
, x is in Nul
A. Since Nul
A is a subspace of
3
, it is closed under scalar multiplication. Thus
28. Let A be the coefficient matrix of the given systems of equations. Since the first system has a
solution, the constant vector
0
1
9
⎡⎤
⎢⎥
=⎢⎥
⎢⎥
⎣⎦
b
is in Col
A. Since Col A is a subspace of
3
, it is closed under
29. a. Since ,A=00
the zero vector is in Col A.
30. Since
()
VW
T=00
, the zero vector
W
0
of W is in the range of T. Let T(x) and T(w) be typical
elements in the range of T. Then since
() ( ) ( ), () ( )
TT T TT
+=+ +
xwxwxw
is in the range of T and
31. a. Let p and q be arbitary polynomials in
2
, and let c be any scalar. Then
( )(0) (0) (0) (0) (0)
() ()()
( )(1) (1) (1) (1) (1)
TTT
++
⎡⎤⎡ ⎤⎡⎤⎡⎤
+= = = + = +
⎢⎥⎢ ⎥⎢⎥⎢⎥
++
⎣⎦⎣ ⎦⎣⎦⎣⎦
pq p q p q
pq p q
pq p q p q
and
32. Any quadratic polynomial q for which
(0) 0=
q
will be in the kernel of T. The polynomial q must
33. a. For any A and B in
22
M×
and for any scalar c,
()()() ( )( )()()
TTTTT
TABABABABABAA BBTATB
+=+++ =++ + =+ ++ = +
and
b. Let B be an element of
22
M×
with
,
T
BB=
and let
1
2
.
AB
=
Then
c. Part b. showed that the range of T contains the set of all B in
22
M×
with
.
T
BB=
It must also be
shown that any B in the range of T has this property. Let B be in the range of T. Then B = T(A) for
some A in
22
.M×
Then
,
T
BAA=+
and
d. Let
ab
Acd
⎡⎤
=
⎢⎥
⎣⎦
be in the kernel of T. Then
() 0
T
TA A A
=+ =
, so
34. Let f and g be any elements in C[0, 1] and let c be any scalar. Then T(f) is the antiderivative F of f
with F(0) = 0 and T(g) is the antiderivative G of g with G(0) = 0. By the rules for antidifferentiation
+FG
will be an antiderivative of
,+
fg
and
()(0)(0)(0)000.+=+=+=
FG F G
Thus
35. Since U is a subspace of V,
V
0
is in U. Since T is linear,
() .
VW
T=00
So
W
0
is in T(U). Let T(x) and
T(y) be typical elements in T(U). Then x and y are in U, and since U is a subspace of V, +xy
is also
4.2 • Solutions 211
36. Since Z is a subspace of W,
W
0
is in Z. Since T is linear,
() .
VW
T=00
So
V
0
is in U. Let x and y be
typical elements in U. Then T(x) and T(y) are in Z, and since Z is a subspace of W,
() ()
TT
+
xy
is
37. [M] Consider the system with augmented matrix
[]
.
Aw
Since
[]
1 0 0 1/95 1/95
0 1 0 39 /19 20 /19 ,
0 0 1 267 / 95 172 / 95
000 0 0
A
−
⎡⎤
⎢⎥
−
⎢⎥
∼
⎢⎥
−
⎢⎥
⎢⎥
w
38. [M] Consider the system with augmented matrix
[]
Aw
. Since
[]
10 10 2
01 20 3
,
00 01 1
A
−−
⎡⎤
⎢⎥
−−
⎢⎥
∼
⎢⎥
⎢⎥
w
39. [M]
a. To show that
3
a
and
5
a
are in the column space of B, we can row reduce the matrices
[]
3
Ba
and
[]
3
Ba
:
212 CHAPTER 4 • Vector Spaces
[]
5
100 10/3
010 26/3
001 4
000 0
B
⎡
⎤
⎢
⎥
−
⎢
⎥
∼
⎢
⎥
−
⎢
⎥
⎢
⎥
⎣
⎦
a
b. We find the general solution of Ax = 0 in terms of the free variables by using the reduced row
echelon form of A given above:
135
(1/3) (10/3)xxx=− −
,
235
(1/3) (26/3)xxx=− +
,
45
4xx=
with
3
x
and
5
x
free. So
⎢
⎢
⎢
⎢
⎣
1
2
1/3 10/3
1/3 26/3
x
x
−−
⎡⎤
⎡
⎤⎡ ⎤
⎢⎥
⎢
⎥⎢ ⎥
−
⎢⎥
⎢
⎥⎢ ⎥
and a spanning set for Nul A is
1/3 10/3
1/3 26/3
⎧⎫
−−
⎡⎤⎡ ⎤
⎪⎪
⎢⎥⎢ ⎥
−
⎪⎪
c. The reduced row echelon form of A shows that the columns of A are linearly dependent and do
not span
4
. Thus by Theorem 12 in Section 1.9, T is neither one-to-one nor onto.
4.3 • Solutions 213
40. [M] Since the line lies both in
12
Span{ , }H=vv
and in
34
Span{ , }K=vv
, w can be written both as
11 2 2
cc+vv
and
33 44
cc+vv
. To find w we must find the c
j
’s which solve
11 22 33 44
cc c c+−−=vvvv0
. Row reduction of
[]
1234
−−
vv v v0
yields
51 2 00 100 10/30
−−
⎡⎤⎡ ⎤
4.3 SOLUTIONS
Notes:
The definition for basis is given initially for subspaces because this emphasizes that the basis
elements must be in the subspace. Students often overlook this point when the definition is given for a
vector space (see Exercise 25). The subsection on bases for Nul A and Col A is essential for Sections 4.5
1. Consider the matrix whose columns are the given set of vectors. This 3 × 3 matrix is in echelon form,
2. Since the zero vector is a member of the given set of vectors, the set cannot be linearly independent
and thus cannot be a basis for
3
. Now consider the matrix whose columns are the given set of
3. Consider the matrix whose columns are the given set of vectors. The reduced echelon form of this
matrix is
132 101
−
⎡⎤⎡⎤
4. Consider the matrix whose columns are the given set of vectors. The reduced echelon form of this
matrix is
228 100
−
⎡⎤⎡⎤
5. Since the zero vector is a member of the given set of vectors, the set cannot be linearly independent
and thus cannot be a basis for
3
. Now consider the matrix whose columns are the given set of
6. Consider the matrix whose columns are the given set of vectors. Since the matrix cannot have a pivot
in each row, its columns cannot span
3
; thus the given set of vectors is not a basis for
3
. The
reduced echelon form of the matrix is
14 10
−
⎡⎤⎡⎤
7. Consider the matrix whose columns are the given set of vectors. Since the matrix cannot have a pivot
in each row, its columns cannot span
3
; thus the given set of vectors is not a basis for
3
. The
reduced echelon form of the matrix is
26 10
−
⎡⎤⎡⎤
8. Consider the matrix whose columns are the given set of vectors. Since the matrix cannot have a pivot
in each column, the set cannot be linearly independent and thus cannot be a basis for
3
. The reduced
echelon form of this matrix is
1020 1020
⎡⎤⎡⎤
9. We find the general solution of Ax = 0 in terms of the free variables by using the reduced echelon
form of A:
10 2 2 10 20
0114 0110.
−− −
⎡⎤⎡⎤
⎢⎥⎢⎥
∼
⎢⎥⎢⎥
4.3 • Solutions 215
1
2
2
1,
x
xx
⎡⎤ ⎡⎤
⎢⎥ ⎢⎥
−
⎢⎥ ⎢⎥
and a basis for Nul A is
2
1.
⎧⎫
⎡⎤
⎪⎪
⎢⎥
−
10. We find the general solution of Ax = 0 in terms of the free variables by using the reduced echelon
form of A:
11 2 1 5 100 2 9
−−
⎡⎤⎡⎤
So
145
29xxx=− +
,
24 5
10xx x=−
,
35
2xx=
, with
4
x
and
5
x
free. So
1
2
29
110
x
x
−
⎡⎤ ⎡⎤ ⎡ ⎤
⎢⎥ ⎢⎥ ⎢ ⎥
−
and a basis for Nul A is
29
110
,.
02
⎧⎫
−
⎡⎤⎡ ⎤
⎪⎪
⎢⎥⎢ ⎥
−
⎪⎪
⎢⎥⎢ ⎥
⎪⎪
⎢⎥⎢ ⎥
⎨⎬
11. Let
[]
132
A
=−
. Then we wish to find a basis for Nul A. We find the general solution of Ax = 0
in terms of the free variables: x = 3y – 2z with y and z free. So
32
x
−
⎡⎤ ⎡⎤ ⎡ ⎤
216 CHAPTER 4 • Vector Spaces
12. We want to find a basis for the set of vectors in
2
in the line 3x + y = 0. Let
[]
31
A
=
. Then we
wish to find a basis for Nul A. We find the general solution of Ax = 0 in terms of the free variables: y
= – 3x with x free. So
1,
3
xx
y
⎡⎤ ⎡ ⎤
==
⎢⎥ ⎢ ⎥
−
⎣⎦ ⎣ ⎦
x
13. Since B is a row echelon form of A, we see that the first and second columns of A are its pivot
columns. Thus a basis for Col A is
24
⎧⎫−
⎡⎤⎡⎤
To find a basis for Nul A, we find the general solution of Ax = 0 in terms of the free variables:
134
65,xxx=− −
234
(5/2) (3/2) ,xxx=− −
with
3
x
and
4
x
free. So
1
65
x
−−
⎡⎤ ⎡ ⎤ ⎡ ⎤
and a basis for Nul A is
65
5/2 3/2
,.
⎧⎫−−
⎡⎤⎡⎤
⎪⎪
⎢⎥⎢⎥
−−
⎪⎪
⎢⎥⎢⎥
⎨⎬
14. Since B is a row echelon form of A, we see that the first, third, and fifth columns of A are its pivot
columns. Thus a basis for Col A is
138
108
,,.
⎧⎫
⎡⎤⎡ ⎤⎡⎤
⎪⎪
⎢⎥⎢ ⎥⎢⎥
⎪⎪
⎢⎥⎢ ⎥⎢⎥