4.6 • Solutions 237
A basis for Row A is the pivot rows of B:
{
}
(2,6, 6,6,3,6),(0,3,0,3,3,0),(0,0,0,0,3,0) .−
To find a
basis for Nul A row reduce to reduced echelon form:
10 3003
−
⎡⎤
6
x
303
010
⎧⎫
−
⎡⎤ ⎡ ⎤⎡ ⎤
⎪⎪
⎢⎥ ⎢ ⎥⎢ ⎥
−
⎪⎪
⎢⎥ ⎢ ⎥⎢ ⎥
4. The matrix B is in echelon form. There are five pivot columns, so the dimension of Col A is 5. There
are five pivot rows, so the dimension of Row A is 5. There is one column without a pivot, so the
equation
A=x0
has one free variable. Thus the dimension of Nul A is 1. A basis for Col A is the
pivot columns of A:
11212
12 12 1
⎧⎫
−−
⎡⎤⎡⎤⎡⎤⎡⎤⎡⎤
⎪⎪
⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥
−−
⎪⎪
⎣⎦⎣⎦⎣⎦⎣⎦⎣⎦
⎩⎭
A basis for Row A is the pivot rows of B:
{
}
(1,1, 2,0,1, 2),(0,1, 1,0, 3, 1),(0,0,1,1, 13, 1),(0,0,0,0,1, 1),(0,0,0,0,0,1) .−− −−− −− −
To find a basis for Nul A row reduce to reduced echelon form:
100 100
010100
000001
⎡⎤
⎢⎥
⎣⎦
The solution to
A=x0
in terms of free variables is
14
xx
=−
,
24
xx
=−
,
34
xx
=−
,
5
0
x
=
,
6
0
x
=
,
238 CHAPTER 4 • Vector Spaces
1
1
⎧⎫
−
⎡⎤
⎪⎪
⎢⎥
−
⎪⎪
⎢⎥
5. By the Rank Theorem,
dimNul 7 rank 7 3 4.AA=− =−=
Since
6. By the Rank Theorem,
dimNul 5 rank 5 2 3.AA=− =−=
Since
7. Yes, Col A =
4
. Since A has four pivot columns,
dimCol 4.A=
Thus Col A is a four-dimensional
No,
8. Since A has four pivot columns,
rank 4,A=
and
dimNul 8 rank 8 4 4.AA=− =−=
9. Since
dimNul 3, rank 6 dimNul 6 3 3.AA A==− =−=
So
dimCol rank 3.AA==
10. Since
dimNul 5, rank 7 dimNul 7 5 2.AA A==− =−=
So
dimCol rank 2.AA==
12. Since
dimNul 2, rank 4 dimNul 4 2 2.AA A==− =−=
So
dimRow dimCol rank 2.AAA===
13. The rank of a matrix A equals the number of pivot positions which the matrix has. If A is either a
14. The dimension of the row space of a matrix A is equal to rank A, which equals the number of pivot
15. Since the rank of A equals the number of pivot positions which the matrix has, and A could have at
16. Since the rank of A equals the number of pivot positions which the matrix has, and A could have at
17. a. True. The rows of A are identified with the columns of
.
T
A
See the paragraph before Example 1.
b . False. See the warning after Example 2.
4.6 • Solutions 239
e . True. See the Numerical Note before the Practice Problem.
18. a. False. Review the warning after Theorem 6 in Section 4.3.
b . False. See the warning after Example 2.
19. Yes. Consider the system as ,A=x0
where A is a
56×
matrix. The problem states that
dimNul 1A=
. By the Rank Theorem,
rank 6 dimNul 5.AA=− =
Thus
dim Col rank 5,AA==
and
20. No. Consider the system as ,A=xb
where A is a
68×
matrix. The problem states that
dimNul 2.A=
By the Rank Theorem,
rank 8 dimNul 6.AA=− =
Thus
dimCol rank 6,AA==
and
21. No. Consider the system as ,A=xb
where A is a
910×
matrix. Since the system has a solution for
22. No. Consider the system as ,A=x0
where A is a
10 12×
matrix. Since A has at most 10 pivot
23. Yes, six equations are sufficient. Consider the system as ,A=x0
where A is a
12 8×
matrix. The
problem states that
dimNul 2.A=
By the Rank Theorem,
rank 8 dimNul 6.AA=− =
Thus
24. Yes, No. Consider the system as ,A=xb
where A is a
76×
matrix. Since A has at most 6 pivot
positions,
rank 6.A≤
By the Rank Theorem,
dim Nul 6 rank 0.AA=− ≥
If
dimNul 0,A=
then the
25. No. Consider the system as ,A=xb
where A is a
10 12×
matrix. The problem states that
dim Nul 3.A=
By the Rank Theorem,
dimCol rank 12 dimNul 9.AA A==− =
Thus Col A will be a
26. Consider the system ,A=x0
where A is a mn× matrix with .mn> Since the rank of A is the
number of pivot positions that A has and A is assumed to have full rank,
rank .An=
By the Rank
27. Since A is an m × n matrix, Row A is a subspace of
n
, Col A is a subspace of
m
, and Nul A is a
subspace of
n
. Likewise since
T
A
is an n × m matrix,
Row
T
A
is a subspace of
m
,
Col
T
A
is a
28. a. Since A is an m × n matrix and dimRow A = rank A,
dimRow A + dimNul A = rank A + dimNul A = n.
29. Let A be an m × n matrix. The system Ax = b will have a solution for all b in
m
if and only if A has a
pivot position in each row, which happens if and only if dimCol A = m. By Exercise 28 b., dimCol A
30. The equation Ax = b is consistent if and only if
[]
rank rankAA=
b
because the two ranks will be
31. Compute that
[]
2222
3333.
⎢
⎣
T
abc
abc a b c
⎡
⎤⎡ ⎤
⎢
⎥⎢ ⎥
=− =− − −
⎢
⎥⎢ ⎥
uv
Each column of
T
uv
is a multiple of u,
32. Note that the second row of the matrix is twice the first row. Thus if v = (1, –3, 4), which is the first
row of the matrix,
33. Let
[]
123
,A=
uu u
and assume that rank A = 1. Suppose that
1
≠u0
. Then
1
{}u
is basis for Col
4.6 • Solutions 241
If
1
=u0
but
2
≠u0
, then similarly
2
{}u
is basis for Col A, since Col A is assumed to be one-
⎡
⎢
⎢
⎢
0
⎡⎤
⎣
⎦
34. Let A be an m × n matrix with of rank r > 0, and let U be an echelon form of A. Since A can be
reduced to U by row operations, there exist invertible elementary matrices
1
,,
p
EE…
with
1
().
p
EEAU⋅⋅⋅ =
Thus
1
1
(),
p
AE EU
−
=⋅⋅⋅
since the product of invertible matrices is invertible. Let
1
1
()
p
EE E
−
=⋅⋅⋅
; then A = EU. Let the columns of E be denoted by
1
,,.
m
…cc
Since the rank of A is
35. [M]
a. Begin by reducing A to reduced echelon form:
1013/20 50 3
0 1 11/2 0 1/2 0 2
−
⎡⎤
⎢⎥
A basis for Col A is the pivot columns of A, so matrix C contains these columns:
7953
4625
.
5752
C
−−
⎡⎤
⎢⎥
−−−
⎢⎥
⎢⎥
=−
242 CHAPTER 4 • Vector Spaces
1013/20 50 3
0111/20 1/20 2
−
⎡⎤
⎢⎥
To find a basis for Nul A row reduce to reduced echelon form, note that the solution to Ax = 0
in terms of free variables is
1357
(13 / 2) 5 3 ,
xxxx
=− − +
2357
(11 / 2) (1 / 2) 2 ,
xxxx
=− − −
010
001
001
⎢⎥
⎢⎥
⎢⎥
−
⎢⎥
⎢⎥
⎣⎦
b. The reduced echelon form of
T
A
is
1000 2/11
0100 41/11
0010 0
−
⎡⎤
⎢⎥
−
⎢⎥
⎢⎥
⎣⎦
so the solution to
T
A
=x0
in terms of free variables is
15
(2/11) ,
xx
=
25
(41/11) ,
xx
=
3
0,
x
=
45
(28/11) ,
xx
=−
with
5
x
free. Thus matrix M is
2/11
41/11
⎡⎤
⎢⎥
⎢⎥
The matrix
T
SR N
⎡
⎤
=
⎣
⎦ is 7 × 7 because the columns of
T
R
and N are in
7
and dimRow A
36. [M] Answers will vary, but in most cases C will be 6 × 4, and will be constructed from the first 4
37. [M] The C and R from Exercise 35 work here, and A = CR.
38. [M] If A is nonzero, then A = CR. Note that
[]
12 n
CR C C C=…rr r
, where
1
r
, …,
n
r
are the
columns of R. The columns of R are either pivot columns of R or are not pivot columns of R.
Consider first the pivot columns of R. The
th
i
pivot column of R is
i
e
, the
th
i
column in the
4.7 SOLUTIONS
Notes:
This section depends heavily on the coordinate systems introduced in Section 4.4. The row
reduction algorithm that produces
cB
P
←
can also be deduced from Exercise 15 in Section 2.2, by row
1. a. Since
112
62=−bcc
and
212
94,=−bcc
1
6
[] ,
2
C
⎡
⎤
=
⎢
⎥
−
⎣
⎦
b
2
9
[] ,
4
C
⎡
⎤
=
⎢
⎥
−
⎣
⎦
b
and
69
.
24
CB
P
←
⎡⎤
=
⎢⎥
−−
⎣⎦
⎡
⎢
⎣
2. a. Since
112
24=− +bcc
and
212
36,=−bcc
1
2
[] ,
4
C
−
⎡
⎤
=
⎢
⎥
⎣
⎦
b
2
3
[] ,
6
C
⎡
⎤
=
⎢
⎥
−
⎣
⎦
b
and
23
.
46
CB
P
←
−
⎡⎤
=
⎢⎥
−
⎣⎦
b. Since
12
23,=+xb b
2
[] 3
B
⎡⎤
=
⎢⎥
x
and
244 CHAPTER 4 • Vector Spaces
5. a. Since
112
4,=−abb
2123
,=− + +abbb
and
32 3
2,=−ab b
1
4
[] 1,
0
B
⎡
⎤
⎢
⎥
=−
⎢
⎥
⎢
⎥
⎣
a
2
1
[] 1,
1
B
−
⎡⎤
⎢⎥
=⎢⎥
⎢⎥
a
b. Since
123
34 ,=+ +xa aa
3
[] 4
1
A
⎡⎤
⎢⎥
=⎢⎥
⎢⎥
⎣⎦
x
and
⎡
⎢
⎢
⎢
⎣
6. a. Since
1123
2,=−+fddd
223
3,=+fdd
and
313
32=− +fdd
,
1
2
[] 1,
1
D
⎡
⎤
⎢
⎥
=−
⎢
⎥
⎢
⎥
⎣
⎦
f
2
0
[] 3,
1
D
⎡⎤
⎢⎥
=⎢⎥
⎢⎥
⎣⎦
f
b. Since
123
22,=− +xf f f
1
[] 2
2
F
⎡⎤
⎢⎥
=−
⎢⎥
⎢⎥
⎣⎦
x
and
7. To find
,
CB
P
←
row reduce the matrix
[]
12 1 2
cc bb
:
8. To find
CB
P
←
, row reduce the matrix
[]
12 1 2
cc bb
:
10 9 8
−
⎡⎤
9. To find
CB
P
←
, row reduce the matrix
[]
12 1 2
cc bb
:
[]
12 1 2
102 3
.
010 1
⎡⎤
∼
⎢⎥
−
⎣⎦
cc bb
10. To find
CB
P
←
, row reduce the matrix
[]
12 1 2
cc bb
:
[]
12 1 2
10 3 1
.
01 20
⎡⎤
∼
⎢⎥
−
⎣⎦
cc bb
11. a. False. See Theorem 15.
b. True. See the first paragraph in the subsection “Change of Basis in
n
.”
12. a. True. The columns of
P
are coordinate vectors of the linearly independent set B. See the
13. Let
222
123
{, , }{12 ,35 4,2 3}Btttttt==−+−++bb b
and let
2
123
{, , }{1,, }.Ctt==cc c
The
C-coordinate vectors of
1
,b
2
,b
and
3
b
are
So
130
252
143
CB
P
←
⎡⎤
⎢⎥
=− −
⎢⎥
⎢⎥
⎣⎦
Let x = –1 + 2t. Then the coordinate vector
[]
B
x
satisfies
246 CHAPTER 4 • Vector Spaces
14. Let
22
123
{, , }{13,2 5,12}Btttt==−+−+bb b
and let
2
123
{, , }{1,, }.Ctt==cc c
The C–coordinate
vectors of
b
,
b
, and
b
are
So
121
012
350
CB
P
←
⎡⎤
⎢⎥
=⎢⎥
⎢⎥
−−
⎣⎦
Let
2
.
t
=x
Then the coordinate vector
[]
B
x
satisfies
15. (a) B is a basis for V
(b) the coordinate mapping is a linear transformation
16. (a)
11
[] []
CB
QQQ
1
1
⎡⎤
⎢⎥
0
⎢⎥
===
⎢⎥
⎢⎥
0
⎢⎥
bb e
#
17. [M]
a. Since we found P in Exercise 34 of Section 4.5, we can calculate that
4.7 • Solutions 247
⎢
⎢
⎢
⎢
⎢
⎢
32 0 16 0 12 0 10
0320240200
0000001
⎡
⎤
⎢
⎥
⎢
⎥
⎢
⎥
⎣
⎦
b. Since P is the change-of-coordinates matrix from C to B,
1
P−
will be the change-of-
coordinates matrix from B to C. By Theorem 15, the columns of
1
P−
will be the C–coordinate
vectors of the basis vectors in B. Thus
2
1
cos (1 cos 2 )
2
tt=+
3
1
cos (3cos cos 3 )
ttt=+
18. [M] The C-coordinate vector of the integrand is (0, 0, 0, 5, –6, 5, –12). Using
1
P−
from the previous
exercise, the B– coordinate vector of the integrand will be
1(0,0,0,5, 6,5, 12) ( 6,55/8, 69/8,45/16, 3,5/16, 3/8)P−−−=− − − −
Thus the integral may be rewritten as
19. [M]
a. If C is the basis
123
{, , },vv v
then the columns of P are
1
[],
C
u
2
[],
C
u
and
3
[].
C
u
So
[]
1231
[],
jC
=uvvvu
and
[][]
123 123
.P=uu u vv v
In the current exercise,
248 CHAPTER 4 • Vector Spaces
b. Analogously to part a.,
[][ ]
123 1 2 3
,P=vv v ww w
so
[]
123
=ww w
[]
1
123
.P
−
vv v
In the current exercise,
20. a.
DB DCCB
PPP
←←←
=
Let x be any vector in the two-dimensional vector space. Since
CB
P
←
is the change-of-coordinates
matrix from B to C and
DC
P
←
is the change-of-coordinates matrix from C to D,
for any vector
[]
B
x
in
2
, and
DB DCCB
PPP
←←←
=
b. [M] For example, let 73
,,
51
B⎧⎫−
⎡⎤⎡ ⎤
⎪⎪
=⎨⎬
⎢⎥⎢ ⎥
−
⎪⎪
⎣⎦⎣ ⎦
⎩⎭
12
,,
52
C
⎧
⎫−
⎡
⎤⎡ ⎤
⎪
⎪
=
⎨
⎬
⎢
⎥⎢ ⎥
−
⎪
⎪
⎣
⎦⎣ ⎦
⎩⎭
and 11
,.
85
D
⎧
⎫−
⎡⎤⎡⎤
⎪
⎪
=
⎨
⎬
⎢⎥⎢⎥
−
⎪
⎪
⎣⎦⎣⎦
⎩⎭
Then we
can calculate the change-of-coordinates matrices:
1 1 1 2 1 0 0 8/3 0 8/3
8552 01114/3 114/3
DC
P
←
−− − −
⎡⎤⎡⎤⎡⎤
∼
⇒
=
⎢⎥⎢⎥⎢⎥
−− − −
⎣⎦⎣⎦⎣⎦
1 1 7 3 1 0 40 / 3 16 / 3 40 / 3 16 / 3
P
−− − −
⎡⎤⎡ ⎤⎡⎤
∼
⇒
=
4.8 • Solutions 249
4.8 SOLUTIONS
Notes:
This is an important section for engineering students and worth extra class time. To spend only
one lecture on this section, you could cover through Example 5, but assign the somewhat lengthy
Example 3 for reading. Finding a spanning set for the solution space of a difference equation uses the
Basis Theorem (Section 4.5) and Theorem 17 in this section, and demonstrates the power of the theory of
Chapter 4 in helping to solve applied problems. This section anticipates Section 5.7 on differential
equations. The reduction of an
th
n
order difference equation to a linear system of first order difference
equations was introduced in Section 1.10, and is revisited in Sections 4.9 and 5.6. Example 3 is the
background for Exercise 26 in Section 6.5.
1. Let
2.
k
k
y=
Then
21
21
2 8 2 2(2 ) 8(2 )
kk k
kkk
yyy
++
++
+−=+ −
Since the difference equation holds for all k,
2k
is a solution.
Let
(4)
k
k
y=−
. Then
21
21
2 8 (4) 2(4) 8(4)
kkk
kkk
yyy ++
++
+−=−+−−−
2. Let
5.
k
k
y=
Then
2
225 5 25(5 )
kk
kk
yy
+
+−=−
Since the difference equation holds for all k,
5
k
is a solution.
Let
(5).
k
k
y=−
Then
2
225 ( 5) 25( 5)
kk
kk
yy +
+−=−−−
3. The signals
2k
and
(4)
k
−
are linearly independent because neither is a multiple of the other; that is,
there is no scalar c such that
2(4)
kk
c=−
for all k. By Theorem 17, the solution set H of the
4. The signals
5
k
and
(5)
k
−
are linearly independent because neither is a multiple of the other; that is,
there is no scalar c such that
5(5)
kk
c=−
for all k. By Theorem 17, the solution set H of the
5. Let
(2).
k
k
y=−
Then
21
21
4 4 (2) 4(2) 4(2)
kkk
kkk
yyy ++
++
+ + =− +− +−
Since the difference equation holds for all k,
(2)
k
−
is in the solution set H.
Let
(2).
k
k
yk=−
Then
21
21
4 4 ( 2)(2) 4( 1)(2) 4(2)
kkk
kkk
yyyk k k
++
++
++=+−++−+−
Since the difference equation holds for all k,
(2)
k
k−
is in the solution set H.
6. Let
2
4cos .
kk
k
y
π
=
Then
2
2
(2)
16 4 cos 16 4 cos
22
kk
kk
kk
yy
ππ
+
+
+⎛⎞
+= +
⎜⎟
⎝⎠
4.8 • Solutions 251
Let
2
4sin .
kk
k
y
π
= Then
2
2
(2)
16 4 sin 16 4 sin
22
kk
kk
kk
yy
ππ
+
+
+⎛⎞
+= +
⎜⎟
⎝⎠
2
(2)
44sin 16sin
22
k
kk
ππ
+
⎛⎞
=+
⎜⎟
⎝⎠
16 4 sin sin
22
k
kk
ππ
π
⎛⎞
⎛⎞
=⋅ + +
⎜⎟⎜⎟
⎝⎠
⎝⎠
7. Compute and row reduce the Casorati matrix for the signals
1,
k
2,
k
and
(2)
k
−
, setting k = 0 for
convenience:
00 0
11 1
12(2) 100
12(2) 010
⎡⎤
−⎡⎤
⎢⎥
⎢⎥
−∼
⎢⎥
8. Compute and row reduce the Casorati matrix for the signals
(1),
k
−
2,
k
and
3,
k
setting k = 0 for
convenience:
000
111
(1) 2 3 100
(1) 2 3 0 1 0
⎡⎤
−⎡⎤
⎢⎥
⎢⎥
−∼
⎢⎥
9. Compute and row reduce the Casorati matrix for the signals
2,
k
2
5cos ,
kk
and
2
5sin
kk
setting k = 0
for convenience:
00 0
11 1
22
25cos05sin0 100
2 5cos 5sin 010
ππ
⎡⎤
⎡⎤
⎢⎥
⎢⎥
∼
⎢⎥
10. Compute and row reduce the Casorati matrix for the signals
(2),
k
−
(2),
k
k−
and
3k
setting k = 0 for
convenience:
000
(2) 0(2) 3 100
⎡⎤
−− ⎡⎤
11. The solution set H of this third-order difference equation has dim H = 3 by Theorem 17. The two
signals
(1)
k
−
and
2
k
cannot possibly span a three-dimensional space, and so cannot be a basis for
H.
13. The auxiliary equation for this difference equation is
22/9 0.rr−+ =
By the quadratic formula
(or factoring), r = 2/3 or r = 1/3, so two solutions of the difference equation are
(2/3)
k
and
(1 / 3)
k
.
14. The auxiliary equation for this difference equation is
2560.rr−+=
By the quadratic formula (or
factoring), r = 2 or r = 3, so two solutions of the difference equation are
2
k
and
3.
k
The signals
2
k
4.8 • Solutions 253
15. The auxiliary equation for this difference equation is
2
620.rr+−=
By the quadratic formula (or
factoring), r = 1/2 or r = –2/3, so two solutions of the difference equation are
(1 / 2)
k
and
(2/3).
k
−
16. The auxiliary equation for this difference equation is
225 0.r−=
By the quadratic formula (or
factoring), r = 5 or r = –5, so two solutions of the difference equation are
5k
and
(5).
k
−
The signals
17. Letting a = .9 and b = 4/9 gives the difference equation
21
1.3 .4 1.
kkk
YYY
++
−+=
First we find a
particular solution
k
YT=
of this equation, where T is a constant. The solution of the equation T –
1.3T + .4T = 1 is T = 10, so 10 is a particular solution to
21
1.3 .4 1
kkk
YYY
++
−+=
. Next we solve the
two-dimensional, so the two linearly independent signals
(.8)
k
and
(.5)
k
form a basis for the
solution space of the homogeneous difference equation by the Basis Theorem. Translating the
solution space of the homogeneous difference equation by the particular solution 10 of the
18. Letting a = .9 and b = .5 gives the difference equation
21
1.35 .45 1.
kkk
YYY
++
−+=
First we find a
particular solution
k
YT=
of this equation, where T is a constant. The solution of the equation
T – 1.35T + .45T = 1 is T = 10, so 10 is a particular solution to
21
1.35 .45 1
kkk
YYY
++
−+=
. Next we
solve the homogeneous difference equation
21
1.35 .45 0.
kkk
YYY
++
−+=
The auxiliary equation for
this difference equation is
21.35 .45 0.rr−+=
By the quadratic formula (or factoring), r = .6 or
r = .75, so two solutions of the homogeneous difference equation are
.6k
and
.75 .
k
The signals
(.6)
k
19. The auxiliary equation for this difference equation is
2410.rr++=
By the quadratic formula,
23r=− +
or
23,r=− −
so two solutions of the difference equation are
(2 3)
k
−+
and
254 CHAPTER 4 • Vector Spaces
20. Let
23a=− +
and
23b=− −
. Using the solution from the previous exercise, we find that
11 2 5000ycacb=+=
and
12
0.
NN
N
ycacb=+=
This is a system of linear equations with variables
1
c
and
2
c
whose augmented matrix may be row reduced:
21. The smoothed signal
k
z has the following values:
1
(957)/37,z=++ =
2
(573)/35,z=++ =
3
(7 3 2) / 3 4,z=++ =
4
(324)/33,z=++ =
5
(246)/34,z=++ =
6
(4 6 5)/3 5,z=++ =
(6 5 7) / 3 6,z=++ =
(5 7 6) / 3 6,z=++ =
(7 6 8) / 3 7,z=++ =
(6 8 10) / 3 8,z=++ =
22. a. The smoothed signal
k
z has the following values:
0210
.35 .5 .35 .35(0) .5(.7) .35(3) 1.4,zyyy=++= ++ =