Another sensible choice for a basis of im(
A
) is
e
1
, e
2
, e
3
.
29. The three column vectors of
A
span all of
R
2
, so that im(
A
) =
R
2
. We can choose any
two of the columns of
A
to form a basis of im(
A
); another sensible choice is
e
1
, e
2
.
30. im(
A
) = span(
e
1
, e
2
)
We can choose
e
1
, e
2
as a basis of im(
A
).
31. The two column vectors of the given matrix
A
are linearly independent (they are not
parallel), so that they form a basis of im(
A
).
137
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Chapter 3
ISM:
Linear Algebra
32. By inspection, the first, third and sixth columns are redundant.
Thus, a basis of the
image consists of the remaining column vectors:
1
0
0
0
,
0
1
0
0
,
0
0
1
0
.
33. im(
A
) = span(
e
1
, e
2
, e
3
), so that
e
1
, e
2
, e
3
is a basis of im(
A
).
34. The fact that
1
2
3
4
is in ker(
A
) means that
A
1
2
3
4
= [
v
1
v
2
v
3
v
4
]
1
2
3
4
=
v
1
+ 2
v
2
+ 3
v
3
+ 4
v
4
= 0, so that
v
4
=

1
4
v
1

1
2
v
2

3
4
v
3
.
35. If
v
i
is a linear combination of the other vectors in the list,
v
i
=
c
1
v
1
+
· · ·
+
c
i

1
v
i

1
+
c
i
+1
v
i
+1
+
· · ·
+
c
n
v
n
, then we can subtract
v
i
from both sides to generate a nontrivial
relation (the coe
ffi
cient of
v
i
will be 1).
Conversely, if there is a nontrivial relation
c
1
v
1
+
· · ·
+
c
i
v
i
+
· · ·
+
c
n
v
n
= 0, with
c
i
= 0,
then we can solve for vector
v
i
and thus express
v
i
as a linear combination of the other
vectors in the list.
36. Yes; we know that there is a nontrivial relation
c
1
v
1
+
c
2
v
2
+
· · ·
+
c
m
v
m
= 0.
Now apply the transformation
T
to the vectors on both sides, and use linearity:
T
(
c
1
v
1
+
c
2
v
2
+
· · ·
+
c
m
v
m
) =
T
(0), so that
c
1
T
(
v
1
) +
c
2
T
(
v
2
) +
· · ·
+
c
m
T
(
v
m
) = 0.
This is a nontrivial relation among the vectors
T
(
v
1
)
, . . . , T
(
v
m
), so that these vectors
are linearly dependent, as claimed.
37. No; as a counterexample, consider the extreme case when
T
is the zero transformation,
that is,
T
(
x
) = 0 for all
x
. Then the vectors
T
(
v
1
)
, . . . , T
(
v
m
) will all be zero, so that
they are linearly dependent.
38. a. Using the terminology introduced in the exercise, we need to show that any vector
v
in
V
is a linear combination of
v
1
, . . . , v
m
. Choose a specific vector
v
in
V
. Since we can
find no more than
m
linearly independent vectors in
V
, the
m
+ 1 vectors
v
1
, . . . , v
m
,
v
will be linearly dependent. Since the vectors
v
1
, . . . , v
m
are independent,
v
must be
redundant, meaning that
v
is a linear combination of
v
1
, . . . , v
m
,
as claimed.
138
ISM:
Linear Algebra
Section 3.2
b. With the terminology introduced in part a, we can let
V
= im [
v
1
· · ·
v
m
]
.
39. Yes; the vectors are linearly independent. The vectors in the list
v
1
, . . . , v
m
are linearly
independent (and therefore nonredundant), and
v
is nonredundant since it fails to be
in the span of
v
1
, . . . , v
m
.
40. Yes; by Fact 3.2.8, ker(
A
) =
{
0
}
and ker(
B
) =
{
0
}
. Then ker(
AB
) =
{
0
}
by Exer
cise 3.1.51, so that the columns of
AB
are linearly independent, by Fact 3.2.8.