SSM:
Linear Algebra
Section 4.1
Chapter 4
4.1
1. Not a subspace since it does not contain the neutral element, that is, the function
f
(
t
) = 0,
for all
t
.
3. This subset
V
is a subspace of
P
2
:
•
The neutral element
f
(
t
) = 0 (for all
t
) is in
V
since
f
0
(1) =
f
(2) = 0.
•
If
f
and
g
are in
V
(so that
f
0
(1) =
f
(2) and
g
0
(1) =
g
(2)), then
(
f
+
g
)
0
(1) = (
f
0
+
g
0
)(1) =
f
0
(1) +
g
0
(1) =
f
(2) +
g
(2) = (
f
+
g
)(2), so that
f
+
g
is in
V
.
•
If
f
is in
V
(so that
f
0
(1) =
f
(2)) and
k
is any constant, then (
kf
)
0
(1) = (
kf
0
)(1) =
kf
0
(1) =
kf
(2) = (
kf
)(2), so that
kf
is in
V
.
If
f
(
t
) =
a
+
bt
+
ct
2
then
f
0
(
t
) =
b
+ 2
ct
, and
f
is in
V
if
f
0
(1) =
b
+ 2
c
=
a
+ 2
b
+ 4
c
=
f
(2), or
a
+
b
+ 2
c
= 0. The general element of
V
is of the
form
f
(
t
) = (

b

2
c
) +
bt
+
ct
2
=
b
(
t

1) +
c
(
t
2

2), so that
t

1
, t
2

2 is a basis of
V
.
5. If
p
(
t
) =
a
+
bt
+
ct
2
then
p
(

t
) =
a

bt
+
ct
2
and

p
(

t
) =

a
+
bt

ct
2
.
Comparing coefficients we see that
p
(
t
) =

p
(

t
) for all
t
if (and only if)
a
=
c
= 0.
The general element of the subset is of the form
p
(
t
) =
bt
.
These polynomials form a subspace of
P
2
, with basis
t
.
7. The set
V
of diagonal 3
×
3 matrices is a subspace of
R
3
×
3
:
a. The zero matrix
0
0
0
0
0
0
0
0
0
is in
V
,
b. If
A
=
a
0
0
0
b
0
0
0
c
and
B
=
p
0
0
0
q
0
0
0
r
are in
V
, then so is their sum
A
+
B
=
a
+
p
0
0
0
b
+
q
0
0
0
c
+
r
.
91
Chapter 4
SSM:
Linear Algebra
c. If
A
=
a
0
0
0
b
0
0
0
c
is in
V
, then so is
kA
=
ka
0
0
0
kb
0
0
0
kc
, for all constants
k
.
9. Not a subspace; consider multiplication with a negative scalar.
I
3
belongs to the set, but

I
3
doesn’t.
11. Not a subspace:
I
3
is in rref, but the scalar multiple 2
I
3
isn’t.
13. Not a subspace: (1
,
2
,
4
,
8
, . . .
) and (1
,
1
,
1
,
1
, . . .
) are both geometric sequences, but their
sum (2
,
3
,
5
,
9
, . . .
) is not, since the ratios of consecutive terms fail to be equal, for example,
3
2
6
=
5
3
.
15. The set
W
of all squaresummable sequences is a subspace of
V
:
•
The sequence (0
,
0
,
0
, . . .
) is in
W
.
•
Suppose (
x
n
) and (
y
n
) are in
W
. Note that the inequality (
x
n
+
y
n
)
2
≤
2
x
2
n
+ 2
y
2
n
holds
for all
n
, since 2
x
2
n
+ 2
y
2
n

(
x
n
+
y
n
)
2
=
x
2
n
+
y
2
n

2
x
n
y
n
= (
x
n

y
n
)
2
≥
0.
Thus
∑
∞
n
=1
(
x
n
+
y
n
)
2
≤
2
∑
∞
n
=1
x
2
n
+ 2
∑
∞
n
=1
y
2
n
converges, so that the sequence (
x
n
+
y
n
) is in
W
as well.
•
If (
x
n
) is in
W
(
so that
∑
∞
n
=1
x
2
n
converges
)
, then (
kx
n
) is in
W
as well, for any constant
k
, since
∑
∞
n
=1
(
kx
n
)
2
=
k
2
∑
∞
n
=1
x
2
n
will converge.
17. Let
E
ij
be the
n
×
m
matrix with a 1 as its
ij
th entry, and zeros everywhere else. Any
A
in
R
n
×
m
can be written as the sum of all
a
ij
E
ij
, and the
E
ij
are linearly independent,
so that they form a basis of
R
n
×
m
. Thus dim(
R
n
×
m
) =
nm
.
19.
a
+
bi
c
+
di
=
a
1
0
+
b
i
0
+
c
0
1
+
d
0
i
The vectors
1
0
,
i
0
,
0
1
,
0
i
form a basis of
C
2
as a
real
linear space, so that dim(
C
2
) =
4.
21. Use Summary 4.1.6. The general element of the subspace is
a
0
0
d
=
a
1
0
0
0
+
d
0
0
0
1
.
Thus
1
0
0
0
,
0
0
0
1
is a basis of the subspace; the
dimension is 2.
92
SSM:
Linear Algebra
Section 4.1
23. Proceeding as in Exercise 21, we find the basis
1
0
0
0
,
0
0
1
0
,
0
0
0
1
; the dimension
is 3.