Chapter 4:
General Vector Spaces
SSM:
Elementary Linear Algebra
82
(d)
False; if a vector space
V
had exactly two
elements, one of them would necessarily be the
zero vector
0
. Call the other vector
u
. Then
0
+
0
=
0
,
0
+
u
=
u
, and
u
+
0
=
u
. Since
−
u
must exist and
−
u
≠
0
, then
−
u
=
u
and
u
+
u
=
0
. Consider the scalar
1
2
. Since
1
2
u
would be an element of
V
,
1
2
=
u0
or
1
2
.
=
uu
If
1
2
,
=
then
11
1
22
.
⎛⎞
==
= + =+=
+
⎜⎟
⎝⎠
u u u000
If
1
2
,
=
then
1
.
+
u u uuu0
Either way we get
u
=
0
which contradicts our
assumption that we had exactly two elements.
(e)
False; the zero vector would be
0
= 0 + 0
x
which
does not have degree exactly 1.
Section 4.2
Exercise Set 4.2
1. (a)
This is a subspace of
3
.
R
(b)
This is not a subspace of
3
,
R
since
121
2
2 2
( ,,) ( ,,) (
, , )
aa
a
a
+=
+
which
is not in the set.
(c)
This is a subspace of
3
.
R
(d)
This is not a subspace of
3
,
R
since for
k
≠
1
()
,
ka c
ka k
c
++ ≠
+ +
so
k
(
a
,
b
,
c
) is not
in the set.
(e)
This is a subspace of
3
.
R
3. (a)
This is a subspace of
3
.
P
(b)
This is a subspace of
3
.
P
(c)
This is not a subspace of
3
P
since
23
01 2
3
ka
ax ax
ax
++
+
is not in the set
for all noninteger values of
k
.
(d)
This is a subspace of
3
.
P
5. (a)
This is a subspace of
.
R
∞
(b)
This is not a subspace of
,
R
∞
since for
k
≠
1,
k
v
is not in the set.
(c)
This is a subspace of
.
R
∞
(d)
This is a subspace of
.
R
∞
7.
Consider
12
(, , )
.
k
k
abc
uv
Thin
1 2 12
02
3
2
(,
,
)
(
,
,
)
.
k k
k
k
+−+
−=
Equating components gives the system
2
2
kk
b
kk c
=
−+
=
or
A
x
=
b
where
01
21
,
A
⎡⎤
⎢⎥
=
−
−
⎣⎦
1
2
,
k
k
⎡
⎤
=
⎢
⎥
⎣
⎦
x
and
.
a
b
c
⎡ ⎤
⎢ ⎥
=
⎢ ⎥
⎣ ⎦
b
The system can be solved simultaneously by
reducing the matrix
2
3
0
0
4
0
2
5
5
0
,
⎡
⎤
⎢
⎥
−
⎢
⎥
−
⎣
⎦
which reduces to
102400
012300
000010
.
⎡
⎤
⎢
⎥
⎢
⎥
⎣
⎦
The
results can be read from the reduced matrix.
(a)
(2, 2, 2) = 2
u
+ 2
v
is a linear combination of
u
and
v
.