81
Chapter 4
General Vector Spaces
Section 4.1
Exercise Set 4.1
1. (a)
12
34
1324
26
(,)(
,)
(,)
(, )
+=−
+
=−+
+
=
uv
3
u
= 3(
−
1, 2) = (0, 6)
(b)
The sum of any two real numbers is a real number. The product of two real numbers is a real number and 0 is
a real number.
(c)
Axioms 1
−
5 hold in
V
because they also hold in
2
.
R
(e)
Let
uu
=
u
with
1
0.
u
≠
Then
2
11
0
(, ) (
, ) .
u
==
≠
3.
The set is a vector space with the given operations.
5.
The set is not a vector space. Axiom 5 fails to hold because of the restriction that
x
≥
0. Axiom 6 fails to hold for
k
< 0.
7.
The set is not a vector space. Axiom 8 fails to hold because
222
()
.
km k m
+≠
+
9.
The set is a vector space with the given operations.
11.
The set is a vector space with the given operations.
23.
Hypothesis
Add
to both sides.
Axiom 3
Axiom 5
Axiom 4
(
)
(
)
[(
)
]
)
]
+=+
++
−=++
−
−
− =++
−
=
uw vw
uw
w vw
w
w
uw w vw w
u0 v0
25.
(1) Axiom 7
(2) Axiom 4
(3) Axiom 5
(4) Axiom 1
(5) Axiom 3
(6) Axiom 5
(7) Axiom 4
True/False 4.1
(a)
False; vectors are not restricted to being directed line segments.
(b)
False; vectors are not restricted to being
n
tuples of real numbers.
(c)
True
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View Full DocumentChapter 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
.
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 Fall '11
 Loveys
 Linear Algebra, Real Numbers, Vector Space, SSM

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