M341 Homework Solution 8
Section 4.1:
4:
It is obviously that the operations satisfy the closure properties. It remains to check eight properties.
For
x
,
y
, and
z
∈
R
, we have
x
⊕
y
= (
x
3
+
y
3
)
1
/
3
= (
y
3
+
x
3
)
1
/
3
=
y
⊕
x
,
and
x
⊕
(
y
⊕
z
) =
[
x
3
+ (
y
3
+
z
3
)
]
1
/
3
=
[
(
x
3
+
y
3
) +
z
3
]
1
/
3
= (
x
⊕
y
)
⊕
z
,
by the commutative law and the associative law of real numbers. Let
0
= 0
∈
R
, then for all
x
∈
R
we have
x
⊕
0
= (
x
3
+ 0
3
)
1
/
3
=
x
=
x
.
For any
x
∈
R
, let
−
x
=
−
x
∈
R
, then
x
⊕ −
x
=
[
x
3
+ (
−
x
)
3
]
1
/
3
= 0
1
/
3
= 0 =
0
.
For
x
,
y
∈
R
and
a, b
∈
R
, we have
a
⊙
(
x
⊕
y
) =
3
√
a
(
x
3
+
y
3
)
1
/
3
= (
ax
3
+
ay
3
)
1
/
3
=
[
(
3
√
ax
)
3
+ (
3
√
ay
)
3
]
1
/
3
= (
a
⊙
x
)
⊕
(
a
⊙
y
)
,
(
a
+
b
)
⊙
x
= (
3
√
a
+
b
)
x
= (
ax
3
+
bx
3
)
1
/
3
=
±
(
3
√
ax
)
3
+ (
3
√
bx
)
3
²
1
/
3
= (
a
⊙
x
)
⊕
(
b
⊙
x
)
,
and
(
ab
)
⊙
x
= (
3
√
ab
)
x
= (
3
√
a
)(
3
√
b x
) =
a
⊙
(
b
⊙
x
)
,
1
⊙
x
= (
3
√
1)
x
=
x
=
x
Therefore,
R
is the vector space under the operations.
6:
Since the zero matrix
O
n
is not in the set of nonsingular
n
×
n
matrices, the set does not have the
identity element for addiction. Therefore the set is not a vector space.
7:
The set
R
is not a vector space under the new operations since it doesn't satisfy the property (8),
because for
x
= 1
∈
R
we have
1
⊙
x
=
0
̸
= 1 =
x
9:
The set
R
2
is not a vector space under the new operations since it does not have the identity element
for addiction. To see this, let
x
= [1
,
1]
∈
R
2
, then we have
x
⊕
y
= [1 +
y
1
,
0]
̸
= [1
,
1] =
x
for any choice of
y
= [
y
1
, y
2
]
∈
R
.
11:
(a) Suppose
B
=
0
. For any
X
1
and
X
2
in
V
and
a
∈
R
, we have
AX
1
=
0
and
AX
2
=
0
, and it
follows
A
(
X
1
+
X
2
) =
AX
1
+
AX
2
=
0
+
0
=
0
,
A
(
a
X
1
) =
a
(
AX
1
) =
a
0
=
0
Therefore the closure properties are satis ed.
Conversely, if the closure properties are satis ed, let
X
∈ V
since the set is nonempty. Then we
have
AX
=
B
and
A
(2
X
) =
B
by
X
and
2
X
are all in
V
. It follows
1
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=
A
(2
X
) =
B
= 2(
AX
) = 2
B
This implies
B
= 2
B
−
B
=
0
.
(b) Properties (1), (2), (5), (6), (7), (8) have already been proven in Theorem 1.3 since
V
is a
subset of
M
n
1
and the set equipped the usual operations as ones used in
M
n
1
.
(c) It is easy to see that the identity element for addiction of
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 Spring '08
 HIETMANN
 Vector Space, Complex number, Eλ

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