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MATH 110: LINEAR ALGEBRA
SPRING 2007/08
PROBLEM SET 2 SOLUTIONS
1.
Let
V
be a vector space over
F
. Let
w
∈
V
be a ﬁxed nonzero vector and
μ
∈
F
be a ﬁxed
nonzero scalar.
(a) Show that the function
f
:
F
→
V
deﬁned by
f
(
λ
) =
λ
w
is injective.
Solution.
If
f
(
λ
1
) =
f
(
λ
2
), then
λ
1
w
=
λ
2
w
, and so (
λ
1

λ
2
)
w
=
0
. By Theorem
1.3
,
we have that
λ
1

λ
2
= 0 since
w
6
=
0
. So
λ
1
=
λ
2
. Hence
f
is injective.
(b) Show that the function
g
:
V
→
V
deﬁned by
g
(
v
) =
μ
v
is bijective.
Solution.
If
g
(
v
1
) =
g
(
v
2
), then
μ
v
1
=
μ
v
2
, and so
μ
(
v
1

v
2
) =
0
. By Theorem
1.3
,
we have that
v
1

v
2
=
0
since
μ
6
= 0. So
v
1
=
v
2
. Hence
g
is injective. Since
μ
6
= 0, it
has a multiplicative inverse
μ

1
∈
F
. Given any
v
∈
V
(codomain), the vector
μ

1
v
∈
V
(domain) has the property that
g
(
μ

1
v
) =
μ
(
μ

1
v
) = (
μμ

1
)
v
= 1
v
=
v
. Hence
g
is
surjective.
(c) Show that the function
h
:
V
→
V
deﬁned by
h
(
v
) =
v
+
w
is bijective.
Solution.
If
h
(
v
1
) =
h
(
v
2
), then
v
1
+
w
=
v
2
+
w
, and adding,

w
, the additve inverse
of
w
to both sides of the equation gives
v
1
=
v
2
. Hence
h
is injective. Given any
v
∈
V
(codomain), the vector
v

w
∈
V
(domain) has the property that
h
(
v

w
) = (
v

w
)+
w
=
v
+ (

w
+
w
) =
v
+
0
=
v
. Hence
h
is surjective.
2.
Let
W
1
and
W
2
be subspaces of a vector space
V
. The
sum
of
W
1
and
W
2
is the subset of
V
deﬁned by
W
1
+
W
2
=
{
w
1
+
w
2
∈
V

w
1
∈
W
1
,
w
2
∈
W
2
}
.
(a) Prove that
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 Spring '08
 GUREVITCH
 Math, Linear Algebra, Algebra, Scalar, Vector Space

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