Alex Gittens
January 19, 2007
ACM 104: HW 1
Exercise 1.
Show that a subset
W
of a vector space
V
is a subspace of
V
if and only if span(
W
) =
W
.
Proof.
This statement is trivially true if
W
=
{
0
}
:
W
is then a subspace of
V
and span(
W
) =
W
.
Assume
W
=
{
0
}
is a subspace of
V
, then clearly
W
⊂
span(
W
), because if
x
∈
W
, then
x
= 1
·
x
∈
span(
W
). If
y
∈
span(
W
), that is, there are
x
i
∈
W
such that
y
=
∑
m
i
=1
a
i
x
i
, then since
W
is closed under
addition and scalar multiplication,
y
∈
W
. This shows span(
W
)
⊂
W
, so span(
W
) =
W
.
Assume
W
=
{
0
}
is a subset of
V
such that span(
W
) =
W
, then if
x, y
∈
W
and
α
is a scalar, then
αx
+
y
∈
span(
W
) =
W
, so
W
is closed under addition and scalar multiplication, therefore
W
is a subspace
of
V
.
Exercise 2.
Let
S
1
and
S
2
be two subsets of a vector space
V
. Prove that
span(
S
1
∩
S
2
)
⊂
span(
S
1
)
∩
span(
S
2
)
.
Give one example to show equality does not hold in general.
Proof.
If
x
∈
span(
S
1
∩
S
2
), then there are
y
1
, . . . , y
m
∈
S
1
∩
S
2
such that
x
=
∑
m
i
=1
a
i
y
i
for some
a
i
∈
R
. But
each
y
i
is in
S
1
and in
S
2
, so
x
∈
span(
S
1
) and
x
∈
span(
S
2
). This shows span(
S
1
∩
S
2
)
⊂
span(
S
1
)
∩
span(
S
2
).
To see equality does not hold in general, take
V
=
R
2
,
S
1
=
{
(0
,
1)
,
(1
,
0)
}
and
S
2
=
{
(1
,
1)
,
(1
,

1)
}
. Then
both
S
1
and
S
2
are bases for V, so span(
S
1
)
∩
span(
S
2
) =
R
2
, but
S
1
∩
S
2
=
∅
, so span(
S
1
∩
S
2
) =
{
0
}
=
R
2
.
Exercise 3.
Let
V
be a vector space having dimension
n
, and let
S
be a subset of
V
such that span(
S
) =
V
.
Prove that there is a subset of
S
that is a basis for
V
. (Note:
S
may be infinite.)
Proof.
Let
β
=
{
v
1
, . . . , v
n
}
be a basis for
V
, then
v
1
, . . . , v
n
are in the span of
S
, so there are
s
1
,
1
,
, . . . , s
1
,m
1
,
s
2
,
1
, . . . , s
2
,m
2
, . . .
,
s
n,
1
, . . . , s
n,m
n
∈
S
such that
v
j
=
∑
m
j
k
=1
a
k
s
j,k
for
j
= 1
, . . . , n
.
Therefore
V
=
span(
s
1
,
1
, . . . , s
n,m
n
).
Since this is a finite set, there is a subset
S
of
{
s
1
,
1
, . . . , s
n,m
n
}
that is a basis
for
V
(by Thm 1.18). Since
S
⊂
S
,
S
contains a basis for
V
.
Exercise 4.
Let
V
and
W
be vector spaces and let
T
:
V
→
W
be a linear transformation.
Suppose
T
is onetoone and
S
is a subset of
V
.
Prove that
S
is linearly independent if and only if
T
(
S
) is linearly
independent.
Proof.
Let
S
be linearly independent. Assume there are
v
1
, . . . , v
n
∈
T
(
S
) and scalars
a
1
, . . . , a
n
such that
∑
n
i
=1
a
i
v
i
= 0. Let
x
i
be such that
v
i
=
Tx
i
, then
n
i
=1
a
i
v
i
=
n
i
=1
a
i
Tx
i
=
T
n
i
=1
a
i
x
i
= 0
⇒
n
i
=1
a
i
x
i
= 0
,
since
T
is onetoone. By the linear independence of
S
, this means
a
i
= 0, for all
i
, which shows
T
(
S
) is
linearly independent.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '10
 Linear Algebra, Vector Space, linearly independent subset, Γj −1

Click to edit the document details