AS3sol/MATH1111/YKL/0809
THE UNIVERSITY OF HONG KONG
DEPARTMENT OF MATHEMATICS
MATH1111: Linear Algebra
Assignment 3 Suggested Solution
1.
(a) No matter whether or not
W
is a subspace,
W
⊂
Span(
W
) since
w
∈
W
⇒
1
·
w
∈
Span(
W
).
”If part”: If
W
is a subspace, then any linear combination of elements of
W
belongs
to
W
. i.e. Span(
W
)
⊂
W
.
∴
W
= Span(
W
).
”Only if” part: Clear by direct checking from definition.
(b) As in part (a), we know
S
⊂
Span(
S
) and Span(
S
) is a subspace.
Want to show:
S
⊂
W
and
W
is a subspace
⇒
Span(
S
)
⊂
W
.
This is also clear, since
w
1
, w
2
,
· · ·
, w
r
∈
S
⇒
w
1
, w
2
,
· · ·
, w
r
∈
W
. Thus,
a
1
w
1
+
a
2
w
2
+
· · ·
+
a
r
w
r
∈
W
because
W
is a subspace.
∴
Span(
S
)
⊂
W
.
2.
(a) Since
U
+
V
is a subspace containing the sets
U
and
V
, by Qn 1 part (b), Span(
U
∪
V
)
⊂
U
+
V
.
Let
x
∈
U
+
V
. Then
x
=
u
+
v
where
u
∈
U
and
v
∈
V
. So
x
is a linear combination
of vectors in
U
∪
V
. i.e.
U
+
V
⊂
Span(
U
∩
V
).
(b) No.
Consider
U
= Span((1
1)
T
),
V
= Span(
e
1
),
W
= Span(
e
2
), all of which are
subspaces of
R
2
.
Then
U
+ (
V
∩
W
) =
U
+
{
0
}
=
U
but (
U
+
V
)
∩
(
U
+
W
) =
R
2
∩
R
2
=
R
2
.
Remark.
U
+ (
V
∩
W
)
⊂
(
U
+
V
)
∩
(
U
+
W
) is always true.
(c) No. Use the counterexample in (b). Now,
U
∩
V
=
{
0
}
and
U
∩
W
=
{
0
}
, but since
V
+
W
=
R
2
, we have
U
∩
(
V
+
W
) =
U
.
3.
(a) dim
X
= 1 as
{
x
}
is a basis for
X
. Similarly, dim
Y
= 1.
dim
W
= 2 as
{
u
,
v
}
is a basis for
W
.
dim(
X
+
W
) = 2 as
X
+
W
=
{
t
+
s
:
t
∈
X, s
∈
W
}
=
{
α
x
+
β
u
+
γ
v
:
α, β, γ
∈
R
}
.
However,
x
= 2
u
+ 2
v
so
X
+
W
= Span(
u
,
v
) where
u
,
v
are linearly independent.
dim(
Y
+
W
) = 3 as
Y
+
W
= Span(
y
,
u
,
v
) and
y
,
u
,
v
are linearly independent.
1
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(b) Claim:
X
∩
W
=
X
.
Proof. 1
◦
X
∩
W
⊂
X
. This is obvious by the definition of intersection.
2
◦
X
⊂
X
∩
W
. Let
f
∈
X
. Then
f
is of the form
f
=
α
x
for some scalar
α
. As
x
= 2
u
+ 2
v
, we see that
f
can be expressed as
f
= 2
α
u
+ 2
α
v
.
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 Fall '08
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 Math, Linear Algebra, Algebra, Vector Space, basis, ej, UNIVERSITY OF HONG KONG DEPARTMENT OF MATHEMATICS

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