Math 121 Homework, Week 3
Michael Von Korﬀ
October 21, 2004
1.6: 31,33; 2.1: 9,14,15,19,35; 2.2: 3,5,13
Problem 1.
1.6 31. Let
W
1
and
W
2
be subspaces of a vector space V having
dimensions m and n, respectively, where
m
≤
n
.
(a) Prove that
dim
(
W
1
∩
W
2
)
≤
n.
(b) Prove that
dim
(
W
1
+
W
2
)
≤
m
+
n
.
Proof.
(a)
∀
v
∈
W
1
∩
W
2
,
by deﬁnition
v
∈
W
2
. So
W
1
∩
W
2
⊂
W
2
. Thus, by
theorem 1.11 (which states that the dimension of a subspace is less than the
dimension of the larger space),
dim
(
W
1
∩
W
2
)
≤
n.
(b) Let
S
1
=
{
u
1
,u
2
,...,u
m
}
be a basis for
W
1
. Let
S
2
=
{
v
1
,v
2
,...,v
n
}
be a
basis for
W
2
. Then
∀
x
∈
W
1
+
W
2
, there exist
w
1
∈
W
1
and
w
2
∈
W
2
such that
w
1
+
w
2
=
x
. Moreover, since
S
1
spans
W
1
and
S
2
spans
W
2
, we can write
w
1
and
w
2
as linear combinations of vectors in
S
1
and
S
2
,
w
1
=
a
1
u
1
+
...
+
a
m
u
m
and
w
2
=
b
1
v
1
+
...b
n
v
n
. So
x
=
w
1
+
w
2
=
a
1
u
1
+
...
+
a
m
u
m
+
b
1
v
1
+
...b
n
v
n
. Thus
x
∈
span
(
S
1
∪
S
2
). So
S
1
∪
S
2
is a set
of size at most m+n (it could be smaller if the two sets have some vectors in
common) that spans
W
1
+
W
2
. But any set spanning a ﬁnitedimensional vector
space must have size greater than or equal to the dimension of the space, so
dim
(
W
1
+
W
2
)
≤
m
+
n
.
Problem 2.
33. (a) Let
W
1
and
W
2
be subspaces of a vector space V such that
V
=
W
1
⊕
W
2
. If
β
1
andβ
2
are bases for
W
1
and
W
2
respectively, show that
β
1
∩
β
2
=
∅
and
β
1
∪
β
2
is a basis for V.
(b) Conversely, let
β
1
and
β
2
be disjoint bases for subspaces
W
1
and
W
2
respec
tively. Prove that if
β
1
∪
β
2
is a basis for V, then
V
=
W
1
⊕
W
2
.
Proof.
(a)
V
=
W
1
⊕
W
2
, so by deﬁnition
W
1
∩
W
2
=
{
0
}
. But
β
1
⊂
W
1
and
β
2
⊂
W
2
, so
β
1
∩
β
2
⊂
W
1
∩
W
2
=
{
0
}
. The zero vector is linearly dependent
with itself and cannot be a basis vector; thus,
β
1
∩
β
2
=
∅
.
Also, we showed in problem 31 that
W
1
+
W
2
is spanned by the union of its
bases. In other words,
β
1
∪
β
2
spans V. To show that
β
1
∪
β
2
is a basis, all we
need to do is show that it is linearly independent.
Let
β
1
=
{
u
1
,u
2
,...,u
m
}
,
β
2
=
{
v
1
,v
2
,...,v
n
}
. Given
a
1
,...,a
m
,b
1
,...,b
n
such that
a
1
u
1
+
...
+
a
m
u
m
+
b
1
v
1
+
...
+
b
n
v
n
= 0, we can write
a
1
u
1
+
...
+
a
m
u
m
=

b
1
v
1

...

b
n
v
n
= 0. The lefthand term is a vector in
W
1
and
1