ma253
,Fa
l
l
2007
— Problem Set
8
Solutions
1.
Problems from the textbook:
a. Section
3
.
3
43.
The point is that if we take a basis
{
~
v
1
,
~
v
2
,...,
~
v
r
}
of
V
and a basis
{
~
w
1
,
~
w
2
~
w
s
}
of
W
,then
{
~
v
1
,
~
v
2
~
v
r
,
~
w
1
,
~
w
2
~
w
s
}
will be a basis of
V
+
W
. To see this, notice Frst that it clearly spans
V
+
W
,sothe
hard part is to show it is independent. ±or that, suppose
c
1
~
v
1
+
···
+
c
r
~
v
r
+
b
1
~
w
1
+
+
b
s
~
w
s
=
~
0.
Then, we get
c
1
~
v
1
+
+
c
r
~
v
r
=(
b
1
~
w
1
+
+
b
s
~
w
s
)
.
In this equation, the expression on the left is an element of
V
and the expression
on the right is an element of
W
. Since they are equal, both expressions must give
(the same) element of
V
∩
W
.But
V
∩
W
=
{
~
0
}
,soinfactweget
c
1
~
v
1
+
+
c
r
~
v
r
=
~
0
and
b
1
~
w
1
+
+
b
s
~
w
s
=
~
0.
Since the original sets were independent, it follows that all the
c
i
and all the
b
j
are
zero, so the big set is still independent.
The upshot is that if
V
∩
W
=
{
~
0
}
,wehave
dim
(
V
+
W
)=
dim
(
V
)+
dim
(
W
)
.
44.
Since
V
∩
W
=
{
~
0
}
, we can conclude from 43 that
V
+
W
has dimension
n
.
Since
V
+
W
⊂
R
n
, it follows that
V
+
W
=
R
n
. Hence every vector
~
x
∈
R
n
can
be written as
~
v
+
~
w
as desired.
To see that this representation is unique, suppose
~
v
+
~
w
=
~
v
1
+
~
w
1
.
Then
~
v

~
v
1
=
~
w
1

~
w,
and the left hand side is in
V
while the right hand side is in
W
.Nowdothesame
as in problem 43.
You don’t have to agree with me, but it’s quicker.