1
Math 110 Homework 3
Partial Solutions
If you have any questions about these solutions, or about any problem
not solved, please ask via email or in office hours, etc.
1.6.23 In this case, by a previous homework problem, we already have that
W
1
⊆
W
2
.
(a) I claim that dim(
W
1
)=dim(
W
2
) if and only if
W
1
=
W
2
if and only
if
v
∈
span(
{
v
1
, . . . , v
k
}
). The first if and only if is straightforward.
For the second, suppose first that
W
1
=
W
2
. Then
v
∈
W
1
. This
proves that
v
is in the span of
{
v
1
, . . . , v
k
}
. On the other hand,
suppose that
v
∈
span(
{
v
1
, . . . , v
k
}
). We will then show that
W
2
⊆
W
1
and thus that they are equal. Let
x
∈
W
2
. Then there exist
scalars
a
1
, . . . , a
k
, a
k
+1
∈
F
such that
x
=
a
1
v
1
+
· · ·
+
a
k
v
k
+
a
k
+1
v
.
But
v
∈
span(
{
v
1
, . . . , v
k
}
), so there are
b
1
, . . . , b
k
∈
F
such that
v
=
b
1
v
1
+
· · ·
+
b
k
v
k
. Then
x
=
a
1
v
1
+
· · ·
+
a
k
v
k
+
a
k
+1
(
b
1
v
1
+
· · ·
+
b
k
v
k
is in span(
{
v
1
, . . . , v
k
}
).
(b) If dim(
W
1
) = dim(
W
2
), then it must be that dim(
W
1
) + 1 =
dim(
W
2
). Choose a basis
β
for
W
1
from among the vectors
{
v
1
, . . . , v
k
}
.
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.
 Winter '07
 Liu
 Math, Linear Algebra, Algebra, Linear Independence, basis, vk

Click to edit the document details