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Wednesday February 25
−
Lecture 20 :
Reducing a spanning family to a basis.
(Refers to section 4.3)
Expectations:
1.
State the Removal Lemma ( The spanning theorem).
2.
Recognize every nontrivial vector space with a finite spanning family has a
basis.
We will consider methods to construct the basis of a vector space.
20.1
Recall
−
If
B
= {
v
1
,
v
2
,
,...,
v
k
} is both
1.
a spanning family of
V
,
and
2.
linearly independent
then we say that
B
is a
basis
of
V
.
20.2
Lemma
−
The
Removal Lemma
.
(
Sometimes called the spanning theorem
) Let
U
=
{
v
1
,
v
2
,...,
v
k
} and
V
= span{
v
1
,
v
2
,...,
v
k
} = span
U
. Suppose the vector
v
1
in
U
= {
v
1
,
v
2
,...,
v
k
} is a linear combination of all the other vectors in that set then span{
v
1
,
v
2
,...,
v
k
} =
span{
v
2
,...,
v
k
}, i.e.,
v
1
can be removed from
U
without affecting the span of the set.
Proof:
•
We are given that
v
1
is a linear combination of the other vectors in
U
.
That is
v
1
=
β
2
v
2
+
β
3
v
3
+.....
+
β
k
v
k
. .
±
Let
w
∈
V
= span{
v
1
,
v
2
,...,
v
k
}. Then, for some scalars
α
1
,
α
2
, , .
..,
α
k
.
w
=
α
1
v
1
+
α
2
v
2
+.....
+
α
k
v
k
=
α
1
(
β
2
v
2
+
β
3
v
3
+.....
+
β
k
v
k
)+
α
2
v
2
+
α
3
v
3
+ .
....+
α
k
v
k
.
=
(
α
1
β
2
+
α
2
)
v
2
+ (
α
1
β
3
+
α
3
)
v
3
+.....
+(
α
1
β
k
+
α
k
2
)
v
k
, a vector in span{
v
2
,...,
v
k
}.
•
So
V
is contained in the span{
v
2
,...,
v
k
}.
•
Hence
V
= span{
v
2
,...,
v
k
}.
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View Full Document•
Since
V
= span{
v
1
,
v
2
,...,
v
k
} then
span{
v
1
,
v
2
,...,
v
k
} =
span{
v
2
,...,
v
k
}.
20.2.2
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 Winter '08
 All
 Vector Space

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