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Economics 204
Lecture 8–Wednesday, August 6, 2008
Chapter 3, Linear Algebra Section 3.1, Bases
Defnition 1
Let
X
be a vector space over a feld
F
.A
linear combination
oF
x
1
,...,x
n
is a vector oF the
Form
y
=
n
X
i
=1
α
i
x
i
where
α
1
,...,α
n
∈
F
α
i
is the
coeﬃcient
oF
x
i
in the linear combination.
IF
V
⊆
X
,sp
an
V
denotes the set oF all linear
combinations oF
V
.
Aset
V
⊆
X
is
linearly dependent
iF there exist
v
1
,...,v
n
∈
X
and
α
1
n
∈
F
not all zero such that
n
X
i
=1
α
i
v
i
=0
V
⊆
X
is
linearly independent
iF it is not linearly dependent.
V
⊆
X
spans
X
iF span
V
=
X
.
A
Hamel basis
(oFten just called a
basis
) oF a vector space
X
is a linearly independent set oF vectors in
X
that spans
X
.
Example:
{
(1
,
0)
,
(0
,
1)
}
is a basis For
R
2
.
{
(1
,
1)
,
(
−
1
,
1)
}
is another basis For
R
2
:
(
x, y
)=
α
(1
,
1) +
β
(
−
1
,
1)
x
=
α
−
β
y
=
α
+
β
x
+
y
=2
α
α
=
x
+
y
2
1
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−
x
=2
β
β
=
y
−
x
2
(
x, y
)=
x
+
y
2
(1
,
1) +
y
−
x
2
(
−
1
,
1)
Since (
x, y
) is an arbitrary element of
R
2
,
{
(1
,
1)
,
(
−
1
,
1)
}
spans
R
2
.I
f(
x, y
)=(0
,
0),
α
=
0+0
2
=0
,β
=
0
−
0
2
so the coeﬃcients are all zero, so
{
(1
,
1)
,
(
−
1
,
1)
}
is linearly independent. Since it is linearly independent
and spans
R
2
, it is a basis.
Example:
{
(1
,
0
,
0)
,
(0
,
1
,
0)
}
is not a basis of
R
3
, because it does not span.
Example:
{
(1
,
0)
,
(0
,
1)
,
(1
,
1)
}
is not a basis for
R
2
.
1(1
,
0) + 1(0
,
1) + (
−
1)(1
,
1) = (0
,
0)
so the set is not linearly independent.
Theorem 2 (1.2’, see Corrections handout)
Let
V
be a Hamel basis for
X
. Then every vector
x
∈
X
has a
unique
representation as a linear combination (with
all
coeﬃcients nonzero) of a Fnite number of
elements of
V
.
(
Aside:
the unique representation of 0 is 0 =
∑
i
∈∅
α
i
b
i
.)
Proof:
Let
x
∈
X
.S
in
c
e
V
spans
X
,wecanwr
ite
x
=
X
s
∈
S
1
α
s
v
s
where
S
1
is Fnite,
α
s
∈
F
,
α
s
6
,
v
s
∈
V
for
s
∈
S
1
. Now, suppose
x
=
X
s
∈
S
1
α
s
v
s
=
X
s
∈
S
2
β
s
v
s
2
where
S
2
is fnite,
β
s
∈
F
,
β
s
6
=0
,and
v
s
∈
V
For
s
∈
S
2
.
Let
S
=
S
1
∪
S
2
, and defne
α
s
=0 For
s
∈
S
2
\
S
1
β
s
s
∈
S
1
\
S
2
Then
0=
x
−
x
=
X
s
∈
S
1
α
s
v
s
−
X
s
∈
S
2
β
s
v
s
=
X
s
∈
S
α
s
v
s
−
X
s
∈
S
β
s
v
s
=
X
s
∈
S
(
α
s
−
β
s
)
v
s
Since
B
is linearly independent, we must have
α
s
−
β
s
,so
α
s
=
β
s
, For all
s
∈
S
.
s
∈
S
1
⇔
α
s
6
⇔
β
s
6
⇔
s
∈
S
2
so
S
1
=
S
2
and
α
s
=
β
s
For
s
∈
S
1
=
S
2
, so the representation is unique.
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This note was uploaded on 08/01/2008 for the course ECON 204 taught by Professor Anderson during the Fall '08 term at University of California, Berkeley.
 Fall '08
 ANDERSON
 Economics

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