3.4. THE DUAL OF A VECTOR SPACE AND MATRICES
177
Lemma 3.4.4.
Let
V
be a finite dimensional vector space over a field
K
.
For each nonzero
v
2
V
, there is a linear functional
f
2
V
such that
f .v/
¤
0
.
Proof.
We know that any linearly independent subset of
V
is contained in
a basis. If
v
is a nonzero vector in
V
, then
f
v
g
is linearly independent.
Therefore, there is a basis
B
of
V
with
v
2
B
. Let
f
be any function from
B
into
K
with
f .v/
¤
0
. By Proposition
3.3.32
,
f
extends to a linear
functional on
V
.
n
Theorem 3.4.5.
If
V
is a finite dimensional vector space, then
W
V
!
V
is a linear isomorphism.
Proof.
We already know that
is linear.
If
v
is a nonzero vector in
V
, then there is an
f
2
V
such that
f .v/
¤
0
, by Lemma
3.4.4
. Thus
.v/.f /
D
f .v/
¤
0
, and
.v/
¤
0
.
Thus
is injective.
Applying Proposition
3.4.2
(c) twice, we have
dim
.V
/
D
dim
.V
/
D
dim
.V /
.
Therefore any injective linear map
from
V
to
V
is necessarily surjective, by Proposition
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 Fall '08
 EVERAGE
 Linear Algebra, Algebra, Matrices, Vector Space, finite dimensional vector, dimensional vector space, Aı

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