38
Proof.
We must show that
B
*
is a basis for
V
*
. First, we show
B
*
is linearly independent.
Suppose
n
X
i
=1
λ
i
v
*
i
= 0
,
the zero linear transformation. Then we have that
n
X
i
=1
λ
i
v
*
i
!
v
j
=
λ
j
= 0
for all
j
= 1
, . . . , n
.
Thus,
B
*
is linearly independent.
We show
B
*
spans
V
*
.
Suppose
ϕ
∈
B
*
. Let
λ
j
=
ϕ
(
v
j
) for
j
= 1
, . . . , n
. Then we have that
ϕ

n
X
i
=1
λ
i
v
*
i
!
v
j
= 0
for all
j
= 1
, . . . , n
. Since a linear transformation is completely determined by its values on
a basis, we have that
ϕ

n
X
i
=1
λ
i
v
*
i
= 0 =
⇒
ϕ
=
n
X
i
=1
λ
i
v
*
i
.
Remark
3.3.4
.
One of the most useful techniques to show linear independence is the “killoff”
method used in the previous proof:
n
X
i
=1
λ
i
v
*
i
!
v
j
=
λ
j
= 0
.
Applying the
v
j
allows us to see that each
λ
j
is zero. We will see this technique again when
we discuss eigenvalues and eigenvectors.
Proposition 3.3.5.
Let
V
be a finite dimensional vector space over
F
. There is a canonical
isomorphism
Ψ:
V
→
V
**
= (
V
*
)
*
given by
v
7→
ev
v
, the linear transformation
V
*
→
F
given by the evaluation map:
ev
v
(
ϕ
) =
ϕ
(
v
)
.
Proof.
First, it is clear that Ψ is a linear transformation as ev
λu
+
v
=
λ
ev
u
+ ev
v
. Hence, we
must show the map Ψ is bijective. Suppose Ψ
v
= 0. Then ev
v
is the zero linear functional
on
V
*
, i.e., for all
ϕ
∈
V
*
,
ϕ
(
v
) = 0. Since
ϕ
is completely determined by where it sends a
basis of
V
, we know that
ϕ
= 0. Hence, the map Ψ is injective. Suppose now that
x
∈
V
**
.
Let
B
=
{
v
1
, . . . , v
n
}
be a basis for
V
and let
B
*
=
{
v
*
1
, . . . , v
*
n
}
be the dual basis. Setting
λ
j
=
x
(
v
*
j
) for
j
= 1
, . . . , n
, we see that if
u
=
n
X
i
=1
λ
i
v
i
,
39
then (
x

ev
u
)
v
*
j
= 0 for all
j
= 1
, . . . , n
.
Once more since a linear transformation is
completely determined by where it sends a basis, we have
x

ev
u
= 0, so
x
= ev
u
and Ψ is
surjective.
Proposition 3.3.6.
Let
V
be a finite dimensional vector space over
F
. Then
V
is isomorphic
to
V
*
, but not canonically.
Proof.
Let
B
=
{
v
1
, . . . , v
n
}
be a basis for
V
. Then
B
*
is a basis for
V
*
. Define a linear
transformation
T
:
V
→
V
*
by
v
i
7→
v
*
i
for all
v
i
∈
B
, and extend this map by linearity.
Then
T
is an isomorphism as there is the obvious inverse map.
3.4
Coordinates
For this section,
V, W
will denote finite dimensional vector spaced over
F
. We now discuss
the way in which all finite dimensional vector spaces over
F
look like
F
n
(nonuniquely). We
then study
L
(
V, W
), with the main result being that if dim(
V
) =
n
and dim(
W
) =
m
, then
L
(
V, W
)
∼
=
M
m
×
n
(
F
) (nonuniquely).
Definition 3.4.1.
Let
V
be a finite dimensional vector space over
F
, and let
B
=
{
v
1
, . . . , v
n
}
be an ordered basis for
V
. The map [
·
]
B
:
V
→
R
n
given by
n
X
i
=1
λ
i
v
i
7→
λ
1
λ
2
.
.
.
λ
n
is called the coordinate map with respect to
B
. We say that [
v
]
B
are coordinates for
v
with
respect to the basis
B
.
Proposition 3.4.2.
The coordinate map is an isomorphism.
Proof.
It is clear that the coordinate map is a linear transformation as [
λu
+
v
]
B
=
λ
[
u
]
B
+[
v
]
B
for all
u, v
∈
V
and
λ
∈
F
. We must show [
·
]
B
is bijective. We show it is injective. Suppose
that [
v
]
B
= (0
, . . . ,
0)
T
. Then since
v
=
n
X
i
=1
λ
i
v
i
for unique
λ
1
, . . . , λ
n
∈
F
, we must have that
λ
i
= 0 for all
i
= 1
, . . . , n
, and
v
= 0. Hence
the map is injective.