Then we claim
[
αT
+
βU
]
A
B
=
α
[
T
]
A
B
+
β
[
U
]
A
B
.
Indeed, we only need to check this formula on the basis vectors
v
1
, . . . , v
n
in
B
. We
must have numbers
a
ij
, b
ij
∈
R
such that
T
(
v
j
) =
m
X
i
=1
a
ij
w
i
,
U
(
v
j
) =
m
X
i
=1
b
ij
w
i
.
51
Now,
(
αT
+
βU
)(
v
j
) =
m
X
i
=1
(
αa
ij
+
βb
ij
)
w
i
,
which shows
[
αT
+
βU
]
A
B
=
α
[
a
ij
] +
β
[
b
ij
] =
α
[
T
]
A
B
+
β
[
U
]
A
B
.
7.2.
Dual spaces.
Given two vector spaces
V
and
W
, we have just seen the vector
space
L
(
V, W
), the vector space of linear transformations from
V
to
W
.
In the
special case that the target
W
=
R
, we obtain something called the dual space to
V
.
Definition 15.
Let
V
be a vector space.
The dual space
V
*
to
V
is the vector
space
V
*
=
L
(
V,
R
)
.
An element
φ
∈
V
*
is a linear function
φ
:
V
→
R
; such a linear map is often
called a
linear functional
. In the special case that
B
=
{
v
1
, . . . , v
n
}
is a basis for
V
, we have a particular set of
n
linear functionals
B
*
=
{
φ
1
, . . . , φ
n
}
, which are
defined by the relation
φ
i
(
v
j
) =
1
i
=
j
0
i
6
=
j.
The main content of the proof of the theorem below is to show that
B
*
is a basis
of
V
*
; it is called the
dual basis
to
B
.
Theorem 39.
Let
V
be a finite dimensional vector space. Then
V
*
is isomorphic
to
V
.
Proof.
We pick a basis
B
=
{
v
1
, . . . , v
n
}
and prove the theorem.
In fact, we al
ready know that
V
is isomorphic to
R
n
, so we only need to prove that
V
*
is also
isomorphic to
R
n
as well. To show this, we only need to show that
V
*
is an
n

dimensional vector space,
i.e.
that
V
*
has a basis with
n
elements. We already
have an excellent candiate for this basis, namely the dual basis
B
*
=
{
φ
1
, . . . , φ
n
}
.
Let
φ
∈
V
*
. We show that there is a unique choice of coefficients
a
1
, . . . , a
n
such
that
φ
=
a
1
φ
1
+
a
2
φ
2
+
· · ·
+
a
n
φ
n
;
as we have seen before, this shows both that
B
*
spans
V
*
(by the existence of a
solution) and that it is linearly independent (by the uniqueness of the solution).
Given our choice of
φ
∈
V
*
, we can pick
a
1
=
φ
(
v
1
)
,
a
2
=
φ
(
v
2
)
,
. . . ,
a
n
=
φ
(
v
n
)
.
Now we compare
φ
to the map
˜
φ
=
a
1
φ
1
+
a
2
φ
2
+
· · ·
+
a
n
φ
n
.
We have, by definition,
φ
(
v
j
) =
a
j
=
˜
φ
(
v
j
), which implies
φ
=
˜
φ
.
Thus we
have represented our arbitrary
φ
∈
V
*
as a linear combination of elements of
B
*
.
Furthermore, if there is some other choice of coefficients
b
1
, . . . , b
n
such that
φ
=
b
1
φ
1
+
· · ·
b
n
φ
n
,
52
then for any
j
= 1
, . . . , n
we have
b
j
= (
b
1
φ
1
+
· · ·
+
b
n
φ
n
)(
v
j
) =
φ
(
v
j
) = (
a
1
φ
1
+
· · ·
+
a
n
φ
n
)(
v
j
) =
a
j
.
Thus the choice of coeffients is unique, completing the proof.
Remark 9.
•
Notice that isomorphism
V
→
V
*
depends quite heavily on the
choice of basis
B
(or, equivalently, on the choice of dual basis
B
*
). If we
change basis in
V
, we will obtain a very different isomorphism
V
→
V
*
.
•
Also notice that in order to construct the dual basis
B
*
, we need to first
pick the basis
B
.
We cannot pick them simultaneously, say choosing the
first two elements of
B
, then the elements of
B
*
, then the rest of
B
. This
is because in order to describe even the first linear functional
φ
1
∈ B
*
, we
need to know what it does to each and every basis vector in
B
.
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 Fall '08
 GUREVITCH
 Linear Algebra, Algebra, Vector Space, aij Aij