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2.6.
2.6. Dual Spaces
1.
(a) F
(linear transformation from V into its ﬁeld of scalars F is called a linear func
tional)
(b) T
(
f
:
F
→
F,
[
f
]
∈
Mat
1
×
1
(
F
))
(c) T
dimV
*
=dim(
L
(
V,F
))=dim
V
dim
F
=dim
V
∴
V
’
V
*
(d) T
For a vector space
V
, we can deﬁne the dual space of
V
i.e. (
L
(
)) =
V
*
Then
V
is the dual space of
V
*
((
V
*
)
*
=
V
)
∴
Every vector space is the dual of some vector space
(e) (example)
V
=
R
2
,F
=
R
β
=
{
e
1
=
±
1
0
¶
,e
2
=
±
0
1
¶
}
:
So
V
*
=
L
(
V,
R
) : 1
×
2 matrices
and
e
*
1
= (1
,
0)
*
2
= (0
,
1)
i.e. β
*
=
{
e
*
1
= (1
,
0)
*
2
= (0
,
1)
}
Now if we deﬁne
T
:
V
→
V
*
by
T
(
e
1
) = (1
,
1)
,T
(
e
2
) = (1
,

1)
Since
{
T
(
e
1
)
(
e
2
)
}
=
T
(
β
) a basis of
V
*
, then clearly
T
is an isomorphism
But
T
(
β
) =
{
T
(
e
1
) = (1
,
1)
6
=
e
*
1
(
e
2
) = (1
,

1)
6
=
e
*
2
} 6
=
β
*
123
PNUMATH
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View Full Document2.6.
(example 2)
V
=
R
, F
=
R
, f
:
V
→
F i.e.V
*
=
F
and
β
*
=
{
1
*
}
(
∵
β
=
{
1
}
)
But
T
:
R
→
R
is an isomorphism
a
7→
2
a T
(
β
) =
{
T
(1)
}
=
{
2
id
} 6
=
β
*
=
{
id
}
(f) T
T
:
V
→
W,T
t
:
W
*
→
V
*
by
T
t
(
g
) =
gT
(
T
t
)
t
: (
V
*
)
*
→
(
W
*
)
*
(g) T
V
’
W
⇔
T
:
V
→
W
: an isomorphism
⇔ ∃
[
T
]
γ
β
: invertible
⇔
([
T
]
γ
β
)
t
= ([
T
t
]
γ
*
β
*
) : invertible
⇔
T
t
:
W
*
→
V
*
: an isomorphism
⇔
V
*
’
W
*
(h) F
f
:
D
n
(
R
)
→
R
by
f
(
g
(
x
)) =
g
0
(
x
)
,
∀
g
(
x
) =
D
n
(
R
)
but in case
g
(
x
) =
x
2
,f
(
g
(
x
)) =
g
0
(
x
) = 2
x
is not in
R
∴
It’s not a linear functional
2.
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 Spring '09
 RUDYAK

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