Let
B
=
{
α
1
, α
2
, α
3
}
be the basis for
C
3
defined by
α
1
= (1
,
0
,

1)
α
2
= (1
,
1
,
1)
α
3
= (2
,
2
,
0)
.
Find the dual basis of
B
.
The first element of the dual basis is the linear function
α
*
1
such
that
α
*
1
(
α
1
) = 1
, α
*
1
(
α
2
) = 0
and
α
*
1
(
α
3
) = 0
. To describe such a function more
explicitly we need to find its values on the standard basis vectors
e
1
,
e
2
and
e
3
.
To do this express
e
1
, e
2
, e
3
through
α
1
, α
2
, α
3
(refer to the solution of Exercise 1
pp.
5455 from Homework 6).
For each
i
= 1
,
2
,
3
you will find the numbers
a
i
, b
i
, c
i
such that
e
1
=
a
i
α
1
+
b
i
α
2
+
c
i
α
3
(i.e. the coordinates of
e
i
relative to the
basis
α
1
, α
2
, α
3
). Then by linearity of
α
*
1
we get that
α
*
1
(
e
i
) =
a
i
. Then
α
*
2
(
e
i
) =
b
i
,
and
α
*
3
(
e
i
) =
c
i
.
This is the answer.
It can also be reformulated as follows.
If
P
is the transition matrix from the standard basis
e
1
, e
2
, e
3
to
α
1
, α
2
, α
3
, i.e.
(
α
1
, α
2
, α
3
) = (
e
1
, e
2
, e
3
)
P
, then
(
P

1
)
t
is the transition matrix from the dual basis
e
*
1
, e
*
2
, e
*
3
to the dual basis
α
*
1
, α
*
2
, a
*
3
, i.e.
(
α
*
1
, α
*
2
, a
*
3
) = (
e
*
1
, e
*
2
, e
*
3
)(
P

1
)
t
.
Note that this problem is basically the change of coordinates problem: e.g.
the value of
α
*
1
on the vector
v
∈
C
3
is the first coordinate of
v
relative to the basis
α
1
, α
2
, α
3
.
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 Spring '10
 Hald,OH
 Differential Equations, Linear Algebra, Algebra, Equations, basis, p1, F3, Dual space, 1 121 3 M

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