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Then
f
r
,the
r
th component of
F
n
(
a
)
,isdefinedas
n−
1
f
r
=
p
(
ω
r
)=
a
k
ω
rk
(6.22)
k
=
0
This computation is equivalent to the following matrix-vector multiplication:
F
n
(
a
)=[
w
ij
]
×
a
(6.23)
where
[
ω
ij
]
is the
n
n
Vandermonde matrix
whose
i
,
j
entry is
ω
ij
,0
×
≤
i
,
j
≤
n
−
1, and
a
is treated as a column vector.
The
n
-point
inverse discrete Fourier transform
F
−
1
:
R
n
R
n
→
is defined as the values
n
of the following polynomial
q
(
x
)
at the inverse
n
th roots of unity:
q
(
x
)=(
f
0
+
f
1
x
+
f
2
x
2
+
+
f
n−
1
x
n−
1
)
/n
···
(6.24)
That is, the inverse DFT maps an
n
-tuple
f
to an
n
-tuple
g
,namely,
F
−
n
(
f
)=
g
,where
g
s
is defined as follows:
n−
1
g
s
=
q
(
ω
−s
)=
1
n
f
l
ω
−ls
(6.25)
l
=
0
This computation is equivalent to the following matrix-vector multiplication:
F
−
n
(
f
)=
1
n
ω
−ij
×
f
Because of the following lemma it is legitimate to call
F
−
1
the
inverse
of
F
n
.
n
LEMMA
6.7.3
For all
a
∈ R
n
,
a
=
F
−
n
(
F
n
(
a
))
.
Proof
Let
f
=
F
n
(
a
)
and
g
=
F
−
n
(
f
)
.Then
g
s
satisfies the following:
n
−
n
−
n
−
1
1
1
g
s
=
1
n
f
l
ω
−ls
=
1
n
a
k
ω
(
k−s
)
l
l
=
0
l
=
0
k
=
0
n
−
n
−
1
1
1
n
ω
(
k−s
)
l
=
a
k
k
=
0
l
=
0
=
a
s
The second equation results from a change in the order of summation. The last follows
from the definition of
n
th roots of unity. It follows that the matrix
[
ω
−ij
/n
]
is the inverse
of
[
ω
ij
]
.
The computation of the
n
-point DFT and its inverse using the naive algorithms suggested
by their definitions requires
O
(
n
2
)
steps. Below we show that a fast DFT algorithm exists for
which only
O
(
n
log
n
)
steps suffice.
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