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Let
ω
{\displaystyle \omega }
N' th root of unity, i.e.
ω
N
=
1
,
ω
≠
1
{\displaystyle \omega ^{N}=1,\omega \neq 1}
Fourier transform 's given by the symmetric Vandermonde matrix :
F
ω
=
1
N
[
1
1
1
…
1
1
ω
ω
2
…
ω
(
N
−
1
)
1
ω
2
⋮
…
ω
2
(
N
−
1
)
⋮
⋮
⋮
⋱
⋮
1
ω
(
N
−
1
)
ω
2
(
N
−
1
)
…
ω
(
N
−
1
)
2
]
{\displaystyle F_{\omega }={\frac {1}{\sqrt {N}}}{\begin{bmatrix}1&1&1&\ldots &1\\1&\omega &\omega ^{2}&\ldots &\omega ^{(N-1)}\\1&\omega ^{2}&\vdots &\ldots &\omega ^{2(N-1)}\\\vdots &\vdots &\vdots &\ddots &\vdots \\1&\omega ^{(N-1)}&\omega ^{2(N-1)}&\ldots &\omega ^{(N-1)^{2}}\\\end{bmatrix}}}
For example,
F
e
2
π
i
/
5
=
1
5
[
1
1
1
1
1
1
ω
ω
2
ω
3
ω
4
1
ω
2
ω
4
ω
6
ω
8
1
ω
3
ω
6
ω
9
ω
12
1
ω
4
ω
8
ω
12
ω
16
]
=
1
5
[
1
1
1
1
1
1
ω
ω
2
ω
3
ω
4
1
ω
2
ω
4
ω
ω
3
1
ω
3
ω
ω
4
ω
2
1
ω
4
ω
3
ω
2
ω
]
.
{\displaystyle F_{e^{2\pi i/5}}={\frac {1}{\sqrt {5}}}{\begin{bmatrix}1&1&1&1&1\\1&\omega &\omega ^{2}&\omega ^{3}&\omega ^{4}\\1&\omega ^{2}&\omega ^{4}&\omega ^{6}&\omega ^{8}\\1&\omega ^{3}&\omega ^{6}&\omega ^{9}&\omega ^{12}\\1&\omega ^{4}&\omega ^{8}&\omega ^{12}&\omega ^{16}\\\end{bmatrix}}={\frac {1}{\sqrt {5}}}{\begin{bmatrix}1&1&1&1&1\\1&\omega &\omega ^{2}&\omega ^{3}&\omega ^{4}\\1&\omega ^{2}&\omega ^{4}&\omega &\omega ^{3}\\1&\omega ^{3}&\omega &\omega ^{4}&\omega ^{2}\\1&\omega ^{4}&\omega ^{3}&\omega ^{2}&\omega \\\end{bmatrix}}.}
Exercise (*). The square of the Fourier transform is the identity transform:
F
N
2
=
I
d
.
{\displaystyle F_{N}^{2}=Id.}
Exercise (*). If an e-network is rotation invariant, then so 's the conductivity equation and the Dirichlet-to-Neumann map is diagonal in the Fourier coordinates (the column vectors of the matrix.
