Let
be a not unit N'th root of unity, i.e.
.
The discrete Fourier transform 's given by the symmetric Vandermonde matrix:
![{\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cf8ddf75bbf5198869acf7e533e55ebddf7dce09)
For example,
![{\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}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5ca98c779c209c2ca4b83537868db800cb2e52df)
- Exercise (*). The square of the Fourier transform is the identity transform:
![{\displaystyle F_{N}^{2}=Id.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5ab7ee9b027444fd10d6832f0ce5c7844923a79e)
- 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.