On 2D Inverse Problems/Fourier coordinates

Let $\omega$ be a not unit N'th root of unity, i.e. ${\omega }^N=1,\omega \ne 1$.

The discrete Fourier transform 's given by the symmetric Vandermonde matrix:

$F_{\omega }=\frac{1}{\sqrt{N}}\left[\begin{array}{ccccc}1& 1& 1& \dots & 1\\ 1& \omega & {\omega }^2& \dots & {\omega }^{(N-1)}\\ 1& {\omega }^2& ⋮& \dots & {\omega }^{2(N-1)}\\ ⋮& ⋮& ⋮& \ddots & ⋮\\ 1& {\omega }^{(N-1)}& {\omega }^{2(N-1)}& \dots & {\omega }^{(N-1{)}^2}\end{array}\right]$

For example,

$F_{e^{2\pi i/5}}=\frac{1}{\sqrt{5}}\left[\begin{array}{ccccc}1& 1& 1& 1& 1\\ 1& \omega & {\omega }^2& {\omega }^3& {\omega }^4\\ 1& {\omega }^2& {\omega }^4& {\omega }^6& {\omega }^8\\ 1& {\omega }^3& {\omega }^6& {\omega }^9& {\omega }^{12}\\ 1& {\omega }^4& {\omega }^8& {\omega }^{12}& {\omega }^{16}\end{array}\right]=\frac{1}{\sqrt{5}}\left[\begin{array}{ccccc}1& 1& 1& 1& 1\\ 1& \omega & {\omega }^2& {\omega }^3& {\omega }^4\\ 1& {\omega }^2& {\omega }^4& \omega & {\omega }^3\\ 1& {\omega }^3& \omega & {\omega }^4& {\omega }^2\\ 1& {\omega }^4& {\omega }^3& {\omega }^2& \omega \end{array}\right].$

Exercise (*). The square of the Fourier transform is the identity transform:

$F_N^2=Id.$

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.