# Inversion of this Vandermonde matrix

by Gerenuk
Tags: inversion, matrix, vandermonde
 P: 1,060 I was trying to expand a three and more parameter functions similarly to the two-parameter case f(x,y)=(f(x,y)+f(y,x))/2+(f(x,y)-f(y,x))/2. Anyway, to do the same for more parameters I need to solve $$\begin{pmatrix} 1 & 1 & 1 & \dotsb & 1\\ 1 & \omega & \omega^2 & \dotsb & \omega^{n-1}\\ 1 & \omega^2 & \omega^4 & \dotsb & \omega^{2(n-1)}\\ 1 & \omega^3 & \omega^6 & \dotsb & \omega^{3(n-1)}\\ \vdots & &&& \vdots \\ 1 & \omega^{n-1} & \omega^{2(n-1)} & \dotsb & \omega^{(n-1)(n-1)} \end{pmatrix}\mathbf{x}= \begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \\ \vdots \\ 0 \end{pmatrix}$$ with $\omega=\exp(2\pi\mathrm{i}/n)$ Is there a closed form expression for x? EDIT: Oh, silly me. I realized it's a discrete Fourier transform. So is this the correct way to expand then? In the 3 parameter case the solution would be $$f(x,y,z)=\frac13(f(x,y,z)+f(y,z,x)+f(z,x,y))+\frac13\left(f(x,y,z)+\ome ga f(y,z,x)+\omega^*f(z,x,y)\right)+\frac13\left(f(x,y,z)+\omega^*f(y,z,x) +\omega f(z,x,y)\right)$$ with $\omega=\exp(2\pi\mathrm{i}/3)$? Now I'm just wondering why I get linear dependent terms when I consider the real part only?
 Sci Advisor PF Gold P: 2,063 You're on the right track. It is convenient to write this equation in matrix notation as $$Wx=b$$ where b is the vector you wrote on the RHS. Since W is full-rank and non-singular, it is invertible and the solution to your problem is $$x=W^{-1}b$$ . You need the inverse of W, but this is easy since we know from the properties of the discrete Fourier transform (DFT) that the individual vector columns of W are independent. Each is an orthonormal basis vector of the DFT. In words, this follows because the component of signal at one frequency (cosine or sine at w_m) is independent to that at every other frequency w_n. In fact, $$W^{-1} = W^{\dagger}$$ that is, W is Hermitian (the dagger is the conjugate transpose operation) and unitary (you get the identity matrix in the next equation) $$WW^{\dagger}=I$$ . You can perform this multiplication explicitly it to see that this is so. Accordingly $$x=W^{\dagger}b$$ and you can write out the components of x explicitly.