# Inversion of this Vandermonde matrix

1. Feb 28, 2010

### Gerenuk

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
$$f(x,y,z)=\frac13(f(x,y,z)+f(y,z,x)+f(z,x,y))+\frac13\left(f(x,y,z)+\omega f(y,z,x)+\omega^*f(z,x,y)\right)+\frac13\left(f(x,y,z)+\omega^*f(y,z,x)+\omega f(z,x,y)\right)$$
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?

Last edited: Feb 28, 2010
2. Feb 28, 2010

### marcusl

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.

Last edited: Feb 28, 2010
3. Mar 1, 2010

### Gerenuk

I just wonder if that way to decompose a multiparameter function makes sense.

It's nicely symmetrical and generalizes to higher dimensions. However it introduces complex number where the initial function might actually be real only. And also I haven't included odd parity permutations of the function arguments...

4. Mar 1, 2010

### marcusl

Sorry, I'm not following your comments. I provided the closed solution to Wx=b, where W is complex, which was the question asked in your first post. Are you looking for something else?

5. Mar 1, 2010

### Gerenuk

I did that exercise to find a way to extend the rule f(x,y)=(f(x,y)+f(y,x))/2+(f(x,y)-f(y,x))/2 to higher dimensions (I didn't explain the connection; just mentioned it in the intro). I assumed some cyclic symmetry for the final form of f(x,y,z)=... and with the help of the discrete Fourier transform (which I didnt recognise at first), I can find some "decomposition".

Now I wasn't sure if the decomposition is useful this way.