Inversion of this Vandermonde matrix

[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
<br /> \begin{pmatrix}<br /> 1 &amp; 1 &amp; 1 &amp; \dotsb &amp; 1\\<br /> 1 &amp; \omega &amp; \omega^2 &amp; \dotsb &amp; \omega^{n-1}\\<br /> 1 &amp; \omega^2 &amp; \omega^4 &amp; \dotsb &amp; \omega^{2(n-1)}\\<br /> 1 &amp; \omega^3 &amp; \omega^6 &amp; \dotsb &amp; \omega^{3(n-1)}\\<br /> \vdots &amp; &amp;&amp;&amp; \vdots \\<br /> 1 &amp; \omega^{n-1} &amp; \omega^{2(n-1)} &amp; \dotsb &amp; \omega^{(n-1)(n-1)}<br /> \end{pmatrix}\mathbf{x}=<br /> \begin{pmatrix}<br /> 1 \\ 0 \\ 0 \\ 0 \\ \vdots \\ 0<br /> \end{pmatrix}<br />
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
<br /> 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?

 

[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.

 

[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...

 

[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?

 

[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.