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Inversion of this Vandermonde matrix

  1. Feb 28, 2010 #1
    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
    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)}
    1 \\ 0 \\ 0 \\ 0 \\ \vdots \\ 0
    with [itex]\omega=\exp(2\pi\mathrm{i}/n)[/itex]
    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)[/tex]
    with [itex]\omega=\exp(2\pi\mathrm{i}/3)[/itex]?

    Now I'm just wondering why I get linear dependent terms when I consider the real part only?
    Last edited: Feb 28, 2010
  2. jcsd
  3. Feb 28, 2010 #2


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    You're on the right track. It is convenient to write this equation in matrix notation as


    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

    [tex]x=W^{-1}b[/tex] .

    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,

    [tex]W^{-1} = W^{\dagger} [/tex]

    that is, W is Hermitian (the dagger is the conjugate transpose operation) and unitary (you get the identity matrix in the next equation)

    [tex]WW^{\dagger}=I [/tex] .

    You can perform this multiplication explicitly it to see that this is so. Accordingly


    and you can write out the components of x explicitly.
    Last edited: Feb 28, 2010
  4. Mar 1, 2010 #3
    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...
  5. Mar 1, 2010 #4


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    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?
  6. Mar 1, 2010 #5
    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.
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