Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    Science Advisor
    Gold Member

    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


    User Avatar
    Science Advisor
    Gold Member

    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.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: Inversion of this Vandermonde matrix
  1. Vandermonde matrix (Replies: 1)

  2. Inverse Matrix (Replies: 2)

  3. Inverse of a matrix (Replies: 4)

  4. Inverse of a matrix (Replies: 3)