Register to reply

Inversion of this Vandermonde matrix

by Gerenuk
Tags: inversion, matrix, vandermonde
Share this thread:
Feb28-10, 08:09 AM
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
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)+\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)[/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?
Phys.Org News Partner Science news on
Experts defend operational earthquake forecasting, counter critiques
EU urged to convert TV frequencies to mobile broadband
Sierra Nevada freshwater runoff could drop 26 percent by 2100
Feb28-10, 09:17 PM
Sci Advisor
PF Gold
P: 2,081
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.
Mar1-10, 06:00 AM
P: 1,060
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...

Mar1-10, 09:50 AM
Sci Advisor
PF Gold
P: 2,081
Inversion of this Vandermonde matrix

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?
Mar1-10, 12:02 PM
P: 1,060
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.

Register to reply

Related Discussions
Vandermonde matrix Linear & Abstract Algebra 1
Vandermonde Matrix - Help! Precalculus Mathematics Homework 2
Finding determinant of Vandermonde matrix Linear & Abstract Algebra 2
Vandermonde matrix Calculus & Beyond Homework 0
Matrix Inversion Precalculus Mathematics Homework 3