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eigenvalues of a circulant matrix

It's hard to argue against the fact that a real symmetric matrix has real
eigenvalues. The eigenvalues of the circulant matrix with first row v are
*polynomials* (not the roots of 1 themselves, unless as Rolf suggested you
start with a vector with all zeros except one 1) in the roots of 1, with
coefficients equal to the entries in v. This is the finite Fourier transform
of v, by the way, and takes real values when the coefficients are real and
symmetric, ie when the matrix is symmetric.

Reid Huntsinger

-----Original Message-----
From: r-help-bounces at stat.math.ethz.ch
[mailto:r-help-bounces at stat.math.ethz.ch] On Behalf Of Globe Trotter
Sent: Monday, May 02, 2005 10:23 AM
To: Rolf Turner
Cc: r-help at stat.math.ethz.ch
Subject: Re: [R] eigenvalues of a circulant matrix
--- Rolf Turner <rolf at math.unb.ca> wrote:
Agreed. As may a circulant matrix if a_i = a_{p-i+2}
The eigenvalues are 4+1*omega+2*omega^2+3*omega^3.
omega=cos(2*pi*k/4)+isin(2*pi*k/4) as k ranges over 1, 2, 3, 4, so the above
holds.

 Bellman may have
No, that is not true: his result can be verified for any circulant matrix,
directly.
Many thanks and best wishes!

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