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The Asymtotic Eiegen Values of a Circulant matrix

  1. May 8, 2012 #1
    Hi,

    The eigenvalues of a circulant matrix are given by:

    [tex]\lambda_n=\sum_{l=0}^Lh_l\exp\left(-j\frac{2\pi}{N}nl\right)[/tex]

    for n=0,1,...N-1. Is it legal to do analysis in asymptiptic sense (as N approaches infinity), in which case:

    [tex]\lambda_1=\cdots=\lambda_N=\sum_{l=0}^Lh_l[/tex]??

    Thanks
     
  2. jcsd
  3. May 8, 2012 #2

    marcusl

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    Gold Member

    I'm not a mathematician so my answer will lack rigor, but here goes:

    I would think the answer is no. The eigenvalues are the Fourier transform or spectrum of the top row (or first column) of the circulant matrix. For the eigenvalues to all be equal defines a constant spectrum, which implies that the first row consists of a one followed by all zeros. Note that this matrix is the identity matrix. If the eigenvalues are instead allowed to vary randomly by a little bit, then the spectrum looks approximately "white" such as you see for Gaussian noise. The row vector is thus a sequence drawn from a Gaussian random variable. Both of these are very special cases, of course, and there is no reason to believe that an arbitrary circulant matrix will resemble them. Hence I believe that your proposition is untrue in general.

    EDIT: Just thought of an obvious counter-example: a matrix of all 1's. No matter how large it gets, it has one non-zero eigenvalue and all rest zeros.
     
    Last edited: May 8, 2012
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