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About Circulant Matrices

  1. Jan 27, 2012 #1
    Hi all,

    I am reading a paper which contains a lot of matrices. Anyway, there is this equation:

    [tex]\|\mathbf{H}_3\mathbf{H}_1\mathbf{e}\|^2=\mathbf{e}^{H}\mathbf{H}_1^{H}\mathbf{H}_3^{H}\mathbf{H}_3\mathbf{H}_1\mathbf{e}[/tex]

    where superscript H means conjugate transpose, and boldface Hs are N-by-N circulant and toeplitz matrices, where the first column is defined as:

    [tex]\begin{array}{ccccccc}h_i(0)&h_i(1)&\cdots &h_i(L)&\mathbf{0}_{1\times N-L-1}\end{array}[/tex]

    and e is some N-by-1 vector. It is claimed that the above equation can be approximated as:

    [tex]\|\mathbf{H}_3\|^2\|\mathbf{H}_1\mathbf{e}\|^2[/tex]

    but the authors did not say how and why? They just claimed that in simulation the mean square error between both of them is tolerable and small. Further it is said that:

    [tex]\|\mathbf{H}_3\|^2=N\sum_{m=0}^L|h_3(m)|^2[/tex]

    are all of that justifiable? and how?

    Thanks in advance
     
  2. jcsd
  3. Jan 28, 2012 #2

    marcusl

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    Evaluate the first claim in terms of the Schwarz inequality
    [tex]||AB||\leq||A||\cdot ||B||[/tex]
    The claim that equality is a good approximation must be specific to this system and the actual quantities involved. (This is further suggested by your quote that they found it to be true in simulation.) I don't see why it would follow generally from properties of circulant matrices.

    You don't state the particular matrix norm used in the 2nd claim. If it is the Frobenius norm
    [tex]||A||_F^2=\sum_{i,j}|a_{i,j}|^2[/tex]
    then your equation follows immediately since all components of a Toeplitz matrix are found in the first column, repeated N times.
     
  4. Jan 29, 2012 #3
    I forgot to include a factor of 1/N in the approximation, and yes, the norm is Frobenius.
     
  5. Jan 29, 2012 #4
    Is there any identity such that:

    [tex]\|\mathbf{A}\mathbf{B}\|^2_F\geq x[/tex]
     
  6. Jan 30, 2012 #5

    marcusl

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    I'm not aware of one, but someone more knowledgeable in math might know.
     
  7. Jan 30, 2012 #6
    Ok, thanks anyway.
     
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