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Lower Bound for a Determinant

  1. Apr 28, 2012 #1
    Hello,

    I have the following determinant:

    [tex]\text{det}\left(\mathbf{A}\mathbf{A}^H\right)[/tex]

    where H denoted complex conjugate transpose, and A is a circulant matrix. I am looking for a lower bound for the above determinant. Is there one?

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

    mfb

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    I think det(AB)=det(A)=det(B) should help to express your determinant via det(A). If something is known about A, it might be possible to evaluate the expression or give some lower bound.
     
  4. Apr 28, 2012 #3
    I do not want to evaluate it, I need a lower bound to see what is the lowest value for the determinant.
     
  5. Apr 28, 2012 #4

    AlephZero

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    The the eigenvectors of a circulant matrix are independent of the matrix and related to the nth roots of unity, so there are some nice ways to write the value of the determinant. See http://en.wikipedia.org/wiki/Circulant_matrix.

    For an arbitrary A the lower bound is zero, which isn't a very interesting result - but maybe you want a lower bound that depends on the elements of A in some way?
     
  6. Apr 28, 2012 #5
    Exactly, that is what I need, especially in term of A's columns.
     
  7. Apr 28, 2012 #6

    mfb

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    The evaluation gives a lower bound. In fact, it is the best possible lower bound.

    The wikipedia article has an explicit formula for the determinant, you can just calculate it and use it as a lower bound. If that needs to much computing power (which I doubt, but I don't know what you are doing), you can produce a lot of lower bounds of variable quality. But in that case, it would be useful to know in which way you need this bound.
     
  8. Apr 28, 2012 #7
    I am working on a wireless communication system, and I need to know the worst performance of the system, which is equivalent to know the "lowest" bound of the above determinant. I do not need any lower bound, I need the lowest bound. A is circulant and toeplitz, so the determinant is strictly greater than zero, and hence , I need the lowest determinant greater than zero, and I need it, if possible, in terms of the the first (or any) row of A, since all contains the same elements. That would ease the analysis.

    Thanks
     
  9. Apr 28, 2012 #8

    Office_Shredder

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    Why is the value for the determinant in the wikipedia article not good enough?
     
  10. Apr 28, 2012 #9
    Again, I do not need to find the determinant itself, I need a lower bound.
     
  11. Apr 29, 2012 #10

    mfb

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    As said before, the determinant IS a lower bound for the determinant of this matrix.
    If you want a lower bound for all possible matrices at the same time, you need additional data about the entries of the matrix. Are they always bounded in some way?


    Do you mean "the highest"? The lowest lower bound is -infinity, which is not what you are looking for I think. The highest lower bound is your determinant.
     
  12. Apr 29, 2012 #11
    [tex]\text{det}\left(\mathbf{A}\mathbf{A}^H\right)\geq f(\mathbf{h})=w[/tex]

    Yeah, I need to find w>0 for all possible realizations of the random matrix A. Further information about A:

    1- A is an N-by-N circulant matrix with first column:

    [tex]\mathbf{h}^T=[h_0\,\,h_1,\cdots,\,h_L,\mathbf{0}_{N-L+1}]^T[/tex]

    2- The entries of h are i.i.d Gasussian random variables with zero-mean and unit variance.

    Does that change any thing?
     
  13. Apr 29, 2012 #12

    mfb

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    I don't think your first equation is what you want:

    f(h)=|det(A)|2 with

    [tex]
    \mathrm{det}(A)
    = \prod_{j=0}^{n-1} (h_0 + h_{n-1} \omega_j + h_{n-2} \omega_j^2 + \dots + h_1\omega_j^{n-1}).[/tex]
    where [itex]\omega_j=\exp \left(\frac{2\pi i j}{n}\right)[/itex].

    But as f(h) depends on h, this bound now depends on the matrix.


    In that case, 0 is the best lower bound you can give. The determinant of your expression can reach arbitrary small values, as it is possible that all entries are 0 (as an example) and therefore the determinant is 0.
     
  14. Apr 29, 2012 #13
    Just throw that exception. I need the determinant that is strictly greater than zero.
     
  15. Apr 30, 2012 #14

    mfb

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    But there is no lower bound larger than 0 with the distribution of hi you gave. In addition to 0, it can reach a value smaller than epsilon for all epsilon > 0.
    It should be possible to evaluate the distribution function and estimate how likely a value smaller epsilon is (for a given epsilon), but it will never be 0.
     
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