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Closed-form determinant of a hermitian banded toeplitz matrix!

  1. Jan 1, 2013 #1
    Hello everyone,

    I found that you're actively discussing math problems here and thought to share my problem with you.

    I have a specially structured complex-valued [itex]n \times n[/itex] matrix, that has only three non-zero constant diagonals (the main diagonal, the [itex]j^{th}[/itex] subdiagonal and the [itex]j^{th}[/itex] superdiagonal), [itex]1 \leq j \leq n-1[/itex]. Moreover, it is a hermitian matrix, where the element composing the superdiagonal is actually the conjugate of that of the subdiagonal. For example, if [itex]n=7, j=3[/itex], the matrix is given by:

    A &=& \left[\begin{array}{ccccccc}
    a &0 &0 &b^* &0 &0 &0 \\
    0 &a &0 &0 &b^* &0 &0 \\
    0 &0 &a &0 &0 &b^* &0 \\
    b &0 &0 &a &0 &0 &b^* \\
    0 &b &0 &0 &a &0 &0 \\
    0 &0 &b &0 &0 &a &0 \\
    0 &0 &0 &b &0 &0 &a
    \end{array} \right]

    [Question:] I want to get the determinant, or the eigenvalues in closed form.

    [Some hints:]
    - It is clear that the determinant will be only a function of [itex]a[/itex], [itex]b[/itex], the shift [itex]j[/itex] and the order of the matrix, [itex]n[/itex].
    - The matrix has the following properties:
    1- It is a sparse Toeplitz matrix, that has only three non-zero diagonals.
    2- It is a hermitian matrix.
    3- It can be regarded as a special banded matrix, with zero diagonals inside the band.
    4- We can also consider it as a diagonally dominant matrix. However, neglecting [itex]b[/itex] may not give a good approximation.
    - A Tridiagonal Toeplitz matrix (for the special case when [itex]j=1[/itex]) already has a known closed form expression for its eigen values, and consequently the determinant which is their direct product. It would be helpful also if we can express this shift in the diagonals as a certain simple operator, and use the known results of the tridiagonal case.

    Any ideas?
    Last edited: Jan 1, 2013
  2. jcsd
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