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On the invertibility of symmetric Toeplitz matrices

  1. Feb 16, 2009 #1
    I am curious if anyone knows conditions on the invertibility of a symmetric Toeplitz matrix. In my research, I have a symmetric Toeplitz matrix with entries coming from the binomial coefficients.

    Any help would be appreciated.

    [6 4 1 0 0]
    [4 6 4 1 0]
    [1 4 6 4 1]
    [0 1 4 6 4]
    [0 0 1 4 6]
  2. jcsd
  3. Feb 16, 2009 #2


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    Gold Member

    In general, a matrix must be nonsingular to possess an inverse. This means that the rows or columns must be linearly independent. A nonsingular matrix cannot have a zero eigenvalue.

    It is known that a bisymmetric (symmetric about each diagonal) matrix such as yours has a bisymmetric inverse, should the inverse exist.
  4. Feb 17, 2009 #3
    I have gone through a standard graduate level linear algebra text and have checked all of the usual methods (finding the eigenvalues, determinant, et c.). This works fine for individual cases, but I am trying to prove this for a family of square matrices.

    In particular, I was hoping someone knew of a paper in which someone showed the symmetric Toeplitz matrices are invertible if they satisfy some simpler condition, e.g., the main diagonal has a certain magnitude.
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