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Math Help - tridiagonal matrix problem

  1. #1
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    tridiagonal matrix problem

    This is a quantum chemistry problem I'm working on; not a homework problem - something I need for an article I'm writing. I don't think knowing the chemistry is important, as at this point it seems to be "just a math problem". I can give some chemical background leading to the formulation of the problem, if that helps.

    The problem is as follows:

    I need to find a general solution to the determinant of an n x n matrix (call it M) with elements M(ij) defined as:

    M(ij) = x, if i = j
    M(ij) = a, if |i-j| = 1 and i+j = 3, 7, 11, ....
    M(ij) = b, if |i-j| = 1 and i+j = 5, 9, 13, ....
    M(ij) = 0 otherwise.

    For example, the M(n = 6) matrix would be:

    x a 0 0 0 0
    a x b 0 0 0
    0 b x a 0 0
    0 0 a x b 0
    0 0 0 b x a
    0 0 0 0 a x

    Basically, the "a's" and "b's" alternate.

    Can the determinant of this matrix be expressed generally as a function of a, b, x and n?
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  2. #2
    Super Member
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    Haven't tried it completely but you might be able to work out a recursive formula, if that should help you
    something like
    D(n) = [+1or-1].x.D(n-1) + [+1or-1].[a^2(or)b^2].D(n-2)
    values within square brackets will depend on 'n', in case of n=6
    D(6) = x.D(5) + a^2.D(4)
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  3. #3
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    Illinois
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    Thanks; if I do it out longhand (work out the determinant for several values of n) I can start to see a pattern, but I'm still having trouble generalizing it.

    I suppose I should write out more clearly what I'm really after. What I really need is a general formula for the roots (values of x, as a function of a, b, and n) when the determinant of M is set equal to zero.

    A simpler version of the problem involves replacing all values of a with b, so that all off-diagonal elements of M are b.

    i.e.

    M(ij) = x if i = j
    M(ij) = b if |i-j| = 1
    M(ij) = 0 otherwise

    A solution for the kth value of x when the determinant is set equal to zero is

    x(k) = -2b*cos(R)

    where

    R = k * pi / (n + 1)

    And k = 0, 1....n.

    In case anyone is interested (to give the problem some relevance), the values of x in these matrices represent the energy states of some model conjugated conducting polymers. Unfortunately, the simpler problem, which appears in many introductory p-chem texts, treats all carbon-carbon bonds as equal in length, which isn't really true. A more accurate representation of most true polymers is that the bonds alternate long and short (hence the alternating a and b terms in the more complex matrix in my original post). I do have the derivation of the solution to the simpler problem, but trying to apply the strategy involved there toward the more complex problem has ran me into a wall; the solution involves breaking down the matrix into its constituent linear equations.

    I do believe the solution to the complex problem is something of the form

    x = +/- SQRT(a^2 + b^2 + 2ab*cos(R))

    But in this case R is a different function of n and I'm not sure of any coefficients.
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