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Determinant of enlarged Correlation Matrix

  1. Dec 14, 2012 #1
    Determinant of "enlarged" Correlation Matrix

    Hi guys,

    I am not a physicist but saw that you guys are actively discussing math problems in this forum. I have the following problem that I've been fighting with for some time now: I have a n-dimensional (n >= 3) correlation matrix with the following special structure

    [itex]C(n, \rho, x) = \left( \begin{array}{c c c c c} 1 & \rho & x & \dots & \rho\\ \rho & 1 & \rho & \dots & \rho\\ x & \rho & 1 & \dots & \rho\\ \vdots & \vdots & \vdots & \ddots & \vdots\\ \rho & \rho & \rho & \dots & 1 \end{array} \right). [/itex]

    I.e. all off-diagonal entries are equal to [itex]\rho[/itex] except for the correlation pair (1, 3). Here is my problem: Let's say I know that the determinant for dimension n is given by [itex]\psi(n)[/itex]. What is the determinant for dimension n + 1? I played around with Mathematica and found that for n = 3

    [itex]\psi(3) = 1 - x^2 - 2 \rho^2 + 2 x \rho^2[/itex]

    (this was the easy part) and for n >= 4, we have the relationship

    [itex]\psi(n) = (\rho - 1) \left( -\psi(n - 1) + (-1)^n (x - 1) (\rho - 1)^{n - 4} \left( \rho (1 + x) - 2 \rho^2 \right) \right)[/itex]

    (this is my problem - how to show this?). Equivalently, we can directly state the formula for the n-th determinant (n >= 3) as

    [itex]\psi(n) = (-1)^n (x - 1) (\rho - 1)^{n - 3} \left( 1 + x + (n - 3) \rho (1 + x) - 2 (n - 2) \rho^2 \right)[/itex]

    but I have no clue of how to get there. Note that by increasing the dimension by one, we basically set

    [itex]C(n, \rho, x) = \left( \begin{array}{c c} C(n - 1, \rho, x) & \rho 1_{\{ (n - 1) \times 1 \}}\\ \rho 1_{\{ 1 \times (n - 1) \}} & 1 \end{array} \right) [/itex]

    Any help is appreciated!
     
  2. jcsd
  3. Dec 14, 2012 #2

    chiro

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    Science Advisor

    Re: Determinant of "enlarged" Correlation Matrix

    Hey LocalVol.

    One question I have for you is are there results for determinants given a partitioned matrix or a matrix which is a tensor product of existing matrices?

    In other words if your final matrix is a tensor product of matrices A X B then is there a result that looks at the determinant of A X B in terms of det(A) and det(B)?

    I don't know enough linear algebra to know if such a result exists.
     
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