There is a relatively simple derivation, that uses three facts from
matrix linear algebra: (a) the sum of the eigenvalues of a positive
definite symmetric matrix is equal to the trace of the matrix (i.e.,
the sum of its diagonal elements, (b) the determinant of such a
matrix is equal to the product of the eigenvalues, and (c) an
intraclass correlation matrix has only two distinct eigenvalues.
For simplicity, suppose that A is a k x k matrix with 1 on the
diagonal and r off the diagonal. (The original problem can be
recovered by rescaling.) A is a correlation matrix with intraclass
structure, provided that r > -1/(k-1). If we multiply the vector of
1's by A, then we recover A, multiplied by L=1+(k-1)r, so that L is
one of the eigenvalues of A. Since there are only two distinct
eigenvalues, and since they must add to the trace of A, the other
(k-1) eigenvalues are all equal to M=1-r. So the determinant of A is
LM^(k-1) = [1+(k-1)r](1-r)^(k-1).
Ron Thisted
At 12:00 AM +0100 9/28/06, allstat automatic digest system wrote:
>Date: Wed, 27 Sep 2006 19:37:29 +0100
>From: "Kim T. Parker" <[log in to unmask]>
>Subject: Re: about the determinant of a symmetric compound matrix
>
>Stefano,
>
>If you divide every element of the matrix by beta, you end up with
>alpha/beta on the diagonals and 1 everywhere else. If we write
>c=alpha/beta, the determinant of the matrix is easier to find - you
>subtract the second row from the first row which gives you only two
>non-zero elements in the first row and thus only two terms in the
>expansion of the determinant. You can then spot some recurrence
>relationships between the k by k matrix and the (k-1) by (k-1)
>cofactors. Following these through gives the determinant as being
>(c-1)^(k-1)*(c+k-1). However, we divided every element in the k rows of
>the original matrix by beta, so we have to multiply our simplified
>determinant by beta^k. Applying this factor, and substituting for c in
>terms of alpha and beta, gives the following answer:
>
>(alpha - beta)^(k-1)*(alpha + beta*(k - 1))
>
>I've checked this up to beta=4 so it looks all right. Other Allstatters
>might spot a flaw, or have a more elegant way of deriving the result.
>
>Best wishes,
>
>Kim Parker
>
>Stefano Sofia wrote:
>
>>Dear Allstat users,
>>probably some of you will be able to help me.
>>
>>I have a square matrix of dimension k by k with alpha on the
>>diagonal and beta everywhee else.
>>This symmetric matrix is called symmetric compound matrix and has the form
>>a( I + cJ),
>>where
>>I is the k by k identity matrix
>>J is the k by k matrix of all ones
>>a = alpha - beta
>>c = beta/a
>>
>>I need to evaluate the determinant of this matrix. Is there any
>>algebric formula for that?
>>
>>thank you for your help
>>Stefano
> >
>>
--
Ronald Thisted, PhD Department of Health Studies
Professor and Chairman The University of Chicago
(773) 834-1242 5841 S Maryland Ave (MC 2007)
[log in to unmask] Chicago, IL 60637
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