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
>
>
--
Dr Kim T Parker
Director of Learning and Teaching
Annex Room 8
Kent Business School
The University of Kent
Canterbury
Kent CT2 7PE
Tel: 01227 823604
Fax: 01227 761187
Email: [log in to unmask]
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