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From: "Roger Stafford" <ellieandrogerxyzzy@mindspring.com.invalid>
Newsgroups: comp.soft-sys.matlab
Subject: Re: Eigenvalues Problem?
Date: Mon, 31 Dec 2007 04:48:40 +0000 (UTC)
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"Michael Nikki" <niqmk@yahoo.com> wrote in message <fl7qqb$1s7
$1@fred.mathworks.com>...
> ........
> So how to get Sk from
> 
> (-1)^n lambda^n + (-1)^(n-1) S1 lambda^(n-1) + (-1)^(n-2) Sk
> > > lambda^(k) + Sn
> 
> It says sub matrix from A (k x k). What the mean of sub
> matrix? Sorry I'm very newbie from this. Thanks for your answer.
--------
  By expanding the expression, det(A-lambda*I), and collecting equal powers 
of lambda, it can be seen that each of your coefficients, Sk, is the "sum of all 
the principal minor determinants of A with k rows", to quote one website.  
That is, Sk is the sum of all the determinants with k rows and k columns that 
can be obtained by eliminating n-k of the diagonal elements of A, along with 
the corresponding row and column of each.  For example, with your 3 x 3 
determinant

 A =
 a b c
 d e f
 g h i

the value of S2 would be the sum of the three determinants

 det([a b;d e]) + det([a c;g i]) + det([e f;h i]) =
 (a*e-b*d)+(a*i-c*g)+(e*i-f*h) ,

which agrees with your earlier expression (if you correct the missing ai.)  
These determinants are all the possible minor determinants of A with one 
(3-2) diagonal element and the associated row and column removed.

  As you can see, for a general k there will be nCk (the number of ways of 
choosing k out of n things) different possible minor determinants forming the 
sum in Sk.  I am not aware of any compact formula for such a sum.  
Computing the characteristic polynomial for a general square matrix of large 
order is therefore a non-trivial task.

Roger Stafford