I haven't thought about how the following might be applied to the difficulty raised by Gene, but it might be worth noting that one of the many ways in which mathematicians conventionally complicate their lives is by defining polynomials over finite fields in an avoidably clumsy fashion. Restricting oneself to linear combinations of powers excludes a lot of functions which ought to be polynomial, but fail to do so on account of the Fermat little theorem x^p = x mod p. Suppose instead polynomials are defined as linear combinations of binomial coefficients, where binomial coefficients are in turn defined via Pascal's triangle. Now --- for example --- linear difference equations with constant coefficients all have polynomial solutions modulo a prime, just as they do over the rationals. Fred Lunnon On 12/3/09, victor miller <victorsmiller@gmail.com> wrote:
What I said is "almost" true. What one really needs to look at is the ring of functions which are generated by all power sums as a subring of the ring of all symmetric functions. As was pointed out, in characteristic 0 (or, more generally anything larger than the degree) they are the same. In small characteristic they are not. There are formulas for the index of the rings (they're related to the theory of modular representations of the symmetric group).
Victor
On Thu, Dec 3, 2009 at 12:13 PM, victor miller <victorsmiller@gmail.com>wrote:
There are actually formulas which work in the "small" characteristic cases. In your case you'd need to look at tr(M^3). This is related to the decoding of BCH codes (where they work out the formulas in characteristic 2).
Victor
On Thu, Dec 3, 2009 at 12:09 PM, Eugene Salamin <gene_salamin@yahoo.com>wrote:
Because of the appearance of n! denominators, all this is true only in characteristic zero. As a counterexample, let M be the 2x2 identity matrix in characteristic 2. The charpoly is (x - 1)^2 = x^2 + 1. But tr(M) = tr(M^2) = 0.
-- Gene
________________________________ From: victor miller <victorsmiller@gmail.com> To: math-fun <math-fun@mailman.xmission.com> Sent: Thu, December 3, 2009 8:39:55 AM Subject: Re: [math-fun] Expansion of charpoly's in terms of traces ?
Yes, The key idea are the Newton Identities ( http://en.wikipedia.org/wiki/Newton's_identities ), which give a way of going from the power sums of the roots of a polynomial to the symmetric functions of the roots. For matrices, the power sums of the eigenvalues are the the traces of the powers of the matrix, and the coefficients of the characteristic polynomial are the symmetric functions of the roots (up to sign). Note that it is only necessary to prove this formula when the matrix has all distinct eigenvalues because this set of matrices is Zariski dense in the set of all matrices (i.e. if there is a polynomial equation giving the characteristic polynomial in terms of the traces of the powers of the matrix which holds for matrices with distinct eigenvalues, it must hold for all matrices).
Victor
On Thu, Dec 3, 2009 at 11:18 AM, Henry Baker <hbaker1@pipeline.com> wrote:
I noticed in Wikipedia's description of eigenvalues, an expansion of the characteristic polynomial of a 3x3 matrix in terms of traces, so I wanted to see if this worked more generally.
For a 2x2 matrix, the charpoly is:
x^2-x*tr(M)+det(A) = x^2-x*tr(M)+ (tr(M)^2-tr(M^2))/2
For a 3x3 matrix, the charpoly is:
x^3-x^2*tr(M)+x*(tr(M)^2-tr(M^2))/2-det(M) =
x^3-x^2*tr(M)+x*(tr(M)^2-tr(M^2))/2-(tr(M)^3+2*tr(M^3)-3*tr(M)*tr(M^2))/6
For a 4x4 matrix, the charpoly is:
x^4-x^3*tr(M)+x^2*(tr(M)^2-tr(M^2))/2-x*(tr(M)^3+2*tr(M^3)-3*tr(M)*tr(M^2))/6+det(M)
Notice that the higher order terms are the same form even though the dimension is higher.
Can we continue this indefinitely?
What is the general term?
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