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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