04/28/2008, 10:04 PM
bo198214 Wrote:If you apply the diagonalization method to the fixed point you get nice triangular matrices with the powers of \( f'(a) \) on the diagonal (the eigenvalues). However if you apply it to 0, you neither have triangular matrices nor are the eigenvalues powers of something (edit: though I should check the latter before asserting).
Hmm, the fixpoints-shifts resulted in similarity scalings with the matrix-method (using powers of the pascalmatrix). Then the eigenvalues should be unchanged.
Gottfried
Gottfried Helms, Kassel

