Gottfried Wrote:Now consider this backways. Using another fixpoint y1 this relation holds again,

y1 = s^y1

V(y1)~ * Bs = V(s^y1)~ = V(y1)~

First row of W1^-1 is V(y1)~

and W0^-1 <> W1^-1 in their first rows and supposedly the same for the whole matrices W0 <> W1 and

W1^-1 * Bs = D1 * W1^-1

is another solution, provided that the form of the first row in W1^-1 is still that of a powerseries.

Sorry, the Eigensystem decomposition for Bs truncated to n is unqiue (up to permutations of eigenvalues), how can there be two solutions???