09/03/2007, 12:46 PM

bo198214 Wrote:Henryk - please excuse the delay. I was a bit exhausted after my search for the eigensystem-solution. Well, I think I've not taken the best wording. What I wanted to say was, that if A is nilpotent, then the series is finite. But A is only nilpotent, if the base is e ( the diagonal is 1 and the diagonal of A-I is zero) - I had in mind, that we were talking about this base. Then the entries of the partial sums up to the power d do not change for rows<d.Gottfried Wrote:With A1 = A-I then log(A) = A1 - A1*A1/2 + A1*A1*A1/3 ....

which is a nice exercise... since it comes out, that A1 is nilpotent and we can compute an exact result using only as many terms as the dimension of A is. For the infinite dimensional case one can note, that the coefficients are constant when dim is increased step by step, only new coefficients are added below the previously last row.

Are you sure about this? For me it rather looks as if they converge.

The Eigenvalues are quite different depending on the truncation.

Even in the case where you can compute the logarithm via the infinite matrix power series, it should depend on where you truncate the matrix.

Call B=S2 - I, where S2 is the matrix of stirling-numbers 2'nd kind. The diagonal of B is zero.

The entries of the partial sums of a series, for instance

I, I+B/1! , I+B/1!+B^2/2!, ...

are constant up to rows

0, 1, 2, , ...

accordingly, since the additional terms do not contribute to that rows due to the nilpotency. This detail was all I wanted to point out; it's surely nothing much important there...

Regards-

Gottfried

Gottfried Helms, Kassel