08/29/2007, 05:28 PM

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.