bo198214 Wrote:If in a triangular matrix A the diagonal are all ones, the eigenvalues of (A-I) are zero. So I have to compute the matrix-logarithm by the ordinary series expressionGottfried Wrote:But it seems as if A must have a diagonal of 1.bo198214 Wrote:Did you anyway already realize that instead ofNo, but it looks very good. I'll give it a deeper look, thanks for the hint!

you can directly use the binomial series for computation?

So probably you have to convert A into a Jordan normal form first as you must do with the logarithm (or how do you compute the matrix logarithm?)

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. So even the infinite case is analytically accessible.

Gottfried

Gottfried Helms, Kassel