Matrix Operator Method

[bo198214]

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-loagarithm by the ordinary series expression
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.

Yes, and the diagonal is 1 in the parabolic case.
Because the matrix \( A-I \) is nilpotent also the binomial series
\( \sum_{k=0}^\infty \left(t\\k\right) (A-I)^k \)
is finite and this is the way Jabotinsky derived the double binomial formula.

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.

Yes, but how do you compute the infinite case. You can not simply do
\( \sum_{n=0}^\infty (-1)^{n+1} \frac{(A-I)^n}{n} \) as this does not converge. So do you compute it as described here via the Jordan form?