Matrix Operator Method

[bo198214]

Did you anyway already realize that instead of
\( \exp(t\cdot\log(A)) \) you can directly use the binomial series for computation?
\( A^t = \sum_{n=0}^\infty \left(t\\n\right) (A-I)^n \)

No, but it looks very good. I'll give it a deeper look, thanks for the hint!

But it seems as if A must have a diagonal of 1.
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?)