I have an exponential matrix:
I now want to calculate $A^4$. And there is this explanation:
I dont understand how this was calculated. This has to imply that $A=0$ in the first place if I am correct and how am I supposed to see that? Does this have something to do with the fact that all 4 eigenvalues to A are 0 somehow (due to highest order of polynomial in $e^{tA}$ and no e term there)?
If$$f(t)=\exp(tA)=\operatorname{Id}+tA+\frac12(tA)^2+\frac1{3!}(tA)^3+\cdots,$$you have $f(0)=\operatorname{Id}$. Also,$$f'(t)=A+tA^2+\frac12t^2A^3+\frac1{3!}t^3A^4+\cdots=A\exp(tA),$$and therefore $f'(0)=A$. More generally, $f^{(n)}(0)=A^n$. But, in your case, since $f(t)$ is a $4\times 4$ matries whose entries are polynomial functions with degree smaller than $4$, $f^{(4)}(0)=0$; in other words, $A^4=0$. It does not follow from this that $A=0$.