The Chinese Remainder Theorem gives a way to compute matrix exponentials.
Indeed, let $A$ be a complex square matrix, put $B:=mathbb C[A]$. This is a Banach algebra, and also a $mathbb C[X]$-algebra ($X$ being an indeterminate). Let $S$ be the set of eigenvalues of $A$, $$mu=prod_{sin S} (X-s)^{m(s)}$$ the minimal polynomial of $A$, and identify $B$ to $mathbb C[X]/(mu)$.
The Chinese Remainder Theorem says that the canonical $mathbb C[X]$-algebra morphism $$Phi:Bto C:=prod_{sin S} mathbb C[X]/(X-s)^{m(s)}$$ is bijective.
Computing exponentials in $C$ is trivial, so the only missing piece in our puzzle is the explicit inversion of $Phi$.
Fix $s$ in $S$ and let $e_s$ be the element of $C$ which has a one at the $s$ place and zeros elsewhere. It suffices to compute $Phi^{-1}(e_s)$. This element will be of the form $$f=frac{mu}{(X-s)^{m(s)}} g mbox{ mod }mu$$ with $f,ginmathbb C[X]$, the only requirement being $$gequivfrac{(X-s)^{m(s)}}{mu}mbox{ mod }(X-s)^{m(s)}$$ (the congruence taking place in the ring of rational fractions defined at $s$). So $g$ is given by Taylor's Formula.
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