Commuting matrices and exponential function Let $A,B$ be $n\times n$ commuting matrices, that is $AB=BA$.
I also know that $\exp(Bt)=X(t)X(0)^{-1}$ where $X$ is the fundamental matrix function.
How can I show that $A\exp(Bt)=\exp(Bt)A$?

For any $n$,
$$A\cdot(I+Bt+\frac12B^2t^2+\frac16B^3t^3+\ldots +\frac1{n!}B^nt^n)=(I+Bt+\frac12B^2t^2+\frac16B^3t^3+\ldots +\frac1{n!}B^nt^n)\cdot A $$
