Commuting matrices and exponential function

Question

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 $A\exp(Bt)=\exp(Bt)A$?

Answer 1

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