Show that $e^{At}B = B e^{At}$ if $AB= BA$ Let $A$ and $B$ be two matrices. How can I show that $e^{At}B = B e^{At}$ if $AB= BA$ and then conclude that $(d/dt) e^{At} e^{Bt} = (A+B) e^{At} e^{Bt}$? I have no idea how to do this. 

By definition
$$e^{At} = \sum_{n=0}^\infty \frac 1{n!} t^n A^n$$
thus term by term
$$B e^{At} = \sum_{n=0}^\infty \frac 1{n!} t^n BA^n = \sum_{n=0}^\infty \frac 1{n!} t^n A^nB = e^{At} B$$
Now apply this lemma for the second part.
