when does ${e^{\left( {A + B} \right)t}} = {e^{At}}{e^{Bt}}$?

Question

is it sufficient that $AB=BA$ to conclude that ${e^{\left( {A + B} \right)t}} = {e^{At}}{e^{Bt}}$ ?

Answer 1

Yes, this is sufficient, because when $AB = BA$ then both sides are limits of the same series. Another way to see this is that by a density argument you can assume that $A$ and $B$ are simultaneously diagonalizable, and then the equality is obvious.

Conversely, when $e^{t(A+B)} = e^{tA} e^{tB}$ for all (or sufficiently small) $t > 0$, then $AB = BA$. This is because the second derivative of both sides at $0$ are $(A+B)^2$ and $A^2 + 2AB + B^2$ respectively.