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

Question

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

See What does $[A,B]=0$ mean in matrix theory?, Show that $e^{t(A+B)} = e^{tA}e^{tB}$ for all $t \in \mathbb{R}$ if, and only if $AB = BA$. — Maximilian Janisch, Feb 12 '22 at 09:26

score 3 · Accepted Answer · answered Feb 12 '22 at 09:14

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.

Please carefully read Enforcement of Quality Standards. – amWhy Feb 15 '22 at 18:57 — amWhy, Feb 15 '22 at 18:57

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

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