# If $\exp(t(A + B)) = \exp(tA) \exp(tB)$ for all $t \geq 0$ then $A,B$ commute

Let $A,B$ be complex valued square matrices. If $\exp(t(A + B)) = \exp(tA) \exp(tB)$ for all $t \geq 0$ then $A,B$ commute.

The converse of this statement can be an easy application of the Cauchy product rule and the binomial theorem.

Note that this statement doesn't hold, if we restrict ourselves to $t = 1$.

So far I have been trying to use the fact, that $A$ and $B$ are infinitesimal generators to the semigroups $\{\exp(tA)\}$ and $\{\exp(tB)\}$ but I have had no success. Do you have any other hints?

Based on the idea of @Did, I came up with the following:

Series expansions give me: $$\sum_{n = 0}^\infty \frac{t^n(A + B)^n}{n!} = I + tA + tB + \frac{t^2(AB + BA)}{2} + \sum_{n = 3}^\infty \frac{t^n(A + B)^n}{n!}$$ and $$\left(\sum_{n = 0}^\infty \frac{t^n(A)^n}{n!} \right) \left(\sum_{n = 0}^\infty \frac{t^n(B)^n}{n!} \right) = I + tA + tB + \frac{t^2A^2}{2} + t^2AB + \frac{t^2B^2}{2} + \sum_{n = 3}^\infty t^n c_n,$$ where $$c_n := \sum_{k = 0}^n \frac{A^k B^{n - k}}{k! n!}.$$

The comparison of both expansions gives $$\frac{t^2(AB + BA)}{2} + \sum_{n = 3}^\infty \frac{t^n(A + B)^n}{n!} = t^2AB + \sum_{n = 3}^\infty t^n c_n.$$ Division by $t > 0$ yields: $$\frac{(AB + BA)}{2} + \sum_{n = 3}^\infty \frac{t^{n-2}(A + B)^n}{n!} = AB + \sum_{n = 3}^\infty t^{n-2} c_n.$$ But I can't quite see, why the two sums $\sum_{n = 3}^\infty \dots$ should go to $0$ for $t \to 0$ yielding the desired equality $$\frac{(AB + BA)}{2} = AB .$$

• Expansions when $t\to0$ up to order $t^2$ show that $$\tfrac12(A+B)^2=AB+\tfrac12A^2+\tfrac12B^2$$ which proves that $AB+BA=2AB$, qed. (And the argument is already on the other page.)
– Did
Oct 30, 2016 at 13:04
• @Did: This seems plausible, but I don't have a rigorous proof of this yet, see my updated post. Oct 30, 2016 at 16:20
• Because the matrix norms of $A$ and $B$ are finite hence one can bound the terms of the two series involved to show that they both converge for every $t$. As an example, use $$c_3=\frac16A^3+\frac12A^2B+\frac12AB^2+\frac16B^3$$ hence $$\|c_3\|\leqslant\frac16\|A\|^3+\frac12\|A\|^2\|B\|+\frac12\|A\|\|B\|^2+\frac16\|B\|^3$$
– Did
Oct 30, 2016 at 21:56
• Another approach is to use the BCH formula Oct 31, 2016 at 16:12
• @el_tenedor sure, use the Weierstrass $M$-test. Oct 31, 2016 at 19:20

Let us write $$T(t)=e^{t(A+B)}$$ and $$S(t)=e^{tA}e^{tB}$$. Then, using (for both) the product rule and (for $$S(t)$$) the fact that the generator commutes with the semigroup,
$$\frac{d}{dt}T(t)=(A+B)T(t),\quad \frac{d}{dt}S(t)=AS(t)+S(t)B$$ $$\frac{d^2}{dt^2}T(t)=(A+B)^2T(t),\quad \frac{d^2}{dt^2}S(t)=A^2S(t)+2AS(t)B+S(t)B^2\tag{1}$$
Since $$T(t)=S(t)$$ for all $$t\geq 0$$, we have $$\frac{d^2}{dt^2}T(t)=\frac{d^2}{dt^2}S(t),\quad\forall\ t\geq 0$$ and thus, from $$(1)$$, $$(AB+BA)S(t)+B^2S(t)=2AS(t)B+S(t)B^2,\quad\forall\ t\geq 0.$$ In particular, for $$t=0$$, $$AB+BA=2AB$$ and the desired result follows.