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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 $$ \frac{(AB + BA)}{2} = AB . $$

Rodrigo de Azevedo
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el_tenedor
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    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 '16 at 13:04
  • @Did: This seems plausible, but I don't have a rigorous proof of this yet, see my updated post. – el_tenedor Oct 30 '16 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 '16 at 21:56
  • Another approach is to use the BCH formula – Ben Grossmann Oct 31 '16 at 16:12
  • What do you mean when you say "the two sums should go to $0$"? Of course they shouldn't go to zero; $\exp(0) = I$. – Ben Grossmann Oct 31 '16 at 16:20
  • @Omnomnomnom: I have updated my post. I am referring to the $\Sigma_{n = 3}^\infty \dots$ terms on both sides. – el_tenedor Oct 31 '16 at 18:52
  • @Omnomnomnom: I would need to prove that both sums $\sum_{n = 3}^\infty \dots$ converge uniformly so I can interchange the limit of the sum with the limit $t \to 0$ to prove my claim. Do you have a proof for this? – el_tenedor Oct 31 '16 at 18:58
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    @el_tenedor sure, use the Weierstrass $M$-test. – Ben Grossmann Oct 31 '16 at 19:20
  • @Omnomnomnom Thanks. Majorizing with $$\sum_{n=3}^\infty \frac{(A + B)^n}{n!}$$, i.e. setting $t=1$, gives uniform convergence on $[0,1]$ which should suffice in my case, right? (similarly with the other sum) – el_tenedor Oct 31 '16 at 21:58
  • That's right${}{}$ – Ben Grossmann Nov 01 '16 at 00:17

1 Answers1

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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.

Remark: This solution follows the hint in Engel's book, page 23.

Pedro
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