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If $A$ and $B$ are $n\times n$ matrices such that $AB = BA$ (that is, $A$ and $B$ commute), show that

$$ e^{A+B}=e^A e^B$$

Note that $A$ and $B$ do NOT have to be diagonalizable.

Julien
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Abudin
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    See my answer in http://math.stackexchange.com/questions/349180/if-ea-and-eb-commute-do-a-and-b-commute/349382#349382 This shows that the converse is not true. – Sungjin Kim Apr 10 '13 at 07:26
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    As a remark, it is actually legitimate to assume that $A$ and $B$ are simultaneously diagonalisable (surprise, surprise!), so the proposition is trivial. But obviously, the reason why we can make such an assumption is way beyond the scope of undergraduate (or even graduate) linear algebra courses. – user1551 Apr 10 '13 at 06:24
  • @user1551 Simultaneous diagonalizability follows from $AB=BA$ already. But the diagonalizable approximation you linked is interesting. – Minh Nguyen Apr 12 '21 at 01:50
  • @MinhNguyen No, that $A$ and $B$ commute does not guarantee that $A$ and $B$ are simultaneously diagonalisable, as they may not be diagonalisable in the first place. Consider e.g. $A=B=\pmatrix{0&1\ 0&0}$. The linked result, however, guarantees that even if none of $A$ and $B$ is diagonalisable, we can still approximate them by a pair of commuting and simultaneously diagonalisable. – user1551 Apr 12 '21 at 01:54
  • @user1551 Oh, I see. That was a mistake – Minh Nguyen Apr 12 '21 at 03:20

2 Answers2

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$$\begin{align*}e^{A}e^{B} &= \left(\sum \frac{A^{n}}{n!}\right)\left(\sum\frac{B^{n}}{n!}\right)\\ &=\sum^{\infty}_{m=0}\sum^{\infty}_{n=0}\frac{A^{m}B^{n}}{m!n!}\\ &=\sum^{\infty}_{l=0}\sum^{l}_{m=0}\frac{A^{m}B^{l-m}}{m!(l-m)!}\\ &=\sum^{\infty}_{l=0}\frac{1}{l!}\sum^{l}_{m=0}\frac{l!}{m!(l-m)!}A^{m}B^{l-m}\\ &=\sum^{\infty}_{l=0}\frac{(A+B)^{l}}{l!}\\ &= e^{A+B}\end{align*}$$

Note:A and B have to commute. Also, I set l=m+n. I did this because we want to use the binomial theorem to simplify this.

amarVashishth
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MaybeALlama
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    Under which conditions are you allowed to reorder the elements of the infinite summation (line 2 -> 3)? – krlmlr Apr 10 '13 at 07:04
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    @krlmlr: A good question. Absolute convergence is enough (for all the matrix entries). It holds in this case, because you can upper bound the entries by the corresponding powers of the matrix norms. The series of these upper bounds then converges absolutely, because they are terms of the real variable exponential series. – Jyrki Lahtonen Apr 10 '13 at 07:11
  • @krlmlr, if what Jyrki said doesn't make sense to you then think of it similar to changing the order of integration. When you do that you have to change the bounds, which I did here as well. I wanted to get it in the form of the binomial theorem, which meant that I had to do the summations in a different way. – MaybeALlama Apr 11 '13 at 00:32
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    @StevenWalton: The comments make sense to me, I just wanted to see why the reordering is allowed at all. – krlmlr Apr 11 '13 at 00:48
  • "Note:A and B have to commute" - which step of the solution does this apply to? Is it a condition of the Binomial Theorem (last step)? – Bill Sep 18 '19 at 21:41
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    What magic brings you from the right side of 4th equal sign to the right side of the 5th? – TVSuchty Jan 28 '21 at 10:41
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Here is a different-ial way, just because it is significantly different from the standard Cauchy product way.

Given a square matrix $M$, the function $X(t):=e^{tM}$ is the unique solution of the linear differential equation: $X'=MX$ and $X(0)=I$.

Now set $X(t):=e^{tA}e^{tB}$ and observe that the factors commute with each other, as well as they commute with $A$ and $B$. It follows that $$ X'(t)=Ae^{tA}e^{tB}+e^{tA}Be^{tB}=(A+B)e^{tA}e^{tB}=(A+B)X(t). $$ And since $X(0)=e^0e^0=I$, it follows from the uniqueness above that $$ X(t)=e^{tA}e^{tB}=e^{t(A+B)}\qquad\forall t\in\mathbb{R}. $$ Set $t:=1$ to get the desired formula.

Julien
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