# Show that $e^{A+B}=e^A e^B$

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.

• 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. Commented Apr 10, 2013 at 6:24
• See my answer in math.stackexchange.com/questions/349180/… This shows that the converse is not true. Commented Apr 10, 2013 at 7:26
• @user1551 Simultaneous diagonalizability follows from $AB=BA$ already. But the diagonalizable approximation you linked is interesting. Commented Apr 12, 2021 at 1:50
• @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. Commented Apr 12, 2021 at 1:54
• @user1551 Oh, I see. That was a mistake Commented Apr 12, 2021 at 3:20

\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.

• Under which conditions are you allowed to reorder the elements of the infinite summation (line 2 -> 3)? Commented Apr 10, 2013 at 7:04
• @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. Commented Apr 10, 2013 at 7:11
• @JyrkiLahtonen I think it is easier to say that the series $e^A$ converges "absolutely" because $\|A^k\|\leq\|A\|^k$. Indeed, this proves that the series is absolutely Cauchy in $M_n(\mathbb{C})$, so it converges absolutely. No need to do it coefficientwise. And anyway, how do you go get $(e^A)_{3,5}$ in terms of the coefficients of $A$? Because we have absolute convergence, any formal power series identity holds. Commented Apr 10, 2013 at 11:14
• @StevenWalton: The comments make sense to me, I just wanted to see why the reordering is allowed at all. Commented Apr 11, 2013 at 0:48
• What magic brings you from the right side of 4th equal sign to the right side of the 5th? Commented Jan 28, 2021 at 10:41

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.