A proof of $=e^{A+B}=e^{A}e^{B}$ using ODEs

Question

I want to prove that $=e^{A+B}=e^{A}e^{B}$ for commuting matrices $A,B$ using differential equations. I found a proof here: LINK

Here is how the proof goes:

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.

Can someone explain to me how the author of this proof gets from $(A+B)X(t)$ to $X(t)=e^{tA}e^{tB}=e^{t(A+B)}$

What is meant by "it follows from uniqueness"?

Answer 1

The commutativity of $A,B$ with the respective exponentials gives the second to last equality. Then set $M=A+B$ and by the original claim, this $X$ is the unique solution to $X'=MX=(A+B)X, X(0)=I$ so that $X=e^{t(A+B)}$, but by assumption we had already set $X=e^{tA}e^{tB}$.

Answer 2

The Picard-Lindelöf's theorem gives you that, for $M\in\mathcal{M}_n(\mathbb{R}),$ the Cauchy's problem $X'=MX, X(0)=\mathrm{Id}_n$ has a unique solution $X:\mathbb{R}\to\mathcal{M}_n(\mathbb{R})$. You show in your proof that both of $t\mapsto e^{t(A+B)}$ and $t\mapsto e^{tA}\cdot e^{tB}$ are solutions, and by uniqueness you give $$\forall t\in\mathbb{R},e^{t(A+B)}=e^{tA}\cdot e^{tB}.$$ Then for $t=1$ you get indeed your relation.

Answer 3

The author checks $X(t)$ satisfies the differential equation $$ X'(t) =(A+B)X(t) $$ with initial condition $X(0)=I$.

Now the quoted theorem asserts that, setting $M = A+B$, this differential equation has a unique solution, which is $$X(t)=X(0)\mathrm e^{tM}=I\mathrm e^{tM}=\mathrm e^{t(A+B)}.$$