Show that $e^{t(A+B)} = e^{tA}e^{tB}$ for all $t \in \mathbb{R}$ if, and only if $AB = BA$. Let A,B real or complex matrixes. Show that $e^{t(A+B)} = e^{tA}e^{tB}$ for all $t \in \mathbb{R}$ if, and only if $AB = BA$.
I demonstrated the reciprocal:
$\Leftarrow )$ The two equations are solutions of $X' = (A+B)X$, $X(0)= I$, because 
$(e^{tA}e^{tB})' = Ae^{tA}e^{tB} + e^{tA}Be^{tB} = Ae^{tA}e^{tB} + Be^{tA}e^{tB} = (A+B)e^{tA}e^{tB}$   and
$(e^t(A+B))' = (A+B)e^{t(A+B)}$. Thus, $e^{t(A+B)} = e^{tA}e^{tB}$.
I have problems to demonstrate the implication.
 A: Let $f(t) = e^{t(A+B)} - e^{tA} e^{tB}$. Then
$$\begin{align}
f'(t) & = (A+B) e^{t(A+B)} - A e^{tA} e^{tB} - e^{tA} B e^{tB} \\
f''(t) & = (A+B)^2 e^{t(A+B)} - A(A e^{tA} e^{tB} + e^{tA} B e^{tB}) - A e^{tA} B e^{tB} - e^{tA} B^2 e^{tB}
\end{align}$$
So if $f(t) \equiv 0$, then $f''(0) = 0 = (A+B)^2 - A^2 - AB - AB - B^2 = BA - AB$. Therefore $AB = BA$.
A: So the proof I've seen of the reverse implication uses the Baker-Campbell-Hausdorff formula. 
The Baker-Campbell-Hausdorff formula says that for $GL_{n}(\mathbb{C})$ or $Gl_{n}(\mathbb{R})$ (you can actually put any Lie group here):
For $X$ and $Y$ close enough to the $0$ matrix, we have
$$\log(e^{X}e^{Y}) = X + Y + \frac{1}{2}[X, Y] + \text{terms involving multiple commutators}$$
where the $\log$ is a local inverse for the exponential function in a neighborhood of the $0$ matrix. 
Now, by choosing $t$ small enough, we can make $tA$ and $tB$ lie in the region where the BCH formula is applicable. Then, by applying $\log$ to both sides of the equation
$$e^{t(A + B)} = e^{tA}e^{tB}$$
we get the expression
$$t(A + B) = t(A + B) + \frac{t^{2}}{2}[A, B] + O(t^{3}).$$
Both sides of this expression are real analytic functions $\mathbb{R} \rightarrow M_{n\times n}(\mathbb{R})$ (resp. $M_{n\times n}(\mathbb{C})$ with nonzero radius of convergence. Hence the only way for them to be equal is if their power series are equal coefficient by coefficient.
Comparing the $t^{2}$ coefficients gives us $[A, B] = 0$ as desired.
