Two matrices $A,B$ such that $e^{(A+B)}=e^{A}e^{B}$ but $AB\neq BA$? So I saw at ODE class the properties of the exponential matrix so we can use it as the  canonical basic solution to linear ODE systems, and the teacher show us that if two matrices $A,B$ hold that $AB=BA$ then $e^{(A+B)t}=e^{At}e^{Bt}$ but said the reciprocal does not necesarily holds. I tried to find two matrices to show a counter example for the reciprocal, but couldn't find them in the matrices with real coefficients.
My question is, such matrices exist? Do i need to consider them with complex coefficients?
Update: I re-checked my notes to see if i made a mistake, and the propertie is without the $t$ involved, or maybe could be seen as $t=1$ (original question: "Two matrices $A,B$ such that $e^{(A+B)t}=e^{At}e^{Bt}$ but $AB\neq BA$?")
 A: If we differentiate twice at $t=0$ the identity $e^{t(A+B)}=e^{tA}e^{tB}$ we get $(A+B)^2=A^2+2AB+B^2$ which entails $AB=BA$. So it seems that should the identity hold for each $t$, then the matrices must commute.
A: In general, if $e^{t(A+B)} = e^{tA} e^{tB}$ is true for all $t$, then $AB = BA$ as demonstrated by other answer. 
However, it is possible for this to be true for specific $t$ even when $AB \ne BA$.
For any matrix $M$ which satisfies $M^2 = -\alpha^2 I$ for some real $\alpha \ne 0$, one can show that
$$e^{tM} = \cos(\alpha t) I + \frac{\sin(\alpha t)}{\alpha} M$$
Now take 
$A = 6\pi\begin{bmatrix} i & 0\\ 0 & -i \end{bmatrix}$
and $B = 8\pi\begin{bmatrix} 0 & 1 \\ -1 & 0\end{bmatrix}$,
it is easy to verify:
$$A^2 = -(6\pi)^2 I,\;\; B^2 = -(8\pi)^2 I\;\;\text{ and }\;\;(A + B)^2 = -(10\pi)^2 I$$
This implies whenever $t = n$ is an integer, one has
$$e^{n(A+B)} = e^{nA} = e^{nB} = I\quad\implies\quad e^{n(A+B)} = e^{nA} e^{nB}$$
In particular, when $n = 1$, we have $e^{A+B} = e^A e^B$.
By direct computation, you will find $A, B$ didn't commute. In fact, they anti-commute.
A: The Baker-Campbell-Hausdorff formula says that
$$
e^{tX}e^{tY}=e^Z
$$
where
$$
Z=tX+tY+\frac{t^2}{2}[X,Y]+\frac{t^3}{12}\big([X,[X,Y]]-[Y,[X,Y]]\big)+\dots
$$
and further terms involve higher powers of $t$ and higher order commutators. A proof of this formula is given in this answer.
This says that $e^{tX}e^{tY}=e^{t(X+Y)}$ if and only if $[X,Y]=0$.
