Is $[X,Y] \neq 0$ the sufficient condition of $e^{X+Y} \neq e^Xe^Y$? We know that if X commutes with Y, where X and Y are $n\times n$ matrices,  then we have
$$e^{X+Y}=e^Xe^Y$$
However, can we conclude that $e^{X+Y} \neq e^Xe^Y$ if X doesn't commute with Y ?
Is there any counterexample ?
Or prove if it is right.
 A: Commutativity is not necessary. Let
$$
A=\pmatrix{0&0\\0&2\pi i},\quad B=\pmatrix{0&1\\0&2\pi i}.
$$
Then $e^A=e^B=e^{A+B}=I$, so
$$
e^Ae^B=e^Be^A=e^{A+B}, \quad \text{but $AB\neq BA$}.
$$
A: Another example in which $[X,Y]\neq 0$ but $e^Xe^Y=e^{X+Y}$.
Let $a\neq b\wedge c\neq0$ real numbers.
$$X=\begin{pmatrix} a&0\\0&b\end{pmatrix}\\Y=\begin{pmatrix} 0&c\\0&0\end{pmatrix}\\
[X,Y]=\begin{pmatrix} 0&c(a-b)\\0&0\end{pmatrix}\neq 0$$
$$e^X=\begin{pmatrix} e^a&0\\0&e^b\end{pmatrix}\\
e^Y=\begin{pmatrix} 1&c\\0&1\end{pmatrix}\\
e^{X+Y}=\begin{pmatrix} e^a&{e^a-e^b\over a-b}c\\0&e^b\end{pmatrix}\\
e^Xe^Y=\begin{pmatrix} e^a&e^ac\\0&e^b\end{pmatrix}$$
Therefore, $e^Xe^Y=e^{X+Y}\Longleftrightarrow (a-b)e^a=e^a-e^b\Longleftrightarrow (a-1)e^a=(b-1)e^b$
But you can easily see that the function $\mathbb{R}\ni x\mapsto (x-1)e^x$ is not injective, so we can find a suitable couple $a\neq b$ to make this equality hold.
An "interesting" fact is that, since the minimum of $(x-1)e^x$ is in $x=0$, this provides counterexamples $X,Y$ arbitrarly close to the zero matrix.
A: Using the identity
\begin{align}
e^{A + B} = e^{A} \, e^{B} \, e^{- \frac{1}{2}[A, B]}
\end{align}
then it is fairly evident that if A and B commute then the identity is reduced to
\begin{align}
e^{A+B} = e^{A} \, e^{B}.
\end{align} 
If A and B do not commute, or have some property of $[A,B]=C$, where C is either another operator, or constant, or etc., then
\begin{align}
e^{A+B} = e^{A} \, e^{B} \, e^{- \frac{1}{2}\, C}.
\end{align}
