multiplication of exponentials for non commuting matrices

Question

Is there any special condition for the following statement to be true for any $n \times n$ matrices?

$$ e^Ae^B=e^{A+B}=e^Be^A $$

Is it always correct to say that

the exponential product equals the sum of the exponentials

and that is why the multiplication of exponentials commutes: because matrix addition is commutative?

Edit

From the answers I understand that

$$e^A e^B \neq e^{A+B}$$

when $A$ and $B$ do not commute.

But if I have two non commuting matrices $A = \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{pmatrix} $ and $B = \begin{pmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}$ like in this example, I see that their exponentials still commute from this computation. (Edit2 - that WolframAlpha result was obviously wrong: those exponentials don't commute and the following inequality is true)

So is there an example of two matrices $A$ and $B$ such that the following holds?

$$e^A e^B \neq e^B e^A $$

Answer 1

This is true if $AB=BA$, that is the matrices commute. If not, this generally does not hold. If $A,B$ do not commute but are Hermitian, then you get the Golden-Thompson inequality: $$tr ~e^{A+B}\le tr ~(e^A e^B)$$ Here, $tr$ denotes matrix trace.

Answer 2

If $A$ and $B$ commute ($AB=BA$) then the equality holds. However, it does not hold in general. Here are several counterexamples.