multiplication of exponentials for non commuting matrices

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?

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

• See also en.wikipedia.org/wiki/… , "A direct application of this identity" – J.G. Apr 13 at 18:07
• @J.G. I think that you pointed me to the answer I was looking for: $e^A e^B = e^{A+B+1/2[A,B]}$ and therefore $e^A e^B = e^B e^A$, could you please confirm it with an answer if that holds true? – Giulio Apr 13 at 19:34
• You realise $[A,\,B]$ is non-commutative, right? – J.G. Apr 13 at 19:35
• @J.G. I was missing that part, thank you: you really answered my question (I guess that's more than a comment), now I see that $e^B e^A = e^A e^B e^{−[A,B]}$ provided—as always—that $[A,B]$ commutes with $A$ and $B$ - as it reads from this appendix – Giulio Apr 13 at 20:21
• By the way, the Kronecker Sum has the property $e^{A \oplus B} =$$e^A \otimes e^B$ as shown in this other question, but I guess it is not commutative. – Giulio Apr 14 at 5:32

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.
• Out of my curiosity, what about the converse? Are there any non-commuting matrices such that their exponentials commute? I think one can only find operators but not matrices such that, e.g., $[A,B] = 2πi$ or (any non zero number times the identity matrix), right? – Giulio Apr 14 at 6:51
If $$A$$ and $$B$$ commute ($$AB=BA$$) then the equality holds. However, it does not hold in general. Here are several counterexamples.
• Out of my curiosity, what about the converse? Are there any non-commuting matrices such that their exponentials commute? I think one can only find operators but not matrices such that, e.g., $[A,B] = 2πi$ or (any non zero number times the identity matrix), right? – Giulio Apr 14 at 6:49