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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?

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

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    $\begingroup$ See also en.wikipedia.org/wiki/… , "A direct application of this identity" $\endgroup$ – J.G. Apr 13 at 18:07
  • $\begingroup$ @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? $\endgroup$ – Giulio Apr 13 at 19:34
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    $\begingroup$ You realise $[A,\,B]$ is non-commutative, right? $\endgroup$ – J.G. Apr 13 at 19:35
  • $\begingroup$ @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 $\endgroup$ – Giulio Apr 13 at 20:21
  • $\begingroup$ By the way, the Kronecker Sum has the property $e^{A \oplus B} = $$e^A \otimes e^B$ as shown in this other question, but I guess it is not commutative. $\endgroup$ – Giulio Apr 14 at 5:32
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This is true if $AB=BA$, that is the matrices commute. If not, this generally does not hold. If $A,B$ do not commute but are Hermitian, then you get the Golden-Thompson inequality: $$tr ~e^{A+B}\le tr ~(e^A e^B)$$ Here, $tr$ denotes matrix trace.

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  • $\begingroup$ 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? $\endgroup$ – Giulio Apr 14 at 6:51
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If $A$ and $B$ commute ($AB=BA$) then the equality holds. However, it does not hold in general. Here are several counterexamples.

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  • $\begingroup$ 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? $\endgroup$ – Giulio Apr 14 at 6:49

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