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.
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4Related: http://math.stackexchange.com/questions/349180/if-ea-and-eb-commute-do-a-and-b-commute – Seirios May 27 '15 at 15:15
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2I was going to provide a long answer, but it appears I am too slow. You may find the paper by Fréchet, Les solutions non commutables de l’équation matricielle $e^{X}e^{Y} = e^{X+Y}$, to be interesting. – Echan May 27 '15 at 15:26
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In fact, there exist non-commuting $3 \times 3$ real matrices $X$ and $Y$ for which $e^{X} = e^{Y} = e^{X+Y} = I$. – Andrew D. Hwang May 27 '15 at 15:26
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@Seirios Thank you for your hint. – William Huang May 27 '15 at 15:53
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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 $$ e^Ae^B=e^Be^A=e^{A+B}, \quad \text{but $AB\neq BA$}. $$
Algebraic Pavel
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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.
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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}
Leucippus
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