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If I have two matrices $A$ and $B$ then I know that $AB$ is not necessarily equal to $BA$. However if I have $exp(tA)exp(tB)$ can I switch them round then? Is $exp(tA)exp(tB) = exp(tB)exp(tA)$? The Taylor expansion suggests not, but I can't find the answer explicitly on google. Thanks for any help.

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    In general, no. I think there's a special case if at least one can be diagonalized. – 2017-01-29
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    @SeanRoberson Ok, thank you. I suppose where I got the idea from originally was I thought $exp(tA)exp(tB)=exp[t(A+B)]$ so then switching them wouldn't make a difference, because $B+A = A+B$, but then you can't treat the exponentials of matrices as I did above. – 2017-01-29
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    This is only true when $A$ and $B$ commute. – 2017-01-29

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Well no, in general. You can observe this, as you said, with a Taylor series.

You know that the matrix exponential can be expressed in the Taylor series definition for the exponential, but only when you can compute all the even and odd orders of the matrix you have. Otherwise you cannot. But suppose we can, then we would have

$$e^{t\mathbb{A}} = \mathbb{1} + t\mathbb{A} + \frac{t}{2}\mathbb{A}^2 + \ldots$$

$$e^{t\mathbb{B}} = \mathbb{1} + t\mathbb{B} + \frac{t}{2}\mathbb{B}^2 + \ldots$$

Hence

$$e^{t\mathbb{A}}e^{t\mathbb{B}} = \left( \mathbb{1} + t\mathbb{A} + \frac{t}{2}\mathbb{A}^2\right)\cdot \left(\mathbb{1} + t\mathbb{B} + \frac{t}{2}\mathbb{B}^2\right) = \mathbb{1} + t(\mathbb{B} + \mathbb{A}) + \frac{t}{2}(\mathbb{B}^2 + \mathbb{A}^2) + \frac{t^2}{4}(4\mathbb{A}\mathbb{B} + 2\mathbb{A}\mathbb{B}^2 + 2\mathbb{A}^2\mathbb{B} + \mathbb{A}^2\mathbb{B}^2)$$

You can easily see that the reversed operation $e^{t\mathbb{B}}e^{t\mathbb{A}}$ would change the order of the matrix products, and you know that in general

$$\mathbb{A}\mathbb{B} \neq \mathbb{B}\mathbb{A}$$

hence yes, order does matter.