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Given the matrix $(I-A)^{-1}$ and $B$, can we compute $e^{A+B}$, where $e^X$ is defined to be $\sum_{i=0}^{\infty} \frac{X^i}{i!}$.

(Note that $A$ and $B$ do not commute, and hence $e^A \cdot e^B \neq e^{A+B}$).

Now I've observed that Laplace transformation might be a useful tool. I've obtained that $$\mathcal{L}[e^{tA+B}](s) ={(sI-A)}^{-1}e^{B}.$$

So is the above (inverse) laplace transformation really useful to compute $e^{A+B}$ from $(I-A)^{-1}$ and $B$? How can I get the resultant $e^{A+B}$ from the Laplace transformation?

Hope anyone who is familiar with linear algebra and Laplace transformation could give me a hand. Thanks!

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Your formula looks way off. If $B=0$, the Laplace transform should be $s\mapsto (sI -A)^{-1}$. –  copper.hat May 11 '12 at 15:10
    
I don't think there is an easy solution here... –  copper.hat May 11 '12 at 16:41
    
I've corrected the laplace equation. Given $(I-A)^{-1}$ and $B$, can we obtain $e^{A+B}$ ? –  John Smith May 11 '12 at 17:02
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Your equation cannot be true, if it were, inverting the result (the Laplace Transform is linear) you have above would yield $e^{At+B} = e^B e^{At}$. Without more conditions on $A,B$, this is a difficult issue. –  copper.hat May 11 '12 at 17:17
    
Sorry, now it is true. I assume that directly computing $e^{(A+B)}$ is difficult, and want to get the result from the known matrices ${(I-A)}^{-1}$ and $B$ (or $e^B$), can I realize it from laplace transform ? Thanks! –  John Smith May 11 '12 at 17:33
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1 Answer 1

up vote 5 down vote accepted

$e^{A+B}$ is not uniquely determined by $e^A$ and $e^B$.

First take $A = \left[ \begin{array}{cc} 0 & -\pi \\\ \pi & 0 \end{array} \right]$ and $B = \left[ \begin{array}{cc} \pi & 0 \\\ 0 & -\pi \end{array} \right]$. Then $A + B$ squares to zero, so we have $$e^A = \left[ \begin{array}{cc} \cos \pi & - \sin \pi \\\ \sin \pi & \cos \pi \end{array} \right] = \left[ \begin{array}{cc} -1 & 0 \\\ 0 & -1 \end{array} \right], e^B = \left[ \begin{array}{cc} e^{\pi} & 0 \\\ 0 & e^{-\pi} \end{array} \right], e^{A+B} = \left[ \begin{array}{cc} 1 + \pi & -\pi \\\ \pi & 1 - \pi \end{array} \right].$$

Now replace $A$ with $\left[ \begin{array}{cc} 0 & - 3\pi \\\ 3\pi & 0 \end{array} \right]$. Then $e^A$ is the same, but now $$A + B = \left[ \begin{array}{cc} \pi & - 3\pi \\\ 3 \pi & - \pi \end{array} \right]$$

has eigenvalues $\pm \pi i \sqrt{7}$, so the eigenvalues of $e^{A+B}$ are different from what they were before.

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The main point here is that modifying the eigenvalues of either $A$ or $B$ by $2 \pi i$ won't change the matrix exponential of either (if they're diagonalizable) but will in general change the eigenvalues of $A + B$ in some more complicated way. –  Qiaochu Yuan May 11 '12 at 15:34
    
Thanks for your answer! Now I've corrected my mistake, and changed the condition. Could you kindly take a look again? –  John Smith May 11 '12 at 17:05
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@John: if you know $(I - A)^{-1}$, then you know $A$. If you also know $B$, then you know $A + B$... –  Qiaochu Yuan May 11 '12 at 17:21
    
I assume that directly computing $e^{(A+B)}$ is difficult, and want to get the result from the known matrices ${(I-A)}^{-1}$ and $B$ (or $e^B$), can I realize it ? Thanks! –  John Smith May 11 '12 at 17:36
    
@John: directly computing $e^{A + B}$ is not difficult. If you only know $e^B$ and not $B$, then the counterexample above works (I didn't change one of the matrices). In any case, setting $B = 0$ your problem is already at least as hard as computing the exponential of a single matrix. I also don't understand what the Laplace transform has to do with this; knowing $e^{tA + B}$ is a lot more information than the other conditions you've given. –  Qiaochu Yuan May 11 '12 at 17:37
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