2
$\begingroup$

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!

  • 1
    Your formula looks way off. If $B=0$, the Laplace transform should be $s\mapsto (sI -A)^{-1}$.2012-05-11
  • 0
    I don't think there is an easy solution here...2012-05-11
  • 0
    I've corrected the laplace equation. Given $(I-A)^{-1}$ and $B$, can we obtain $e^{A+B}$ ?2012-05-11
  • 1
    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.2012-05-11
  • 0
    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!2012-05-11
  • 0
    Exactly the same reasoning applies, the above would imply $e^{At+B} = a^{At} e^B$, which is not true in general. Basically your formula is wrong, in general, and there is no easy fix.2012-05-11

1 Answers 1

5

$e^{A+B}$ 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.

  • 1
    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.2012-05-11
  • 0
    Thanks for your answer! Now I've corrected my mistake, and changed the condition. Could you kindly take a look again?2012-05-11
  • 1
    @John: if you know $(I - A)^{-1}$, then you know $A$. If you also know $B$, then you know $A + B$...2012-05-11
  • 0
    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!2012-05-11
  • 0
    @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.2012-05-11
  • 0
    yeah, I know computing the exponential of a matrix is difficult. That's why I want to avoid $exp(\cdot)$ calculation while obtain the result $exp(A+B)$ from the previously given $A$, $exp(A)$, ${(I-A)}^{-1}$, $B$, $exp(B)$. How can I realize this? Originally, I want to use inverse Laplace transform to get $exp(A+B)$ from R.H.S of my laplace equation. Can I yield the result in this way?2012-05-11
  • 0
    @John: basically I do not think this is possible (in a way that isn't equivalent to just directly computing a matrix exponential).2012-05-11