0
$\begingroup$

Let $$ M = \begin{pmatrix}A& B\\C& D\end{pmatrix}. $$ Is there a general formula for $e^M$ in terms of the sub-matrices $A$, $B$, $C$ and $D$? If not, can anything useful be said about the properties of $e^M$, given properties of its component blocks?

I've tried a few different approaches but haven't got anywhere, mostly because the formula $e^{A + B}=e^Ae^B$ only works if $A$ and $B$ commute.

If it helps, I'm most interested in the case where $M$ is a symmetryic real matrix, and an 'infinitesimal stochastic' matrix in particular. (An infinitesimal stochastic matrix is a symmetric real matrix whose column sums are all zero, and which has no negative off-diagonal elements.) If there are other special sub-classes of matrices for which something useful can be said here, I would be interested to know that.

$\endgroup$
  • $\begingroup$ the case where $M$ is a real symmetric matrix, or where the sub-matrices are? $\endgroup$ – Thoth Oct 23 '15 at 5:33
  • $\begingroup$ Since any matrix $M$ can be represented in block form as indicated, it is unclear why any special expression for the exponential of $M$ ought to be expected. $\endgroup$ – hardmath Oct 23 '15 at 5:39
  • $\begingroup$ @Thoth the case where $M$ is - I've clarified in the question $\endgroup$ – Nathaniel Oct 23 '15 at 7:01
  • $\begingroup$ @hardmath that's a good point. (I was coming from the point of view of "it would be really useful to me if there was a formula for this," not "I have some reason to expect there would be a formula for this.") I've edited the question to ask whether there's anything else that can be said about $e^M$, given properties of its component blocks. $\endgroup$ – Nathaniel Oct 23 '15 at 7:02
  • 1
    $\begingroup$ I doubt you can say much about $A,B,C,D$, but if $M$ is symmetric then it has an eigen-decomp as $M=P^TDP$ and so I think you have $e^M=P^Te^DP$. I don't know if that helps any. $D$ contains the (possibly absolute value or squared can't remember) eigenvalues of $M$, but unless the top right or bottom left block is the zero block, there is no relationship known between the eigenvalues of the blocks and the eigenvalues of $M$. $\endgroup$ – Thoth Oct 23 '15 at 8:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.