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Nov 28, 2016 at 9:29 vote accept N. Gast
Nov 25, 2016 at 11:01 comment added Federico Poloni Note also that one can compute column $jj$ of $M^{-1}$ by solving the Lyapunov equation $AX+X^*A=E$, where $E$ is the matrix that has a 1 in position $j,j$ and zeros elsewhere. The needed integral is the $ii$ entry of its solution $X$.
Nov 24, 2016 at 21:23 comment added Anthony Quas I see now. That's very cool.
Nov 24, 2016 at 20:52 comment added N. Gast I will verify it tomorrow but this seems to work.
Nov 24, 2016 at 20:08 comment added Will Sawin @AnthonyQuas As long as the real parts of the eigenvalues of $A$ are negative, yes. If $A$ is has eigenvalues $\lambda_i$, then $M$ has eigenvalues $\lambda_i+\lambda_j$, which all have negative real parts, so the inverse exists.
Nov 24, 2016 at 18:42 comment added Anthony Quas @WillSawin: does M^{-1} exist?
Nov 24, 2016 at 18:09 comment added Will Sawin @N.Gast Sorry, that was wrong, is this clearer?
Nov 24, 2016 at 18:09 comment added Will Sawin @AnthonyQuas That was a stupid mistake, fixed now.
Nov 24, 2016 at 18:08 history edited Will Sawin CC BY-SA 3.0
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Nov 24, 2016 at 16:39 comment added Anthony Quas I think this is problematic because each of the summands is of rank $n$, so the sum is of rank at most $2n$. For $n>2$, $M$ will not be invertible.
Nov 24, 2016 at 16:18 comment added N. Gast Thanks for this answer. I am not sure to understand why the two summands would commute and why $e^{Mt}_{ij,kl}$ can be expressed in terms of $e^{At}$ (with these indices, it seems false).
Nov 24, 2016 at 10:34 history answered Will Sawin CC BY-SA 3.0