Timeline for Integral of the entrywise square of the exponential of a matrix
Current License: CC BY-SA 3.0
12 events
when toggle format | what | by | license | comment | |
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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 |
added 210 characters in body
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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 |