Timeline for When the integral of the product of two matrix exponentials is singular?
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
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| Aug 8, 2016 at 7:25 | comment | added | nadia | Yes, I am aware of that but the commutator of $A$ and $B$ is not multiple of the identity matrix. | |
| Aug 7, 2016 at 21:12 | comment | added | Andreas Rüdinger | I would just like to remark that using the Baker-Campbell-Hausdorff formula (e.g. en.wikipedia.org/wiki/…) you can convert the integrand into the form $\exp(tf(A,B))$. This could be helpful at least in special cases, e.g. if $[A,B]$ is proportional to the unit matrix so that the Baker-Campbell-Hausdorff formula consists of only three terms. | |
| Aug 5, 2016 at 10:58 | history | edited | nadia | CC BY-SA 3.0 |
added 9 characters in body
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| Aug 5, 2016 at 10:58 | comment | added | nadia | I mean a singular matrix, thanks Fedor! | |
| Aug 5, 2016 at 10:50 | comment | added | Fedor Petrov | You mean that the value of this integral is a singular matrix? Usually "singular integral" means something very different. | |
| Aug 5, 2016 at 8:51 | review | First posts | |||
| Aug 5, 2016 at 8:56 | |||||
| Aug 5, 2016 at 8:42 | history | asked | nadia | CC BY-SA 3.0 |