Timeline for Asymptotic expansion square root matrix
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
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| Feb 27, 2015 at 15:13 | comment | added | user68601 | Thank you Carlo for the reply. As far as I understand, that solution coincides with the one I posted above, i.e. $\frac{1}{2}\underline\gamma^{ra}c_{as}$ and which, regrettably, is not working (in the sense that it does not solve the equation $\underline\gamma_{ra}\underline c_{as}+\underline c_{ra}\underline\gamma_{as}=c_{rs}$). Any ideas? | |
| Feb 27, 2015 at 14:38 | comment | added | Carlo Beenakker | mathoverflow.net/questions/193905/… | |
| Feb 27, 2015 at 13:28 | comment | added | user68601 | That's the whole point. Of course I realize that if I plug back the (wrong) solution in $\underline\gamma_{ra}\underline c_{as}+\underline c_{ra}\underline\gamma_{as}$ I do not get $c_{rs}$, but why symmetry of all the matrices involved does not help to simplify the calculations? | |
| Feb 27, 2015 at 12:55 | comment | added | Federico Poloni | Are you sure about $2\underline\gamma_{ra}\underline c_{as}=c_{rs}$? That's not the same as $\underline\gamma_{ra}\underline c_{as}+\underline c_{ra}\underline\gamma_{as}=c_{rs}$, or $\underline{\gamma}\underline{c}+\underline{c}\underline{\gamma}=c$ in matric notation. | |
| Feb 27, 2015 at 11:35 | comment | added | user68601 | All matrices involved are positive definite and real valued. As you told me and as I have verified the solution is guaranteed to be symmetric as well, therefore the equation should reduce to $2\underline\gamma_{ra}\underline c_{as}=c_{rs}$ and the solution should be $\underline c_{rs}=\frac{1}{2}\underline\gamma^{ra}c_{as}$, where $\underline\gamma^{rs}$ is the inverse of $\underline\gamma_{rs}$. Am I right or Am I missing something here? I'm asking because this was my first (and raw) solution to the problem, but whenever I use such a result to develop further expansions I get strange results. | |
| Feb 27, 2015 at 9:53 | comment | added | Federico Poloni | I mean positive or nonnegative definite, i.e., $v^T\gamma v \geq 0$ for each vector $v$. | |
| Feb 27, 2015 at 9:31 | comment | added | user68601 | Thank you Federico for the reply, I am going to see the Lyapunov equation right now! Yes, $c$ is symmetric (I am 100% sure about it, as I have an explicit expression). What do you mean by definite? If I understand correctly, I can tell you that I have the expressions for both $\gamma$ and $c$. | |
| Feb 27, 2015 at 9:15 | comment | added | Federico Poloni | Is $c$ symmetric? If I decipher correctly your indices, what you have is a Lyapunov equation with a symmetric $A$, so the solution should be guaranteed to be symmetric. Also, are $\gamma$ and $c$ definite? | |
| Feb 27, 2015 at 9:05 | review | First posts | |||
| Feb 27, 2015 at 9:16 | |||||
| Feb 27, 2015 at 9:02 | history | asked | user68601 | CC BY-SA 3.0 |