Timeline for Singular values of $X$ in $AX+XA=C$?
Current License: CC BY-SA 4.0
21 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 30, 2019 at 18:00 | history | tweeted | twitter.com/StackSciComp/status/1178731429917798408 | ||
| Sep 15, 2019 at 6:47 | answer | added | Federico Poloni | timeline score: 2 | |
| Sep 13, 2019 at 17:22 | vote | accept | Yaroslav Bulatov | ||
| Sep 12, 2019 at 6:43 | answer | added | Federico Poloni | timeline score: 2 | |
| Sep 12, 2019 at 6:40 | comment | added | Federico Poloni | @YaroslavBulatov The quick decay in the singular values of X is a very common phenomenon in Lyapunov equations; there is nothing strange about it and it is not a side-effect of singularity (but rather of the fact that $C$ is low-rank). | |
| Sep 11, 2019 at 19:52 | comment | added | Brian Borchers | There's simply no recent to expect the singular values of $X$ to be determined from this equation when the equation doesn't have a unique solution. | |
| Sep 11, 2019 at 19:19 | history | edited | Yaroslav Bulatov | CC BY-SA 4.0 |
added 43 characters in body
|
| Sep 11, 2019 at 19:05 | comment | added | Yaroslav Bulatov | @BrianBorchers I've added "Background" section where I explain where these matrices come from | |
| Sep 11, 2019 at 19:04 | history | edited | Yaroslav Bulatov | CC BY-SA 4.0 |
added 1336 characters in body
|
| Sep 11, 2019 at 18:21 | comment | added | Brian Borchers | I'm afraid you need to back up and ask a different question about how to deal with the Lyapunov equation when $A$ is singular. Start by explaining where you Lyapunov equation came from. | |
| Sep 11, 2019 at 18:02 | history | edited | Yaroslav Bulatov | CC BY-SA 4.0 |
added 169 characters in body
|
| Sep 11, 2019 at 17:46 | comment | added | Yaroslav Bulatov | @BrianBorchers Thanks for pointing out non-uniqueness, it seems lyapunov_solve(A, 2*A) doesn't quite do the right thing here, it gives me matrices of norm of the order 100, where I was expecting something of norm 1 | |
| Sep 11, 2019 at 16:54 | history | edited | Yaroslav Bulatov | CC BY-SA 4.0 |
added 268 characters in body
|
| Sep 11, 2019 at 16:50 | history | edited | Anton Menshov♦ | CC BY-SA 4.0 |
added 57 characters in body; edited tags; edited title
|
| Sep 11, 2019 at 16:47 | history | edited | Yaroslav Bulatov | CC BY-SA 4.0 |
added 92 characters in body
|
| Sep 11, 2019 at 16:46 | comment | added | Yaroslav Bulatov | Right, I'm expecting some non-uniqueness in X, does this imply non-uniqueness in singular values of X? I tried adding small identity matrix to A, C and the singular values seem to cluster around a constant value, with a rapid fall-off in the tail (added plot). I wonder if this fall-off is the side-effect of regularization | |
| Sep 11, 2019 at 16:40 | comment | added | Brian Borchers | You understand that there are nonuniqueness problems when A and C aren't positive definite, right? You may need to regularize the Lyapunov equation. | |
| Sep 11, 2019 at 16:36 | comment | added | Yaroslav Bulatov | The lyapunov equation solver is the slow part, several times slower than SVD. It prints warnings about numerical stability and perturbing coefficients, probably not expecting A,C to be low-rank | |
| Sep 11, 2019 at 16:34 | comment | added | Brian Borchers | What are your time constraints? Are you using an optimized BLAS/LAPACK library? | |
| Sep 11, 2019 at 16:33 | comment | added | Brian Borchers | Which part is too slow- the solution of the Lyapunov equation or the SVD that follows? Or, are they taking about equal amounts of time? | |
| Sep 11, 2019 at 16:08 | history | asked | Yaroslav Bulatov | CC BY-SA 4.0 |