Timeline for Unique coordinates (solutions) in a single Gauss-Seidel iteration
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 27, 2012 at 15:21 | comment | added | usero | $L$ is a weighted Laplacian matrix (negative off-diagonal entries, with rows(collumns) summing to $1$ residing on its diagonal). This is why I say $y$ is centered. | |
| Mar 27, 2012 at 14:03 | comment | added | Jed Brown | This tells me nothing at all. You haven't said what $L$ is so the setting is just obfuscation; any $x_0 := y$ and $b := Ly$ can have this form. | |
| Mar 27, 2012 at 13:42 | comment | added | usero | The initial iterate would be $x_0=y$, from $Ax=Ly$, the main linear system. Note that the initial solution might have non-unique entries. | |
| Mar 27, 2012 at 13:38 | history | edited | Jed Brown | CC BY-SA 3.0 |
don't implicitly skip the affine part
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| Mar 27, 2012 at 13:29 | comment | added | Jed Brown | @usero What conditions do you intend to impose on the initial iterate? | |
| Mar 27, 2012 at 11:32 | comment | added | Aron Ahmadia | Are you sure that you've got the right definition of unique? Can you point us to this proof? | |
| Mar 27, 2012 at 9:15 | comment | added | usero | There exist a proof that a solution to the problem that I've shown to be equivalent to a single GS iteration yields unique $d-$dimensional solution. So, either the proof is wrong, or my reduction is, which is to be investigated. That is why I wonder if, perhaps, a specific visiting order might provide guarantees. | |
| Mar 26, 2012 at 22:51 | comment | added | Paul | @usero: Out of curiosity, what do you expect to gain from knowledge of the uniqueness of the first iteration vectors? What application is this for? | |
| Mar 26, 2012 at 20:25 | comment | added | Aron Ahmadia | Where is it claimed that uniqueness is guaranteed? This is a very strange definition of unique... | |
| Mar 26, 2012 at 20:11 | comment | added | usero | Good point. Because it $is$ claimed on some places that the uniqueness is guaranteed. I wanted to prove that, perhaps, under certain visiting order, one might really guarantee uniqueness. | |
| Mar 26, 2012 at 20:01 | comment | added | Jed Brown | No, there is no reasonable way to guarantee that. Why would you care whether two entries in the same vector happened to be the same or not? | |
| Mar 26, 2012 at 19:57 | comment | added | usero | I don't think you understood it correctly. Suppose a system $$Ax=Ly$$ is given, with the unknown $x$, and suppose I want to perform a single iteration of GS, where an iteration implies that all $x_i$ (all rows of $X$) are visited once, in some order. Now, given that the initial guess $x_0$ from which I start GS is $x_0=y$, I want to know if there are guarantees that all new $x_i$ (so, rows on the resulting matrix, after the first iteration) are different (there are no coincidences). I hope this clarifies the question. | |
| Mar 26, 2012 at 19:43 | history | answered | Jed Brown | CC BY-SA 3.0 |