Timeline for Find Determinant/or Row Reduce parameter dependent matrix
Current License: CC BY-SA 3.0
24 events
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| Jul 3, 2012 at 23:21 | comment | added | tau1777 | I do believe I have the problem of my initial matrix being close to singular. How can I get the second to last singular value instead of the very last. I've tried SingularValueList[mat /. \[Kappa] -> x,-2]]], but I don't think it is working. | |
| Jul 3, 2012 at 2:50 | comment | added | tau1777 | @Jens, I did not mean to get to far off topic, since this is a Mathematica related forum. I guess my main Mathematica related question that remains is with the SingularValueList command? As I was saying in "6th-Edit" I'm still a little stuck about how to interpret the results plot. Other than that I guess you are saying that Det[mat] should be zero, because we are using a singular value, but it is not b/c Mathematica is calculating the Det differently than it calculated the singular values. Thanks again for all your help. | |
| Jul 3, 2012 at 2:36 | comment | added | Jens | This doesn't sound like the remaining problems are related to Mathematica, but to the setup of your matrix problem. I can't think of anything to say about that, except that it doesn't sound numerically stable. Many matrix decomposition methods attempt to renormalize the rows and/or columns before doing the decomposition, whereas that may not be possible for the determinant calculation. That could explain the difference in the results. | |
| Jul 2, 2012 at 22:07 | history | edited | tau1777 | CC BY-SA 3.0 | added 1287 characters in body |
| Jul 1, 2012 at 6:26 | history | edited | Mr.Wizard | CC BY-SA 3.0 | added 12 characters in body |
| Jun 30, 2012 at 21:00 | comment | added | tau1777 | @Jens, WOW. I officially think you are a Mathematica wizard. Sorry to have bothered you about this again, it seems like I really need a better understanding of what the SingularValueList command does and what SVD does in general. I will have to present these results to my advisor so knowing exactly what my code does will be very important. But more importantly I would really like to know what my code is doing. If you have any suggestions about where to learn about this material, I would be happy to know. Thanks again. | |
| Jun 30, 2012 at 20:56 | history | edited | tau1777 | CC BY-SA 3.0 | added 191 characters in body |
| Jun 30, 2012 at 20:46 | comment | added | Jens | Again a stab in the dark: in your Edit 4, the jumps indicate that some of the small singular values are thrown away occasionally and then picked up again. You could try replacing SingularValueList[m] by SingularValueList[m, Tolerance -> 0] which will ensure that all singular values are kept. | |
| Jun 30, 2012 at 19:51 | history | edited | tau1777 | CC BY-SA 3.0 | added 2648 characters in body |
| Jun 28, 2012 at 17:59 | history | edited | tau1777 | CC BY-SA 3.0 | added 243 characters in body |
| Jun 28, 2012 at 16:54 | comment | added | rcollyer | Your example matrix does not appear as if it can be rewritten that way. But, it may be separable into multiple updates. That said, it isn't trivial, and I don't know if it can be done automatically, or a case-by-case analysis is required. I think the latter is more likely. If you can make use of it, I'd be interested in hearing about it. | |
| Jun 28, 2012 at 16:51 | comment | added | tau1777 | @TomD, Yes there should be finite roots, as I'm really try to reproduce research. Also I did get finite roots for a small version of my problem. Making the problem much larger is where the trouble starts. | |
| Jun 28, 2012 at 16:50 | comment | added | tau1777 | @rcollyer Thanks for the suggestion. I'm not sure if I can rewrite my problem like this, since there are terms like $\kappa$a(1,1) etc, but this is certainly something I will look into. | |
| Jun 28, 2012 at 16:45 | comment | added | rcollyer | Finding the determinant of a symbolic matrix is very, very slow, and it gets bad fast. On my machine, I cap out at $11\times 11$ matrices (which take 4 secs) before Mathematica begins to chew up a lot of memory. Since you have a numerical matrix, $M$ plus a symbolic matrix of a single variable, $\kappa A$, it may be worthwhile to see if you can recast $A$ as a rank-1 or rank-l ($l\le n$) update. Then you can use the matrix determinant lemma. As this is non-trivial, I opted to just comment instead of providing an answer. | |
| Jun 28, 2012 at 15:41 | answer | added | Daniel Lichtblau | timeline score: 6 | |
| Jun 28, 2012 at 7:32 | comment | added | user1066 | Are you sure kappa has finite roots? Plot[Det[t2], {\[Kappa], -10, 10}, AxesOrigin -> {0, 0}] (or Solve[Det[t2] == 0, \[Kappa]]) | |
| Jun 28, 2012 at 7:01 | comment | added | tau1777 | @nightowl, I just checked with $\kappa$ instead of "x" it did not make a difference. I didn't think it would since I defined the function "f" it should have been able to take any variable as it input. Thanks for help, at least that part is troubleshot. | |
| Jun 28, 2012 at 6:43 | comment | added | night owl | Should it be FindRoot[f[\[Kappa]],{\[Kappa],.5}]? There was an x. Could this have generated the error messages? | |
| Jun 28, 2012 at 5:17 | history | edited | tau1777 | CC BY-SA 3.0 | added 1825 characters in body |
| Jun 28, 2012 at 3:22 | history | edited | Verbeia | CC BY-SA 3.0 | added 6 characters in body; edited tags |
| Jun 28, 2012 at 3:20 | history | edited | tau1777 | CC BY-SA 3.0 | added 318 characters in body |
| Jun 28, 2012 at 3:14 | history | edited | tau1777 | CC BY-SA 3.0 | added 318 characters in body |
| Jun 28, 2012 at 1:56 | answer | added | Jens | timeline score: 10 | |
| Jun 28, 2012 at 1:41 | history | asked | tau1777 | CC BY-SA 3.0 |