Speeding up lsqlin to find the base of a matrix

2 次查看（过去 30 天）
Tintin Milou
Tintin Milou 2022-8-4
评论： Bruno Luong 2022-8-7
Hello,
I have to solve the following problem over and over again for slightly different values of mu.
n = lsqlin(mu-eye(J),zeros(J,1),[],[],ones(1,J),1,[],[],[]);
The matrix mu has columns sum up to 1, values are between zero and 1 and the diagonal elements are typically above 0.9. There are only very few 0 elements in mu (although many of them are 'close' to 0, e.g. 1e-4). J is equal to 400.
You can do the same calculation using
n = null(mu-eye(J),1e-10);
n = n/sum(n);
but that's not any faster. Are there any ideas on how to speed that up? Since the solution n does not change much in the various calls, I thought about providing an initial guess, but the interior algorithm does not accept any initial guesses.
Here's a sample matrix (with J=5)
0.980472260884484 0.0169062020634941 1.31828882462499e-05 0.00712276192859210 0.00734667253541008
0.0122127547484456 0.972782440495672 1.15814051875989e-05 0.00808693210831310 0.00830831290909974
3.60311943991737e-08 4.68217920214649e-08 0.999953080612125 1.05940998862873e-07 3.71871063642989e-05
0.00445881693357994 0.00660554154798086 1.18490160022582e-05 0.976444331730883 0.0148592752450766
0.00285613140229620 0.00370576907106219 1.03060784389177e-05 0.00834586829121426 0.969448552204049
Thanks!
  3 个评论
显示 1更早的评论
John D'Errico
John D'Errico 2022-8-4
There are at least a couple of other ways I could describe to solve for the null space of a matrix. It would be easier if you would provide a sample matrix to play with, and compare methods, without needing to describe in detail how to solve it for each method.
Tintin Milou
Tintin Milou 2022-8-4
The different mu are not available at the same time. I'll provide a sample matrix to play with.

请先登录，再进行评论。

请先登录，再回答此问题。

采纳的回答

Bruno Luong
Bruno Luong 2022-8-4
Can you try this:
A = mu-eye(size(A));
[Q,R,p] = qr(A,'vector');
n = [R(1:end-1,1:end-1)\R(1:end-1,end); -1];
n(p) = n/sum(n)
  10 个评论
显示 8更早的评论
Tintin Milou
Tintin Milou 2022-8-7
Update: Before calling fsolve for the outer loop, I now add an fsolve on a small-scale problem to get a reasonable initial guess for the complete problem. With this improved initial guess, the code runs more smoothly and the initial suggestion by Bruno Luong is the best solution I have found. Thanks again!
Bruno Luong
Bruno Luong 2022-8-7
Thanks for the update. It is puzzled me that the outer loop is that sensitive to numerical error.

请先登录，再进行评论。

更多回答（0 个）

类别

在 Help Center 和 File Exchange 中查找有关 Linear Least Squares 的更多信息

标签

产品


版本

R2022a

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by