Timeline for Convergence of iterative algorithm.
Current License: CC BY-SA 2.5
9 events
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| Feb 4, 2010 at 20:45 | comment | added | Steve Huntsman | It does, I could do A(1,[42,48]) = 1 instead of A(1,42) = 1 and A(1,48) = 1, for instance. There are even more succinct ways to do it for large examples. | |
| Feb 4, 2010 at 20:36 | comment | added | Mariano Suárez-Álvarez | Please tell me matlab has a less verbose way of entering sparse matrices! :) | |
| Feb 4, 2010 at 20:35 | comment | added | Steve Huntsman | However, the rank of A in this example is 4 rather than 6, so there are certainly some uniqueness issues. | |
| Feb 4, 2010 at 20:31 | history | edited | Steve Huntsman | CC BY-SA 2.5 |
changed "invert a matrix" to "solve a matrix equation"
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| Feb 4, 2010 at 20:30 | comment | added | Steve Huntsman | BTW, the commands above are related to the specific example in the question. | |
| Feb 4, 2010 at 20:30 | comment | added | Steve Huntsman | Fair enough, I wasn't thinking all the way through. But using this index I can do the following in MATLAB... A = zeros(56); A(1,42) = 1; A(1,48) = 1; A(2,48) = 1; A(2,22) = 1; A(3,42) = 1; A(3,14) = 1; A(4,48) = 1; A(4,14) = 1; A(5,22) = 1; A(6,42) = 1; A(6,22) = 1; A(6,14) = 1; ...and then pA=pinv(A);sparse(pA.*(abs(pA)>10^-10)) gives a pretty tractable psuedoinverse. | |
| Feb 4, 2010 at 19:37 | comment | added | Harald Hanche-Olsen | But the linear system above is underdetermined, is it not? And then, the problem of finding x from z is overdetermined, or so it seems to me. So this is not a great way to decompose the problem. | |
| Feb 4, 2010 at 19:01 | history | answered | Steve Huntsman | CC BY-SA 2.5 |