Timeline for Factor matrix ${\bf A}$ into the product ${\bf B}{\bf C}$ where ${\bf C}$ has no negative entries and ${\bf B}$ has few non-zero entries
Current License: CC BY-SA 3.0
15 events
| when toggle format | what | by | license | comment | |
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| S Jun 29, 2016 at 18:06 | history | suggested | David G. Stork | CC BY-SA 3.0 |
Fixed a typo
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| Jun 29, 2016 at 17:51 | review | Suggested edits | |||
| S Jun 29, 2016 at 18:06 | |||||
| S Jun 29, 2016 at 0:15 | history | edited | Pedro Lauridsen Ribeiro | CC BY-SA 3.0 |
fixed typo in the subscripts, a few aesthetic and formatting improvements
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| S Jun 29, 2016 at 0:15 | history | suggested | David G. Stork | CC BY-SA 3.0 |
fixed typo in the subscripts
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| Jun 28, 2016 at 23:53 | review | Suggested edits | |||
| S Jun 29, 2016 at 0:15 | |||||
| Jun 14, 2016 at 21:54 | vote | accept | David G. Stork | ||
| Apr 6, 2016 at 16:37 | comment | added | David G. Stork | If ${\bf A}$ has non-vanishing determinant, then its $n$ eigenvectors span an $n$-dimensional space. Once the $k$ non-negative or non-positive mutually independent rows have been eliminated, the remaining $n-k$ rows still be mutually independent may be spannable by $n-k$ non-negative or non-positive rows. If not, what condition upon ${\bf A}$ will ensure that it can so be spanned. | |
| Apr 6, 2016 at 0:29 | comment | added | Will Sawin | @DavidG.Stork I thought of those too. If that improvement always worked, then it is easy to prove that it gives the optimal method. However, then the method requires me to write $v_1$ as a linear combination of an all positive vector $u$ and a subset of the $v_2,\dots,v_n$ - the subset consisting of vectors that have both positive and negative entries. Unfortunately it's not obvious that this is possible. | |
| Apr 5, 2016 at 21:44 | comment | added | David G. Stork | I upvoted your answer, @Will Sawin, because it is a step in the right direction. However, there are a number of very quick improvements that I've found. For instance, if any row $i$ in ${\bf A}$ has only non-negative entries, then a single $+1$ in the $i,i$ entry of ${\bf B}$ with the $i$ row of ${\bf C}$ set to $i$ row of ${\bf A}$ reduces the cost. Likewise if the row of ${\bf A}$ has a row with non-positive entries, simply set the $i,i$ entry of ${\bf B}$ to $-1$, and so forth. I must say, I'm surprised such factorization (apparently) doesn't appear in the scholarly literature. | |
| Apr 2, 2016 at 3:09 | comment | added | Will Sawin | @DavidG.Stork Yes. Of course it is off by a factor of at most $2$. I thought a simple modification of my argument gives the optimum but that turns out not to be the case. | |
| Apr 2, 2016 at 2:48 | comment | added | David G. Stork | And the $2n$ entries need not be optimal, as my simple example shows... right? | |
| Apr 2, 2016 at 0:32 | comment | added | Will Sawin | @DavidG.Stork Right. | |
| Apr 1, 2016 at 21:27 | comment | added | David G. Stork | Will Shawn (@Will Shawn): Your third line should be $\nu_1 = cu + a_2 \nu_2 + \ldots a_n \nu_n$. Right? | |
| Apr 1, 2016 at 16:03 | comment | added | David G. Stork | Thanks, @Will Shawn. Let me try coding this up and seeing if it works on a few simple cases and how sensitive it might be to changes in entries to ${\bf C}$. (This will take a two or three days...) | |
| Apr 1, 2016 at 3:15 | history | answered | Will Sawin | CC BY-SA 3.0 |