Timeline for Singular Value Decomposition more columns than rows
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 17, 2021 at 19:17 | answer | added | Ângelo Polotto | timeline score: 0 | |
| May 9, 2014 at 9:58 | review | First posts | |||
| May 9, 2014 at 10:00 | |||||
| May 8, 2014 at 11:08 | comment | added | joidegn | you changed the first comment. The second one is similar to what I accepted as the right answer by Brian Borchers below. Thanks | |
| May 6, 2014 at 14:57 | vote | accept | joidegn | ||
| May 6, 2014 at 13:17 | comment | added | ttnphns | You should take into account that only eigenvectors corresponding to nonzero singular values make sense. So, the trailing eigenvectors of V, corresponding to zero singular values, can be safely set to zero. If so, they may be not computed or not shown by a function. | |
| May 6, 2014 at 13:09 | answer | added | Brian Borchers | timeline score: 6 | |
| May 6, 2014 at 11:42 | comment | added | joidegn | @ttnphns: yes you are right. My question is with respect to the right eigenvectors $V^*$. I should have been more precise I guess. | |
| May 6, 2014 at 11:42 | history | edited | joidegn | CC BY-SA 3.0 |
added 1 character in body
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| May 6, 2014 at 11:02 | comment | added | ttnphns | When you do SVD of a nXp matrix, you normally get 3 matrices as the result: left eigenvectors U (nXn), right eigenvectors V (pXp), diagonal matrix of singular values S (nXp). Some implementations of the function may cut-off empty rows or columns of S. | |
| May 6, 2014 at 9:54 | history | asked | joidegn | CC BY-SA 3.0 |