Timeline for Calculating a generalized inverse (Moore–Penrose pseudoinverse)
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 18, 2016 at 14:43 | comment | added | TommasoF | Is there any kind of proof of your answer? I am very interested in it. | |
| Feb 28, 2014 at 13:38 | comment | added | Nathaniel Johnston | @Thomas -- Yep, if you don't normalize then the $\mathbf{v}_j$'s become nicer, but at the expense that the form of $L_n^+$ becomes slightly messier. Explicitly, if $\mathbf{w_j} = \sqrt{(2j-n)(2j-n-1)}\mathbf{v}_j$ (i.e., the all-integer eigenvector) then $L_n^+ = \sum_{j=1}^n \frac{1}{j(2j-n)(2j-n-1)}\mathbf{w}_j\mathbf{w}_j^*$. | |
| Feb 28, 2014 at 13:10 | comment | added | MthQ | Ok, I understand now. I guess I should not normalize in order to get nicer expression. Thank you very much | |
| Feb 28, 2014 at 13:07 | vote | accept | MthQ | ||
| Feb 28, 2014 at 12:05 | comment | added | Nathaniel Johnston | @Thomas -- I coded my answer above in MATLAB to double-check that it works for $n \leq 10$. The eigenvectors above do consist entirely of integers, up to being normalized (i.e., up to being divided by $\sqrt{(2j-n)(2j-n-1)}$, which is the norm of the integer vector). | |
| Feb 28, 2014 at 9:37 | comment | added | MthQ | Are you sure about those eigenvectors? When I compute the eigenvectors of $L_n$, they completely consist of integers. Am I not getting something? | |
| Feb 27, 2014 at 19:22 | history | answered | Nathaniel Johnston | CC BY-SA 3.0 |