Timeline for Existence of generalized inverse-like operator
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 18, 2017 at 12:38 | vote | accept | PThomasCS | ||
| May 18, 2017 at 0:18 | answer | added | Luca Ghidelli | timeline score: 7 | |
| May 17, 2017 at 23:24 | comment | added | PThomasCS | That was to relate it to the Moore-Penrose Pseudoinverse. The $\star$ operator might also be applied to non-square matrices, but you are right that the condition we need only considers square matrices. | |
| May 17, 2017 at 22:48 | comment | added | Robert Israel | So why did you say $\star \;: M(m,n, \mathbb R) \to M(n,m,\mathbb R)$? | |
| May 17, 2017 at 22:12 | comment | added | PThomasCS | @RobertIsrael Yes, $A$ must be square. If it helps, we can even assume that $A$ is symmetric. | |
| May 17, 2017 at 21:40 | comment | added | Robert Israel | If $B^T A B$ makes any sense, $A$ must be a square matrix. | |
| May 17, 2017 at 20:44 | comment | added | PThomasCS | @მამუკაჯიბლაძე Moore-Penrose pseudoinverse does not satisfy this. For example, if B=[1 1; 1 2] and A = [1 0; 0 0]. | |
| May 17, 2017 at 19:59 | comment | added | მამუკა ჯიბლაძე | See Moore-Penrose pseudoinverse | |
| May 17, 2017 at 19:33 | history | edited | PThomasCS | CC BY-SA 3.0 |
The transpose suggested in a comment was actually on the wrong B term.
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| May 17, 2017 at 19:32 | history | edited | PThomasCS | CC BY-SA 3.0 |
Added that the operator flips the dimensions of the matrix.
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| May 17, 2017 at 19:26 | history | asked | PThomasCS | CC BY-SA 3.0 |