Timeline for Is the solution of this linear system always positive definite?
Current License: CC BY-SA 3.0
11 events
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| May 30, 2020 at 23:09 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Jan 31, 2020 at 23:01 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Oct 3, 2019 at 22:02 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Jun 5, 2019 at 21:02 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| May 6, 2019 at 20:49 | answer | added | Christopher Hillenbrand | timeline score: 1 | |
| Jun 22, 2012 at 21:02 | comment | added | user11870 | I see. But if so, I want to point out that the left-hand side is not $J_r\circ S$. | |
| Jun 22, 2012 at 20:47 | comment | added | Noah Stein | @wmmiao: I believe Will Sawin's point was to consider what happens if the "arbitrary orthogonal matrix" you mentioned was the identity, making $P$ the first $r$ columns of the identity. | |
| Jun 22, 2012 at 20:41 | comment | added | user11870 | I believe the answer of the question is true based on random generated numerical tests. Numerical tests also show that it should be true if the term $I_r$ on the right-hand side is replaced by any diagonal matrix with positive diagonal entries. | |
| Jun 22, 2012 at 20:36 | comment | added | user11870 | Assumption: $1<r<n$. So $P$ cannot be the identity matrix. If so, we cannot have $n^2-n\geq r^2$ in the last line but two. | |
| Jun 22, 2012 at 20:29 | comment | added | Will Sawin | What if your orthogonal matrix is the identity matrix? Then the left side is just $J_r \circ S$, which cannot equal $I_r$. | |
| Jun 22, 2012 at 19:29 | history | asked | user11870 | CC BY-SA 3.0 |