Timeline for Neighborhood of an orthogonal matrix
Current License: CC BY-SA 4.0
13 events
| when toggle format | what | by | license | comment | |
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| Oct 16, 2020 at 1:03 | 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 18, 2020 at 0:09 | 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. | |
| May 19, 2020 at 18:08 | comment | added | user114668 | It’s the case you mentioned where multiple maxima are in the same column. The upper limit $a=1/\sqrt{2}$ also gives a matrix $A$ where it’s not immediately clear to me what rotation produces a valid $B$. | |
| May 19, 2020 at 17:44 | comment | added | Lo Celso | @student Could you elaborate more on your comment? Thanks! | |
| May 19, 2020 at 9:15 | comment | added | user114668 | $$\left( \begin{array}{cccc} a & \sqrt{\frac{1}{2}-a^2} & \frac{1}{2} & -\frac{1}{2} \\ a & \sqrt{\frac{1}{2}-a^2} & -\frac{1}{2} & \frac{1}{2} \\ -\sqrt{\frac{1}{2}-a^2} & a & \frac{1}{2} & \frac{1}{2} \\ -\sqrt{\frac{1}{2}-a^2} & a & -\frac{1}{2} & -\frac{1}{2} \\ \end{array} \right) \;, \quad \frac{1}{2} < a < \sqrt{\frac{1}{2}}$$ | |
| May 18, 2020 at 23:05 | answer | added | Ben Grossmann | timeline score: 1 | |
| May 18, 2020 at 22:24 | comment | added | Lo Celso | @Omnomnomnom I have thought about that, but there exists a counterexample since the max of two rows may occur at the same column (I got an explicit 4$\times$4 counterexample but probably too large for a comment). | |
| May 18, 2020 at 21:55 | comment | added | Ben Grossmann | A naive constructive approach that comes to mind: of course, a permutation matrix $P$ is a matrix that satisfies the requirements of $B$ except that it might fail to be in a sufficiently small neighborhood of $A$. If we look at the path $p:[0,1] \to \Bbb R^{n \times n}$ defined by $p(t) = (1-t)A + tP$ and project onto $O(n)$ (via the "orthogonal procrustes" polar decomposition method), then it must hold that $p(t)$ eventually satisfies the necessary requirements. Perhaps it is possible to guarantee that this happens for sufficiently small $t$ for the right choice of $P$. | |
| May 15, 2020 at 3:09 | comment | added | DSM | I am sorry, I assumed you wanted only orthogonality, and not orthonormality. In that case, my comment does not mean much. | |
| May 14, 2020 at 16:09 | comment | added | Lo Celso | @DSM Why would this be an orthogonal matrix though? The length of every row would be $1+\epsilon$ so each row would not be a unit vector. | |
| May 14, 2020 at 5:11 | comment | added | DSM | Why not consider $(1+\epsilon)A$. This would also be orthogonal with $\max(\vec{b}_i)>\max(\vec{a}_i)$ for every $i$, and can be made to lie within an arbitrary small ball around $A$. Hope I follow your question correctly. | |
| May 13, 2020 at 21:06 | review | First posts | |||
| May 13, 2020 at 21:10 | |||||
| May 13, 2020 at 21:04 | history | asked | Lo Celso | CC BY-SA 4.0 |