Timeline for When is this matrix singular?
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 24, 2014 at 4:42 | comment | added | Hans | @Suvrit: oh, right. | |
| Feb 24, 2014 at 1:58 | comment | added | Suvrit | $\sin(x-y)=s_xc_y-c_xs_y$, which is at most rank-2. | |
| Feb 24, 2014 at 1:55 | answer | added | Lev Borisov | timeline score: 2 | |
| Feb 24, 2014 at 1:50 | history | edited | Hans | CC BY-SA 3.0 |
Added a question.
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| Feb 24, 2014 at 1:36 | comment | added | Hans | @Suvrit: This condition falls in the simple scenarios mentioned in question 1). Still, in this case, why is $A$ always rank 2? | |
| Feb 24, 2014 at 0:56 | comment | added | Hans | @LevBorisov: Yes, I have tried $2\times 2$ case with $\phi_1=\phi_2=0$. The singularity condition is the trivial $\omega_2=\pm\omega_1+q\pi$ for some integer $q$, similar to the ones listed in the question. Perhaps I should find a way to use some symbolic manipulation software to simplify the determinant of the general $3\times 3$ cases. | |
| Feb 24, 2014 at 0:20 | comment | added | Suvrit | If $\omega_j=1$, $\phi_j=-t_j$, then $A_{jk}=\sin(t_k-t_j)$, which means that $A$ is a rank-2 matrix, and hence singular (for arbitrary $t_k$). I guess you are trying to get necessary and sufficient conditions which will be much harder... | |
| Feb 23, 2014 at 22:06 | comment | added | Lev Borisov | Have you tried $2x2$ and $3x3$ cases? This may give you a sense if this is feasible as stated. | |
| Feb 23, 2014 at 18:01 | history | undeleted | Hans | ||
| Feb 23, 2014 at 18:00 | history | deleted | Hans | via Vote | |
| Feb 23, 2014 at 17:58 | history | asked | Hans | CC BY-SA 3.0 |