Timeline for A short question about the DFT matrix
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| S Mar 27, 2013 at 14:30 | vote | accept | Olivier Leveque | ||
| Mar 27, 2013 at 14:08 | vote | accept | Olivier Leveque | ||
| S Mar 27, 2013 at 14:30 | |||||
| Mar 27, 2013 at 14:08 | vote | accept | Olivier Leveque | ||
| Mar 27, 2013 at 14:08 | |||||
| Mar 26, 2013 at 23:26 | answer | added | Chris Godsil | timeline score: 6 | |
| Mar 26, 2013 at 21:59 | comment | added | Suvrit | ;-) duh! I somehow interpreted it to mean magnitude of the nonzeros! | |
| Mar 26, 2013 at 17:57 | comment | added | Steve Huntsman | I should have said "$N$th roots of unity" above. | |
| Mar 26, 2013 at 17:55 | comment | added | Steve Huntsman | Let $M$ a matrix of the form you describe, with $M_{jk} = C e^{i\omega_{jk}}$ and $C > 0$. Unitarity implies that $C^2 \sum_\ell e^{i(\omega_{\ell k} - \omega_{\ell j})} = \delta_{j k} = C^2 \sum_\ell e^{i(\omega_{j \ell} - \omega_{k \ell})}$. Taking $j = k$ gives that $C = N^{-1/2}$, where $N = \dim M$. For $j \ne k$, $\sum_\ell e^{i(\omega_{j \ell} - \omega_{k \ell})} = 0$. The only way this can happen is if the angles $\omega_{j \ell} - \omega_{k \ell}$ ``balance out''. Roots of unity are a particularly nice way for this to happen, but as Mark Meckes points out, not the only one. | |
| Mar 26, 2013 at 17:43 | answer | added | Mark Meckes | timeline score: 4 | |
| Mar 26, 2013 at 17:25 | comment | added | Steve Huntsman | The identity matrix has lots of zeros. | |
| Mar 26, 2013 at 17:20 | comment | added | Suvrit | what about the identity matrix? | |
| Mar 26, 2013 at 17:12 | history | asked | Olivier Leveque | CC BY-SA 3.0 |