Timeline for Finiteness of "novel variance" from a kernel on a compact space [closed]
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 16, 2014 at 18:50 | history | closed |
Tom LaGatta Chris Godsil Stefan Kohl♦ user9072 Noah Stein |
Needs details or clarity | |
| Apr 3, 2014 at 1:26 | vote | accept | Tom LaGatta | ||
| Apr 3, 2014 at 1:09 | review | Close votes | |||
| Apr 16, 2014 at 18:50 | |||||
| Apr 3, 2014 at 1:02 | comment | added | Tom LaGatta | Voting to close my own question. | |
| Apr 3, 2014 at 0:46 | comment | added | Alexander Shamov | Then I don't have a guess what you are computing in the case of more than 2 vectors. | |
| Apr 3, 2014 at 0:40 | comment | added | Tom LaGatta | Thanks, Alexander. I clearly have screwed something up in the definition. The 2 vector case is supposed to be $\|i_1\|^2 + \|i_2\|^2 - \langle i_1, i_2 \rangle$, which is symmetric. | |
| Apr 3, 2014 at 0:29 | answer | added | Alexander Shamov | timeline score: 1 | |
| Apr 3, 2014 at 0:22 | comment | added | Alexander Shamov | On the other hand, it does depend on the order in which you take the $i_n$'s. For instance, even for 2 vectors, say, $i_1$ and $i_2$, your sum of squared norms will be $\Vert i_1 \Vert^2 + \Vert i_2 \Vert^2 - \frac{\langle i_1,i_2 \rangle}{\Vert i_1 \Vert^2}$, which is not symmetric in $i_1,i_2$. | |
| Apr 3, 2014 at 0:20 | comment | added | Alexander Shamov | What do you mean by "doesn't depend on the choice of basis"? It seems that you are not choosing a basis. | |
| Apr 3, 2014 at 0:01 | history | asked | Tom LaGatta | CC BY-SA 3.0 |