Timeline for Maximum number of orthonormal vectors contained in an open cone
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| Mar 1, 2012 at 17:21 | vote | accept | Jesús Álvarez | ||
| Mar 1, 2012 at 17:06 | comment | added | Robert Israel | For the maximum number with $p=2$ to be $\ge 3$ you need the right side to be at most $(2/3) \|u\|^2$, not $\|u\|^2/\sqrt{2}$. I suspect you're thinking about $\|u\|^2/2$. | |
| Mar 1, 2012 at 17:00 | answer | added | Robert Israel | timeline score: 3 | |
| Mar 1, 2012 at 16:16 | comment | added | Uday | oops! I interpreted the 'number of orthonormal vectors' as dimension. I thought that is the natural thing to do. In the convex cone case, the answer must be p (dimension of the range). | |
| Mar 1, 2012 at 16:06 | comment | added | Jesús Álvarez | About the motivation, I arrived to this question studying growth conditions for the eigenvalues of self-adjoint operators with discrete spectrum. Suggestions about this kind of study are also welcome. As an example, it is easy to see that the maximum number is 1 if $p=1$, and it is $\ge 3$ if $p=2$. The dimension of U is not mentioned in the question, but its definition uses a linear subspace of finite dimension. | |
| Mar 1, 2012 at 15:31 | comment | added | Uday | Is it a convex cone? Otherwise, what do you mean by dimension of cone? | |
| Mar 1, 2012 at 15:02 | comment | added | Anthony Quas | and the motivation for the question is?... | |
| Mar 1, 2012 at 13:42 | history | edited | Jesús Álvarez | CC BY-SA 3.0 |
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| Mar 1, 2012 at 12:21 | history | asked | Jesús Álvarez | CC BY-SA 3.0 |