Timeline for spectrum of the compression of a selfadjoint operator
Current License: CC BY-SA 3.0
7 events
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| S Mar 23, 2016 at 17:52 | history | suggested | CommunityBot | CC BY-SA 3.0 |
Fixed a typo
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| Mar 23, 2016 at 17:29 | review | Suggested edits | |||
| S Mar 23, 2016 at 17:52 | |||||
| Nov 23, 2010 at 4:16 | vote | accept | Paul Z | ||
| Nov 23, 2010 at 4:16 | |||||
| Nov 22, 2010 at 15:38 | comment | added | Denis Serre | This is exactly what I mean. In my example, $e_{pTp}(-\infty,t)$ is the eigenspace (a line) of $pTp$ associated with the eigenvalue $-\sqrt{1+a^2}$, whereas $e_{T}(-\infty,t)$ is the eigenspace (a line) of $T$ associated with the eigenvalue $-1$. These lines turn out to be distinct, so there is no inequality between the projectors. | |
| Nov 22, 2010 at 15:25 | comment | added | Paul Z | But $e_{pTp}(-\infty, t)$ $\le$ $e_T(-\infty, t)$ does not imply that the range of $e_{pTp}(-\infty, t)$ is invariant under $T$. | |
| Nov 22, 2010 at 15:22 | comment | added | Paul Z | Denis, Thanks for your answer. You guess is correct. I am sorry I should clarify my notations. If we write $T=\int_R \lambda d(e_{\lambda}) $, then $e_T(-\infty, t)$ means the projections corresponding to unique spectral decomposition of the operator $T$. Similarly, for $e_{pTp}(-\infty, t)$. And $e_{pTp}(-\infty, t)$ $\le$ $e_T(-\infty, t)$ means the range of the first projection is included in the range of the second projection. | |
| Nov 22, 2010 at 7:12 | history | answered | Denis Serre | CC BY-SA 2.5 |