Timeline for Jordan form of compact operator
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 24, 2011 at 20:26 | vote | accept | Alexander | ||
| Nov 21, 2011 at 21:01 | vote | accept | Alexander | ||
| Nov 24, 2011 at 20:26 | |||||
| Nov 21, 2011 at 21:00 | vote | accept | Alexander | ||
| Nov 21, 2011 at 21:00 | |||||
| Nov 21, 2011 at 17:14 | comment | added | Florian | The point is that for a nonzero eigenvalue, its eigenspace can be separated, and on the remaining space the operator no longer has this eigenvalue. A countable repetition of this process yields a decomposition into the various eigenspaces and a remaining space on which there are no nonzero eigenvalues, and at this point you are stuck. So it doesn't help you too much that $A^0$ just has one eigenvalue. | |
| Nov 21, 2011 at 7:58 | comment | added | Alexander | I choose the operator $A^0$ such that it has non-zero eigenvalues. | |
| Nov 20, 2011 at 21:03 | comment | added | Robert Israel | One fact about compact operators that is true in Hilbert space but not in all Banach spaces is that compact operators can be approximated (in norm) by finite-rank operators (see <en.wikipedia.org/wiki/Approximation_property>). | |
| Nov 20, 2011 at 20:52 | comment | added | Robert Israel | I don't know what you expect to say about nonnormal compact operators on a Hilbert space. They too can have no nonzero eigenvalues. A useful example might be $T$ defined on $\ell^2$ by $(T x)_n = x_{n+1}/n$. Note that in this case the union of the kernels of powers of $T$ is dense in $\ell^2$. | |
| Nov 20, 2011 at 20:00 | history | answered | Florian | CC BY-SA 3.0 |