Timeline for Can an operator have Exp(z) as its characteristic "polynomial"?
Current License: CC BY-SA 3.0
14 events
| when toggle format | what | by | license | comment | |
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| Feb 9, 2015 at 11:48 | comment | added | Wolfgang | Hi John, have you seen my recent question? mathoverflow.net/questions/196030/… | |
| Oct 14, 2011 at 16:30 | vote | accept | John Wiltshire-Gordon | ||
| Oct 13, 2011 at 23:15 | comment | added | George Lowther | @John: See my answer. I include a reference for the product expansion. | |
| Oct 13, 2011 at 23:14 | answer | added | George Lowther | timeline score: 15 | |
| Oct 13, 2011 at 1:25 | comment | added | George Lowther | @John: I think it is true in general. Diagonalization is not necessary. You can still decompose $\mathcal{H}$ into the orthonal sum of a sequence of finite dimensional subspaces preserved by $T$, and the orthogonal complement $V$ on which $T$ (followed by orthogonal projection onto $V$) is quasinilpotent. So, by Yemon Choi's comment below, $T$ projected onto $V$ is traceless, as are its powers. So, the calculation of the determinant reduces to the increasing limit over finite dimensional spaces, and the product follows from the finite dimensional case. | |
| Oct 13, 2011 at 0:49 | comment | added | John Wiltshire-Gordon | @George Lowther I think such an expression is true when $T$ can be diagonalized (by the spectral theorem, for instance). It seems less likely in general. | |
| Oct 13, 2011 at 0:36 | comment | added | George Lowther | I think it's true that $f(z)={\rm Det}(I+zT)$ can be written as $\prod\_n(1+z\lambda\_n)$ where $\lambda\_n$ are the eigenvalues of $T$ (I just say think, because I haven't really looked at Fredholm determinants before). Then, if $f$ is nowhere zero, we have $\lambda\_n=0$, so $f(z)=1$. Therefore, the answer to the question is no. | |
| Oct 12, 2011 at 21:46 | answer | added | kap44 | timeline score: 7 | |
| Oct 11, 2011 at 23:25 | comment | added | Yemon Choi | @ARupinski: wouldn't the Fredholm determinant of a quasinilpotent trace-class operator be an entire, nowhere-vanishing function? | |
| Oct 11, 2011 at 23:08 | history | edited | John Wiltshire-Gordon | CC BY-SA 3.0 |
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| Oct 11, 2011 at 13:58 | comment | added | John Wiltshire-Gordon | @András Bátkai What do you mean? | |
| Oct 11, 2011 at 5:26 | comment | added | András Bátkai | Well, it does have zeros... | |
| Oct 11, 2011 at 0:45 | comment | added | ARupinski | Since $e^z$ has no zeros, it seems that such a $T$ either cannot exist or must be extremely weird to begin with. | |
| Oct 10, 2011 at 23:19 | history | asked | John Wiltshire-Gordon | CC BY-SA 3.0 |