Timeline for Can an operator have Exp(z) as its characteristic "polynomial"?
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| Oct 12, 2011 at 22:53 | comment | added | George Lowther | Ok. That proves that $T$ is quasinilpotent (all eigenvalues are zero). Then, I think, $\Lambda^nT$ is also quasinilpotent and ${\rm Det}(I+zT)=1$, giving a negative result to the question. | |
| Oct 12, 2011 at 22:28 | comment | added | kap44 | If you consider the Taylor expansion in the finite-dimensional case in terms of eigenvalues, the $n$-th coefficient will be a function of degree $n$ with a symmetrical expression for eigenvalues. This means that $Trace T^{2n}=0$. | |
| Oct 12, 2011 at 22:24 | comment | added | George Lowther | or are these the same thing? | |
| Oct 12, 2011 at 22:21 | comment | added | George Lowther | Don't you mean ${\rm Trace}(\Lambda^nT^2)=0$? | |
| Oct 12, 2011 at 22:19 | comment | added | kap44 | If so, maybe my first argument that $T^2=0$ is wrong; I got it from the finite-dimensional situation. | |
| Oct 12, 2011 at 22:07 | comment | added | Yemon Choi | If $T^2$ is a quasinilpotent trace class operator, then it has trace zero (Lidskii). So all powers of $T^2$ have trace zero. Does this imply that $\Tr(\wedge^{2k} T) \neq 1/(2k!)$? | |
| Oct 12, 2011 at 22:03 | history | edited | kap44 | CC BY-SA 3.0 |
added 21 characters in body
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| Oct 12, 2011 at 21:46 | history | answered | kap44 | CC BY-SA 3.0 |