Timeline for Vanishing Trace
Current License: CC BY-SA 2.5
9 events
| when toggle format | what | by | license | comment | |
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| Jan 24, 2011 at 0:29 | vote | accept | Andre | ||
| Jan 22, 2011 at 2:05 | comment | added | Martin Argerami | Thanks a lot, Mikael! I should start thinking before writing... | |
| Jan 21, 2011 at 22:02 | comment | added | Mikael de la Salle | @Martin: How do you define the trace of an operator which is not of trace class? For you second remark, what I meant is that if the compression to a closed subspace $K$ of a trace-zero operator has trace zero, then so has its compression to the orthogonal of $K$. This is just because of the equality $Tr(PaP+(1-P)a(1-P))=Tr(a)$, valid for any projection $P$. | |
| Jan 21, 2011 at 19:40 | comment | added | Martin Argerami | Mikael, I'm confused with a couple of things in your answer. First you seem to imply that an operator with zero trace is trace-class (and thus compact), which is not the case. Second, in your maximality argument in the second paragraph you say that the compression of a trace-zero operator by a projection is still trace-zero, which again I think is not true. | |
| Jan 20, 2011 at 9:40 | history | edited | Mikael de la Salle | CC BY-SA 2.5 |
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| Jan 20, 2011 at 8:28 | comment | added | Mikael de la Salle | Bill: unless I miss something, I really believe that my argument proves that the condition $Tr(a)=0$ implies that the numerical range of $a$ is closed. See my edit. | |
| Jan 20, 2011 at 7:45 | history | edited | Mikael de la Salle | CC BY-SA 2.5 |
added 750 characters in body; added 3 characters in body
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| Jan 19, 2011 at 23:10 | comment | added | Bill Johnson | Mikael, you only get that $0$ is in the closure of the convex hull of the $\lambda_n$'s. IIRC, the numerical range need not be closed. For those who don't know: in finite dimensions Mikael's argument works and is the standard proof that an $n$ by $n$ matrix that has trace zero is unitarily equivalent to a matrix that has zero diagonal. Of course, here simple induction is used. | |
| Jan 19, 2011 at 10:13 | history | answered | Mikael de la Salle | CC BY-SA 2.5 |