Timeline for The generalization of commutative property of orthogonal projectors on a subspace to the whole space
Current License: CC BY-SA 3.0
10 events
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| Nov 16, 2017 at 17:38 | comment | added | Christian Remling | @DominiqueUnruh: Yes, I was sloppy. I thought originally for $n$ projections $v$ would still have to be in the sum of the now $2^n$ summands formed with the spaces we project on, but it's not so clear how that would follow or if it's true. | |
| Nov 16, 2017 at 17:37 | history | edited | Christian Remling | CC BY-SA 3.0 |
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| Nov 16, 2017 at 12:35 | comment | added | Dominique Unruh | I do not think the generalization to many projectors works with the same argument. Say $P,Q,R$ pairwise commute on $V$. Then, using your proof, we get $\hat P,\hat Q$ such that $\hat P=P$ on $V$ and $\hat Q=Q$ on $V$, and $\hat P,\hat Q$ commute. Now we apply want to apply the argument to the next pair, say $P,R$. But if we apply it to $P,R$, we get a different $\hat P$ that might not commute with $\hat Q$ any more. Alternatively, we may try to apply the argument to $\hat P,R$. But that doesn't work because we don't know whether $\hat PRv=R\hat Pv$ for all $v\in V$. | |
| Nov 16, 2017 at 12:26 | comment | added | Nik Weaver | Oh, I see. Well, Halmos's writing is unusually clear. | |
| Nov 16, 2017 at 3:51 | comment | added | Christian Remling | @NikWeaver: The article I linked to says they'll call it the Halmos 2 projections theorem because Halmos's presentation is so well written, but many others had this or similar versions before... | |
| Nov 16, 2017 at 3:23 | comment | added | Nik Weaver | I had no idea this was due to Halmos! I always considered it folklore. | |
| Nov 16, 2017 at 0:16 | history | undeleted | Christian Remling | ||
| Nov 16, 2017 at 0:16 | history | edited | Christian Remling | CC BY-SA 3.0 |
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| Nov 15, 2017 at 22:56 | history | deleted | Christian Remling | via Vote | |
| Nov 15, 2017 at 22:54 | history | answered | Christian Remling | CC BY-SA 3.0 |