Timeline for Prove that the inner product space of 2x2 traceless Hermitian matrices form a Hilbert space [closed]
Current License: CC BY-SA 3.0
16 events
| when toggle format | what | by | license | comment | |
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| Nov 1, 2016 at 16:01 | review | Reopen votes | |||
| Nov 1, 2016 at 17:42 | |||||
| Nov 1, 2016 at 15:45 | history | edited | Annie | CC BY-SA 3.0 |
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| S Nov 1, 2016 at 13:36 | history | unlocked | CommunityBot | ||
| S Nov 1, 2016 at 13:36 | history | locked | CommunityBot | ||
| S Nov 1, 2016 at 13:36 | history | closed |
Javier user36790 auden glS Jon Custer |
Not suitable for this site | |
| Oct 31, 2016 at 12:44 | answer | added | Valter Moretti | timeline score: 2 | |
| Oct 31, 2016 at 12:01 | history | edited | ACuriousMind♦ |
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| S Oct 31, 2016 at 11:40 | history | suggested | user130529 | CC BY-SA 3.0 |
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| Oct 31, 2016 at 11:13 | answer | added | user130529 | timeline score: 0 | |
| Oct 31, 2016 at 11:01 | comment | added | Valter Moretti | "A vector space is a hilbert space iff every cauchy sequence converges in the vector space itself." This definition is not complete: "A vector space equipped with a scalar product is a Hilbert space iff every Cauchy sequence - defined with respect to the norm associated to the scalar product - converges in the vector space itself." | |
| Oct 31, 2016 at 10:59 | comment | added | Valter Moretti | The question is meaningless if you do not define a symmetric real scalar product on that space. An important point is that we are speaking about a real Hilbert space, since a complex combination of Hermitian matrices is not Hermitian. | |
| Oct 31, 2016 at 10:53 | review | Suggested edits | |||
| S Oct 31, 2016 at 11:40 | |||||
| Oct 31, 2016 at 10:30 | review | Close votes | |||
| Nov 1, 2016 at 13:36 | |||||
| Oct 31, 2016 at 10:11 | comment | added | Mass | Math SE would be the better for this question. | |
| Oct 31, 2016 at 10:09 | comment | added | andypea | Is it isomorphic to a simpler space? | |
| Oct 31, 2016 at 10:06 | history | asked | Annie | CC BY-SA 3.0 |