Timeline for Does ${\rm Tr}(l(a)+l(b)) ={\rm Tr}(l(\alpha_1 a +\alpha_2 b ))$ imply that $l(a)l(b)=r(a)r(b)=0$?
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
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| Apr 14, 2023 at 10:17 | comment | added | user92646 | @GeraldEdgar I see! THX | |
| Apr 14, 2023 at 10:17 | history | edited | user92646 | CC BY-SA 4.0 |
edited title
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| Apr 14, 2023 at 10:16 | comment | added | Gerald Edgar | It is best not to include displayed math inside a title. | |
| Apr 14, 2023 at 10:13 | vote | accept | user92646 | ||
| Apr 14, 2023 at 10:07 | answer | added | Stefaan Vaes | timeline score: 3 | |
| Apr 14, 2023 at 9:52 | comment | added | user92646 | @StefaanVaes THX Stefaan, you are right about that. | |
| Apr 14, 2023 at 9:51 | history | edited | user92646 | CC BY-SA 4.0 |
added 5 characters in body
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| Apr 14, 2023 at 8:55 | comment | added | Stefaan Vaes | If you require $\text{Tr}(l(a)+l(b)) = \text{Tr}(l(\alpha_1 a + \alpha_2 b))$ for really all $\alpha_1,\alpha_2$, then also for $\alpha_1 = \alpha_2 = 0$, which forces $a=b=0$. You probably want to modify the question? | |
| Apr 14, 2023 at 7:20 | history | asked | user92646 | CC BY-SA 4.0 |