Timeline for Eigenvalues come in pairs
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
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| Jun 17, 2022 at 9:57 | history | became hot network question | |||
| Jun 17, 2022 at 9:55 | comment | added | Pritam Bemis | @DenisSerre thanks. Based on Carlo Beenakker's answer, this is true as $A_1^{-1} = UA_1U$. Thus $A_1^{-T}=VA_1^{-1}V=VUA_1 UV.$ Thus $J=VU.$ | |
| Jun 17, 2022 at 9:50 | comment | added | Denis Serre | Likely, $A_1$ and $A_2$ belong to a classical group $G$, defined by the identity $A^TJA=J$ for some invertible $J$ (which you must find). Then $JAJ^{-1}=A^{-T}$ tells you that the spectrum if invariant under $\lambda\mapsto\lambda^{-1}$. | |
| Jun 17, 2022 at 9:41 | vote | accept | Pritam Bemis | ||
| Jun 17, 2022 at 8:48 | answer | added | Carlo Beenakker | timeline score: 13 | |
| Jun 17, 2022 at 7:42 | history | edited | Pritam Bemis | CC BY-SA 4.0 |
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| Jun 17, 2022 at 1:57 | history | asked | Pritam Bemis | CC BY-SA 4.0 |