Timeline for Why is the set of Hermitian matrices with repeated eigenvalue of measure zero?
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
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| Apr 14, 2022 at 0:52 | comment | added | Duchamp Gérard H. E. | (+1) Good and simple argument. To make it unshaky, I expect that (a) you write $𝑆\setminus \{𝐴\}$ as the cartesian product of a sphere centered at $A$ and the open half line and (b) you use Fubini formula to conclude. | |
| Apr 13, 2022 at 22:26 | comment | added | Mizar | @LSpice good points; I corrected my answer. | |
| Apr 13, 2022 at 22:26 | history | edited | Mizar | CC BY-SA 4.0 |
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| Apr 13, 2022 at 22:25 | comment | added | LSpice | I guess your symmetric matrix is real, so as also to be Hermitian; but I am confused by "any complex line through $A$". The space of Hermitian matrices is naturally a real, but not a complex, vector space; how do you define a complex line? | |
| Apr 13, 2022 at 22:24 | comment | added | Guido Li | shaky argument, but yes ;) | |
| Apr 13, 2022 at 22:21 | vote | accept | Guido Li | ||
| Apr 13, 2022 at 22:15 | history | answered | Mizar | CC BY-SA 4.0 |