Timeline for Find the determinant of a matrix given the determinant of all $p\times p$ sub-matrices?
Current License: CC BY-SA 4.0
12 events
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| Feb 24, 2022 at 8:16 | comment | added | Denis Serre | @FrançoisBrunault . Thanks a lot, François ! | |
| Feb 23, 2022 at 16:42 | comment | added | François Brunault | I corrected a typo: $\Lambda^p(K)$ should be $\Lambda^p(K^n)$. | |
| Feb 23, 2022 at 16:41 | history | edited | François Brunault | CC BY-SA 4.0 |
Corrected typo: Lambda^p(K) -> Lambda^p(K^n)
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| Feb 21, 2022 at 16:20 | vote | accept | Mathew George | ||
| Feb 21, 2022 at 15:14 | comment | added | Denis Serre | @LSpice. Thanks a lot ! | |
| Feb 21, 2022 at 15:13 | history | edited | Denis Serre | CC BY-SA 4.0 |
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| Feb 21, 2022 at 14:56 | comment | added | LSpice |
TeX note: In case it is of interest, you might find that $\displaystyle\sum_{\rho \in \mathfrak S_4; \rho(1) < \rho(2), \rho(3) < \rho(4)}$ looks better as $\displaystyle\sum_{\substack{\rho \in \mathfrak S_4 \\ \rho(1) < \rho(2), \rho(3) < \rho(4)}}$ \sum_{\substack{\rho \in \mathfrak S_4 \\ \rho(1) < \rho(2), \rho(3) < \rho(4)}} (or even an index of summation on three lines, trading more vertical space for horizontal space).
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| Feb 21, 2022 at 14:55 | history | edited | LSpice | CC BY-SA 4.0 |
Link to comment
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| Feb 21, 2022 at 14:33 | history | edited | Denis Serre | CC BY-SA 4.0 |
added 16 characters in body
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| Feb 21, 2022 at 14:17 | comment | added | Ilya Bogdanov | This can also be proved by induction using generalized Laplace expansion. | |
| Feb 21, 2022 at 10:18 | history | edited | Denis Serre | CC BY-SA 4.0 |
added 367 characters in body
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| Feb 21, 2022 at 10:08 | history | answered | Denis Serre | CC BY-SA 4.0 |