Timeline for answer to Special determinant formula by dharr

11 events

when toggle format	what		by	license	comment
Jan 19, 2025 at 0:38	history	edited	dharr	CC BY-SA 4.0	add minor clarification
Dec 29, 2024 at 6:01	history	edited	dharr	CC BY-SA 4.0	fix minor error
Dec 29, 2024 at 2:17	comment	added	dharr		For the inverse, it's not clear what form you want, which presumably depends on what you want to do with it.. Now that you have the determinant, the simplest is just the adjugate divided by the determinant. If you want a formula in terms of the eigenvectors and eigenvalues, $M(mk)^{-1}=U^{T} \Lambda ^{-1}U$ (different $U$ than in my answer), then finding the last eigenvectors is non-trivial and involves roots of a cubic.
Dec 28, 2024 at 21:19	history	edited	dharr	CC BY-SA 4.0	typo
Dec 28, 2024 at 20:29	history	edited	LSpice	CC BY-SA 4.0	Minor tidying, while this is on the front page
Dec 28, 2024 at 19:58	history	edited	dharr	CC BY-SA 4.0	correct typo
Dec 28, 2024 at 19:47	history	edited	dharr	CC BY-SA 4.0	Replaced with complete proof
Dec 25, 2024 at 14:01	comment	added	Honza		Nice observation by @dharr. PS In addition to a determinant, I'll also need the inverse.
Dec 24, 2024 at 16:52	history	edited	dharr	CC BY-SA 4.0	typo
Dec 24, 2024 at 6:52	comment	added	dharr		I think I can prove this but will need to work with the eigenvectors, which will be fiddly, so may take a few days,
Dec 24, 2024 at 6:46	history	answered	dharr	CC BY-SA 4.0