Timeline for Products of adjugate matrices
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot | replaced http://mathoverflow.net/ with https://mathoverflow.net/ | |
| Aug 19, 2015 at 13:02 | comment | added | user91684 | @ darij grinberg , Yes, I need a trick for the first step. In fact, the required result about the $(i,j)$ entry of the studied matrix $M$ is a formal equality (using only addition and multiplication) of the form $M_{i,j}=\det(S)P_{i,j}((s_{i,j}),(a_{i,j}))$ where $P_{i,j}$ is a polynomial. Where would be the obstruction that prevents this is true in my polynomial ring ? | |
| Aug 19, 2015 at 12:28 | comment | added | darij grinberg | Getting from a polynomial ring over $\mathbb{Z}$ to an arbitrary commutative ring is easy: You can always take the evaluation homomorphism at the entries of your matrices. The hard part is getting from a field to a polynomial ring over $\mathbb{Z}$. It is here that you probably want to use a form of Gauss's lemma. | |
| Aug 19, 2015 at 12:18 | history | edited | user91684 | CC BY-SA 3.0 | added 1259 characters in body |
| Aug 15, 2015 at 4:22 | comment | added | user91684 | Yes Darij, you are right. | |
| Aug 14, 2015 at 21:08 | comment | added | darij grinberg | +1. Nice trick in here! But you can spare yourself the algebraic geometry orgy by generalizing: For any $n\times n$-matrix $S$ and any alternating $n\times n$-matrix $A$, the entries of $\left(\operatorname{adj} S\right) \cdot A \cdot \left(\operatorname{adj}\left(S^T\right)\right)$ are divisible by $\det S$. The irreducibility of $\det S$ is much more elementary here, and instead of $\operatorname{adj} S = avv^T$ you now need $\operatorname{adj} S = vw^T$ (which is also easier to prove and less reliant on the base field). | |
| Aug 14, 2015 at 14:58 | history | edited | user91684 | CC BY-SA 3.0 | added 223 characters in body |
| Aug 14, 2015 at 10:00 | comment | added | user91684 | In fact the rank is 0 or 1 | |
| Aug 14, 2015 at 9:59 | comment | added | user91684 | It is true for any matrix ! | |
| Aug 13, 2015 at 23:06 | comment | added | user54031 | Why is the rank of adj S = 1 for det S = 0? | |
| Aug 13, 2015 at 21:48 | history | answered | user91684 | CC BY-SA 3.0 |