Timeline for Can we recover all $k$-minors of a square matrix from some of them?
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 11, 2020 at 5:08 | vote | accept | Asaf Shachar | ||
| Dec 11, 2019 at 8:37 | answer | added | Arnaud Mortier | timeline score: 3 | |
| Nov 18, 2019 at 13:32 | comment | added | Asaf Shachar | @EricCanton Yeah, I am assuming that $A$ is invertible. I included the discussion about non-degeneracy conditions as a motivation for why requiring invertibility. | |
| Nov 17, 2019 at 3:12 | comment | added | Eric Canton | Why do you need the non-degeneracy condition? Aren't you assuming $A$ is invertible? | |
| Nov 16, 2019 at 13:14 | answer | added | Libli | timeline score: 7 | |
| Nov 15, 2019 at 15:00 | comment | added | Kapil | (Another thinking out loud attempt!) The matrix defines a section of $S^{*}\otimes Q$ on $G_s\times G_q$ where $S$ is the universal sub-bundle on the Grassmannian $G_s$ of rank $k$ sub-spaces and $Q$ is the universal quotient bundle on the Grassmannian $G_q$ of rank $k$ quotient spaces. Such a section is determined by its values at finitely many points. However, you are not asking for the value, rather only the value of its image in $\det(S)^*\otimes\det(Q)$. Moreover, the points are "pre-determined" since you are only using points determined by basis vectors. | |
| Nov 15, 2019 at 9:26 | comment | added | Gro-Tsen | (Just thinking out loud.) I feel it might be worth generalizing the question to the case where you delete $k$ columns and $ℓ$ rows, with the convention that the “determinant” of a non-square matrix is simply the wedge product of its columns (say). So one sub-question becomes: how many wedge products of $n-k$ among $n$ column vectors do you need to recover them all? (The Plücker relations certainly have something to say here.) And another: now what if we also delete some entries? | |
| Nov 15, 2019 at 9:03 | history | asked | Asaf Shachar | CC BY-SA 4.0 |