Timeline for Determining consistency of a general overdetermined linear system
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 8, 2014 at 4:52 | vote | accept | Travis Willse | ||
| Oct 3, 2014 at 9:42 | history | bounty awarded | Travis Willse | ||
| Sep 29, 2014 at 6:51 | comment | added | Travis Willse | Thanks for these comments, Ewan, the outstanding points are very clear to me now. As for finding a Groebner basis this seems out of reach without another insight, as I don't have a characterization of the ideal of obstructions (for general $m$) besides the definition. I'll attempt some variations of your argument for $m = 3$ (which works just as well for $m = n + 1$ I think), but perhaps this is harder than it seems. | |
| Sep 29, 2014 at 5:44 | comment | added | Ewan Delanoy | @Travis "This argument must be modified significantly when $m>n+1$ as in that case there are multiple determinant obstructions" Indeed, I’m not even certain that there are no other obstructions in this case. What is needed, probably, is showing that the set of determinental obstructions is a Groebner basis with respect to your favorite monomial ordering. | |
| Sep 29, 2014 at 5:35 | comment | added | Ewan Delanoy | @Travis "Why does $R\in I$ imply that $R(\phi(w))=0$" ? Because $\phi(w)\in V(I)$ by construction of $\phi$ ; what $\phi$ does is choose the unique value of $b_3$ so that $b$ will indeed be a linear combination of the columns of $A$. Then, by definition of $V(I)$, all the polynomials of $I$ must be zero on $V(I)$. | |
| Sep 29, 2014 at 5:30 | comment | added | Ewan Delanoy | @Travis "we must assume that $p\geq 1$, but if it is not, by symmetry we can replace $B_3$ with any generator that does occur in $P$" Indeed, or you can use another argument I’ve just inserted in my answer, see above. | |
| Sep 29, 2014 at 5:28 | history | edited | Ewan Delanoy | CC BY-SA 3.0 | added 103 characters in body |
| Sep 29, 2014 at 3:22 | comment | added | Travis Willse | Thanks for this, Ewan. In order to apply the lemma, we must assume that $p \geq 1$, but if it is not, by symmetry we can replace $B_3$ with any generator that does occur in $P$. Why does $R \in I$ imply that $R(\phi(w)) = 0$? Also, it seems like this argument must be modified significantly when $m > n + 1$ (rather than equal as in this case), as in that case there are multiple determinant obstructions---or is there an obvious and easy modification I'm not seeing? | |
| Sep 28, 2014 at 8:37 | history | answered | Ewan Delanoy | CC BY-SA 3.0 |