Timeline for How many minors I need to check to conclude all minors will vanish ?
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| Nov 1, 2010 at 18:43 | comment | added | Gerhard Paseman | Here is an example to show that most minors need to be checked. Let k = m - 1. Take the subset C of R^m which lies on the intersection of S^k with a k-dimensional plane, both centered at the origin. You can now successively pick points from C so that no k of them lie in a sub-vector space of dimension less than k. The coordinates of these points can be the columns of such a matrix: every mxm minor will have rank m-1. Gerhard "Ask Me About System Design" Paseman, 2010.11.01 | |
| Nov 1, 2010 at 18:08 | comment | added | Gerhard Paseman | Also, in the first sentence of your first paragraph, it is not clear that all the minors need to be checked. For an mxn matrix, with m < n, if the first minor you check belongs to an mxm submatrix with rank anything other than m-1, then you use the information to narrow the search, as you either have a negative answer or a set of m-1 columns with rank less than m-1, which indicates several other mxm minors which vanish. The examples given show that all columns need to be checked, not that all minors need to be checked. Gerhard "Ask Me About System Design" Paseman, 2010.11.01 | |
| Nov 1, 2010 at 14:58 | comment | added | Vagabond | Just as an aside I was curious about generalized vandermonde matrices whose determinants vanishes. As I was surfing through the literature quite at random I came across some argument in a paper which goes like this: its a $3x4$ matrix with rows $({a_{i}}^{j_k}- s^{j_k}),i=1,..,3;j=1,..,4$. Then it says note for all minors to vanish, it is enough to show that two of them vanish. See page 17 of this paper Tropical secant graphs of monomial curves arxiv.org/pdf/1005.3364 | |
| Nov 1, 2010 at 14:52 | comment | added | Vagabond | @shieikraisinrollbank @ Steven Sam Thanks to both of you for being so helpful. I must admit as I lack the necessary background it would take me some ( I guess a lot actually) to understand all the argument. I will try now to pick up the necessary background and figure out things for myself. Any suggestion for where to begin ? | |
| Nov 1, 2010 at 14:13 | comment | added | Steven Sam | @sheikraisinrollbank: well it seems that we both benefited from this discussion. I've updated my answer. | |
| Nov 1, 2010 at 14:12 | history | edited | Steven Sam | CC BY-SA 2.5 |
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| Nov 1, 2010 at 11:01 | comment | added | Sheikraisinrollbank | I really think you should modify the "Your questions amounts..." bit to make it more accurate. As it is it's misleading and confusing. The way you put it in the third paragraph "Another question..." makes it seem like you've already answered the OP's question when really you're punting! | |
| Nov 1, 2010 at 11:01 | comment | added | Sheikraisinrollbank | I see, thanks for the clarification. On the other hand, it really seems that the whole point here is that the ideal is radical: the linear independence of the minors follows already from Rennemo's observation and doesn't need any rep. theory (incidentally, this also shows I said something wrong about the minors: the evaluation at points functionals do span the whole dual, but are not equal to it). | |
| Oct 29, 2010 at 17:56 | comment | added | Steven Sam | @shiekraisinrollbank: I am aware that the minors can't take arbitrary values independent of one another. This answer was to the original question of when a matrix has rank $< r$. Since the ideal generated by the $r \times r$ minors generate a radical ideal and they linearly independent, they form a minimal generating set. Hence some proper subset of them vanishing cannot imply that the other ones vanish. That's all I am saying. I am not saying anything about the possible values they can take. | |
| Oct 29, 2010 at 14:37 | comment | added | Sheikraisinrollbank | @Steven Sam: The determinantal ideal being radical doesn't help here; it's just not possible for the minors to take on any set of values, independent of one another. For instance, in the 2 by 4 case you can't have $m_{12}=0=m_{13}$ and $m_{14}=1=m_{23}$. The point really is what I wrote above: the relevant linear functionals in this case are the evaluation functionals at points, and these don't separate the span of the minors. | |
| Oct 29, 2010 at 13:23 | comment | added | Steven Sam | @sheikraisinrollbank I certainly overlooked something, you are right. For example, the functions $x^2, xy, y^2$ are certainly linearly independent but the vanishing of $x^2, y^2$ (over a field) implies that $xy = 0$. But this ideal is not radical, so I should emphasize that determinantal ideals are radical, but I didn't want to get into why that is true. One could also cheat and say that $xy \ne 0$ when $x^2 = y^2 = 0$ for certain commutative rings $k$ :) | |
| Oct 29, 2010 at 11:19 | comment | added | Sheikraisinrollbank | @Steven Sam: The only linear maps that are relevant for the question are the ones coming from evaluation at a matrix, and these do not span the full dual space of the linear span of the minors. Certainly the Plucker relations show that despite the fact that the minors are linearly independent, it is not true that they can take on arbitrary values independent of one another. So the phrase "Your question amounts..." at the beginning of the second paragraph should be modified. | |
| Oct 29, 2010 at 5:10 | comment | added | Vagabond | @Steven Sam Thanks a ton for explaining and explaining it so well. Specially for writing in such a manner which even a person like me who has no prior acquaintance with many of the concepts can follow to a reasonable degree. | |
| Oct 28, 2010 at 15:52 | comment | added | Steven Sam | @David: "dependent" makes more sense than "independent", thanks. | |
| Oct 28, 2010 at 15:51 | history | edited | Steven Sam | CC BY-SA 2.5 |
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| Oct 28, 2010 at 15:22 | comment | added | David E Speyer | Also, I would emphasize the first sentence of your third paragraph more strongly. If you want to compute the rank of a matrix in practice (and you have exact entries, dealing with round off error is a whole different subject) then the right algorithm is Gaussian elimination, not computing minors. | |
| Oct 28, 2010 at 15:20 | comment | added | David E Speyer | In the first sentence of your second paragraph, I think "independent" should read "dependent". | |
| Oct 28, 2010 at 13:56 | history | answered | Steven Sam | CC BY-SA 2.5 |