Timeline for Extremal properties of the determinant for matrices with entries in a fixed subset of $[-1,1]^{n^2}$?
Current License: CC BY-SA 3.0
11 events
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| Dec 28, 2011 at 19:03 | comment | added | Gerhard Paseman | It may be possible to do something similar for the three ones case, although not as slick. Using the form (a-d)(b-f) - (a-c)(b-e), one might get a nice amount of cancellation between the two products. (I think 4/9 is possible.) If you manage to get 1/2 for this, that will just leave two more cases to establish a slick upper bound on a(3). Gerhard "Ask Me About System Design" Paseman, 2011.12.28 | |
| Dec 26, 2011 at 19:11 | comment | added | Will Orrick | To tie up one loose end, the 1/4 bound can be put on completely solid ground. Let the five non-1 elements be $-1\le a\le b\le c\le d\le e\le 1$ and let $w=b-a$, $x=c-b$, $y=d-c$, $z=e-d$. We show that $wy\le1/4$ or $xz\le1/4$. We have $w+x+y+z\le2$. Now if $w+y\le1$ then $wy\le1/4$. Therefore, if $wy>1/4$ then $w+y>1$ and $x+z<1$ and so $xz<1/4$. Still looking for a good upper bound in cases where there are less than four 1s... | |
| Dec 23, 2011 at 21:30 | comment | added | Will Orrick | Thanks for the explanation. I see now that I was mistakenly allowing only the four elements that share a row or column with the 2x2 block to move. Once the fifth element is in play things are much more constrained. I can't see a way to beat the 1/4 bound in that case. | |
| Dec 23, 2011 at 20:28 | comment | added | Gerhard Paseman | As support for the 1/4 bound though, I am thinking of spaces between the unknowns as (1/2 + delta), (1/2 + epsilon), (1/2 + zeta), and (1/2 - delta -epsilon - zeta). Granted I can't choose any two of these spaces and multiply them together, but for many configurations I get a usable product of less than 1/4, and if pressed for a proof, I can probably guarantee something like 3/8. This is still better than n^(n/2) for an upper bound. Gerhard "Ask Me About System Design" Paseman, 2011.12.23 | |
| Dec 23, 2011 at 20:21 | comment | added | Gerhard Paseman | Even if the 1/4 turns out to be a bad value, for the four ones case I am convinced that a(3)< 1/2, and I hope to encourage someone to develop a cheap upper bound like a(n) < 1 or even a(n) > a(n+1) with the ideas I've suggested. I feel as you do a(3) is not much larger than 9/50. Gerhard "Ask Me About Uneducated Guessing" Paseman, 2011.12.23 | |
| Dec 23, 2011 at 20:16 | comment | added | Gerhard Paseman | It is a "shoot from the lip" estimate that is likely incorrect, but the idea is as follows. Since there are four ones, there are 5 unknowns distributed in [-1,1), guaranteeing that at least two of them are less than 1/2 distance apart. Although it may be possible to get five arranged so that any two pairs from the five will form a product greater than 1/4, it seemed hard to produce such a case. Anyway, it is clear to me that from the five entries one can get a product less than 1/2, and 1/4 may be likely. Gerhard "Fires From Lip And Hip" Paseman, 2011.12.23 | |
| Dec 23, 2011 at 18:25 | comment | added | Will Orrick | Gerhard, would you elaborate on the derivation of the 1/4 upper bound in the case where the multiset contains four 1s? If you put the four 1s in a 2x2 block, giving determinant D=(a-b)(c-d), it seems to me that you can find values of a, b, c, and d in [-1,1] such that the minimum value of |D| over all permutations of a, b, c, and d is larger than 1/4, but perhaps I'm misunderstanding the argument. By the way, I am now strongly convinced that the lower bound of 0.1859138 for a(3) cannot be improved. | |
| Dec 23, 2011 at 8:15 | history | edited | Gerhard Paseman | CC BY-SA 3.0 |
new remarks; added 4 characters in body
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| Dec 23, 2011 at 2:07 | comment | added | Gerhard Paseman | Indeed Will, there should be five: I conflated two classes when I should not have. I will edit later and include some more remarks. Gerhard "Ask Me About System Design" Paseman, 2011.12.22 | |
| Dec 22, 2011 at 12:46 | comment | added | Will Orrick | @ Gerhard: Shouldn't there be five essentially different forms? In my answer I give three different forms (two of my four matrices differ only in the placement of the non-1 elements). In addition to these three, there is the matrix with three 1s in a row and three 1s in a column, and the matrix with three 1s in a row and two rows each containing a single 1 in different columns. | |
| Dec 6, 2011 at 22:38 | history | answered | Gerhard Paseman | CC BY-SA 3.0 |