Timeline for The determinant of the hadamard product of two matrices
Current License: CC BY-SA 3.0
6 events
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| Jun 3, 2011 at 4:41 | comment | added | fedja | Ah, that was the ambiguity of English then: I parsed "an explicit way of picking the xis such that I get nonzero determinant no matter what the dis and R are" as "there exists X such that for all D and R" instead of the correct "for all D,R, there exists X". | |
| Jun 2, 2011 at 19:49 | comment | added | Anadim | I'm sorry for not making this explicit :). All elements of ${\bf R}$ are $m$-th roots of unity. On another note, I now have a probabilistic argument on how to choose the $x_i$ elements such that the determinant is nonzero: Let your $x_i$s being drawn uniformly at random from a finite field. Then by expanding the determinant of this matrix using the Leibniz formula, you get a nonzero polynomial in your $x_i$ entries. Using the Schwartz–Zippel lemma, you obtain that this polynomial has a nonzero solution with arbitrarily high probability for arbitrarily large finite fields. | |
| Jun 2, 2011 at 17:10 | comment | added | fedja | Erm... isn't it true that in a finite field a certain power of every element is 1? I am pretty certain that you didn't mean this trivial counterexample but if you put some effort into stating the problem correctly, the result may be that somebody will really think of it instead of posting stupid comments ;) | |
| May 31, 2011 at 3:27 | comment | added | Anadim | Correct. I want this determinant to be nonzero over a finite field (I don't mind the order), so I thought that a lower bound given by the determinants of the two matrices could be useful. So, the exact question I am interested in is if there is an explicit way of picking the $x_i$s such that I get nonzero determinant no matter what the $d_i$s and ${\bf R}$ are? | |
| May 31, 2011 at 1:13 | comment | added | fedja | It can easily be $0$: I do not see anything that prevents you from taking $x_j$ to be roots of unity and $r_{ij}=x_j^{-d_i}$, giving you the matrix of all ones. Why don't you just ask exactly what you are interested in? | |
| May 30, 2011 at 21:23 | history | asked | Anadim | CC BY-SA 3.0 |