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We know that the determinant of a Hadamard product of two positive semidefinite matrices $|{\bf A}\circ{\bf B}|$ is greater than or equal to $|{\bf A}||{\bf B}|$. Are there any general results on arbitrary matrices or something specific on other classes of matrices?

I am specifically interested in the determinant of $|{\bf V}\circ {\bf R}|$ where the $i$-th row of ${\bf V}$ is $\left[x_1^{d_i} \;x_2^{d_i}\;\ldots \;x_N^{d_i}\right]$ and ${\bf R}$'s elements are all roots of unity.

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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? – fedja May 31 '11 at 1:13
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? – Anadim May 31 '11 at 3:27
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 ;) – fedja Jun 2 '11 at 17:10
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. – Anadim Jun 2 '11 at 19:49
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". – fedja Jun 3 '11 at 4:41

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