Gene therapy! Yes, for n = 1 and n = 2, the answer is +/- 1. When I listed the subsets in an obvious "binary" order, the matrix I got had the contradiagonal and everything below it all 1. This suggested a proof sketch for all n, as follows: Arrange the subsets so that all the elements containing 1 are listed first, followed by all the ones that don't. The resulting matrix is all 1 on and above the main contradiagonal, because subsets containing 1 have nonempty intersection. Now the "sketch" part: it was very easy to use standard determinant-invariant operations to clear the other triangle of all 1s, after which one could just read off the determinant as 1. On Fri, Dec 7, 2018 at 3:32 PM Eugene Salamin via math-fun < math-fun@mailman.xmission.com> wrote:
The hypothesis that the determinant is zero can be checked, and possibly refuted, by an explicit calculation for small n.
-- Gene
On Friday, December 7, 2018, 12:24:43 PM PST, Dan Asimov < dasimov@earthlink.net> wrote:
I never got a few things right in the statement of this problem, even on the second try. But now it's right:
----- A2 Let S_1, S_2, ..., S_(2^n−1) be the nonempty subsets of {1,2,...,n} in some order, and let M be the (2^n−1)×(2^n−1) matrix whose (i,j)th entry m(i,j) = 0 if S_i ∩ S_j is empty; m(i,j) = 1 otherwise.
Calculate the determinant of this matrix. -----
Allan's point that if the order doesn't matter, the determinant must be 0 also occurred to me. But maybe they're looking for an answer of the form
±[expression]
??? In any case, if the answer doesn't turn out to be too trivial, this is a very cool problem.
—Dan
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