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Let $M$ be a $2^n \times 2^n$ matrix over real number field, where the rows and columns are indexed by subsets of $[n] := {1,2,\ldots,n}$, and defined as follows, $ M_{A, B} = 1 $ if $A \subseteq B$; $M_{A, B} = -1$ if $B \subsetneq A$; $M_{A, B}$ can take arbitrary value over $\mathbb{R}$.

In words, $M$ is a matrix with some missing entries. Is there some result lower bounds the rank of matrix $M$, for example, $rk(M) \ge n^{\Omega(\log n)}$.

edited Feb 8 at 0:50

asked Feb 7 at 0:26

jsliyuan
315●5

	What does "undefined" mean? Do you need the lower bound for all matrices with some prescribed values? – Ilya Bogdanov Feb 7 at 15:16
	"undefined" = could take any value. Lower bound means the lower bound for all such matrices – jsliyuan Feb 8 at 0:43
	Do you have constructions that allow reducing the rank so much?? – S. Sra Feb 8 at 3:57
	(it seems to be easy to reduce the rank $2^{n/2}$, but one needs something more clever to reduce it even further...) – S. Sra Feb 8 at 4:00
	I think I have constructions achieving this bound. But I don't know how to prove the lower bound. – jsliyuan Feb 8 at 7:26

Aaron Meyerowitz · Accepted Answer · 2013-02-09 05:24:32Z UTC

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Amazingly (to me) the rank can be as low as $n$. Simply define $M_{A,B}=1$ when $|A| \le |B|$ and $M_{A,B}=-1$ when $|A| \gt |B|.$ This is consistent with the previous requirements and makes the $A$ row depend only on $|A|.$

answered Feb 9 at 5:24

Aaron Meyerowitz
11.6k●1●10●31

Rank of a matrix with missing entries

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