Problem setting

Consider a set $ S = \big\{ 1,2,\cdots,n \big\}$. Now consider $k$ equal-sized subsets $S_i \subset S$ s.t of size $\big|S_i\big|=n' \;\forall i$.

Consider a $k\times k$ matrix $M$ s.t $\;M_{ij} = \big|S_i\cap S_j\big|\big/n'$, size of the intersection of the $i^{th}, j^{th}$ subset divided by $n'$.


Does $M_{ij}$ lie in the polytope of zero-one matrices? Paraphrasing the question - Can we express every $M_{ij}$ as a convex combination of zero-one matrices?

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    $\begingroup$ It should be assumed that $n' \le k$, since otherwise the diagonal entries of $M$ are larger than one. $\endgroup$ – Andrew Morgan Sep 23 '16 at 21:07
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    $\begingroup$ The convex hull of zero-one matrices consists exactly of the matrices whose entries are in $[0,1]$. So I think the answer to your question is "yes". $\endgroup$ – Sasho Nikolov Sep 23 '16 at 21:49

Sasho already gave you a yes/no answer, but here's an actual convex combination for you: If $B_\ell$ is the $k \times k$ matrix which is $1$ when $|S_i \cap S_j| \ge \ell$ and zero otherwise, then $M = \sum_{\ell=1}^{n'} \frac{1}{n'} B_\ell$.

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    $\begingroup$ This has a similar flavour of $ E[X] = \int P(X>x) $ $\endgroup$ – Vivek Bagaria Sep 23 '16 at 23:31
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    $\begingroup$ In case it wasn't obvious, let me add that the convex hull of zero-one vectors in $\mathbb{R}^d$ is $[0,1]^d$. So there is nothing specific in this answer about $k \times k$ matrices, which are the special case of $d=k^2$ (up to an obvious bijection between $k$ by $k$ matrices and $\mathbb{R}^{k^2}$). $\endgroup$ – Sasho Nikolov Sep 24 '16 at 19:08

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