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Given a matrix $M\in\Bbb F_2^{n\times n}$, define its Hadamard rank $h(M)$ to be the minimum number of rank $\leq2$ matrices in $\Bbb F_2^{n\times n}$ with Hadamard product (that is, the entry-wise product $\circ$) equal to $M$. That is, $$h(M)=\min\{k:\exists M_1,\dots,M_k\mid\max_i\mathrm{rk}(M_i)\le 2,\;M_1\circ M_2\circ\dots \circ M_k=M\}.$$

From Arnaud Mortier's argument, $h(M)\leq n$.

Is there a geometric meaning behind the Hadamard rank $h(M)$?

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    $\begingroup$ Am I correct that the corank of the 3 by 3 identity is 2? Where does this notion come from? $\endgroup$ – Dirk Jan 12 '18 at 16:35
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    $\begingroup$ It's possibly a natural notion but not a natural terminology, as usually corank means something distinct: en.wikipedia.org/wiki/Corank. Even for those not familiar with the classical definition, "co" suggest something such as dual or orthogonal, which is not reflected in your definition. Would you suggest another name? $\endgroup$ – YCor Jan 12 '18 at 16:40
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    $\begingroup$ I think (and am sure @YCor would agree) that there's no question that there's many ways of being "dual to rank", but a concept that not only is not dual to rank in any obvious way, but (per your question (4)) doesn't even have any obvious relation to the rank, surely at least faces an uphill battle to justify the name 'corank'. $\endgroup$ – LSpice Jan 12 '18 at 16:46
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    $\begingroup$ How about 2-rank? $\endgroup$ – Arnaud Mortier Jan 12 '18 at 16:56
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    $\begingroup$ Maybe Hadamard rank? $\endgroup$ – მამუკა ჯიბლაძე Jan 12 '18 at 16:58
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A matrix full of $1 $'s except on one particular row has rank at most $2$. Hence $1.$ and $2 $. are true.

[Deleted contents regarding 3. as there was a mistake in my spreadsheet - $ A\circ A $ is generically invertible only when $ n\leq 3 $.] $ $

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    $\begingroup$ Of course this should be a comment but the comments section is being used for other purposes. $\endgroup$ – Arnaud Mortier Jan 12 '18 at 17:22
  • $\begingroup$ it is ok we can delete comments and move to discussion except yours. $\endgroup$ – Brout Jan 12 '18 at 17:30
  • $\begingroup$ How does 2 follow? $\endgroup$ – Dirk Jan 12 '18 at 19:00
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    $\begingroup$ Am I mistaken? If you take $M_i$ to coincide with $M$ on the $i$-th row and full of ones everywhere else, then $M$ is the Hadamard product $M_1\circ\ldots\circ M_n$... right? $\endgroup$ – Arnaud Mortier Jan 12 '18 at 19:13
  • $\begingroup$ @ArnaudMortier sorry corrected 4. to what I had in mind. However what do you mean by no two columns are linearly dependent and rank $2$? $\endgroup$ – Brout Jan 12 '18 at 19:18

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