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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
    Commented Jan 12, 2018 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
    Commented Jan 12, 2018 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
    Commented Jan 12, 2018 at 16:46
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    $\begingroup$ How about 2-rank? $\endgroup$
    – Arnaud Mortier
    Commented Jan 12, 2018 at 16:56
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    $\begingroup$ Maybe Hadamard rank? $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Jan 12, 2018 at 16:58

1 Answer 1

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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
    Commented Jan 12, 2018 at 17:22
  • $\begingroup$ it is ok we can delete comments and move to discussion except yours. $\endgroup$
    – Turbo
    Commented Jan 12, 2018 at 17:30
  • $\begingroup$ How does 2 follow? $\endgroup$
    – Dirk
    Commented Jan 12, 2018 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
    Commented Jan 12, 2018 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 $2$? $\endgroup$
    – Turbo
    Commented Jan 12, 2018 at 19:18

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