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I have a matrix $A\in\{0,1\}^{d\times n}$ and $rank(A)=d,d<n$, and another matrix $X\in \mathbb{R}^{d\times n}$, but I do not know the rank of $X$. What can we say about the rank of their Hadamard product $rank(A\odot X)$

And what if $rank(A)=k, k<d$?

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    $\begingroup$ I would say that you can say nothing about it. Consider $A$ a matrix with an identity block of size $d$ and $X=A$, their product would have the same rank. Then start replacing the $1$s by $0$s and you end up with a sequence of matrices with ranks $d-1, d-2, \cdots 0$. $\endgroup$ – nicoguaro May 9 at 2:22
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    $\begingroup$ Conversely, start with the matrix $A$ that is all ones. It has rank 1. The mask it with a matrix $X$ that is the identity matrix, and the rank of the Hadamard product is full. $\endgroup$ – Wolfgang Bangerth May 9 at 16:36
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For Hadamard product, you have (source):

$$ \text{rank}(A\odot X) \leq \text{rank}(A) \text{rank}(X) $$

This is as tight as you can go even with knowing $\text{rank}(X)$. Without knowing anything about $X$, there is certainly nothing you can say. (See the examples that nicoguaro and Wolfgang Bangerth brought in the comments section).

Now, imagine you know $\text{rank}(X)=r$, and $\text{rank}(X)\geq1$ (I think the without loss of generality clause is applicable here).

  1. If $\text{rank}(A)=d$, then the inequality above is totally useless, as it would not bound anything: $d\cdot r \geq d$, which is the same conclusion you would get from just looking at $A\odot X \in \mathbb{F}^{d\times n}, d<n$.
  2. If $\text{rank}(A)=r<d$, depending on what you know about $X$ can give you something, but still not too exciting.

Hadamard products are tricky, and a lot of things one expects from them (based on prior knowledge of the regular matrix-matrix product) do not hold.

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