I have a matrix $$A\in\{0,1\}^{d\times n}$$ and $$rank(A)=d,d, and another matrix $$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?

• 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 $d-1, d-2, \cdots 0$. – nicoguaro May 9 '19 at 2:22
• 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. – Wolfgang Bangerth May 9 '19 at 16:36

$$\text{rank}(A\odot X) \leq \text{rank}(A) \text{rank}(X)$$
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 $$A\odot X \in \mathbb{F}^{d\times n}, d.
2. If $$\text{rank}(A)=r, depending on what you know about $$X$$ can give you something, but still not too exciting.