# Hadamard (element-wise multiplication) product rank

I am having some problems on understanding an inequality regarding the rank of the Hadamard product (element-wise product). I have $$B=A\circ A$$ where $$A$$ is a $$n\times r$$ matrix, and $$\circ$$ is the Hadamard product, so $$B$$ must also be a $$n\times r$$ matrix, so, according to my intuition, $$rank(B)\le \min (n,r)$$, but I have read that the actual upper bound for the rank is as follows: $$rank(B)\le {d+1 \choose 2}$$, where $$d$$ is the rank of $$A$$, but I am confused since the value of $${d+1 \choose 2}$$ is greater than $$d$$ so it means that the Hadamard product is increasing the rank of the original matrix $$A$$, but this sounds weird for me since we are not increasing the amount of rows and columns, so the upper bound for $$B$$ must be the same as for $$A$$. Thanks for any explanation about this.

• My I ask where did you read the upper bound $rank(B) \leq \binom{d+1}{2}$? Is there some references? Jan 16, 2023 at 17:42

The value $$\binom{d+1}{2}$$ is only greater than the maximal possible rank of $$B$$ if $$A$$ has sufficiently high rank. Not every $$m\ \times r$$ matrix has the maximal possible rank of $$\min\{m,r\}$$.
The point of the inequality is that if $$A$$ has a small enough rank so that $$\binom{d+1}{2} < \min\{n,r\}$$, then $$\operatorname{rank}(B) \leq \binom{d+1}2$$ gives us a non-trivial upper bound. In the case that $$\binom{d+1}{2} \geq \min\{n,r\}$$, this inequality doesn't give us any information about the rank of $$B$$ because, as you noted, we already know that $$\operatorname{rank}(B) \leq \min\{m,r\}$$.