Let $Q$ be a orthogonal matrix, i.e. $QQ^T=I$
I know there exist upper bound of the rank of Hadamard product ($\operatorname{rank}(A \circ B) < \operatorname{rank}(A) \times \operatorname{rank}(B)$)
But is there any lower bound of the rank of its Hadamard square, $\operatorname{rank}(Q \circ Q)$?
Best Answer
If $H$ is a Hadamard matrix then $H\circ H$ is the all-ones matrix, with rank 1. So there's your lower bound :-(
Also, if $Q$ has rank $d$, the rank of $Q\circ Q$ is at most $\binom{d+1}2$. (Here $Q$ need not be orthogonal.)