Rank of Hadamard product with random matrices

I do research in statistics and am not sure whether the following is considered research level or not in mathematics. If it isn't, I'm happy because that means the answer is probably known and I can use it. If it is, any ideas for how to solve the problem are appreciated.

Suppose $Y$ is a real, $n \times k$ random matrix from a continuous distribution and that $Q$ is an $n\times n$ projection matrix of rank $q \leq n$. Let $C$ be a real, $n \times k$ matrix with strictly positive entries. What can be said about the rank of $(QY)\circ C$? Here, $\circ$ denotes the elementwise, or Hadamard, product.

I'm mostly interested in the nontrivial cases where $Q\neq I_n$ and the $C_{ij}$ are not all unity. My intuition tells me the rank is unchanged by the Hadamard product but I can't prove it, or find a proof in the literature.

• $\operatorname{rank}(QY \circ C) < \operatorname{rank}(QY) \times \operatorname{rank}(C)$ --- I don't think there is anything more to say in general. – Carlo Beenakker Aug 25 '17 at 7:31
• Also, the rank can indeed change via the Hadamard product. $A = \begin{bmatrix}1 & 2 \\ 2 & 1\end{bmatrix}$ and $B = \begin{bmatrix}2 & 1 \\ 1 & 2\end{bmatrix}$ both have rank $2$, but their Hadamard product has rank 1. – Nathaniel Johnston Aug 26 '17 at 21:21
• That's certainly true @NathanielJohnston. I was sloppy in my statement. I meant probabilistically, i.e. if $QY$ has rank $q$ almost surely, which I believe is true, then my intuition would say $(QY) \circ C$ also has rank $q$ almost surely. The intuition being that scaling r.v.s doesn't change the dependence between them, loosely speaking. Except it does of course, unless talking about correlation or some other standardized measure of dependence. – ekvall Aug 26 '17 at 23:54