Hadamard product decomposition with lower rank matrices

Question

Given integers $k$ and $l$ and a matrix $A$ of rank $kl$, can we always find a matrix $B$ of rank $k$ and a matrix $C$ of rank $l$, such that $A$ is the Hadamard product of $B$ and $C$, namely $A=B \odot C$?

For example, when $k=2$ and $l=2$ and $A$ is the 4 by 4 identity matrix,

\begin{equation*} A = \begin{bmatrix} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1\\ \end{bmatrix} \end{equation*}

we can find $B$ of rank 2 and $C$ of rank 2:

\begin{equation*} B = \begin{bmatrix} 1 & 1 & 0 & 0\\ 1 & 1 & 0 & 0\\ 0 & 0 & 1 & 1\\ 0 & 0 & 1 & 1\\ \end{bmatrix}, C = \begin{bmatrix} 1 & 0 & 1 & 0\\ 0 & 1 & 0 & 1\\ 1 & 0 & 1 & 0\\ 0 & 1 & 0 & 1\\ \end{bmatrix} \end{equation*}

satisfying $A=B \odot C$.

Can we find $B$ and $C$ in the general case for any given $A$ and $k$, $l$?

Answer 1

There is a new paper published in May 2024 claiming that it cannot be done for arbitrary matrices. They did not give a specific example, but only provided high level intuitions. But their argument looked pretty convincing. https://link.springer.com/article/10.1007/s10618-024-01033-y