I am looking for a real matrix A of rank r with non-negative entries with the following property :
For every complex matrix B such that $B\circ \overline B=A$, the rank of B is strictly greater than r.
Here what I denote $B\circ \overline B$ is the matrix with entries equal to the square of the absolute value of the entries of $B$ (element-wise operations). In the case of real matrices, this would be the square element-wise of $B$, denoted $B^{\circ 2}$.
I solve below the case where B is restricted to be a real matrix , but would be interested in solving the problem where B can be a complex matrix, or showing that it has no solution.
Consider $A=\left(\begin{matrix} 2 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \\ \end{matrix}\right),$ it is clear that $A$ has rank 2. Now consider all real matrices B such that $A=B^{\circ 2}$, these matrices can be written, up to arbitrary signs on each entries, as $B=\left(\begin{matrix} \sqrt{2} & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \\ \end{matrix}\right),$ and these matrices all have rank 3, which solves the problem.
However if we allow B to be a complex matrix, there are infinitely many matrices such that $B\circ \overline B=A$, since the previous matrix elements can have any phase. In particular $B=\left(\begin{matrix} 1+i & 1 & 1 \\ 1 & 0 & 1 \\ i & 1 & 0 \\ \end{matrix}\right)$ is such a matrix with rank 2, so the property doesn't hold for $A$.