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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 $A$.

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    $\begingroup$ As it's written, your $A \neq B^{\circ 2}$, and $A$ is invertible. $\endgroup$ – Matthew Leingang Feb 20 at 18:25
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    $\begingroup$ I think he means $A = [[2 1 1], [1 0 1], [1 1 0]]$ instead $\endgroup$ – MoonKnight Feb 20 at 18:35
  • $\begingroup$ @MoonKnight I think you're right. For a minute I was calculating that $B^{\circ 2}$ was invertible, but I had it wrong. $\endgroup$ – Matthew Leingang Feb 20 at 18:44
  • $\begingroup$ Sorry for the typo, thanks for noticing it, it is now fixed. $\endgroup$ – cestyx Feb 20 at 19:45

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