# Rank of the element-wise square of a matrix

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 $$A$$.

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