TITLE: The minimum rank of a matrix with a given pattern of zeros
QUESTION [5 upvotes]: For real matrices $A=(a_{ij})$ and $B=(b_{ij})$ of the same size, I write $A\prec B$ if $a_{ij}=0$ whenever $b_{ij}=0$.
If 
  $$ B = \begin{pmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \end{pmatrix}^{\otimes10}, $$
then the matrix 
  $$ A = \begin{pmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ -1 & 0 & 1 \end{pmatrix}^{\otimes10} $$ 
satisfies $\mathrm{rk}(A)<\mathrm{rk}(B)$ and $I\prec A\prec B$ (where $I$ is the identity matrix of order $3^{10}$) . Does there exist a square matrix $C$ of order $3^{10}$ such that $\mathrm{rk}(C)<\mathrm{rk}(A)$ and $I\prec C\prec A$?

REPLY [4 votes]: As Misha Muzychuck has observed, the answer is "no": since 
  $$ \begin{pmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ -1 & 0 & 1 \end{pmatrix} $$ contains a non-degenerate upper-triangular submatrix of size $2$, the matrix $A$ contains a non-degenerate upper-triangular submatrix of size $2^{10}$, whence $\mathrm{rk}(C)\ge 2^{10}=\mathrm{rk}(A)$ for any matrix $C$ with $I\prec C\prec A$.