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    $\begingroup$ Another sufficient condition is that $A$ has 3 identical/proportional lines (or columns), as then, for each submatrix $B$, either $B$ or $\bar B$ has 2 identical/proportional lines and thus is singular. In this case, the rank of $A$ can be as much as $n-2$. We may still consider this as a trivial case, so I guess the (most interesting part of the) question is whether such a matrix exists with rank $n-1$. $\endgroup$ Commented Jan 3 at 19:34
  • $\begingroup$ @Wolfgang Rank $n-1$ is possible with a zero row or column. I suppose you mean to ask if rank $n-1$ is possible without a zero row or column? $\endgroup$ Commented Jan 4 at 6:44
  • $\begingroup$ As a class of examples that contains both Lchencz's and Wolfgang's: If for some $k$ the matrix contains a set of $k$ rows or $k$ columns whose span has dimension strictly less than $k/2$, then the matrix has the desired property. $\endgroup$ Commented Jan 4 at 8:57
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    $\begingroup$ They can be both zero. I revise it into Det(B)Det(\bar B)=0. @Rodrigo de Azevedo $\endgroup$ Commented Jan 5 at 0:29
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    $\begingroup$ Here is a sufficient condition that includes Kevin P. Costello's condition and Wolfgang's second condition: If $A$ has an $a \times b$ submatrix of rank $r< \frac{ a+ b -n}{2}$ then $A$ has this property. Indeed, a $c \times d$ submatrix of an $m \times m$ invertible matrix has rank at least $c+d-m$. If $B$ and $\bar{B}$ are invertible than adding this formula applied to $B$ to this formula applied to $\bar{B}$ gives $2r \geq a+b-n$. $\endgroup$ Commented Jan 6 at 20:16