Worst case difference in rank by column-row swapping Given a matrix $m\in\{-1,+1\}^{n\times n}$. Consider $m^\sigma$ to be collection of all matrices obtained from $m$ by permuting rows and columns.
Consider $\mathscr{M}[m^\sigma]$ to be collection of all $n\times n$ matrices obtained from matrices in $m^\sigma$ by swapping an equal number of rows for an equal number of columns of same indices.
As an example, say you pick row/column indices $i$ and $j$. Then you include matrix where you first replace $i$th row with transpose of $i$th column and vice versa followed by similar operation on $j$th row and column.
What is worst case difference between least rank and largest rank of any matrix in $\mathscr{M}[m^\sigma]$?
Can we say anything about their ratios (such as bound based on some intrinsic property of the matrix)?
I am guessing there is a constant $c\in[1,4]$ such that ratio of largest rank to least rank is bounded above by $c$. My guess is $c=4$.
 A: Sorry for the multiple deletions and edits; I hope this partial answer will serve as partial penance.
Up to switching some rows and columns, your operation is 
$$\pmatrix{A&B\cr C&D\cr}\mapsto \pmatrix{A^T&C^T\cr B^T&D\cr}$$
where $A$ and $D$ are square.
Edited to add:  Since you still haven't clarified what "swapping" means, it's possible that $A^T$ should be $A$ above.  Fortunately, the argument to follow works either way.
I claim that if $A$ is invertible, then the worst-case ratio is $3$.  To see this, put $X=CA^{-1}B$, and note that the ranks of the original and transformed matrices are
$$rk(A)+rk(D-X)\qquad\hbox{and}\qquad rk(A)+rk(D-X^T)$$
Note that $rk(D-X)$ and $rk(D-X^T)$ are both bounded below by $|rk(D)-rk(A)|$ and above by $rk(D)+rk(A)$.    Treating the cases $rk(D)\le rk(A)$ and $rk(D)\ge rk(A)$ separately, it follows in either case that 
$${rk(A)+rk(D-X)\over rk(A)+rk(D-X^T)}\le 3$$
This settles the case where $A$ is invertible, and does not use the assumption that all entries are $\pm 1$.  Of course at the opposite extreme, when $A=0$, your ratio is equal to $1$.
A: Take a matrix of order $n$, with $n$ even, so that every entry is $+1$ except for the upper right quarter, which is a nonsingular matrix of order $n/2$.  The rank is either $n/2$ or $n/2+1$.  Now replace the first $n/2$ rows by the first $n/2$ columns.  The new matrix has rank $1$.  However, if the last $n/2$ rows are replaced by the first $n/2$ columns, the rank remains the same. So there is no uniform bound on the ratio you seek.
On the other hand, if the original matrix has full rank $n$, the new matrix has rank at least $n/2$, so it may be possible to save something by tightening the conditions. Either put a lower bound on the original rank or an upper bound on the number of rows replaced.
