choosing a square matrix to have a product with one 1 und other 0's Let $A$ be a $m\times n$ real matrix with maximal rank. Let $i\in\\{1,\dots,m\\}$, $j\in\\{1,\dots,n\\}$. I'm curious if it is possible (for any choice of $i,j$) to find a square matrix $B$ such that (at least) one of matrices $AB$, $BA$ has "1" on $(i,j)$ position und zeros on all other positions.

Yes, it is always possible.

Suppose that $m \leq n$. Let $e_i$ denote the column vector in $\Bbb R^m$ whose $i$th entry is a $1$ and whose other entries are $0$. Then since $A$ has maximal rank, the system $$ Ax = e_i $$ has a solution. Consider any such solution, and call this solution $x$. Let $B$ be the matrix with $m$ columns such that the $j$th column is $x$, and all other columns are just $0$s. Then we find that $AB$ will be an $m \times n$ matrix with a $1$ in the $i,j$ position, and $0$s everywhere else.

For the case of $n \leq m$, it suffices to take the previous solution and note that $$ (AB)^T = B^TA^T $$