Consider $n$ convex polytopes $S_1, \cdots, S_n$ and a set of matrices $W$ such that each matrix $A\in W$, we have that the $i$-th row of $A$ is a member of $S_i$. (In general $W$ is infinite.)

Furthermore, given a vector $e$, I want a finite set $F$ of vectors from $W$ and $e$, such that for any $x,y$

$\{xAe \mid A\in W\} = \{yAe \mid A\in W\}$ iff $x \cdot f =y\cdot f$ for each $f\in F$.

Is there a simple way to come up with such an $F$?

  • 1
    $\begingroup$ You really couldn't think of a more descriptive title?? $\endgroup$ Sep 18, 2014 at 14:39
  • $\begingroup$ I am sorry, but I do not know how to classify this question. Do you have any suggestion? $\endgroup$
    – user157989
    Sep 18, 2014 at 15:44
  • $\begingroup$ Are all the polytopes of the same dimension? Are the matrices of the same dimension? Is $W$ a set of matrices or a set of vectors? How can $W$ be infinite? Where do $x$ and $y$ live? What is the meaning of $xAe$? $\endgroup$ Sep 18, 2014 at 15:52
  • $\begingroup$ the polytopes are not necessarily of the same dimension. The matrices in W are of the same dimension. W is a set of matrices. Because each matrix $A\in W$ could have different rows from different convex polytopes. $x,y$ are row vectors and $e$ is a column vector, this explains $xAe$ $\endgroup$
    – user157989
    Sep 18, 2014 at 15:57
  • $\begingroup$ I'm confused. The left equality seems to be an equality of intervals of the reals ($x,y,e$ are fixed and the $A$s vary over a convex set), depending on $x,y$. But the right equalities are saying that $x-y$ is in some linear subspace. It is not clear to me at all that the left equality only depends on $x-y$, or that it is a linear condition even in $(x,y)$. $\endgroup$ Sep 18, 2014 at 17:12


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