0
$\begingroup$

Suppose $A$ be a skew-symmetric matrix whose entries are positive numbers. A perturbation of $A$, $A'$, is obtained by adding another skew-symmetric matrix whose entries are positive integers.

My question is for all such $A$ is there any perturbation $A'$ such that all Pfaffian minors of $A'$ are positive?

By definition a Pfaffian minor of a skew symmetric matrix $A$ is a Pfaffian of a submatrix of $A$ consisting of rows and columns indexed by $i_1, i_2, ..., i_{2t}$ for some $i_1 < i_2 < ... < i_{2t}$.

$\endgroup$
  • $\begingroup$ By positive you mean positive above the diagonal right? But doesn't this question just come down to "is there a matrix $B$ all of whose Pfaffians are nonnegative"? Because, if there is, we can choose $t$ large enough that $tB-A$ is positive, and then $tB$ is a perturbation of $A$ by your definition. I assume you left something out of the conditions. $\endgroup$ – David E Speyer Dec 6 '16 at 18:28
  • $\begingroup$ I really meant for a fixed definition of pfaffian, the pfaffian (as a number) is positive for all those minors. $\endgroup$ – SiOn Dec 6 '16 at 18:38
  • $\begingroup$ Okay. But your definition of perturbation is that $A'$ is a perturbation of $A$ if $A'-A$ is nonnegative, with no hypothesis that $A'-A$ be small. So, take any positive $B$ and then, for any $A$, if we take $t$ sufficiently large, $tB$ will be a perturbation of $A$. Is that what you wanted to ask for? $\endgroup$ – David E Speyer Dec 6 '16 at 18:44
  • $\begingroup$ Sorry I didn't get your question. Perturbation is done by skew symmetric matrices with positive integers entires while $A$ could be any skew symmetric matrix with positive real entries. $\endgroup$ – SiOn Dec 6 '16 at 18:51
  • 1
    $\begingroup$ If you have $A' = A + t D$, then Pfaffians of $A'$ will have as their leading terms as $t\to \infty$ the Pfaffians of $D$ times the appropriate power of $t$. So positivity for $A'$ follows from that of $D$. $\endgroup$ – Lev Borisov Dec 13 '16 at 20:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.