For $A$ and $B$ being matrices of the same dimension and $B$ being rank $1$, one knows that $det(A+tB)$ is a linear polynomial in $t \in \mathbb{R}$. Hence by Taylor series it follows that $det(A + tB) = (1 + t\partial_p)det(A+pB) \vert _{p=0}$. Now (coincidentally!?) it so happens that this operator $(1+ t\partial_z)$ is an operator that preserves real-stability of complex multi-variable polynomials. [for proof see : https://blogs.princeton.edu/sas/2014/10/05/lectures-2-and-3-proof-of-kadison-singer-1/ ]

  • Is this a freak coincidence for rank-1 matrices (like $B$) of is there some generalization of it?

For example one could take a rank-2 matrix $B$ and similarly argue (using matrix determinant lemma and Shermann-Morrison formula) that $det(A + tB)$ is a quadratic polynomial in $t \in \mathbb{R}$. Hence by running a similar argument as above one would get the operator, $(1+t\partial_z + \frac{t^2}{2}\partial_z^2)$. But this does not seem to be an operator which preserves real-stability (at least it doesn't seem to satisfy that its symbol polynomial (i.e $1 -tz + \frac{t^2}{2}z^2$) is real stable (for all $t$) as gleaned from the results here, http://annals.math.princeton.edu/2009/170-1/p14)

[..or am I reading this Borcea-Branden paper wrong?..]

  • Is there a hack around this which generalizes beyond rank $=1$?

A related MO discussion, Differential operators that preserve real-rootedness

  • $\begingroup$ Any $2 \times 2$ matrix has rank $2$, and real $2$ matrices can have non-real eigenvalues. Why would you expect real stability to be preserved? $\endgroup$ – Robert Israel Feb 19 '15 at 18:41
  • $\begingroup$ @RobertIsrael I am kind of looking for the converse - say $B$ is a rank-2 matrix (and $A$ and $B$ are of the same dimension) then can there be a differential operator (in $z$) which acts on $det(A+zB)$ and gives another real-stable polynomial provided $det(A+zB)$ was real stable to begin with? $\endgroup$ – user6818 Feb 19 '15 at 19:55
  • $\begingroup$ @RobertIsrael I used the Taylor series idea to motivate how such an operator might be found...(it workes for rank-1 but not higher) $\endgroup$ – user6818 Feb 19 '15 at 19:56

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.