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
