I have read in this paper http://dx.doi.org/10.4007/annals.2015.182.1.8 that for any real stable polynomial $p \in \mathbb{R}[z_1,z_2]$ of degree exactly $d$, there exists $d \times d$ symmetric positive semidefinite matrices $A$ and $B$ and a symmetric matrix $C$ such that \begin{equation*} p(z_1,z_2)=\pm \det (z_1A+z_2B+C). \end{equation*} Here real stable just means that $p$ is stable and has real coefficients. In the same paper it is stated that: For any $z_1,z_2>0$, $z_1A+z_2B$ is positive definite (*). They argue by contradiction, and say that if (*) does not hold, there is a non-zero vector in the null-space of both $A$ and $B$. This I get. Then they claim that this implies that the degree of $p$ must be less then $d$, arriving at a contraction. My question is: Why is this last implication true?
I guess that the idea is to relate the determinant expression to the coefficients of the highest degree monomials of $p$. I tried expanding $p(z_1,z_2)=\det(z_1A+z_2B+C)$ using the definition involving permutations, and if I am not mistaking, this gives something of the form $\det(A)z_1^d+\det(B)z_2^d+\sum_{i=1}^{d-1}k_iz_1^{d-i}z_2^{i}$ (plus monomials in $z_1,z_2$ of degree less than $d$) for some coefficients $k_i$ that depend on the matrices $A, B$ and $C$. I know that $\det(A)=0=\det(B)$, but don't we also need to know that all the $k_i$'s are zero in order to conclude that $p$ has degree less than $d$? I tried to find expressions for the $k_i$'s involving $\det(A), \det(B)$, but without luck...
Is there some sort of formula for these coefficients that could be usefull? Or is there a better/easier way to see why (*) is true? Chances are that I am overlooking something obvious, and any kind of help would be appreciated.
Thank you in advance.
If $A$ and $B$ have a common non-zero vector in the nullspace, then $p(z_1,z_2)=\pm \det (z_1A+z_2B+C)$ indeed has degree less than $d$. We can see this as follows:
Suppose that $x$ is a unit-vector and that $Ax = Bx = 0$. Let $U$ be an orthogonal matrix whose first column is $x$. Note (or show) that $U^TAU$ and $U^TBU$ have zeros in their first row and column. We see that $$ \det (z_1A+z_2B+C) = \det (U^T[z_1A+z_2B+C]U) \\ = \det(z_1[U^TAU] + z_2[U^TBU] + U^TCU). $$ Since the matrix $M(z_1,z_2) = z_1[U^TAU] + z_2[U^TBU] + U^TCU$ has no variables in its first row, we can conclude that its determinant has (total) degree at most $d-1$.