Handelman's positivstellensatz for symmetric matrix-valued polynomials

Question

For certain classes of sets $S \subseteq \mathbb{R}^n$, there exist algebraic characterizations of real valued polynomials $p: \mathbb{R}^n \rightarrow \mathbb{R}$ that are positive on $S$. Several of these algebraic characterizations, known as positivstellensatz, are listed here in Wikipedia.

Some of these positivstellensatze can be generalized for matrix polynomials, i.e. a function $P$ from $\mathbb{R}^n$ to the set of all $r \times r$ symmetric matrices whose entries are polynomials. In this case, $P$ is understood to be positive in $S$ if $P(x)$ is positive definite for all $x \in S$.

For instance, in [1], Pólya's positivstellensatz for homogeneous polynomials is generalized to homogeneous matrix polynomials and in [2], Putinar's positivstellensatz is also generalized to matrix polynomials. Stengle's and Schweighofer's positivstellensatze are also generalized to matrix polynomials in [3].

Does Handelman's positivstellensatz have also a matrix polynomial version?

[1]: Scherer, C. W. "Relaxations for robust linear matrix inequality problems with verifications for exactness." SIAM Journal on Matrix Analysis and Applications 27(2) (2005): 365-395.

[2]: Scherer, C. W. and Hol, C. W. J. "Matrix sum-of-squares relaxations for robust semi-definite programs." Mathematical programming 107(1-2) (2006): 189-211.

[3]: Công-Trình, Lê. "Some Positivstellensätze for polynomial matrices." arXiv preprint arXiv:1403.3783 (2014)

Answer 1

The obvious generalization is false: Consider $A=\begin{pmatrix} 1+x &y \\y & 1-x \end{pmatrix}$ which is a linear matrix polynomial. It defines a compact set in the plane, namely the unit disc. The algebra $\mathbb{R}[A]$ of all polynomial expressions in $A$ is contained in $\mathbb{R}(x,y)[A]$ which is a two dimensional $\mathbb{R}(x,y)$-vector space since $A^2=2A+x^2+y^2-1$. Therefore $\mathbb{R}[A]$ can not contain all of these three $\mathbb{R}(x,y)$-linear independent matrices: $\begin{pmatrix}1 & 0 \\ 0 & 2 \end{pmatrix}$, $\begin{pmatrix}2 & 0 \\ 0 & 1 \end{pmatrix}$, $\begin{pmatrix}2 & 1 \\ 1 & 2 \end{pmatrix}$ which are all positive definite on the unit disc.

Edit: So Handelman's statement is the following: If the linear forms $l_1, \ldots, l_r$ define a compact set and if the polynomial $f$ is strictly positive on this set, then $f$ can be expressed as a sum of products of $l_1, \ldots, l_r$ with positive coefficients.

Now replace "(linear) polynomials" by "(linear) matrix polynomials" and you get a generalization to matrix polynomials which is false. The matrix $A$ plays hereby the role of $l_1, \ldots, l_r$.