This is (quite obviously) inspired by [this question.](https://mathoverflow.net/questions/104645/descartes-rule-of-signs-for-noncommutative-polynomia) Let $C\_i$ be symmetric positive definite matrices. Then is it true that there is exactly one symmetric positive definite $X$ such that 
$F(X) = X^n - \sum\_{i=0}^n C\_i \circ X^i = 0$, where $\circ$ denotes the Schur (component-wise) product (and exponentiation is with respect to that same product) Notice that unlike in the inspiring question, the Schur product is commutative.



A conjecturally easier version of matrix Descartes rule of signs.
I suspect you mean "is it true that there is exactly one symmetric $X$ such that $F(X) = X^n - \sum\_{i=0}^{n-1} C\_i \circ X^i = 0$", which is more in line with the inspiration question. If you meant something else, I'll just delete this later today.


In that case, the answer is no. Take 
$$
C\_2 = 
\begin{bmatrix}
1 & 2 \\\
2 & 5
\end{bmatrix}
\quad
C\_1 = 
\begin{bmatrix}
1 & 1 \\\
1 & 2
\end{bmatrix}
\quad
C\_0 = 
\begin{bmatrix}
2 & -2 \\\
-2 & 24
\end{bmatrix}
$$
Then we have four polynomials defining the entries in the matrix $X$ (really three by symmetry). The two diagonal entries are unique by Descartes rule of signs, being the real solutions to $z^3 -z^2 -z - 2 =0$, and $z^3 -5z^2 -2z - 24 =0$, so $2$ and $6$ - but the off diagonal entries are the solutions to $z^3 - 2z^2 -z +2 = 0$, so they can be $2, 1$ or $-1$. Each of these gives a positive definite matrix. 


