Recently, while studying certain notions of "averaging" a set of input matrices, I obtained a nonlinear polynomial in matrix variables. A simple example is

\begin{equation*} \mathcal{G}(X) := X^n - \sum_{i=0}^{n-1} (C_iX^i + X^iC_i), \end{equation*}

where the $C_i$ are symmetric positive definite matrices.

If all the terms above were scalars, then Descartes' Rule of Signs tells us that $\mathcal G$ has exactly one positive root. This led me to wonder if a similar "rule" is also known for the above case.

Does there exist a unique symmetric positive definite solution to $\mathcal{G}(X)=0$?

EDIT If the $C_i$ commute with each other, we can simplify matters as follows. Since the $C_i$ commute, they can be simultaneously diagonalized by the same orthogonal matrix $U$; thus, let $C_i=U\Lambda_i U^T$. Suppose for a moment that $\mathcal{G}(X)=0$ does have a solution. Then, pre and post multiplying by $U^T$ and $U$, respectively, we see that this solution must satisfy

\begin{eqnarray*} U^TX^nU &=& \sum_{i=0}^{n-1} (U^TC_iUU^TX^iU + U^TX^iUU^TC_iU),\\\\ Y^n &=& \sum_{i=0}^{n-1} (\Lambda_iY^i + Y^i\Lambda_i), \end{eqnarray*} where $Y=U^TXU$. Now, if we pick $Y$ to be diagonal, then we see that indeed, for each diagonal entry we have a separate polynomial that has a unique positive root. Hence, we have a unique diagonal matrix $Y$. But as Mark Sapir alerted me in a comment below, it seems that having a unique diagonal $Y$ (and thus possibly non-diagonal $X=UYU^T$), does not yet rule out the possibility of other solutions.

Update After some hours of struggle due to the pressure of having posted my question on MO, under an additional assumption, I think I could show that the answer is "yes" (actually algorithmic). But my answer is not yet clean enough to be put on MO. If I manage to clean it up, I'll update my question.

Recently, while studying certain notions of "averaging" a set of input matrices, I obtained a nonlinear polynomial in matrix variables. A simple example is

\begin{equation*} \mathcal{G}(X) := X^n - \sum_{i=0}^{n-1} (C_iX^i + X^iC_i), \end{equation*}

where the $C_i$ are symmetric positive definite matrices.

If all the terms above were scalars, then Descartes' Rule of Signs tells us that $\mathcal G$ has exactly one positive root. This led me to wonder if a similar "rule" is also known for the above case.

Does there exist a unique symmetric positive definite solution to $\mathcal{G}(X)=0$?

EDIT If the $C_i$ commute with each other, we can simplify matters as follows. Since the $C_i$ commute, they can be simultaneously diagonalized by the same orthogonal matrix $U$; thus, let $C_i=U\Lambda_i U^T$. Suppose for a moment that $\mathcal{G}(X)=0$ does have a solution. Then, pre and post multiplying by $U^T$ and $U$, respectively, we see that this solution must satisfy

\begin{eqnarray*} U^TX^nU &=& \sum_{i=0}^{n-1} (U^TC_iUU^TX^iU + U^TX^iUU^TC_iU),\\\\ Y^n &=& \sum_{i=0}^{n-1} (\Lambda_iY^i + Y^i\Lambda_i), \end{eqnarray*} where $Y=U^TXU$. Now, if we pick $Y$ to be diagonal, then we see that indeed, for each diagonal entry we have a separate polynomial that has a unique positive root. Hence, we have a unique diagonal matrix $Y$. But as Mark Sapir alerted me in a comment below, it seems that having a unique diagonal $Y$ (and thus possibly non-diagonal $X=UYU^T$), does not yet rule out the possibility of other solutions.

Update After some hours of struggle due to the pressure of having posted my question on MO, under the additional assumption that each term in the sum is positive definite, the answer is "yes". But this assumption seems to be way too strong...

Recently, while studying certain notions of "averaging" a set of input matrices, I obtained a nonlinear polynomial in matrix variables. A simple example is

\begin{equation*} \mathcal{G}(X) := X^n - \sum_{i=0}^{n-1} (C_iX^i + X^iC_i), \end{equation*}

where the $C_i$ are symmetric positive definite matrices.

If all the terms above were scalars, then Descartes' Rule of Signs tells us that $\mathcal G$ has exactly one positive root. This led me to wonder if a similar "rule" is also known for the above case.

Does there exist a unique symmetric positive definite solution to $\mathcal{G}(X)=0$?

EDIT If the $C_i$ commute with each other, we can simplify matters as follows. Since the $C_i$ commute, they can be simultaneously diagonalized by the same orthogonal matrix $U$; thus, let $C_i=U\Lambda_i U^T$. Suppose for a moment that $\mathcal{G}(X)=0$ does have a solution. Then, pre and post multiplying by $U^T$ and $U$, respectively, we see that this solution must satisfy

\begin{eqnarray*} U^TX^nU &=& \sum_{i=0}^{n-1} (U^TC_iUU^TX^iU + U^TX^iUU^TC_iU),\\\\ Y^n &=& \sum_{i=0}^{n-1} (\Lambda_iY^i + Y^i\Lambda_i), \end{eqnarray*} where $Y=U^TXU$. Now, if we pick $Y$ to be diagonal, then we see that indeed, for each diagonal entry we have a separate polynomial that has a unique positive root. Hence, we have a unique diagonal matrix $Y$. But as Mark Sapir alerted me in a comment below, it seems that having a unique diagonal $Y$ (and thus possibly non-diagonal $X=UYU^T$), does not yet rule out the possibility of other solutions.

Recently, while studying certain notions of "averaging" a set of input matrices, I obtained a nonlinear polynomial in matrix variables. A simple example is

\begin{equation*} \mathcal{G}(X) := X^n - \sum_{i=0}^{n-1} (C_iX^i + X^iC_i), \end{equation*}

where the $C_i$ are symmetric positive definite matrices.

If all the terms above were scalars, then Descartes' Rule of Signs tells us that $\mathcal G$ has exactly one positive root. This led me to wonder if a similar "rule" is also known for the above case.

Does there exist a unique symmetric positive definite solution to $\mathcal{G}(X)=0$?

EDIT If the $C_i$ commute with each other, we can simplify matters as follows. Since the $C_i$ commute, they can be simultaneously diagonalized by the same orthogonal matrix $U$; thus, let $C_i=U\Lambda_i U^T$. Suppose for a moment that $\mathcal{G}(X)=0$ does have a solution. Then, pre and post multiplying by $U^T$ and $U$, respectively, we see that this solution must satisfy

\begin{eqnarray*} U^TX^nU &=& \sum_{i=0}^{n-1} (U^TC_iUU^TX^iU + U^TX^iUU^TC_iU),\\\\ Y^n &=& \sum_{i=0}^{n-1} (\Lambda_iY^i + Y^i\Lambda_i), \end{eqnarray*} where $Y=U^TXU$. Now, if we pick $Y$ to be diagonal, then we see that indeed, for each diagonal entry we have a separate polynomial that has a unique positive root. Hence, we have a unique diagonal matrix $Y$. But as Mark Sapir alerted me in a comment below, it seems that having a unique diagonal $Y$ (and thus possibly non-diagonal $X=UYU^T$), does not yet rule out the possibility of other solutions.

Update After some hours of struggle due to the pressure of having posted my question on MO, under an additional assumption, I think I could show that the answer is "yes" (actually algorithmic). But my answer is not yet clean enough to be put on MO. If I manage to clean it up, I'll update my question.

Recently, while studying certain notions of "averaging" a set of input matrices, I obtained a nonlinear polynomial in matrix variables. A simple example is

\begin{equation*} \mathcal{G}(X) := X^n - \sum_{i=0}^{n-1} (C_iX^i + X^iC_i), \end{equation*}

where the $C_i$ are symmetric positive definite matrices.

If all the terms above were scalars, then Descartes' Rule of Signs tells us that $\mathcal G$ has exactly one positive root. This led me to wonder if a similar "rule" is also known for the above case.

Does there exist a unique symmetric positive definite solution to $\mathcal{G}(X)=0$?

Note: Obviously, if the $C_i$ commute with each other, we can simultaneously diagonalize everything and apply the rule of signs to conclude uniqueness. I'm hoping the same conclusion also holds for the general case.

Recently, while studying certain notions of "averaging" a set of input matrices, I obtained a nonlinear polynomial in matrix variables. A simple example is

\begin{equation*} \mathcal{G}(X) := X^n - \sum_{i=0}^{n-1} (C_iX^i + X^iC_i), \end{equation*}

where the $C_i$ are symmetric positive definite matrices.

If all the terms above were scalars, then Descartes' Rule of Signs tells us that $\mathcal G$ has exactly one positive root. This led me to wonder if a similar "rule" is also known for the above case.

Does there exist a unique symmetric positive definite solution to $\mathcal{G}(X)=0$?

EDIT If the $C_i$ commute with each other, we can simplify matters as follows. Since the $C_i$ commute, they can be simultaneously diagonalized by the same orthogonal matrix $U$; thus, let $C_i=U\Lambda_i U^T$. Suppose for a moment that $\mathcal{G}(X)=0$ does have a solution. Then, pre and post multiplying by $U^T$ and $U$, respectively, we see that this solution must satisfy

\begin{eqnarray*} U^TX^nU &=& \sum_{i=0}^{n-1} (U^TC_iUU^TX^iU + U^TX^iUU^TC_iU),\\\\ Y^n &=& \sum_{i=0}^{n-1} (\Lambda_iY^i + Y^i\Lambda_i), \end{eqnarray*} where $Y=U^TXU$. Now, if we pick $Y$ to be diagonal, then we see that indeed, for each diagonal entry we have a separate polynomial that has a unique positive root. Hence, we have a unique diagonal matrix $Y$. But as Mark Sapir alerted me in a comment below, it seems that having a unique diagonal $Y$ (and thus possibly non-diagonal $X=UYU^T$), does not yet rule out the possibility of other solutions.