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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...

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This may be a stupid question, but while it is obvious that $\mathcal{G}(X)$ is symmetric when $X = I,$ it is not so clear it is ever symmetric otherwise, unless the system is very special. Am I missing something? –  Igor Rivin Aug 13 '12 at 21:53
    
@Igor: I think $X$ is supposed to be symmetric too. Then, since $C_i$ are symmetric, the sum is also symmetric. –  Mark Sapir Aug 13 '12 at 21:59
    
Aha, I see, I was careless reading the equation! –  Igor Rivin Aug 13 '12 at 23:15
    
@Suvrit: Could you explain why if all $C_i$ pairwise commute, you can diagonalize everything? You cannot assume that $X$ commutes with $C_i$. –  Mark Sapir Aug 14 '12 at 1:14
    
@Mark: Hmm...if all the $C_i$ commute, we diagonalize them, and then we see that there exists a unique diagonal matrix that satisfies the equation. But you are right, just because we have a unique diagonal matrix, does not mean that the solution to the original equation is unique. I'll edit the question to work in these details. Thanks. –  Suvrit Aug 14 '12 at 7:35
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