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This is (quite obviously) inspired by this question.this question. 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.

This is (quite obviously) inspired by this question. 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.

This is (quite obviously) inspired by this question. 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.

fixed the statement.
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Igor Rivin
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This is (quite obviously) inspired by this question. Let $C_i$ be symmetric positive definite matrices. Then is it true that there is exactly one symmetric positive definite $X$ such that $F(X) = X^n - \sum_{i=0}^n C_i \circ X^i$ is positive definite$F(X) = X^n - \sum_{i=0}^n C_i \circ X^i = 0$, 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.

This is (quite obviously) inspired by this question. Let $C_i$ be symmetric positive definite matrices. Then is it true that there is exactly one symmetric $X$ such that $F(X) = X^n - \sum_{i=0}^n C_i \circ X^i$ is positive definite, 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.

This is (quite obviously) inspired by this question. 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.

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Igor Rivin
  • 96.4k
  • 11
  • 153
  • 366

A conjecturally easier version of matrix Descartes rule of signs.

This is (quite obviously) inspired by this question. Let $C_i$ be symmetric positive definite matrices. Then is it true that there is exactly one symmetric $X$ such that $F(X) = X^n - \sum_{i=0}^n C_i \circ X^i$ is positive definite, 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.