MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
up vote 7 down vote accepted

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.

share|cite|improve this answer
That's the right interpretation! – Igor Rivin Aug 14 '12 at 13:12
I fixed the question. – Igor Rivin Aug 14 '12 at 14:44

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.