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I have a system of $n$ quadratic equations with $n$ unknowns. It can be written as

$diag(x)Ax=1$

$x$ is an $n$-vector, $A$ is $n\times n$, real, symmetric and positive definite, the diagonal elements of $A$ are strictly positive, other elements of $A$ are arbitrary, $1$ is a vector of ones and $diag(x)$ is just s shorthand for taking a vector and putting it on the diagonal of a matrix, the rest of the entries are all zeros.

I am interested in the following question:

  1. How many strictly positive solutions exist (in general)?

Any help (examples etc). is greatly appreciated!

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  • $\begingroup$ Do the entries of $A$ have known signs, or can they be arbitrary? $\endgroup$ – Federico Poloni Mar 3 '16 at 19:24
  • $\begingroup$ The diagonal elements are strictly greater than zero, for the rest they can be arbitrary. $\endgroup$ – Andy Mar 3 '16 at 19:35
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Your system says $(Ax)_i = x_i^{-1}$. Suppose you had two distinct, positive solutions $x, y$. Then

$$ \eqalign{(x-y)^T A (x-y) &= \sum_{i} (x-y)_i (A (x-y))_i\cr & = \sum_i (x_i-y_i)(x_i^{-1} - y_i^{-1})\cr &= - \sum_{i} \dfrac{(x_i - y_i)^2}{x_i y_i} < 0} $$ which contradicts the assumption that $A$ is positive definite. So there can only be at most one positive solution.

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  • $\begingroup$ Is there always at one positive solution? I think so. $\endgroup$ – Andy Mar 3 '16 at 20:56

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