1
$\begingroup$

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!

$\endgroup$
2
  • $\begingroup$ Do the entries of $A$ have known signs, or can they be arbitrary? $\endgroup$ Mar 3, 2016 at 19:24
  • $\begingroup$ The diagonal elements are strictly greater than zero, for the rest they can be arbitrary. $\endgroup$
    – Andy
    Mar 3, 2016 at 19:35

1 Answer 1

5
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ Is there always at one positive solution? I think so. $\endgroup$
    – Andy
    Mar 3, 2016 at 20:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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