Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. 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

I have a linear system $Ax=b$, where

  • $A$ is symmetric, positive semidefinite, and positive. $A$ is a variance-covariance matrix.
  • vector $b$ has elements $b_1>0$ and the rest $b_i<0$, for all $i \in \{2, \dots, N\}$.

Prove that the first component of the solution is positive, i.e., $x_1>0$.

Does anybody have any idea?

share|cite|improve this question
    
Thanks. What about if A is positive definite? Is in that case first component positive? I can also "normalize" diagonal elements to be equal to 1 since they are variances, and rest of elements to be in interval [0,1], which would then be correlation coefficients. – Branka Jul 3 '12 at 11:01

I don't think $x_1$ must be positive.

A counter example might be a positive definite matrix $A = [1 \space -0.2 ; \space -0.2 \space 1]$ with its inverse matrix $A^{-1}$ having $A_{11}, A_{12} > 0$.

- Edit: Sorry. A counter example might be a normalized covariance matrix

$ A= \left( \begin{array}{ccc} 1 & 0.6292 & 0.6747 & 0.7208 \\ 0.6292 & 1 & 0.3914 & 0.0315 \\ 0.6747 & 0.3914 & 1 & 0.6387 \\ 0.7208 & 0.0315 & 0.6387 & 1 \end{array} \right) $.

share|cite|improve this answer
    
IT seems that A is not only restricted to be positive definite, but also must have positive elements. – leonbloy Jul 5 '12 at 11:53
    
Thanks. Fixed it. – tatterdemalion Jul 5 '12 at 12:36

Your Answer

 
discard

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.