From: israel@math.ubc.ca (Robert Israel)
Subject: Re: Matrix Inequality
Date: 11 Aug 2000 20:34:20 GMT
Newsgroups: sci.math.research
Summary: [missing]

In article <3992C6E9.2F2D494E@stat.unibe.ch>,
Enkelejd Hashorva  <enkelejd.hashorva@stat.unibe.ch> wrote:

>is there a characterization for symmetric  matrices $\Sigma$ of size $d
>\times d$ such that
>there exists a vector $\x  \in R^d$ with strictly positive elements, so
>that

>$$ \Sigma \x > (0, \cdots 0)$$
>holds.

I assume that by \x > \y for vectors \x,\y you mean x_i > y_i for all i.

There is such a vector \x if and only if the only vector \y >= 0 with
\Sigma \y <= 0 is \y = 0.  
This is related to duality in linear programming, and it holds more
generally for rectangular matrices with \Sigma \y replaced by \Sigma^T
y.  See e.g. V. Chvatal, "Linear Programming", exercise 16.10.

>When $\Sigma$ is positive definite for example, it is easy to show that
>for some $\x^*$
>$$\Sigma \x^* \ge (0, \cdots 0)$$

>I wonder if the inequality can be made strict at least for this example.

Yes, by the criterion above.  If \y >= 0 and \Sigma \y <= 0 then 
\y^T \Sigma \y <= 0, and if \Sigma is positive definite this is only
possible for \y = 0.

A slight extension of this: if \Sigma is positive semidefinite and
the only \y with \Sigma \y = 0 and \y >= 0 is \y = 0, then again
your \x exists. 

Robert Israel                                israel@math.ubc.ca
Department of Mathematics        http://www.math.ubc.ca/~israel 
University of British Columbia            
Vancouver, BC, Canada V6T 1Z2