Linear system with positive semidefinite matrix 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?

Someone bumped up this old question today. For sake of having an answer, we will see that the problem statement arises from a classical property of $M$-matrices.
Let $D=\operatorname{diag}(1,-1,\ldots,-1),\ M=DAD+tI,\ y=Dx$ and $q=Db+ty$. Note that $y_1$ has the same sign as $x_1$ and $DAD$ is positive semidefinite. Pick a sufficiently small $t>0$. Then $q$ is positive and $M$ is positive definite (hence nonsingular). However, as $M=DAD+tI$, all off-diagonal entries of $M$ are negative. This makes $M$ an $M$-matrix. Now the problem boils down to the following known property of $M$-matrices:

Suppose $M$ is a matrix whose eigenvalues have positive real parts and its off-diagonal entries are negative. Then $M^{-1}>0$ (entrywise). Consequently, if $My=q>0$, then $y>0$.

Proof. See e.g. Horn and Johnson's Topics in Matrix Analysis. The usual proof is very easy. By the given assumptions on $M$, when $\alpha>0$ is sufficiently large, $P=\alpha I - M$ is nonsingular and (entrywise) positive and hence
$$
M^{-1}=\frac1\alpha\left(I-\frac1\alpha P\right)^{-1}
=\frac1\alpha\left(I+\frac1\alpha P+\frac1\alpha^2 P^2+\ldots\right)>0.
$$
