TITLE: What kind of matrix $A$ satisfies $Ax\geq 0\Rightarrow x\geq 0$?
QUESTION [7 upvotes]: $A\in \mathbb{R}^{n\times n}$ is an n-by-n matrix. $x=(x_1,x_2,\ldots ,x_n)\in \mathbb{R}^n$ is a vector. $x\geq 0$ means $x_i\geq 0,\forall i$. 
Q1: When $A$ satisfies what conditions, $\forall x\in\mathbb{R}^n ,x\geq 0\Rightarrow Ax\geq 0$?
Q2: When $A$ satisfies what conditions, $\forall x\in\mathbb{R}^n ,Ax\geq 0\Rightarrow x\geq 0$?
Q1 is solved by Martin Argerami(The matrix's entries are all positive). But what are the matrices in Q2?
In Q2, the answer is not "entries all positive". The counterexample is $A=\begin{pmatrix} 1 & 1 \\ 1 & 1\end{pmatrix}$, $A\begin{pmatrix} -1 \\ 2 \end{pmatrix}\geq 0$.
Let's see an example $A=\begin{pmatrix} 2 & -1 \\ -1 & 2 \end{pmatrix}$.$A\begin{pmatrix} x \\ y \end{pmatrix} \geq 0\Rightarrow x\geq 0,y\geq 0$. So $A$ is in answer to Q2. What general properties of matrix $A$ would ensure Q2 be satisfied?
Another obvious class of matrix in Q2 is diagnoal matrices with positive diagonal entries.
Q2 is settled by Robert Israel($A^{-1}$ has entries all positive).

REPLY [7 votes]: For $Ax \ge 0$ implies $x \ge 0$, a necessary and sufficient condition is that $A$ is invertible and all elements of $A^{-1}$ are nonnegative.
Proof: If all elements of $A^{-1}$ are nonnegative, then $Ax \ge 0$ implies $x = A^{-1} (A x) \ge 0$.  
Conversely, if $A$ is not invertible, there is $v \ne 0$ with $Av = 0$, and either $v$ or $-v$ has a negative entry but $Av = A(-v) = 0 \ge 0$.
If $A$ is invertible and $(A^{-1})_{ij} < 0$, then take $x = A^{-1} e_j$ so that $x_i < 0$, but $A x = e_j \ge 0$.
EDIT: to the added question about diagonally dominant matrices, the answer is no.  For example, $$\pmatrix{2& 1\cr 0& 2\cr}$$ is diagonally dominant, but its inverse has a negative entry.