What kind of matrix $A$ satisfies $Ax\geq 0\Rightarrow x\geq 0$? $A\in \mathbb{R}^{n\times n}$ 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).
 A: 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.
A: If you question is which matrices $A$ will make $Ax\geq0$ for every $x\geq0$, then the answer is all matrices such that all their entries are nonnegative (we can write that as $A\geq0$). 
Indeed, if $Ax\geq0$, for all $x$, then let $e_j=(0,\ldots,0,1,0,\ldots,0)$ where the $1$ is in the $j^{\rm th}$ place. Then $0\leq Ae_j=(a_{1j},a_{2j},\ldots,a_{nj})$, so every entry in the column $j$ of $A$ is non-negative. As $j$ was arbitrary, $A\geq0$. 
Conversely, if $A\geq0$ and $x\geq0$, then the $j^{\rm th}$ entry of $Ax$ is $\sum_{k}a_{jk}x_k$. As every term in the sum is non-negative, so is the sum. So every entry of $Ax$ is non-negative, i.e. $Ax\geq0$. 
A: Let $A_i= (A_{i1}, \cdots ,A_{in})$ the $i=1,\dots, n $ line of $A$. 
Then multiplication matrix ruler $Ax=(\langle A_{i1},x \rangle , \cdots ,\langle A_{in},x \rangle)$. Here $\langle , \rangle$ is the inner product of $\mathbb{R}$. Then $Ax\geq 0$ if, only if, $\langle A_{i1},x \rangle \geq 0, \cdots ,\langle A_{in},x \rangle \geq 0. $
