matrix with all rows positive I am thinking about a problem in a different area than linear algebra, but I came across a matrix with sum of entries of all rows positive, i.e. a matrix $A$ such that $\sum_{j} A_{ij}>0$ for all $i$. Are there any interesting properties of such a matrix (anything about the determinant, eigenvalues, etc)? 
 A: Partial answer:
Take the matrix $\left( \begin{array}{cc}
1 & 0 \\
1 & 0  \end{array} \right)$, it's determinant is $0$. 
Take the identity matrix $\left( \begin{array}{cc}
1 & 0 \\
0 & 1  \end{array} \right)$, it's determinant is $1$. 
Take the matrix $\left( \begin{array}{cc}
0 & 1 \\
1 & 0  \end{array} \right)$, it's determinant is $-1$.
So the you can't guarantee anything about the determinant. I wouldn't expect you could say anything definitive about the eigenvalues in this case either so, since you can't tell if it has any non-zero, or any negative positive etc. The first matrix has eigenvalues $1$ and $0$, and the second both equal to $1$ and The last has eigenvalues $1$ and $-1$.
You can certainly say the sum of two such matrices has the same property, but the product however does not, e.g. 
$\left( \begin{array}{cc}
2 & -1 \\
0 & 1  \end{array} \right) 
\left( \begin{array}{cc}
-1 & 2 \\
3 & 0  \end{array} \right) =
\left( \begin{array}{cc}
-5 & 4 \\
3 & 0  \end{array} \right) $
A: If for any row
$$|a_{ii}|>\sum_{j=1,j\ne i}^n |a_{ij}|$$
Then the matrix is called diagonally dominated and can be proved to be invertible. 
Moreover, if $a_{ii}>0$, then it can be shown to be positive definite, i.e. all eigenvalues are positive.
