A particular subgroup of the general linear group Say a $n \times n$ matrix with real coefficients $a_{ij}$ has property P if $\sum_j a_{ij} >0, \forall i$. Say a group (or subgroup) of matrices has property P if every element has property P.
What would be the biggest subgroup of the general linear group $\text{GL}_ n(\mathbb{R})$ having property P ?
I'm alread having some problems for the $2 \times 2$ case. I found that matrices of the form $\begin{bmatrix}
a & 0 \\
0 & b\end{bmatrix}$ or $\begin{bmatrix}
0 & a \\
b & 0\end{bmatrix}$ with strictly positive coefficients work, obviously, as well as matrices of the form $\begin{bmatrix}
1 & a \\
0 & 1+a\end{bmatrix}$ ($a > 0$), but I don't see a general way of doing it.
 A: Do I get it right, you want the sum of the elements in each row to be positive. So you are considering linear maps which take the vector $[1,1, \dots, 1]^{t}$ to a vector with positive entries.
My idea about constructing a biggish subgroup of maps like that would be to take a basis $e_1, \dots, e_n$ of the underlying vector space $V$ which has $e_1 = [1,1, \dots, 1]^{t}$ as its first element. Then define a subgroup $G$ of $\operatorname{GL}(V)$ by
$$
G = \left\{ g \in GL(V) : \text{$g(e_1) = a e_{1}$,  for some $a > 0$} \right\}.
$$
In the case $n = 2$ you get $[1,1]^{t} \mapsto [a,a]^{t}$, and say $[0,1]^{t} \mapsto [b,c]^{t}$, with $a > 0$ and $b \ne c$ not both zero. Then with respect to the standard basis you get the group of matrices
$$
G = \left\{ \begin{bmatrix} a-b&b\\a-c&c\end{bmatrix}
:
\text{$a > 0$, $b \ne c$,  $b$ and $c$ not both zero} \right\}.
$$
However, your first two examples do not fit in. I believe this might mean there is no unique maximal subgroup with respect this property. Although this is no conclusive evidence, note, for instance, that the matrices
$$
\begin{bmatrix}-1&2\\-2&3\end{bmatrix},
\qquad
\begin{bmatrix}-1&3\\-2&3\end{bmatrix}
$$
satisfy the condition, while their product
$$
\begin{bmatrix}-3&3\\-4&3\end{bmatrix}
$$
does not.
