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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.

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What you describe is not a subgroup. – Tobias Kildetoft Feb 13 '13 at 16:30
Also, given that you had a sum in the definition, your last example of such a matrix would only require $a>-1$. – Joe Tait Feb 13 '13 at 16:33
@Tobias I don't think that OP is asking whether the set of all $A\in GL_n(\mathbb{R})$ with $\sum_j a_{ij}>0$ is a subgroup. I think he is asking for a characterization of subgroups for which that property holds. – Alexander Gruber Feb 13 '13 at 17:47
I can see how confusing this can be, I will try to reformulate... – JeanThiviers Feb 18 '13 at 8:57
Do we know a characterization of those matrices $A$ such that the cyclic group generated by $A$ has property $P$? I think that would be a useful starting point. An interesting question actually. – Jyrki Lahtonen Feb 18 '13 at 11:12

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.

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