TITLE: A matrix with all entries being positive, each row has the same sum.
QUESTION [6 upvotes]: A matrix $A$ with all entries being positive, each row has the same sum, that is, $\sum_j a_{ij}=c>0$. Show that up to constant multiplier, the eigenvector $(1,\cdots,1)'$ is the unique eigenvector with all entries being positive.
Let $x=(x_1,\cdots,x_n)'$ being the eigenvector of $A$, with eigenvalue $\mu>0$, $x_i$ being not identically the same, How to derive a contradiction?

REPLY [4 votes]: We can start from here. Let's take an eigenvector x with all positive entries and let $m=min_i x_i$ and $M=max_i x_i$ the minimum and maximum of the components of the eigenvector. Than:
$\lambda x_i=\sum_j a_{i,j}x_j$ [1]
$Cm \le \lambda x_i \le CM$ [2]
for every $i$. Since this is valid for every $i$ it must be valid for $x_i=m$ and $x_i=M$, so that we have $C\le \lambda$ from the left inequality and $C \ge \lambda$ from the right. This shows that $\lambda=C$.
Now we can look back at the system [1] where $x_i=m$ with $\lambda=\sum_j a_{i,j}$:
$m \sum_j a_{i,j}=\sum_j a_{i,j}x_j$
, that is:
$\sum_j a_{i,j} (x_j-m)=0$
,which shows that we must have $x_j=m$ for each $j$ since all entries $a_{i,j}$ are positive.