TITLE: Prove that if the sum of each row of A equals s, then s is an eigenvalue of A.
QUESTION [9 upvotes]: Consider an $n \times n$ matrix $A$ with the property that the row sums all equal the same number $s$. Show that  $s$ is an eigenvalue of $A$. [Hint: Find an eigenvector]
My attempt:
By definition: $Ax = sx$ which implies that $(A - sI)x = 0$
$s$ is an eigenvalue for $A$ iff $\det(A - sI)  = 0$
When you do $A - sI$ the sum of each row is now $0$. I think that's important but I don't know what it means. So this is where I'm stuck

REPLY [4 votes]: What can you deduce about the linear dependence/independence of the columns of $A-sI$?