TITLE: When does an eigenvector of a matrix contain only a constant?
QUESTION [7 upvotes]: When I compute the eigenvectors of a certain matrix, the first of them is composed entirely of a single constant.
What properties of a matrix lead to this result?
Update
By "a vector composed entirely of a constant" I mean n repetitions of a constant comprising a length-n vector.

REPLY [8 votes]: Take an $n \times n$ matrix $A$, and suppose that $v$ is an eigenvector of $A$, with all entries of $v$ equal to a constant $k$. Naturally, $k \ne 0$.  Let $\lambda$ be the eigenvalue of $A$ that has $v$ as an eigenvector. If $(b_1, b_2, \dots, b_n)$ is any row of $A$, then by the definition of eigenvalue and eigenvector, we have 
$$kb_1+kb_2+\cdots +kb_n=\lambda k,$$
from which we conclude that $b_1+b_2+\cdots+b_n=\lambda$.  It follows that each row sum of the matrix is equal to $\lambda$.
Conversely, suppose that all row sums of $A$ are equal to $\sigma$. Let $v$ be the vector with all entries equal to $1$.  Then $Av$ is a vector with all entries equal to $\sigma$, which means that $v$ is an eigenvector of $A$ with eigenvalue $\sigma$.
Thus $A$ has an eigenvector with all entries equal if and only if all row sums of $A$ are equal.

REPLY [2 votes]: Exercise: a matrix has an eigenvector with all its entries equal if and only if all rows of the matrix have the same sum.