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
Try the vector of all 1's. That'll do it.
If this is not clear, then think about it this way. Let $$e_1 = \left( \begin{array}{c}1\\0\\ \vdots \\ 0\end{array} \right), e_2 = \left( \begin{array}{c}0\\1\\ \vdots \\ 0\end{array} \right), ..., e_n = \left( \begin{array}{c}0\\0\\ \vdots \\ 1\end{array}\right)$$
The operation $Ae_1$ is the first column of $A$. $Ae_2$ is the second column. Thus $A(e_1 + e_2)$ is the vector that results from the addition of the first and second columns. The vector of all 1's is given by $e_1 + e_2 + \cdots + e_n$. Applying $A$ to this vector yields $$Ae_1 + Ae_2 + \cdots + Ae_n$$ which is the vector that results from summing all of the columns.
This means $A ( 1, 1, ..., 1)^T = (s, s, ..., s)^T$ and we are done.