Prove that if the sum of each row of $A$ equals $s$, then $s$ is an eigenvalue of $A$. Let $A$ be an $n \times n$ matrix.
$i)$Prove that if the sum of each row of $A$ equals $s$, then $s$ is an eigenvalue of $A$.
$ii)$Prove that if the sum of each column of $A$ equals $s$, then $s$ is an eigenvalue of $A$.
I think that being an eigenvalue of $A$ implies that $sv=Av$ for some vector $v$. Furthermore, I know that $[a_i] = s$ if we let $a_i$ denote the i-th row of $A$. However, I do not seem to be able to find a link between these two facts. Could anyone please help me out?
 A: HINT: Calculate $Av$ when $v=(1,1,\ldots ,1,1)^t$, what can you say?
A: There is a simple simple proof beind this,
Let A = $
        \begin{pmatrix}
        a & b & c \\
        d & e & f \\
        g & h & i \\
        \end{pmatrix}
$ such that (a+b+c) = (d+e+f) = (g+h+i) = s (say).
Now AX = λX , X ≠ 0.
So for calculating its eigenvalue simply observe the det (A-λI):
det(A-λI) = $
        \begin{vmatrix}
        a - λ & b & c \\
        d & e - λ & f \\
        g & h & i- λ \\
        \end{vmatrix}
$ = 0
Thus,
$
        \begin{vmatrix}
        a + b +c - λ & b & c \\
        d + e + f  - λ  & e - λ & f \\
        g+h+i - λ & h & i- λ \\
        \end{vmatrix}
$ = 0
So,
 $
        \begin{vmatrix}
        s - λ & b & c \\
        s  - λ  & e - λ & f \\
       s- λ & h & i- λ \\
        \end{vmatrix}
$ = 0
And hence,
$
        \begin{vmatrix}
        s - λ
        \end{vmatrix}*\begin{vmatrix}1 & b & c \\1 & e - λ & f \\1 & h & i- λ \\
        \end{vmatrix}
$ = 0.
Hence, we conclude that s is an eigenvalue of A. Similarly, we prove when sum of each column is constant all over the matrix.
