Prove that if the sum of each row of $A$ equals $s$, then $s$ is an eigenvalue of $A$.

Solution 1:

HINT: Calculate $Av$ when $v=(1,1,\ldots ,1,1)^t$, what can you say?

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.