# 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?

• The easiest way to start is to find the eigenvector. Think of a vector $x$ such that the $i$-th entry of $Ax$ is the $i$-th row sum of $A$. Commented Mar 31, 2013 at 16:38

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

• That every element in $Av$ is equal to $s$ I think? So that would indeed mean that $s$ is an eigenvalue in this case. However, how does this proof the general case? Commented Mar 31, 2013 at 17:10
• @mause: What do you mean by the general case ? The HINT says that $v$ is an eigen vector with corresponding eigen value $s$, thats what you wanted to show. Commented Mar 31, 2013 at 17:17
• Yes, but what I mean is, don't I have to show that this holds for arbitraty $v$? Because your hint only shows that this is true if you choose this particular $v$. Commented Mar 31, 2013 at 17:32
• @mause:You don't have to show for arbitrary $v$. The definition of eigen value of a matrix is: $\lambda$ is an eigen value of $A$ if $\textit{there exists}$ a vector $v$ such that $v\ne 0$ and $Av=\lambda v$. Commented Mar 31, 2013 at 17:37
• @mause If $Av=sv$ for any arbitrary $v$, then $A=sI$. You don't think that $sI$ is the only matrix whose row sums are equal to $s$, do you? Commented Mar 31, 2013 at 18:10

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.

• Sorry for bringing this question back, but what operation did you perform to get from step 1 to 2? Commented Apr 26, 2018 at 15:05
• I have used C1 = C1 + C2 + C3. Commented Jul 22, 2018 at 12:34
• I still don't get it, why C1 = C1 + C2 + C3 should be equal. Is there some theorem I don't know that says this should be true Commented Jan 9, 2023 at 8:12