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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?

Martin Sleziak
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dreamer
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    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 $A$. – Chris Godsil Mar 31 '13 at 16:38

2 Answers2

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

pritam
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    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? – dreamer Mar 31 '13 at 17:10
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    @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. – pritam Mar 31 '13 at 17:17
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    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 $v$. – dreamer Mar 31 '13 at 17:32
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    @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 $Av=\lambda v$. – pritam Mar 31 '13 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? – user1551 Mar 31 '13 at 18:10
  • @pritam I see your point now :). Thank you for your help :). – dreamer Mar 31 '13 at 18:20
  • @user1551 I see now that you're right indeed. Thanks! – dreamer Mar 31 '13 at 18:21
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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 $AX = \lambda X , X \neq 0$. So to calculate its eigenvalue simply observe the $\det (A-\lambda I)$:

$\det (A-\lambda I) = \begin{vmatrix} a - λ & b & c \\ d & e - λ & f \\ g & h & i- λ \\ \end{vmatrix} = 0$

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.

Cloud Walker
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Amit Bendkhale
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