Problem: Let $A$ be an $n\times n$ matrix over $\mathbb{R} $ (or $\mathbb{C} $ ). Suppose that the sum of each row of $A$ is $\lambda $ and sum of each column of $A$ is $\mu$. Then show that $\lambda $ is an eigenvalue of $A$ and $\mu $ is an eigenvalue of $A^T$
Solution: Given that the sum of each row of $A$ is $\lambda $. Let $\mathbf{v} =(1,1,1\dots,1)^t$. We claim that $\lambda $ is an eigenvalue of $A$ with eigenvector $\mathbf{v} $. Note that \begin{align*} A \mathbf{v} & = \begin{pmatrix} a_{11} + a_{12} + \dots + a_{1n} \\ a_{21} + a_{22} + \dots + a_{2n} \\ \vdots \\ a_{n1} + a_{n2} + \dots + a_{nn} \\ \end{pmatrix} = \begin{pmatrix} \lambda \\ \lambda \\ \lambda \\ \lambda \\ \end{pmatrix} = \lambda \mathbf{v}. \end{align*} Thus, $\lambda $ is an eigenvalue of $A$ with eigenvector $\mathbf{v} $.
If sum of each column of $A$ is $\mu $, then the sum of each row of $A^t$ will be $\mu $ and hence $\mu $ is an eigenvalue of $A^t$ with an eigenvector $\mathbf{v} $.