Proposition: Let $A=[a_{ij}]_{n\times n}$ be a non-zero matrix where $a_{ij}\in\mathbb{R}$. If $\sum_{j=1}^na_{ij}=M$ for all $i=1,2,\cdots, n$ then at least one of the eigenvalues of $A$ is equal to $M$.
Proof: Let $x=\begin{bmatrix}1\\ 1\\ \vdots \\ 1\end{bmatrix}$
Then, we have $Ax=Mx$. Hence, $M$ is the eigenvalue of the matrix $A$.
In fact, it is also true if the sum of the columns of a matrix is equal because the eigenvalues of $A^T$ and $A$ are equal.
Is there any other proof? Thanks in advance.
Consider the matrix $A-MI$,if you prove that $A-MI$ is not invertible then it shows that $M$ is an eigenvalue of $A$.
To show that $A-MI$ is not invertible you can prove that $rank(A-MI) < n$, which follows easily from the fact that sum of entries of each row equals zero and there exits a non-trivial linear combination for $A-MI$ rows to equal zero hence $A-MI$ rows are linearly dependent so $rank(A-MI)<n$