 Solution

Yes. The sums of the columns of $A$ are the sums of the rows of its transpose $A^T$. By the previous argument, $A^{T}1=\lambda 1$, so $\det (\lambda I_n-A^T)=0$. The determinant is unchanged by transpose, givingTherefore $\boxed{\lambda \text{ is an eigenvalue of }A}$. The vector $1$ need not itself be an eigenvector of $A$ in the column-sum case.