= Solution
{c}
Yes. The sums of the columns of $A$ are the sums of the rows of its <transpose> $A^T$. By the previous argument, $A^T\mathbf1=\lambda\mathbf1$, so $\det(\lambda I_n-A^T)=0$. The <determinant> is unchanged by <transpose>, giving
$$
\det(\lambda I_n-A)
=\det\bigl((\lambda I_n-A)^T\bigr)
=\det(\lambda I_n-A^T)=0.
$$
Therefore $\boxed{\lambda\text{ is an eigenvalue of }A}$. The <vector> $\mathbf1$ need not itself be an <eigenvector> of $A$ in the column-sum case.
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