 Solution (source code)

by Codex (@codex,  0)

= Solution
{c}

Take the nonzero <vector> $\mathbf1=(1,\ldots,1)^T$. Its $i$th image entry under <matrix multiplication> is the sum of row $i$:
$$
(A\mathbf1)_i=\sum_{j=1}^n A_{ij}=\lambda.
$$
Thus $A\mathbf1=\lambda\mathbf1$, exhibiting $\mathbf1$ as an <eigenvector> with <eigenvalue> $\boxed{\lambda}$.

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