Let $A$ be an $n\times n$ matrix and $a_{ij}$ be the entries of $A$ ($i,j \in \{1, \dots, n \}$).
If $$\sum_{i=1}^n\sum_{j=1}^n a_{ij} = n$$ and $$ a_{ij} \geq 0 \qquad (i,j \in \{1, \dots, n \})$$ then prove
(a) $|\det A| \leq 1$
(b) If $|\det A| = 1$ and $\lambda\in\mathbb C$ is an eigenvalue of $A$, then $|\lambda| = 1$.
Proving part (a) in the case $A$ is a diagonal matrix is easy, but I'm not sure how to go about the general case. Any hint or suggestion of a lemma/theorem/etc would be appreciated.
Not sure how I would go about part (b), but I imagine it would follow from the result obtained in (a).
Here is a different approach via Hadamard Determinant Inequality.
Let column $j$ of $A$ be given by $\mathbf a_j$
$\Big(\det\big(A^T A\big)\Big)^\frac{1}{n}$
$ \leq \Big(\prod_{j=1}^n \big \Vert \mathbf a_j\big \Vert_2^2\Big)^\frac{1}{n}=\Big(\prod_{j=1}^n \big \Vert \mathbf a_j\big \Vert_2\Big)^\frac{2}{n}$
$ \leq \Big(\frac{1}{n}\sum_{j=1}^n \big\Vert \mathbf a_j\big\Vert_2\Big)^2$
$\leq \Big(\frac{1}{n}\sum_{j=1}^n \big\Vert \mathbf a_j\big\Vert_1\Big)^2= \Big(\frac{1}{n}\sum_{j=1}^n\sum_{i=1}^n a_{ij}\Big)^2 =1^2=1$
$\implies \vert \det A\vert \leq 1$
where the inequalities are (i) Hadamard Determinant Inequality, (ii) $\text{GM}\leq \text{AM}$, (iii) Triangle Inequality
with equality iff the columns of $A$ are mutually orthogonal, per (i), each column has the same norm, per (ii)-- which must be 1 since this implies $\prod_{j=1}^n \big \Vert \mathbf a_j\big \Vert_2^2 =1$-- and finally each column is $\propto$ a standard basis vector, per (iii). That is, equality occurs iff $A$ is composed of $n$ linearly independent standard basis vectors aka iff $A$ is a Permutation Matrix (which, being orthogonal, necessarily has all eigenvalues on the unit circle).