Let $A = \left(a_{i,j} \right) \in \mathrm{M}\left( n\times n, \mathbb{R} \right)$ be an orthogonal matrix with $A_{i,j} \geq 0$ for all $i, j \in \left\{ 1, \dots, n \right\}$. Prove that $A$ is a permutation matrix.
My approach was to use $A A^T = E $ but that did not work out.
Hint You're on the right track. Taking the $(i, j)$ entry of the equation $E = A^T A$ characterizing orthogonal matrices $A$ gives
$$\delta_{ij} = E_{ij} = (A^T A)_{ij} = \sum_{k = 1}^n (A^T)_{ik} A_{kj} = \sum_{k = 1}^n A_{ki} A_{kj} .$$ By definition, the lattermost quantity is the dot product of the $i$th and $j$th columns of $A$ regarded as vectors in $\Bbb R^n$. For $i \neq j$, we have $0 = \sum_{k = 1}^n A_{ki} A_{kj}$, and since each product $A_{ki} A_{kj}$ is nonnegative, we must have $A_{ki} A_{kj} = 0$ for all $k$.
Now, show that this implies that for any $k$ exactly one entry of the $k$th row of $A$ is $1$ and the rest are $0$.
Assume any row of $A$ has two non-zero entries, say $a_{ij_1}>0$ and $a_{ij_2}>0$ for $j_1\neq j_2.$ Then the entry in position $(j_1,j_2)$ of $AA^T$ must be positive, a contradiction to $AA^T=I$ ($I$ being the identity matrix). Thus, each row of $A$ has at most one (in fact, exactly one, as rows of all zeros are impossible) non-zero entry, and so has each column. It is also clear now that each of these numbers must be one. Thus, $A$ is a permutation matrix.
