Non-negative, unitary, symmetric matrix is a permutation matrix

Question

I have a hypothesis that if a matrix is (edit: elementwise) non-negative, unitary and symmetric then it must be a permutation matrix. Does anyone have a proof of this, or otherwise a counter example?

I know that this must be true if the matrix is also traceless, is it also true when it is not traceless?

Answer 1

You don't need symmetry. The assumptions are that $AA'=I$ and $A\ge 0$. For each $i,j,k$ we must have $a_{ik} a_{jk}=0$ whenever $i\ne j$, so for each $k$ only one of the $a_{ik}$'s can be nonzero. Its square must be 1 because of the orthogonality condition, so it equals 1, and then you are done.