Minimal spectral radius of a primitive matrix Given the set of all primitive matrices of dimensions $m$ by $m$ that are non-negative and integer - which one is the matrix with the minimal spectral radius?
Edit (according to the first comment):
The matrix without the red '1' is an irreducible matrix with minimal (1) eigenvalue (in order to be irreducible it must have at least one '1' in each row). This matrix is not primitive, because of the locations of the '1's. In order to get a primitive matrix, we must add at least one '1'. The matrix below is indeed primitive with only one additional '1' (proof is needed...). Addition of more '1's will result in a larger spectral radius (for non negative matrices, $A \ge B$ implies $\rho(A) \ge \rho(B)$). The question is:
how do we know that this specific location of the red '1' will give the minimum?

(This is too long for a comment.) I don't know the answer, but it follows from Perron-Frobenius theorem that if $A$ is a primitive matrix with integer entries, then its spectral radius must be bounded by its minimum row/column sum and maximum row/column sum. Therefore, I am convinced that the minimiser is a $0-1$ matrix with at most two nonzero entries on every column and every row. My own guess is that when $m\ge3$, the minimiser is given by the following:
$$
A=\pmatrix{0&1&\color{red}{1}\\ &\ddots&\ddots\\ &&\ddots&\ddots\\ &&&\ddots&1\\ 1&&&&0},
$$
i.e. it is obtained by adding a $1$ to the $(1,3)$-th entry of a permutation matrix (in case $m=2$, use modulo arithmetic, i.e. add a $1$ to the $(1,1)$-th entry instead). When $m\to\infty$, we have $\rho(A)\to1$.
