I've read in a number of places (eg here) that if a matrix $M$ is Markov (aka stochastic), then $|\det(M)|\leq 1$, with equality $|\det(M)|=1\iff$ $M$ is a permutation matrix. It is fairly easy to show the first part of this result, based on the eigenvalues of $M$ being of magnitude 1 or less, but I'm struggling with the second part.
If anyone could provide a proof (or reference) of this result, I would be much obliged.
Best Answer
If we denote entries in the matrix $M$ in $k$ row by $a_{k,1},a_{k,2},\ldots,a_{k,n}$ then $a_{k,i} \ge 0$ and $\sum_{i=1}^{i=n}a_{k,i} = 1$ so $a_{k,i} \le1$.
By Hadamard's inequality $$\det(M) \le \prod_{i=1}^{i=n}\sum_{k=1}^{k=n} a_{k,i}^{2}$$ This can be $1$ only if $a_{k,i} = 1$ for any $k$. In this case $M$ is a permutation matrix.