This is a followup to a previous question where a nice counter example came up to the proposition "stochastic matrices can only have trivial Jordan forms".
This question looks at the more strict case of doubly stochastic matrices:
$${\bf M} \in \mathbb [0,1]^{n\times n} : \cases{ \displaystyle \sum_{i}{\bf M}_{ij} = 1, \forall j\\\displaystyle \sum_{j}{\bf M}_{ij} = 1, \forall i}$$
Can we find a counter example for those too?
Yes. As $1$ is always a semi-simple eigenvalue of a doubly stochastic matrix, the smallest-sized counterexample is $3\times3$, and here is a random counterexample. Consider $$ M=\frac1{15}\pmatrix{ 4&4&7\\ 6&6&3\\ 5&5&5}. $$ Since $Mu=0$ for $u=(1,-1,0)^T$ and $Mv=u$ for $v=(0,-5,5)^T$, $M$ has a $2\times2$ nilpotent Jordan block in its Jordan form.