Can doubly stochastic matrices have non trivial Jordan forms?

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.
• @mathreadler Sure. $tM+(1-t)I$ would be such a counterexample for any $0<t<1$. – user1551 Sep 4 '17 at 12:37
• @Omnomnomnom There wasn't any trick. As $1$ is a semi-simple eigenvalue, if there is any 3x3 counterexample $M$ at all, it has to be a convex combination of $I_3$ and a doubly stochastic matrix similar to $1\oplus J_2(0)$. So it suffices to consider a singular $M$ and a natural choice is to perturb the all-one matrix in a minimal way, i.e. to try $$M=\frac1{3c}\pmatrix{c-d&c-d&c+2d\\c+d&c+d&c-2d\\c&c&c}$$ with $c\ge|2d|>0$. Now any such $c,d$ in this range will give you a valid counterexample because $M(1,-1,0)^T=0$ and $\frac cdM(0,-1,1)^T=(1,-1,0)^T$. – user1551 Sep 4 '17 at 16:17