Are there constructive examples of doubly stochastic matrices (whose rows and columns all sum up to $1$ and contain only non-negative entries) that are not diagonalizable?



Non-diagonalizable doubly stochastic matrices
Sure. For example:


$$A = \begin{pmatrix}
5/12 & 5/12 & 1/6 \\
1/4 & 1/4 & 1/2 \\ 
1/3 & 1/3 & 1/3 
\end{pmatrix}$$


Note that 
$$A \begin{pmatrix} 0 \\ 1 \\ -1 \end{pmatrix} = \begin{pmatrix} 1/4 \\ -1/4 \\ 0 \end{pmatrix} \ \mbox{and} \ A^2 \begin{pmatrix} 0 \\ 1 \\ -1 \end{pmatrix} = 0.$$
This shows that $A$ is not diagonalizable, as, for diagonalizable matrices, $A$ and $A^2$ have the same kernel.




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Now, let me explain how to find this. Let $w$ be the all ones vector. The condition that $A$ is doubly stochastic is that $Aw =w$ and $A^T w = w$ (ignoring positivity for now). For any nonzero vector $v \in \mathbb{R}^n$, we have $Av =v$ and $A^T v = v$ if and only if $Av=v$ and $A$ sends $v^{\perp}$ into itself. This equivalence is obvious for $v=e\_1$, and the truth of the statement is preserved by orthogonal changes of coordinate, so it is true for any nonzero $v$.


So, I wanted $Aw=w$ and $A$ to preserve $w^{\perp}$. So, if $A$ is going to be non-diagonalizable, it has to have a nontrivial Jordan block on $w^{\perp}$. So I tried making $A$ be of the form $\left( \begin{smallmatrix} 0 & c \\ 0 & 0 \end{smallmatrix} \right)$ in the basis $\begin{pmatrix} 1 & -1 & 0 \end{pmatrix}^T$, $\begin{pmatrix} 0 & 1 & -1 \end{pmatrix}^T$. At first I tried this with $c=1$, but some of the entries came out negative. So I redid it with a smaller value of $c$. (I knew this had to work because, when $c=0$, you get $A = \left( \begin{smallmatrix} 1/3 & 1/3 & 1/3 \\ 1/3 & 1/3 & 1/3 \\ 1/3 & 1/3 & 1/3 \end{smallmatrix} \right)$ so, by continuity, for $c$ small enough I had to get nonnegative entries.


