TITLE: Non-diagonalizable doubly stochastic matrices
QUESTION [15 upvotes]: 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?

REPLY [9 votes]: Here's a 2-parameter family of examples. 
Let $$A=\pmatrix{a&b&1-a-b\cr2-2a-3b&3a+4b-2&1-a-b\cr a+3b-1&3-3a-5b&2a+2b-1\cr}$$ If $a$ and $b$ are chosen so that all the entries are between 0 and 1, then this is a doubly-stochastic matrix. It has the repeated eigenvalue $3a+3b-2$, which is not 1 if $a+b\ne1$. The matrix $A-(3a+3b-2)I$ has rank 2, provided $a+2b\ne1$. So if $a$ and $b$ are chosen in accordance with the stated conditions, this is a non-diagonalizable doubly-stochastic matrix. 
I found this by letting $A(1,1)=a$, $A(1,2)=b$, $A(2,1)=a-\lambda$, $A(2,2)=b+\lambda$, letting the other entries be what they have to be to make all row- and column-sums 1, and then choosing $\lambda$ to make the trace $1+2\lambda$, so the eigenvalues have to be $\lambda$ repeated and 1. 
EDIT: If, for some reason, you want the matrix to be singular, then just take $\lambda=0$. Then you get the one-parameter family $$A=\pmatrix{a&(2/3)-a&1/3\cr a&(2/3)-a&1/3\cr1-2a&2a-(1/3)&1/3\cr}$$ Now if you choose $a=1/4$ you get (essentially) David Speyer's example.