Product and convex combination of two stochastic matrices Let $K_1$ and $K_2$ be two $N \times N$ stochastic matrices (hence non-negative and rows adding up to one) with zeros on the diagonal.  If $\alpha \in (0,1)$, is it possible to have 
$$K_1 K_2 = \alpha K_1 + (1-\alpha)K_2?$$
 A: Let $$A=\pmatrix{0&.5&0&0&.5\cr.5&0&.5&0&0\cr0&.5&0&.5&0\cr0&0&.5&0&.5\cr.5&0&0&.5&0\cr}{\rm\ and\ }B=\pmatrix{0&0&.5&.5&0\cr0&0&0&.5&.5\cr.5&0&0&0&.5\cr.5&.5&0&0&0\cr0&.5&.5&0&0\cr}$$ Then $AB=.5A+.5B$. 
A: EDITED: Federico Poloni points out that this fails to satisfy one of the criteria in the original posting in that this does not give a non-trivial convex combination of the two matrices. 
Set
$$
K_1=\begin{pmatrix}0&\tfrac 12&0&\tfrac 12\\
\tfrac 12&0&\tfrac 12&0\\
0&\tfrac 12&0&\tfrac 12\\
\tfrac 12&0&\tfrac 12&0\end{pmatrix}
$$
and
$$
K_2=\begin{pmatrix}0&0&1&0\\0&0&0&1\\1&0&0&0\\0&1&0&0\end{pmatrix}
$$
Then $K_2K_1=K_1$.
I think of $K_2$ as a rotation of the square by $\pi$ and $K_1$ as a Markov transition to one of your neighbours on the square. Then $K_2K_1$ is a Markov transition to one neighbour, followed by a flip that sends you to the opposite neighbour. 
A: Let $\mathcal{D}_n$ denote the set of doubly stochastic matrices. Then via a simple optimization routine, we can find several numerical examples. For instance,
\begin{equation*}
 K_1 = \begin{bmatrix}
  0 &    0 &    0&   1\\
  0 &    0 &    1&   0\\
  1 &    0 &    0&   0\\
  0 &    1 &    0&   0
\end{bmatrix},\qquad K_2=\text{argmin}_{K_2\in \mathcal{D}_n}\|K_1K_2-\alpha K_1 - (1-\alpha)K_2\|.
\end{equation*}
For instance, I used the following CVX code:
 n=4; % k1 = set as above
 cvx_begin
  variable k2(n,n)
  minimize norm(k1*k2-0.3*k1-0.7*k2)
  subject to
   k2 >= 0
   sum(k2,2)==ones(n,1)
   sum(k2,1)==ones(1,n)
 cvx_end

I obtained an objective value of $\sim 3\times 10^{-9}$. You can experiment with this code to construct several other examples, including perhaps more rational looking ones.
