Markov Chain Converging in Single Step

Question

I have a markov kernel K. From this I find the invariant probability $\pi$. The question is to design a "dream" matrix K*, that converges in one step. Such that $\lambda_{SLEM}=0$ (second largest eigen-value modulus). I am not sure how to go about designing the dream matrix. Any pointers to literature/pointers would be welcome

Answer 1

Consider three states $A,B,C$. We have that $A$ almost surely goes to itself and no other states. Let $B$ almost surely go to $A$ and to no other states. Let $C$ go to $A$ with probability $0.5$ and to $B$ with probability $0.5$.

The transition matrix may be seen as $M = \left( \begin{array}{ccc} 1 & 0 & 0 \\ 1 & 0 & 0 \\ 0.5 & 0.5 & 0 \end{array} \right) $

The eigenvalues of the associated transition matrix $M$ are $1,0,0$. The powers of the transition matrix $M$ are $M^2, M^3, \dots$, these just give us the matrix $A$ defined by

$A= \left( \begin{array}{ccc} 1 & 0 & 0 \\ 1 & 0 & 0 \\ 1 & 0 & 0 \end{array} \right) $

This converges in just one step.

EDIT:

Suppose we didn't know how to get there. We seek a transition matrix with an existing limiting distribution, then the Perron-Frobenius theorem tells us the spectral radius of this matrix must be $1$ with unique multiplicity one and all other eigenvalues smaller. So simply pick a matrix of order three by three with nine coefficients to be found and calculate what you need to satisfy those conditions.