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Im trying to come to grips with what the physical interpretation of a non diagonalisable Markov Matrix means in terms of what we can deduce about it having a limiting distribution/ what potential limiting distributions would be. Please refer to this question: Example of a Markov chain transition matrix that is not diagonalizable? where the Answer shows a transition matrix, $P$ that is recurrent and defective (i.e. non diagonalisable).

This is interesting because I like to think of the fact that for non defective $P$, I can express any initial valid distribution vector $\mathbf{x}^0$, (satisfying $\sum_i x_i^0 = 1$ ) as a linear combination of the left eigenvectors of $P$. This is nice because $\mathbf{x}^{\infty} = \lim_{n\to \infty} \mathbf{x}^0P^n$ will then no longer contain any components of those eigenvectors whose corresponding eigenvalues $|\lambda|<1$. However if $P$ is defective we can't necessarily express any initial distribution in terms of the eigenvectors and hence can't draw any conclusions of what the limiting distributions will look like, if they exist. Is this the end of the story or is there something else I'm missing.

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For an ergodic chain it still converges to a stationary distribution. By Perron-Frobenius theorem $1$ is a simple eigenvalue, so all other eigenvalues hold $|\lambda_i| < 1$. Now you can build an orthogonal basis of generalized eigenvectors. Though it's a bit more involved, other terms [those eigenvectors of other than $1$ eigenvalues] will disappear in limit. See Rosenthal, Jeffrey S. "Convergence rates for Markov chains." Siam Review 37.3 (1995): 387-405 for more details.

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