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Let $A$ be an $n \times n$ matrix with real, non-negative entries. Assume $A$ is primitive, meaning there exists an integer $k$ such that $A^k>0$ (here the inequality means all entries in $A$ are positive). Also, assume $A$ is row stochastic, meaning that the the entries of each row sum to 1.

An alternate way of stating the above is to say: let $A$ be the transition probability matrix of an irreducible, aperiodic Markov chain.

Is $A$ diagonalizable? Thank you in advance for any proofs or counterexamples.

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For example, consider

$$ \left[ \begin {array}{ccc} 0&{\frac {49}{72}}&{\frac {23}{72}} \\ 1/2&1/6&1/3\\ 1&0&0\end {array} \right] $$

which is not diagonalizable (the eigenvalue $-5/12$ has algebraic multiplicity $2$ but geometric multiplicity $1$).

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  • $\begingroup$ In this case, $A^k > 0$ when $k \geq 3$ $\endgroup$
    – Ben Grossmann
    Sep 19, 2014 at 4:35
  • $\begingroup$ Also, neat! Out of my own curiosity, is there some clever way you managed to come across this example? $\endgroup$
    – Ben Grossmann
    Sep 19, 2014 at 4:38
  • $\begingroup$ Thanks! In addition to being primitive and row stochastic, some transition probability matrices have the form where each row has as an entry either $0$ or $(1/i)$ i-times. For example: $\begin{pmatrix} 1/2 & 0 & 1/2 \\ 1/3 & 1/3 & 1/3 \\ 0 & 1 & 0 \end{pmatrix}$. It seems a bit far-fetched, but if we add this additional constraint on $A$, is there any reason to believe $A$ will then be diagonalizable? $\endgroup$
    – John
    Sep 19, 2014 at 5:10
  • $\begingroup$ The basic methodology was to select the basic form of the matrix subject to parameters, and adjust the parameters to make the characteristic polynomial have discriminant $0$. $\endgroup$
    – Robert Israel
    Sep 19, 2014 at 20:02
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    $\begingroup$ Try $$\pmatrix{1/2 & 1/2 & 0 & 0\cr 0 & 0 & 1/2 & 1/2\cr 1/2 & 0 & 0 & 1/2\cr 0 & 1/2 & 1/2 & 0\cr}$$ where eigenvalue $0$ has algebraic multiplicity $2$ and geometric multiplicity $1$. $\endgroup$
    – Robert Israel
    Sep 19, 2014 at 20:06

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