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Let $P = [p_{ij}]_{1 \leqslant i,j \leqslant m} \geqslant 0$ a primitive and irreducible matrix. And $P^k > 0$ for some $k.$ Prove that $ P^{k+i} > 0, i =1,2, \dots.$

I have used a hint suggested by N.S below and wrote this proof what do you think?

By the irreducibility of $P \geqslant 0$ there exists a permutation matrix $M$ (an identity matrix whose rows and columns have been reordered) such that $MP$ can be written in the form $ MP = \begin{pmatrix} P_{11} & 0 \\ P_{21} & P_{22} \end{pmatrix}$ where $P_{11}$ and $P_{22}$ are square matrices. This implies $ P = M^{-1}\begin{pmatrix} P_{11} & 0 \\ P_{21} & P_{22} \end{pmatrix}.$ The matrix $P$ has at least a non-zero element (which is in fact positive) in each row, thus we use this fact in the proof below

We have $P^k >0.$ At $i=1$ we have $P^{k+1} = P P^k > 0.$ We assume at $i$ that we have $P^{k+i} >0.$ Thus at $i+1$ we have $ P^{k+i+1} = PP^{k+i} > 0.$ This is correct for all $i = 1,2, \dotsm$

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Haven't you mixed up reducibilty and irreducibility in the above? –  Geoff Robinson Jul 12 '12 at 19:09
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1 Answer

Hint Prove first that any row of $P$ has a non-zero element. Then

$P^{k+i+1}=PP^{k+i}$

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Well, its N.S due to the irreducibility right? –  Zizo Dec 10 '11 at 15:30
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