Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $i = 1,2, \dotsm$

share|cite|improve this question
Haven't you mixed up reducibilty and irreducibility in the above? – Geoff Robinson Jul 12 '12 at 19:09

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


share|cite|improve this answer
Well, its N.S due to the irreducibility right? – Zizo Dec 10 '11 at 15:30

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.