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$$ \left[ \begin{array}{@{}ccccc@{}} 0.9& 0.1& 0& 0& 0& 0& \\ 0& 0.9& 0.1& 0& 0& 0& \\ 0& 0& 0.9& 0& 0& 0.1& \\ 0& 0& 0& 0.9& 0.1& 0& \\ 0& 0& 0& 0.1& 0.9& 0& \\ 0.1& 0& 0& 0& 0& 0.9& \\ \end{array} \right] $$

I think this matrix is irreducible and primitive but after I looked properties of irreducible i was confused and don't know how can to prove that the matrix irreducible and primitive

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    $\begingroup$ State the definitions and explain your progress toward understanding. $\endgroup$ Mar 17, 2013 at 17:54
  • $\begingroup$ Yes you are right that is no good fits on it but I only interested according this matrix.my program give me a Peron matrix and I try to understand the matrix is irreducible and primitive $\endgroup$
    – Toto
    Mar 17, 2013 at 18:07

3 Answers 3

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As every state here is pathwise connected to any other state, the matrix is irreducible. Also, as every state is accessible from itself, it has to be aperiodic and hence the matrix is primitive.

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  • $\begingroup$ Thanks for helps all of you to answered the question it helped me a lot! $\endgroup$
    – Toto
    Mar 17, 2013 at 22:54
  • $\begingroup$ Hi User1551 i am so sorry i made a mistake to show my matrix wrong. about it i had some problem in my program. i corrected the matrix now and according to my understand it is also primitive am i right?Thank you so much .Regards $\endgroup$
    – Toto
    Mar 25, 2013 at 18:43
  • $\begingroup$ @Toto No. The modified matrix is reducible, hence not primitive. You can see that states 1,2,3,6 and states 4,5 form two connected components. Therefore the matrix is reducible. Each of the two connected components gives rise to a primitive submatrix, though. $\endgroup$
    – user1551
    Mar 25, 2013 at 19:24
  • $\begingroup$ Thank you very much ,it means so much for me.Best Regards $\endgroup$
    – Toto
    Mar 25, 2013 at 19:28
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You have to look at the powers of your matrix. A matrix is primitive when for some power $k\in \mathbb{N}_0$, all entries of the matrix are strictly positive. A matrix is irreducible when for each entry, there is a power $k\in \mathbb{N}_0$ such that that entry for the power of the matrix is strictly positive. Primitive matrices are always irreducible but not the other way around.

Take for instance a $n\times n$-matrix with $0$ on the diagonal, $1$ everywhere else. That matrix is primitive. It is also therefore irreducible. Check this for yourself by computing powers.

On the other hand a $n\times n$-matrix with 1's on the $(k,k+1)$ for $k\in{1,\ldots,n}$ and $(n,1)$ entries of the matrix but zeros everywhere else is only irreducible. Check again by computing powers.

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  • $\begingroup$ +1 for enlightening us about the meaning of primitive and irreducible in this context :-) $\endgroup$ Mar 17, 2013 at 18:14
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As for seeing the matrix is primitive, recall the definition. A matrix is primitive if there is $k \gt 0$ such that $P^k$ has all strictly positive entries. As my friend said Primitive matrices are always irreducible!

When we draw the directed graph corresponding to the matrix, then your matrix is irreducible if it is possible to travel from any node to any other node and Your example not fails this condition because as it is possible to travel away from node.

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