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This is a follow-up questions to posts:

Stationary distribution for directed graph

Stationary distribution for different types of graph

The definition of stationary distribution in wikipediaSteady-state analysis and limiting distributions

Are stationary distributions of graphs with every properties(for example directed or undirected, strongly connected or sparse, periodic or aperiodic)proportional with eigenvector corresponding to eigenvalue 1 or this property is only satisfied for strongly connected aperiodic graph?

If not, what is the difference for each case(I mean for example for Aperiodic or for periodic etc)?

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Cross posted mathoverflow.net/questions/122779/… – Byron Schmuland Feb 24 at 15:50
@ByronSchmuland :) what do you mean by this comment?! Is there any problem to post in 2 site in my account. I just want to consult different scientists. – Fatime Feb 24 at 17:19
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It is just polite to let others know when you crosspost. Imagine someone who spends a lot of time and effort to answer your question, only to find that a complete solution already exists on another site. How will that person feel? If you must crosspost, at least add that information to your question so that everybody knows. – Byron Schmuland Feb 24 at 18:39
@ Byron Schmuland. Thanks again :). I insert the link of my question in other sites from now. – Fatime Feb 24 at 18:50
@ Byron Schmuland. Is my question is so bad that has negative vote? what is the reason? – Fatime Feb 24 at 18:53
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1 Answer

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The same Wikipedia article you link to gives the answer. This is true for any Markov chain with finite state space.

In other words, the stationary distribution π is a normalized (meaning that the sum of its entries is 1) left eigenvector of the transition matrix associated with the eigenvalue 1.

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What matrix? What eigenvector? What Matlab computation? Please add all of this to your post and ask the question clearly. We want to see all the numbers. Nobody can help you if you are hiding important information. – Byron Schmuland Feb 24 at 20:35
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Add this to the body of your question. Not in the comments. – Byron Schmuland Feb 24 at 20:36
@FatimeRahman Did you calculate the left eigenvector or the right eigenvector? Wikipedia clearly tells you which of these is correct.... – Byron Schmuland Feb 24 at 20:48
@ Byron Schmuland.As I understand,It must be the left but the problem as I mentioned above is that the result is not the same as an Example.Is there a point? – Fatime Feb 24 at 21:15
@FatimeRahman Did you calculate the left eigenvector or the right eigenvector? – Byron Schmuland Feb 24 at 21:19
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