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$\begingroup$

Of course the term "non-negative" is entry-wise.

I know the fact that if this matrix is a stochastic matrix, that is, the sum of each of its rows is 1, then it has a stationary probability vector $\pi$, which is of course non-negative.

But what if it is just a general non-negative matrix? Does it again always have a non-negative eigenvector?

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    I see. You deleted your previous question on this. The answer is yes.2012-09-19
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    @WillJagy: Yeah. Sorry about that. I tried really hard to analyze it, and did some experiments, but just can't see why it is true.2012-09-19
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    [Try this](http://en.wikipedia.org/wiki/Perron%E2%80%93Frobenius_theorem).2012-09-19
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    @did: Oh! Thanks. How silly I was. That's just the theorem that proved the existence of stationary probability vector.2012-09-19
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    This is knpwn as the Perron_frobenius theorem, there is a WP page: http://en.wikipedia.org/wiki/Perron%E2%80%93Frobenius_theorem2012-09-19
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    @Did Please consider converting your comment into an answer, so that this question gets removed from the [unanswered tab](http://meta.math.stackexchange.com/q/3138). If you do so, it is helpful to post it to [this chat room](http://chat.stackexchange.com/rooms/9141) to make people aware of it (and attract some upvotes). For further reading upon the issue of too many unanswered questions, see [here](http://meta.stackexchange.com/q/143113), [here](http://meta.math.stackexchange.com/q/1148) or [here](http://meta.math.stackexchange.com/a/9868).2013-06-16

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Try the WP page on Perron-Frobenius theorem.