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I am currently trying to understand a proof for the above, which states that, in other words, there exists a unique $\overrightarrow{v}$, such that $\overrightarrow{v}P = \overrightarrow{v}$ for the transition matrix $P$.

He first shows that there exists a non-zero $\overrightarrow{v}$, which satisfies $\overrightarrow{v}P = \overrightarrow{v}$.

Next, he proves that either $\forall v \in \overrightarrow{v} (v \text{ is non-negative})$ or $\forall v \in \overrightarrow{v} (v \text{ is non-positive})$ by contradiction. However, I don't understand what's the contradiction in this part of the proof. Attached is the proof for your reference.

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  • $P=\begin{bmatrix}0 & 1 \ 1 & 0\end{bmatrix}$ doesn't have a stationary distribution $P^2=I$ and hence $P^k=\begin{cases} P & \text{ if } k \text{ is odd}\I & \text{ if } k \text{ is even} \end{cases}$. They're missing the aperiodic property! :p – N8tron May 28 '18 at 05:40
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    @N8tron They are assuming that each transition probability is strictly greater than zero (line -5 up in the proof) – AnyAD May 28 '18 at 06:01
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    I don't think that they have reached a contradiction, the way the proof is written. – AnyAD May 28 '18 at 06:06
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    @AnyAD nice catch! I don't think they're condition is correct that $P$ is irreducible if the entries are strictly positive unless this author includes aperiodic as part of their irreducibility criteria. Generally being irreducible corresponds to the underlying weighted digraph being strongly connected and aperiodic means that the GCD of the cycle lengths will be one. However, both conditions combined are equivalent to the matrix being primitive that is $P^k$ has strictly positive entries for some $k$. – N8tron May 28 '18 at 06:19
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    Here's a better proof, you'll have to deal with the spectral radius $\rho(A)$ in your case it's just one, but Gelfands formula is still pretty handy in the proof. http://www.math.cornell.edu/~web6720/Perron-Frobenius_Hannah%20Cairns.pdf – N8tron May 28 '18 at 06:41

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