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If $A$ is a Markov matrix, all the columns of $A^k$ approaches the steady state vector $x_\textrm{ss}$. How can you prove this?

So far I was trying to understand with the decomposition.

$A^k = S \Lambda^k S^{-1} = c[x_1, 0 , ..., 0]S^{-1} = c[x_1, x_1, ..., x_1] $ where $x_1$ is the first column of $S$ or an eigenvector of $A^k$ associated with $\lambda = 1$, which is the steady state vector, $x_\textrm{ss}$. Here $c$ is just a constant so that the components of $cx_1$ sum to 1.

I can't figure out the last equality.

Although the suggested question is very similar to my question, I think my question focuses more on why other columns of $A^k$ which does not have to do with "probability vector" mentioned in the other question. So I don't think this is an exact duplicate.

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    You need to add some conditions for this to be true. For instance, the columns of $A=\pmatrix{0&1\\1&0}$ don’t converge. – 2017-02-20
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    This is a consequence of the Perron-Frobenius theorem. If the hypothesis are fulfilled, $A^k$ converge to a rank-$1$ matrix whose columns are given by the only eigenvector for the $\lambda=1$ eigenvalue, i.e. the stationary distribution. – 2017-02-20
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    However, you need that $\lambda=1$ is a simple eigenvalue and the only eigenvalue on the boundary of the unit circle. – 2017-02-20
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    @Sheljohn I thought so, too, at first, but this question is asking about the columns of the matrix powers, while the other question is asking about the product of powers of the matrix and a probability vector. The two questions are, of course, related. – 2017-02-21
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    @JackD'Aurizio Could you elaborate more on why $A^k$ converges to a rank-1 matrix? – 2017-02-21
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    @hwchung: The proposed duplicate answers the question you pose in the first line of the body of your Question. Much of what you say later in your post is confused. "I can't figure out the last equality." Please be more specific as to what equality you are struggling with. Note that if $A$ is stochastic, so are the powers of $A$. Also what you want to prove needs additional hypotheses, as the earlier Commenters point out. In all I think the older Question would be worth your study. – 2017-02-21

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