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There is a difference equation of Markov process.

$y_{1}=0.9y_{0}+0.2z_{0}$
$z_{1}=0.1y_{0}+0.8z_{0}$

Let matrix A =$\begin{pmatrix} 0.9 & 0.2 \\ 0.1 & 0.8 \end{pmatrix}$

Then from det(A-$\lambda$I)=0
$\lambda_{1}=1$ and $\lambda_{2}=0.7$

A=$S\lambda S^{-1}$=$\begin{pmatrix} \frac{2}{3} & \frac{1}{3} \\ \frac{1}{3} & -\frac{1}{3} \end{pmatrix}$ $\begin{pmatrix} 1 & 0 \\ 0 & 0.7 \end{pmatrix}$ $\begin{pmatrix} 1 & 1 \\ 1 & -2 \end{pmatrix}$

Why is this? I think the eigenvectors are $x_{1}$=(2 1) and $x_{2}$=(1 -1) but can't understand why the eigenvector matrix S is formed like that.

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    Sorry. I meant $\(2/\sqrt{5},1/\sqrt{5}\)$.2012-11-14

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Yes, $(2,1)$ and $(1,-1)$ are eigenvectors. So are $(2a,a)$ and $(b,-b)$ for any nonzero $a$ and $b$. For reasons that are not clear to me, whoever wrote what you are quoting chose to use $a=b=1/3$, hence eigenvectors $(2/3,1/3)$ and $(1/3,-1/3)$, hence $S=\pmatrix{2/3&1/3\cr1/3&-1/3\cr}$ It seems a strange choice, but there is nothing wrong with it.