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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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    Note that if $x_1$ is an eigenvector, then $ax_1,\quad a \neq 0$ is also an eigenvector with the same eigenvalue.2012-11-14
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    Um...sorry I can't follow your explanation. How their norm could be 1? root(4/9+1/9)=root(5/9)? Something wrong with my calculation...2012-11-14
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    I changed my previous comment. My point is that when writing an eigenvector, it is standard to normalize it. In this case (2,1) will become $(\frac{2}{\sqrt{5}}, \frac{2}{\sqrt{5}})$. As for $(2/3,1/3)$ the norm is $\sqrt{5}/3$. So when u divide the vector by this, the 3 cancels and you get $(\frac{2}{\sqrt{5}}, \frac{2}{\sqrt{5}})$.2012-11-14
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    S in any sense is not unique.2012-11-14
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    Sorry. I meant $\(2/\sqrt{5},1/\sqrt{5}\)$.2012-11-14

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