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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. –  Gautam Shenoy Nov 14 '12 at 7:36
    
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... –  ssku Nov 14 '12 at 7:42
    
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 $(\frac{2}{\sqrt{5}}, \frac{2}{\sqrt{5}})$. –  Gautam Shenoy Nov 14 '12 at 7:50
    
S in any sense is not unique. –  Gautam Shenoy Nov 14 '12 at 7:51
    
Sorry. I meant $\(2/\sqrt{5},1/\sqrt{5}\)$. –  Gautam Shenoy Nov 14 '12 at 8:18
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1 Answer 1

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.

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