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$M = \begin{bmatrix}0.95&0.10&0.10\\0.05&0.80&0.05\\0.00&0.10&0.85 \end{bmatrix}$

$v_0 = \begin{bmatrix}x_0\\y_0\\z_0 \end{bmatrix}$

$v_0$ represents the initial state of a market, whereas M describes the switching behavior of consumers. Apparently we can describe this as $v_{n+1}=M \cdot v_n$.

a) Let $u$ be an eigenvector of M corresponding to the value $\lambda$. Prove that the sequence constructed by: $v_n=\lambda^n \cdot u$ satisfies the recursive relation for $n=0,1,2,...$

I don't know how to do this.... I guess I have to make the following substitution: $v_{n+1}=M \cdot \lambda^n \cdot u$ But I don't know how to proceed....

And the second part of the question looks even more difficult:

b) Let $v$ be an eigenvector of M corresponding to $\mu \neq \lambda$ and let &c& and $d$ be some numbers. Prove that the sequence $w_0,w_1,w_2,...$constructed by: $w_k=c \cdot \lambda^n \cdot u + d \cdot \mu^n \cdot v$ also satisfies the recursive relation for $n=0,1,2,...$

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    If you prefer round brackets this is pmatrix .2012-11-08

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Hint: You have to use the fact that $M$ defines a linear map $\mathbb{R}^3\to \mathbb{R}^3$. Hence $M(\lambda x)=\lambda Mx$ and $M(x+y)=Mx+My$.

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    let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/6376/discussion-between-julian-kuelshammer-and-bob)2012-11-08