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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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    You can get the brackets with `bmatrix` instead of `matrix`.2012-11-08
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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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    So if I proof M is a linear mapping then, $$v_{n+1}=M \cdot \lambda^n \cdot u$$ is ofcourse on of the points we are mapping to, but I don't see how this is useful...2012-11-08
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    Second hint: What is $Mu$ if $u$ is an eigenvector?2012-11-08
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    The inner product will be zero, but I don't want it to be if I want to satisfy the relation..2012-11-08
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    I don't see an inner product somewhere in your question.2012-11-08
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    No, I know that, but that's the only thing I could come up with to answer your question2012-11-08
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    Recall the definitions, e.g. how is an eigenvector defined. Without knowing the definitions you can't normally answer mathematical questions.2012-11-08
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    Oh yeah ofcourse, I know that $M \cdot u=\lambda \cdot u$2012-11-08
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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