# Proofs with sequences, eigenvalues and matrices

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

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

$v_0$ 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. – Brian M. Scott Nov 8 '12 at 22:09
If you prefer round brackets this is pmatrix . – Julian Kuelshammer Nov 8 '12 at 22:11

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 $M(x+y)=Mx+My$.
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... – Bob Nov 8 '12 at 22:13
Second hint: What is $Mu$ if $u$ is an eigenvector? – Julian Kuelshammer Nov 8 '12 at 22:17