Q:

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 $n=0,1,2,...$

A:

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 $M(x+y)=Mx+My$.

