Sequence of matrices

Question

Suppose that you have parameters $\alpha, \beta \in \mathbb{R}$ and a matrix $A_{0} \in \mathbb{R}^{2 \times 2}$ where $A_{0}$ is a rotational matrix given like below

$$ A_{0} = \begin{bmatrix} \cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta) \end{bmatrix} $$

Now suppose that

$$ A_{i} = \begin{align}\begin{cases} \alpha A_{i-1} & i \textrm{ is even } \\ (1+\beta) A_{i-1} & i \textrm{ is odd } \end{cases} \end{align}$$

What can you say about the behavior of $A_{i}$ given the parameters $\alpha, \beta$?

Answer 1

\begin{align} A_i &= \begin{cases} \alpha A_{i-1} & i \text{ is even}\\ (1+\beta) A_{i-1} & i \text{ is odd} \end{cases}\\ &= \begin{cases} \alpha(1+\beta) A_{i-2} & i \text{ is even}\\ (1+\beta)\alpha A_{i-2} & i \text{ is odd} \end{cases} = (1+\beta)\alpha A_{i-2}\\ &= \begin{cases} \alpha(1+\beta)\alpha A_{i-3} & i \text{ is even}\\ (1+\beta)\alpha(1+\beta) A_{i-3} & i \text{ is odd} \end{cases}\\ \vdots\\ &=\begin{cases} [\alpha(1+\beta)]^{i/2} A_{0} & i \text{ is even}\\ (1+\beta)[\alpha(1+\beta)]^{(i-1)/2} & i \text{ is odd} \end{cases} \end{align}

Answer 2

$A_{2i}=\alpha^i(1+\beta)^iA_0$ and $A_{2i+1}=\alpha^i(1+\beta)^{i+1}A_0$.