Let ${a_{n + 1}} = \frac{{{a_n} - 1}}{{{a_n} + 1}}.$ How can we found general term of this sequence? Let


*

*$\left\{ {{a_n}} \right\}$ be a sequence.

*$\alpha \in \mathbb{R} .$

*$a_1=\alpha$

*${a_{n + 1}} = \frac{{{a_n} - 1}}{{{a_n} + 1}}.$


How can we found general term of this sequence?
 A: Assume that $a_n=\frac{p_n}{q_n}$ is represented by the vector $v_n=(p_n,q_n)^T\in\mathbb{R}^2$. Then 
$$ a_{n+1} = \frac{p_n-q_n}{p_n+q_n} $$
is represented by the vector
$$ v_{n+1}=\begin{pmatrix}p_{n+1}\\ q_{n+1}\end{pmatrix} = \begin{pmatrix}1 & -1 \\ 1 & 1\end{pmatrix}v_n$$
so we may solve the problem by finding an explicit expression for $M^n$ where $M=\begin{pmatrix}1 & -1 \\ 1 & 1\end{pmatrix}$.
Since the eigenvalues of $M$ are $1+i$ and $1-i$, by diagonalizing $M$ we get that
$$ p_n = A(1+i)^n + B(1-i)^n,\qquad q_n = C(1+i)^n+D(1-i)^n $$
for a set of constants $A,B,C,D$ that depend on $p_0,q_0,p_1,q_1$. By imposing $p_1=\alpha,q_1=1$, $p_2=\alpha-1$, $q_2=\alpha+1$ it is easy to find $A,B,C,D$, then the general expression for $a_n$:
$$ a_n = \color{red}{\frac{(1-i\alpha)+(\alpha-i)(-i)^n}{(1+i\alpha)(-i)^n-(\alpha+i)}}.$$
We may notice that since $M^4=-4I$, $\;\color{red}{a_n=a_{n+4}}$, so the whole sequence is made by the terms $\alpha,\frac{\alpha-1}{\alpha+1},-\frac{1}{\alpha}$ and $\frac{1+\alpha}{1-\alpha}$ only.
