$~A:=\text{matrix} ~\rightarrow~\lim_{n\to\infty}A^{n}=?~;~$How should I approach against it first? $$A:=\begin{pmatrix} \alpha&1-\alpha\\1-\beta&\beta\\\end{pmatrix}$$
$$\left(0<\alpha,\beta<1\right)~~\wedge~~\left(\alpha+\beta\neq 1\right)$$
$$\underbrace{\lim_{n\to\infty}A^{n}}_{\text{What can I do?}}$$
The problem didn't specified whether$~n~$is an natural number.
I think this problem is of a quite typical problem.
$$\det\left(B-\lambda I\right)$$
$$=\det\left( \begin{matrix} \alpha-\lambda&1-\alpha\\ 1-\beta&\beta-\lambda\\ \end{matrix}\right)$$
$$=\det\left(\left(\alpha-\lambda\right)\left(\beta-\lambda\right)-\left(1-\alpha\right)\left(1-\beta\right)\right)$$
$$=\det\left(\left(\lambda-\alpha\right)\left(\lambda-\beta\right)-\left(\alpha-1\right)\left(\beta-1\right)\right)$$
$$=\det\left(\lambda^{2}-\left(\alpha+\beta\right)\lambda+\alpha\beta-\left(\alpha\beta-\alpha-\beta+1\right)\right)$$
$$=\det\left(\lambda^{2}-\left(\alpha+\beta\right)\lambda+\alpha\beta-\alpha\beta+\alpha+\beta-1\right)$$
$$=\det\left(\lambda^{2}-\left(\alpha+\beta\right)\lambda+\left(\left(\alpha+\beta\right)-1\right)\right)$$
$$\lambda=\frac{\left(\alpha+\beta\right)\pm\sqrt{\left(\alpha+\beta\right)^{2}-4\left(\left(\alpha+\beta\right)-1\right)}}{2}$$
About inside the square root.
$$\left(\alpha+\beta\right)^{2}-4\left(\left(\alpha+\beta\right)-1\right)$$
$$=\left(\alpha+\beta\right)^{2}-4\left(\alpha+\beta\right)+4$$
$$=\left(\left(\alpha+\beta\right)-2\right)^{2}\geq0$$
$$\therefore~~~\lambda=\frac{\left(\alpha+\beta\right)\pm\sqrt{\left(\left(\alpha+\beta\right)-2\right)^{2}}}{2}$$
$$=\frac{\left(\alpha+\beta\right)\pm\sqrt{\left(2-\left(\alpha+\beta\right)\right)^{2}}}{2}$$
$$=\frac{\left(\alpha+\beta\right)\pm\left|2-\left(\alpha+\beta\right)\right|}{2}$$
Since$~\alpha+\beta<2~$is held,
$$\lambda=\frac{\left(\alpha+\beta\right)\pm\left(2-\left(\alpha+\beta\right)\right)}{2}~~\leftarrow~~\text{Removed operator of absolute value}$$
$$\lambda^{+}=\frac{\left(\alpha+\beta\right)+\left(2-\left(\alpha+\beta\right)\right)}{2}$$
$$=1$$
$$\lambda^{-}=\frac{\left(\alpha+\beta\right)-\left(2-\left(\alpha+\beta\right)\right)}{2}$$
$$=\frac{\left(\alpha+\beta\right)-2+\left(\alpha+\beta\right)}{2}$$
$$=\frac{2\left(\alpha+\beta\right)-2}{2}$$
$$=\left(\alpha+\beta\right)-1$$
$$\therefore~~~\lambda=1,\underbrace{\left(\alpha+\beta\right)-1}_{\neq0}$$

$$p_{A}\left(x\right)=x^{2}-sx+s-1$$
$$n\in\mathbb{N}_{\geq2}$$
$$q\left(x\right)=n-2~\text{degree polynomial}$$
$$a_{n},b_{n}\in\mathbb{R}$$
$$\underbrace{x^{n}=q\left(x\right)p_{A}\left(x\right)+a_{n}x+b_{n}}_{\text{I've been struggling to derive it}}$$
$$x^{n}=q\left(x\right)p_{A}\left(x\right)+a_{n}x+b_{n}$$
$$=\left\{\text{const}_{n-2}x^{n-2}+\sum_{i=0}^{n-3?}\text{const}_{i}x^{i}\right\}\left(x^{2}-sx+s-1\right)+a_{n}x+b_{n}$$
About above, at least, I can understand that RHS of the above equation is n degree polynomial but unable to prove that other$~x^{i}~~\leftrightarrow~~i\in\mathbb{N}\setminus\left\{n-2\right\}~$disappears.
I think as$~n~$is greater than 2, then any const takes zero hence$~a_n, b_n~$is always zero except as n is 2.
 A: The characteristic polynomial $p_A$ of $A$ is given by
\begin{eqnarray}
p_A(x)&=&(x-\alpha)(x-\beta)-(1-\alpha)(1-\beta)\cr
&=&x^2-sx+s-1
\end{eqnarray}
where $s=\alpha+\beta$.  The eigenvalues are therefore
\begin{eqnarray}
\lambda&=&\frac{s \pm \sqrt{s^2-4(s-1)}}{2} \cr
&=&\frac{s \pm \sqrt{(s-2)^2}}{2} \cr
&=&\frac{s \pm (2-s)|}{2} \cr
&=&\frac{(1\mp1)s \pm 2}{2} \cr
\lambda_1&=&s-1 \in (-1,1)\cr
\lambda_2&=&1
\end{eqnarray}
For any positive integer $n \ge 2$, there exist a polynomial $q$ of degree $n-2$ and real coefficients $a_n, b_n$  such that
$$\tag{1}
x^n=q(x)p_A(x)+a_nx+b_n
$$
Since $p_A(\lambda)=0$, we have
\begin{eqnarray}
a_n+b_n&=&1\cr
a_n\lambda_1+b_n&=&\lambda_1^n
\end{eqnarray}
i.e.
\begin{eqnarray}
a_n&=&\frac{\lambda_1^n-1}{\lambda_1-1}\cr
b_n&=&\frac{\lambda_1^n-\lambda_1}{1-\lambda_1}
\end{eqnarray}
Using (1), we have
$$
A^n=a_nA+b_nI
$$
Because $|\lambda_1|<1$, we have  $\displaystyle \lim_{n\to \infty}\lambda_1^n=0$
and
$$
\lim_{n\to \infty}a_n=\frac{-1}{\lambda_1-1}, \quad 
\lim_{n\to \infty}b_n=\frac{-\lambda_1}{1-\lambda_1}
$$
Hence
$$
\lim_{n\to \infty}A^n=\frac{-1}{\lambda_1-1}A+\frac{\lambda_1}{\lambda_1-1}I=
\left(
\begin{array}[cc]
1\frac{\beta-1}{\alpha+\beta-2}& \frac{\alpha-1}{\alpha+\beta-2}\cr
\frac{\beta-1}{\alpha+\beta-2} & \frac{\alpha-1}{\alpha+\beta-2}
\end{array}
\right).
$$
