# $~A:=\text{matrix} ~\rightarrow~\lim_{n\to\infty}A^{n}=?~;~$How should I approach against it first

exponentiationlimitslinear algebrasystems of equations

$$A:=\begin{pmatrix} \alpha&1-\alpha\\1-\beta&\beta\\\end{pmatrix}$$
$$\left(0<\alpha,\beta<1\right)~~\wedge~~\left(\alpha+\beta\neq 1\right)$$
$$\underbrace{\lim_{n\to\infty}A^{n}}_{\text{What can I do?}}$$
The problem didn't specified whether$$~n~$$is an natural number.

I think this problem is of a quite typical problem.

$$\det\left(B-\lambda I\right)$$

$$=\det\left( \begin{matrix} \alpha-\lambda&1-\alpha\\ 1-\beta&\beta-\lambda\\ \end{matrix}\right)$$
$$=\det\left(\left(\alpha-\lambda\right)\left(\beta-\lambda\right)-\left(1-\alpha\right)\left(1-\beta\right)\right)$$
$$=\det\left(\left(\lambda-\alpha\right)\left(\lambda-\beta\right)-\left(\alpha-1\right)\left(\beta-1\right)\right)$$
$$=\det\left(\lambda^{2}-\left(\alpha+\beta\right)\lambda+\alpha\beta-\left(\alpha\beta-\alpha-\beta+1\right)\right)$$
$$=\det\left(\lambda^{2}-\left(\alpha+\beta\right)\lambda+\alpha\beta-\alpha\beta+\alpha+\beta-1\right)$$
$$=\det\left(\lambda^{2}-\left(\alpha+\beta\right)\lambda+\left(\left(\alpha+\beta\right)-1\right)\right)$$
$$\lambda=\frac{\left(\alpha+\beta\right)\pm\sqrt{\left(\alpha+\beta\right)^{2}-4\left(\left(\alpha+\beta\right)-1\right)}}{2}$$
$$\left(\alpha+\beta\right)^{2}-4\left(\left(\alpha+\beta\right)-1\right)$$
$$=\left(\alpha+\beta\right)^{2}-4\left(\alpha+\beta\right)+4$$
$$=\left(\left(\alpha+\beta\right)-2\right)^{2}\geq0$$

$$\therefore~~~\lambda=\frac{\left(\alpha+\beta\right)\pm\sqrt{\left(\left(\alpha+\beta\right)-2\right)^{2}}}{2}$$

$$=\frac{\left(\alpha+\beta\right)\pm\sqrt{\left(2-\left(\alpha+\beta\right)\right)^{2}}}{2}$$

$$=\frac{\left(\alpha+\beta\right)\pm\left|2-\left(\alpha+\beta\right)\right|}{2}$$

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.

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 $$\lim_{n\to \infty}A^n=\frac{-1}{\lambda_1-1}A+\frac{\lambda_1}{\lambda_1-1}I= \left( \begin{array}[cc] 1\frac{\beta-1}{\alpha+\beta-2}& \frac{\alpha-1}{\alpha+\beta-2}\cr \frac{\beta-1}{\alpha+\beta-2} & \frac{\alpha-1}{\alpha+\beta-2} \end{array} \right).$$