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手动推导如下公式。

证明：

1. 首先将如下矩阵对角化：
   $$\left\{\begin{array}{cc}1-a& a\\ b& 1-b\end{array}\right\}$$

(1)求如下矩阵的特征值：
$$\left\{\begin{array}{cc}1-a& a\\ b& 1-b\end{array}\right\}\left\{\begin{array}{c}{x}_1\\ x_2\end{array}\right\}=\lambda \left\{\begin{array}{c}{x}_1\\ x_2\end{array}\right\}==>$$
$$\left|\begin{array}{cc}1-a-\lambda & a\\ b& 1-b-\lambda \end{array}\right|=0==>$$
$$(1-a-\lambda )(1-b-\lambda )-ab=0==>$$
$$\begin{gathered} {\lambda }^2+(a+b-2)\lambda +(1-a-b)=0==> \\ \lambda =\frac{(2-a-b)+-\sqrt{(a+b-2{)}^2-4(1-a-b)}}{2}= \\ \frac{(2-a-b)+-(a+b)}{2}=(1)or(1-a-b) \end{gathered}$$

(2)求得正交特征向量

$$\left|\begin{array}{cc}-a& a\\ b& -b\end{array}\right|\left|\begin{array}{c}{x}_1\\ x_2\end{array}\right|=0==>x_1=1,x_2=1$$

$$\left|\begin{array}{cc}b& a\\ b& a\end{array}\right|\left|\begin{array}{c}{x}_1\\ x_2\end{array}\right|=0==>x_1=a,x_2=-b$$

也即：
$$A=P^{-1}\Lambda P=\left\{\begin{array}{cc}\frac{1}{\sqrt{2}}& \frac{a}{\sqrt{a^2+b^2}}\\ & \\ \frac{1}{\sqrt{2}}& \frac{-b}{\sqrt{a^2+b^2}}\end{array}\right\}\left\{\begin{array}{cc}1& 0\\ & \\ 0& 1-a-b\end{array}\right\}\left\{\begin{array}{cc}\frac{1}{\sqrt{2}}& \frac{1}{\sqrt{2}}\\ & \\ \frac{a}{\sqrt{a^2+b^2}}& \frac{-b}{\sqrt{a^2+b^2}}\end{array}\right\}$$

$$\begin{gathered} A^n=P^{-1}{\Lambda }^{n}P=\left\{\begin{array}{cc}\frac{1}{\sqrt{2}}& \frac{a}{\sqrt{a^2+b^2}}\\ & \\ \frac{1}{\sqrt{2}}& \frac{-b}{\sqrt{a^2+b^2}}\end{array}\right\}\left\{\begin{array}{cc}1& 0\\ & \\ 0& (1-a-b{)}^n\end{array}\right\}\left\{\begin{array}{cc}\frac{1}{\sqrt{2}}& \frac{1}{\sqrt{2}}\\ & \\ \frac{a}{\sqrt{a^2+b^2}}& \frac{-b}{\sqrt{a^2+b^2}}\end{array}\right\}= \\ \left\{\begin{array}{cc}\frac{1}{2}+\frac{a^2(1-a-b{)}^2}{a^2+b^2}& \frac{1}{2}+\frac{-ab(1-a-b{)}^2}{a^2+b^2}\\ & \\ \frac{1}{2}+\frac{-ab(1-a-b{)}^2}{a^2+b^2}& \frac{1}{2}+\frac{b^2(1-a-b{)}^2}{a^2+b^2}\end{array}\right\} \end{gathered}$$