例子 快速计算矩阵的幂

对于矩阵

$$A=\left(\begin{array}{rr}1& 2\\ -1& 4\end{array}\right)$$

我们可以很容易地算知

$${\mathbf{A}}^2=\left(\begin{array}{ll}-1& 10\\ -5& 14\end{array}\right),{\mathbf{A}}^3=\left(\begin{array}{ll}-11& 38\\ -19& 46\end{array}\right),{\mathbf{A}}^4=\left(\begin{array}{ll}-49& 130\\ -65& 146\end{array}\right),.$$

但是,当让我们算 ${\mathbf{A}}^{100}$ 时,如果直接计算,计算量势必会很大,不可取. 那么, 是不是存在一种简单易行的方法呢? 我们在随后的两节中将给出一种计算这类题目的简便方法. 单考虑上面的矩阵, 我们可以找到一个可逆矩阵 (求法参见本节末)

$$P=\left(\begin{array}{ll}2& -3\\ 1& -3\end{array}\right)$$

使得

$$\begin{array}{rl}{\mathbf{P}}^{-1}\mathbf{AP}& =\left(\begin{array}{rr}1& -1\\ \frac{1}{3}& -\frac{2}{3}\end{array}\right)\left(\begin{array}{rr}1& 2\\ -1& 4\end{array}\right)\left(\begin{array}{ll}2& -3\\ 1& -3\end{array}\right)\\ & =\left(\begin{array}{ll}2& 0\\ 0& 3\end{array}\right)\\ & =\mathbf{B}.\end{array}$$

这样, $A=PB{P}^{-1}$ ,则有

$$\begin{array}{rl}{A}^k& =\underset{k\text{ times}}{\underbrace{(PB{P}^{-1})(PB{P}^{-1})\cdots (PB{P}^{-1})}}\\ & =P{B}^k{P}^{-1}\\ & =P\left(\begin{array}{cc}{2}^k& 0\\ 0& 3^k\end{array}\right)P^{-1}.\end{array}$$

于是, 我们可以容易算出

$$\begin{array}{rl}{A}^{100}& =\left(\begin{array}{cc}2& -3\\ 1& -3\end{array}\right)\left(\begin{array}{cc}{2}^{100}& 0\\ 0& 3^{100}\end{array}\right)\left(\begin{array}{cc}1& -1\\ \frac{1}{3}& -\frac{2}{3}\end{array}\right)\\ & =\left(\begin{array}{ll}{2}^{101}-3^{100}& -2^{101}+2\cdot 3^{100}\\ 2^{100}-3^{100}& -2^{100}+2\cdot 3^{100}\end{array}\right).\end{array}$$