是生成这些数的矩阵关系. 矩阵的特征值与特征向量, $\mathbf{A}$ , 计算如下

$$\frac{1}{2} + \frac{\sqrt{5}}{2} \text{ with eigenvector } X = \begin{bmatrix} \frac{1}{2} + \frac{\sqrt{5}}{2} \ 1 \end{bmatrix} \text{ and}$$

$$\frac{1}{2} - \frac{\sqrt{5}}{2} \text{ with eigenvector } Y = \begin{bmatrix} \frac{1}{2} - \frac{\sqrt{5}}{2} \ 1 \end{bmatrix}$$

矩阵形式可以反复应用:

$$\begin{bmatrix} F_{k+1} \ F_k \end{bmatrix} = \mathbf{A} \begin{bmatrix} F_k \ F_{k-1} \end{bmatrix} = \mathbf{A}^2 \begin{bmatrix} F_{k-1} \ F_{k-2} \end{bmatrix} = \mathbf{A}^3 \begin{bmatrix} F_{k-2} \ F_{k-3} \end{bmatrix} = \dots = \mathbf{A}^{k-1} \begin{bmatrix} F_2 \ F_1 \end{bmatrix}$$

由于特征值互不相同, 特征向量是线性无关的, so any 向量可以写成 $aX + bY$ 对于某些常数 $a, b$ 特别是

$$\begin{bmatrix} F_2 \ F_1 \end{bmatrix} = \begin{bmatrix} 1 \ 1 \end{bmatrix} = \frac{1}{\sqrt{5}}(\frac{1}{2} + \frac{\sqrt{5}}{2})X - \frac{1}{\sqrt{5}}(\frac{1}{2} - \frac{\sqrt{5}}{2})Y$$

Since $\mathbf{A}^{k-1}X = (\frac{1}{2} + \frac{\sqrt{5}}{2})^{k-1}X$ and $\mathbf{A}^{k-1}Y = (\frac{1}{2} - \frac{\sqrt{5}}{2})^{k-1}Y$

$$\begin{bmatrix} F_{k+1} \ F_k \end{bmatrix} = \mathbf{A}^{k-1} \begin{bmatrix} 1 \ 1 \end{bmatrix} = \frac{1}{\sqrt{5}}(\frac{1}{2} + \frac{\sqrt{5}}{2})^k X - \frac{1}{\sqrt{5}}(\frac{1}{2} - \frac{\sqrt{5}}{2})Y$$

于是我们可以从第二行推出

$$F_k = \frac{1}{\sqrt{5}}(\frac{1}{2} + \frac{\sqrt{5}}{2})^k - \frac{1}{\sqrt{5}}(\frac{1}{2} - \frac{\sqrt{5}}{2})^k$$

并且这个公式生成斐波那契数.

Example 5.46

求下列矩阵的特征值与特征向量

$$\mathbf{A} = \begin{bmatrix} \cos \theta & -\sin \theta \ \sin \theta & \cos \theta \end{bmatrix}$$

Solution Now

$$\begin{aligned} |\lambda I - \mathbf{A}| &= \begin{vmatrix} \lambda - \cos \theta & \sin \theta \ -\sin \theta & \lambda - \cos \theta \end{vmatrix} \ &= \lambda^2 - 2\lambda \cos \theta + \cos^2 \theta + \sin^2 \theta = \lambda^2 - 2\lambda \cos \theta + 1 \end{aligned}$$