施密特正交化

公式

$$\left\{ \begin{aligned} &\beta_1 = \alpha_1 \\ &\beta_2 = \alpha_2 - \frac{(\alpha_2, \beta_1)}{(\beta_1,\beta_1)} \beta_1 \\ &\beta_3 = \alpha_3 - \frac{(\alpha_3, \beta_1)}{(\beta_1,\beta_1)} \beta_1 - \frac{(\alpha_3, \beta_2)}{(\beta_2,\beta_2)} \beta_2 \\ \end{aligned} \right.$$

推导

令

$$\left\{ \begin{aligned} &\beta_1 = \alpha_1 \\ &\beta_2 = \alpha_2 + k \beta_1 \\ &\beta_3 = \alpha_3 + m_1 \beta_1 + m_2 \beta_2 \\ \end{aligned} \right.$$

已知 $\beta_1,\beta_2,\beta_3$ 两两正交，即：

$$\left\{ \begin{aligned} &\beta_1^{\mathrm{T}} \beta_2 = 0 & ①\\ &\beta_1^{\mathrm{T}} \beta_3 = 0 & ②\\ &\beta_2^{\mathrm{T}} \beta_3 = 0 & ③\\ \end{aligned} \right.$$

由 ① 得：

$$\begin{aligned} & \beta_1^{\mathrm{T}} \beta_2 = 0 \\ \Rightarrow & \beta_1^{\mathrm{T}} (\alpha_2+k \beta_1) = 0 \\ \Rightarrow & \beta_1^{\mathrm{T}} \alpha_2 + k \beta_1^{\mathrm{T}} \beta_1 = 0 \\ \Rightarrow & k = -\frac{\beta_1^{\mathrm{T}} \alpha_2}{ \beta_1^{\mathrm{T}} \beta_1} \end{aligned}$$

由 ② 得：

$$\begin{aligned} & \beta_1^{\mathrm{T}} \beta_3 = 0 \\ \Rightarrow & \beta_1^{\mathrm{T}} (\alpha_3 + m_1 \beta_1 + m_2 \beta_2) = 0 \\ \Rightarrow & \beta_1^{\mathrm{T}} \alpha_3 + m_1 \beta_1^{\mathrm{T}} \beta_1 + m_2 \beta_1^{\mathrm{T}} \beta_2 = 0 \\ \Rightarrow & \beta_1^{\mathrm{T}} \alpha_3 + m_1 \beta_1^{\mathrm{T}} \beta_1 = 0 \\ \Rightarrow & m_1 = -\frac{\beta_1^{\mathrm{T}} \alpha_3}{ \beta_1^{\mathrm{T}} \beta_1} \end{aligned}$$

由 ③ 得：

$$\begin{aligned} & \beta_2^{\mathrm{T}} \beta_3 = 0 \\ \Rightarrow & \beta_2^{\mathrm{T}} (\alpha_3 + m_1 \beta_1 + m_2 \beta_2) = 0 \\ \Rightarrow & \beta_2^{\mathrm{T}} \alpha_3 + m_1 \beta_2^{\mathrm{T}} \beta_1 + m_2 \beta_2^{\mathrm{T}} \beta_2 = 0 \\ \Rightarrow & \beta_2^{\mathrm{T}} \alpha_3 + m_2 \beta_2^{\mathrm{T}} \beta_2 = 0 \\ \Rightarrow & m_2 = -\frac{\beta_2^{\mathrm{T}} \alpha_3}{ \beta_2^{\mathrm{T}} \beta_2} \end{aligned}$$

一种避开正交化的小技巧

该技巧源自李正元数学全书（线性代数部分是尤承业老师写的）。

假设从 $(A-E)x=0$ 得到方程 $x_1 + 2x_2 - 2x_3 = 0$，该方程显然有两个线性无关的解。

先求出第一个特征向量为 $\alpha_1 = (0,1,1)^{\mathrm{T}}$。为保证与 $\alpha_1$ 正交，第二个特征向量先设为 $\alpha_2 = (c,1,-1)^{\mathrm{T}}$，然后代入到方程中，解得 $c=-4$，所以轻松求得与 $\alpha_1$ 正交的 $\alpha_2 = (-4,1,-1)^{\mathrm{T}}$。

拓展

以上推导的思路还可以应用于下题中：

【例】设 $\alpha_1,\alpha_2,\alpha_3$ 线性无关，$A\alpha_1=\alpha_1, A\alpha_2=2(\alpha_1+\alpha_2), A\alpha_3=3(\alpha_1+\alpha_2+\alpha_3)$，求 $A$ 的特征值和特征向量。

【解】待定系数法：

$$\begin{aligned} & A(\alpha_1 + m\alpha_2 + n\alpha_3) = \alpha_1 + 2m(\alpha_1+\alpha_2) + 3n(\alpha_1+\alpha_2+\alpha_3) \\ \Rightarrow & \lambda (\alpha_1 + m\alpha_2 + n\alpha_3) = \alpha_1 + 2m(\alpha_1+\alpha_2) + 3n(\alpha_1+\alpha_2+\alpha_3) \\ \Rightarrow & \lambda \alpha_1 + m \lambda \alpha_2 + n \lambda \alpha_3 = (1+2m+3n)\alpha_1 + (2m+3n)\alpha_2 + 3n\alpha_3 \end{aligned}$$

得到以下方程组：

$$\left\{ \begin{aligned} &1+2m+3n = \lambda & ①\\ &2m+3n = m \lambda & ②\\ &3n = n \lambda & ③\\ \end{aligned} \right.$$

由 ③ 可分为两种情形：

（1）$n \neq 0, \lambda = 3$

将 $\lambda = 3$ 代入 ①、②：

$$\left\{ \begin{aligned} &1+2m+3n = 3 & ①\\ &2m+3n = 3m & ②\\ \end{aligned} \right.$$

解得：

$$\left\{ \begin{aligned} &m = \frac{2}{3} \\ &n = \frac{2}{9} \\ \end{aligned} \right.$$

代入到最初的式子可得：

$$\begin{aligned} & A(\alpha_1 + m\alpha_2 + n\alpha_3) = \alpha_1 + 2m(\alpha_1+\alpha_2) + 3n(\alpha_1+\alpha_2+\alpha_3) \\ \Rightarrow & A(\alpha_1 + \frac{2}{3} \alpha_2 + \frac{2}{9} \alpha_3) = \alpha_1 + \frac{4}{3}(\alpha_1+\alpha_2) + \frac{2}{3}(\alpha_1+\alpha_2+\alpha_3) \\ \Rightarrow & A(\alpha_1 + \frac{2}{3} \alpha_2 + \frac{2}{9} \alpha_3) = 3(\alpha_1 + \frac{2}{3} \alpha_2 + \frac{2}{9} \alpha_3) \\ \end{aligned}$$

（2）$n = 0$

将 $n = 0$ 代入 ①、②：

$$\left\{ \begin{aligned} &1+2m = \lambda & ①\\ &2m = m \lambda & ②\\ \end{aligned} \right.$$

由 ② 又分为两种情形：

（2-1）$m \neq 0, \lambda = 2$

$\lambda = 2$ 代入 ① 得：$1+2m=2$，解得 $m=\frac{1}{2}$

将已求得结果代入最初式子得：

$$A (\alpha_1 + \frac{1}{2} \alpha_2) = 2 (\alpha_1 + \frac{1}{2} \alpha_2)$$

（2-2）$m = 0$

$m = 0$ 代入 ① 得：$\lambda = 1$

将已求得结果代入最初式子得：$A \alpha_1 = \alpha_1$

（3）总结

所以 $A$ 的特征值为 $3,2,1$

对应的特征向量为 $\alpha_1 + \frac{2}{3} \alpha_2 + \frac{2}{9} \alpha_3，\alpha_1 + \frac{1}{2} \alpha_2，\alpha_1$