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：

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

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