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 \(\alpha_1 + \frac{2}{3} \alpha_2 + \frac{2}{9} \alpha_3，\alpha_1 + \frac{1}{2} \alpha_2，\alpha_1\)
