施密特正交化
文章目录
- 公式
- 推导
- 拓展
公式
{ β 1 = α 1 β 2 = α 2 − ( α 2 , β 1 ) ( β 1 , β 1 ) β 1 β 3 = α 3 − ( α 3 , β 1 ) ( β 1 , β 1 ) β 1 − ( α 3 , β 2 ) ( β 2 , β 2 ) β 2 \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. ⎩ ⎨ ⎧β1=α1β2=α2−(β1,β1)(α2,β1)β1β3=α3−(β1,β1)(α3,β1)β1−(β2,β2)(α3,β2)β2
推导
令
{ β 1 = α 1 β 2 = α 2 + k β 1 β 3 = α 3 + m 1 β 1 + m 2 β 2 \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. ⎩ ⎨ ⎧β1=α1β2=α2+kβ1β3=α3+m1β1+m2β2
已知 β 1 , β 2 , β 3 \beta_1,\beta_2,\beta_3 β1,β2,β3 两两正交，即：
{ β 1 T β 2 = 0 ① β 1 T β 3 = 0 ② β 2 T β 3 = 0 ③ \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. ⎩ ⎨ ⎧β1Tβ2=0β1Tβ3=0β2Tβ3=0①②③
由 ① 得：
β 1 T β 2 = 0 ⇒ β 1 T ( α 2 + k β 1 ) = 0 ⇒ β 1 T α 2 + k β 1 T β 1 = 0 ⇒ k = − β 1 T α 2 β 1 T β 1 \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} ⇒⇒⇒β1Tβ2=0β1T(α2+kβ1)=0β1Tα2+kβ1Tβ1=0k=−β1Tβ1β1Tα2
由 ② 得：
β 1 T β 3 = 0 ⇒ β 1 T ( α 3 + m 1 β 1 + m 2 β 2 ) = 0 ⇒ β 1 T α 3 + m 1 β 1 T β 1 + m 2 β 1 T β 2 = 0 ⇒ β 1 T α 3 + m 1 β 1 T β 1 = 0 ⇒ m 1 = − β 1 T α 3 β 1 T β 1 \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} ⇒⇒⇒⇒β1Tβ3=0β1T(α3+m1β1+m2β2)=0β1Tα3+m1β1Tβ1+m2β1Tβ2=0β1Tα3+m1β1Tβ1=0m1=−β1Tβ1β1Tα3
由 ③ 得：
β 2 T β 3 = 0 ⇒ β 2 T ( α 3 + m 1 β 1 + m 2 β 2 ) = 0 ⇒ β 2 T α 3 + m 1 β 2 T β 1 + m 2 β 2 T β 2 = 0 ⇒ β 2 T α 3 + m 2 β 2 T β 2 = 0 ⇒ m 2 = − β 2 T α 3 β 2 T β 2 \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} ⇒⇒⇒⇒β2Tβ3=0β2T(α3+m1β1+m2β2)=0β2Tα3+m1β2Tβ1+m2β2Tβ2=0β2Tα3+m2β2Tβ2=0m2=−β2Tβ2β2Tα3
拓展
以上推导的思路还可以应用于下题中：
【例】设 α 1 , α 2 , α 3 \alpha_1,\alpha_2,\alpha_3 α1,α2,α3 线性无关， A α 1 = α 1 , A α 2 = 2 ( α 1 + α 2 ) , A α 3 = 3 ( α 1 + α 2 + α 3 ) A\alpha_1=\alpha_1, A\alpha_2=2(\alpha_1+\alpha_2), A\alpha_3=3(\alpha_1+\alpha_2+\alpha_3) Aα1=α1,Aα2=2(α1+α2),Aα3=3(α1+α2+α3)，求 A A A 的特征值和特征向量。
【解】待定系数法：
A ( α 1 + m α 2 + n α 3 ) = α 1 + 2 m ( α 1 + α 2 ) + 3 n ( α 1 + α 2 + α 3 ) ⇒ λ ( α 1 + m α 2 + n α 3 ) = α 1 + 2 m ( α 1 + α 2 ) + 3 n ( α 1 + α 2 + α 3 ) ⇒ λ α 1 + m λ α 2 + n λ α 3 = ( 1 + 2 m + 3 n ) α 1 + ( 2 m + 3 n ) α 2 + 3 n α 3 \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} ⇒⇒A(α1+mα2+nα3)=α1+2m(α1+α2)+3n(α1+α2+α3)λ(α1+mα2+nα3)=α1+2m(α1+α2)+3n(α1+α2+α3)λα1+mλα2+nλα3=(1+2m+3n)α1+(2m+3n)α2+3nα3
得到以下方程组：
{ 1 + 2 m + 3 n = λ ① 2 m + 3 n = m λ ② 3 n = n λ ③ \left\{ \begin{aligned} &1+2m+3n = \lambda & ①\\ &2m+3n = m \lambda & ②\\ &3n = n \lambda & ③\\ \end{aligned} \right. ⎩ ⎨ ⎧1+2m+3n=λ2m+3n=mλ3n=nλ①②③
由 ③ 可分为两种情形：
（1） n ≠ 0 , λ = 3 n \neq 0, \lambda = 3 n=0,λ=3
将 λ = 3 \lambda = 3 λ=3 代入 ①、②：
{ 1 + 2 m + 3 n = 3 ① 2 m + 3 n = 3 m ② \left\{ \begin{aligned} &1+2m+3n = 3 & ①\\ &2m+3n = 3m & ②\\ \end{aligned} \right. {1+2m+3n=32m+3n=3m①②
解得：
{ m = 2 3 n = 2 9 \left\{ \begin{aligned} &m = \frac{2}{3} \\ &n = \frac{2}{9} \\ \end{aligned} \right. ⎩ ⎨ ⎧m=32n=92
代入到最初的式子可得：
A ( α 1 + m α 2 + n α 3 ) = α 1 + 2 m ( α 1 + α 2 ) + 3 n ( α 1 + α 2 + α 3 ) ⇒ A ( α 1 + 2 3 α 2 + 2 9 α 3 ) = α 1 + 4 3 ( α 1 + α 2 ) + 2 3 ( α 1 + α 2 + α 3 ) ⇒ A ( α 1 + 2 3 α 2 + 2 9 α 3 ) = 3 ( α 1 + 2 3 α 2 + 2 9 α 3 ) \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} ⇒⇒A(α1+mα2+nα3)=α1+2m(α1+α2)+3n(α1+α2+α3)A(α1+32α2+92α3)=α1+34(α1+α2)+32(α1+α2+α3)A(α1+32α2+92α3)=3(α1+32α2+92α3)
（2） n = 0 n = 0 n=0
将 n = 0 n = 0 n=0 代入 ①、②：
{ 1 + 2 m = λ ① 2 m = m λ ② \left\{ \begin{aligned} &1+2m = \lambda & ①\\ &2m = m \lambda & ②\\ \end{aligned} \right. {1+2m=λ2m=mλ①②
由 ② 又分为两种情形：
（2-1） m ≠ 0 , λ = 2 m \neq 0, \lambda = 2 m=0,λ=2
λ = 2 \lambda = 2 λ=2 代入 ① 得： 1 + 2 m = 2 1+2m=2 1+2m=2，解得 m = 1 2 m=\frac{1}{2} m=21
将已求得结果代入最初式子得：
A ( α 1 + 1 2 α 2 ) = 2 ( α 1 + 1 2 α 2 ) A (\alpha_1 + \frac{1}{2} \alpha_2) = 2 (\alpha_1 + \frac{1}{2} \alpha_2) A(α1+21α2)=2(α1+21α2)
（2-2） m = 0 m = 0 m=0
m = 0 m = 0 m=0 代入 ① 得： λ = 1 \lambda = 1 λ=1
将已求得结果代入最初式子得： A α 1 = α 1 A \alpha_1 = \alpha_1 Aα1=α1
（3）总结
所以 A A A 的特征值为 3 , 2 , 1 3,2,1 3,2,1
对应的特征向量为 α 1 + 2 3 α 2 + 2 9 α 3 ， α 1 + 1 2 α 2 ， α 1 \alpha_1 + \frac{2}{3} \alpha_2 + \frac{2}{9} \alpha_3，\alpha_1 + \frac{1}{2} \alpha_2，\alpha_1 α1+32α2+92α3，α1+21α2，α1