文章目录
- 正交化方法
- 示例
- 矩阵正交化
正交化方法
设 R n R^n Rn 中线性无关组 a 1 , a 2 , a 3 , … , a n a_1,a_2,a_3,\dots,a_n a1,a2,a3,…,an，令
β 1 = α 1 β 2 = α 2 − [ α 2 β 1 ] ∣ ∣ β 1 ∣ ∣ β 1 β 3 = α 3 − [ α 3 β 1 ] ∣ ∣ β 1 ∣ ∣ β 1 − [ α 3 β 2 ] ∣ ∣ β 2 ∣ ∣ β 2 β n = α 3 − [ α n β 1 ] ∣ ∣ β 1 ∣ ∣ β 1 − ⋯ − [ α n β n − 1 ] ∣ ∣ β n − 1 ∣ ∣ β n − 1 \begin{aligned} & \beta_1 = \alpha_1 \\ & \beta_2 = \alpha_2 - {\frac{[\alpha_2\beta_1]}{||\beta_1||}} \beta_1 \\ & \beta_3 = \alpha_3 - {\frac{[\alpha_3\beta_1]}{||\beta_1||}} \beta_1 - {\frac{[\alpha_3\beta_2]}{||\beta_2||}} \beta_2 \\ & \beta_n = \alpha_3 - {\frac{[\alpha_n\beta_1]}{||\beta_1||}} \beta_1 - \cdots - {\frac{[\alpha_n\beta_{n-1}]}{||\beta_{n-1}||}} \beta_{n-1} \end{aligned} β1=α1β2=α2−∣∣β1∣∣[α2β1]β1β3=α3−∣∣β1∣∣[α3β1]β1−∣∣β2∣∣[α3β2]β2βn=α3−∣∣β1∣∣[αnβ1]β1−⋯−∣∣βn−1∣∣[αnβn−1]βn−1
该方法称为施密特正交化（Gram–Schmidt process）
[ x , y ] [x, y] [x,y] 为向量的内积， ∣ ∣ x ∣ ∣ = [ x , x ] ||x||=[x,x] ∣∣x∣∣=[x,x]
[ x , y ] = x 1 y 1 + x 2 y 2 + ⋯ + x n y n [x, y] = x_1y_1 + x_2y_2 + \cdots + x_ny_n [x,y]=x1y1+x2y2+⋯+xnyn
示例
将向量组
α 1 = ( 1 , 1 , 0 , 0 ) T , α 2 = ( 1 , 0 , 1 , 0 ) T α 3 = ( − 1 , 0 , 0 , 1 ) T , α 4 = ( 1 , − 1 , − 1 , 1 ) T \begin{align} & \alpha_1=(1,1,0,0)^T,\alpha_2=(1,0,1,0)^T \\ & \alpha_3=(-1,0,0,1)^T,\alpha_4=(1,-1,-1,1)^T \\ \end{align} α1=(1,1,0,0)T,α2=(1,0,1,0)Tα3=(−1,0,0,1)T,α4=(1,−1,−1,1)T
标准正交化
解： 先正交化
β 1 = ( 1 , 1 , 0 , 0 ) T β 2 = ( 1 , 0 , 1 , 0 ) T − 1 2 ( 1 , 1 , 0 , 0 ) T = 1 2 ( 1 , − 1 , 2 , 0 ) T β 3 = ( − 1 , 0 , 0 , 1 ) T + 1 2 ( 1 , 1 , 0 , 0 ) T + 1 6 ( 1 , − 1 , 2 , 0 ) T = 1 3 ( − 1 , 1 , 1 , 3 ) T β 4 = ( 1 , − 1 , − 1 , 1 ) T − 0 − 0 − 0 = ( 1 , − 1 , − 1 , 1 ) T \begin{aligned} & \beta_1 =(1,1,0,0)^T \\ & \beta_2 = (1,0,1,0)^T-\frac{1}{2}(1,1,0,0)^T = \frac{1}{2}(1,-1,2,0)^T \\ & \beta_3 = (-1,0,0,1)^T + \frac{1}{2}(1,1,0,0)^T + \frac{1}{6}(1,-1,2,0)^T = \frac{1}{3}(-1,1,1,3)^T \\ & \beta_4 = (1,-1,-1,1)^T-0-0-0=(1,-1,-1,1)^T \end{aligned} β1=(1,1,0,0)Tβ2=(1,0,1,0)T−21(1,1,0,0)T=21(1,−1,2,0)Tβ3=(−1,0,0,1)T+21(1,1,0,0)T+61(1,−1,2,0)T=31(−1,1,1,3)Tβ4=(1,−1,−1,1)T−0−0−0=(1,−1,−1,1)T
再标准化
β 1 = 1 2 ( 1 , 1 , 0 , 0 ) T β 2 = 1 6 ( 1 , − 1 , 2 , 0 ) T β 3 = 1 2 3 ( − 1 , 1 , 1 , 3 ) T β 4 = 1 2 ( 1 , − 1 , − 1 , 1 ) T \begin{aligned} & \beta_1 = \frac{1}{\sqrt2} (1,1,0,0)^T \\ & \beta_2 = \frac{1}{\sqrt6} (1,-1,2,0)^T \\ & \beta_3 = \frac{1}{2\sqrt3} (-1,1,1,3)^T \\ & \beta_4 = \frac{1}{2} (1,-1,-1,1)^T \end{aligned} β1=21(1,1,0,0)Tβ2=61(1,−1,2,0)Tβ3=231(−1,1,1,3)Tβ4=21(1,−1,−1,1)T
矩阵正交化
A = ( 0 1 1 − 1 1 0 − 1 1 1 − 1 0 1 − 1 1 1 0 ) A = \begin{pmatrix} 0 & 1 & 1 & -1 \\ 1 & 0 & -1 & 1 \\ 1 & -1 & 0 & 1 \\ -1 & 1 & 1 & 0 \\ \end{pmatrix} A= 011−110−111−101−1110
求一正交矩阵 P P P，使 P T A P P^{T}AP PTAP 成对角形。
解：
∣ A − λ E ∣ = ∣ − λ 1 1 − 1 1 − λ − 1 1 1 − 1 − λ 1 − 1 1 1 − λ ∣ = ∣ 1 − λ 1 − λ 1 − λ 1 − λ 1 − λ − 1 1 1 − 1 − λ 1 − 1 1 1 − λ ∣ = ( 1 − λ ) ∣ 1 1 1 1 1 − λ − 1 1 1 − 1 − λ 1 − 1 1 1 − λ ∣ = ( 1 − λ ) ∣ 1 1 1 1 0 − λ − 1 − 2 0 0 − 2 − λ − 1 0 0 2 2 1 − λ ∣ = ( 1 − λ ) 2 ( λ 2 + 2 λ − 3 ) = ( λ − 1 ) 3 ( λ + 3 ) \begin{aligned} & |A-\lambda E| ~=~ \begin{vmatrix}-\lambda & 1 & 1 & -1 \\ 1 & -\lambda & -1 & 1 \\ 1 & -1 & -\lambda & 1 \\ -1 & 1 & 1 & -\lambda \\ \end{vmatrix} ~=~ \begin{vmatrix} 1-\lambda & 1-\lambda & 1-\lambda & 1-\lambda \\ 1 & -\lambda & -1 & 1 \\ 1 & -1 & -\lambda & 1 \\ -1 & 1 & 1 & -\lambda \\ \end{vmatrix} \\\\\\ & =~ (1-\lambda) \begin{vmatrix} 1 & 1 & 1 & 1 \\ 1 & -\lambda & -1 & 1 \\ 1 & -1 & -\lambda & 1 \\ -1 & 1 & 1 & -\lambda \\ \end{vmatrix} ~=~ (1-\lambda) \begin{vmatrix} 1 & 1 & 1 & 1 \\ 0 & -\lambda-1 & -2 & 0 \\ 0 & -2 & -\lambda-1 & 0 \\ 0 & 2 & 2 & 1-\lambda \\ \end{vmatrix} \\\\\\\ & =~ (1-\lambda)^2(\lambda^2+2\lambda-3) = (\lambda-1)^3(\lambda+3) \end{aligned} ∣A−λE∣ = −λ11−11−λ−111−1−λ1−111−λ = 1−λ11−11−λ−λ−111−λ−1−λ11−λ11−λ = (1−λ) 111−11−λ−111−1−λ1111−λ = (1−λ) 10001−λ−1−221−2−λ−121001−λ = (1−λ)2(λ2+2λ−3)=(λ−1)3(λ+3)
求得 λ 1 = λ 2 = λ 3 = 1 , λ 4 = − 3 \large \lambda_1=\lambda_2=\lambda_3=1, \lambda_4=-3 λ1=λ2=λ3=1,λ4=−3 ，
把 λ 1 = 1 \lambda_1=1 λ1=1 （3重）带入齐次方程组，得
A − E = ( − 1 1 1 − 1 1 − 1 − 1 1 1 − 1 − 1 1 − 1 1 1 − 1 ) = ( 1 − 1 − 1 1 0 0 0 0 0 0 0 0 0 0 0 0 ) A - E = \begin{pmatrix}-1 & 1 & 1 & -1 \\ 1 & -1 & -1 & 1 \\ 1 & -1 & -1 & 1 \\-1 & 1 & 1 & -1 \\ \end{pmatrix}= \begin{pmatrix} 1 & -1 & -1 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ \end{pmatrix} A−E= −111−11−1−111−1−11−111−1 = 1000−1000−10001000 { x 1 = x 2 + x 3 − x 4 x 2 = x 2 x 3 = x 3 x 4 = x 4 = > x 2 ( 1 1 0 0 ) + x 3 ( 1 0 1 0 ) + x 4 ( − 1 0 0 1 ) \begin{cases} x_1 = x_2 + x_3 - x_4 \\ x_2 = x_2 \\ x_3 = ~~~~~~~~~x_3 \\ x_4 = ~~~~~~~~~~~~~~~~~~x_4 \\ \end{cases} => x_2 \begin{pmatrix} 1 \\ 1 \\ 0 \\ 0 \end{pmatrix}+ x_3 \begin{pmatrix} 1 \\ 0 \\ 1 \\ 0 \end{pmatrix} + x_4 \begin{pmatrix} -1 \\ 0 \\ 0 \\ 1 \end{pmatrix} ⎩ ⎨ ⎧x1=x2+x3−x4x2=x2x3= x3x4= x4=>x2 1100 +x3 1010 +x4 −1001
得出基础解系 ζ 1 , ζ 2 , ζ 3 \zeta_1,\zeta_2,\zeta_3 ζ1,ζ2,ζ3 ，
ζ 1 = ( 1 1 0 0 ) , ζ 2 = ( 1 0 1 0 ) , ζ 3 = ( − 1 0 0 1 ) \zeta_1 = \begin{pmatrix} 1 \\ 1 \\ 0 \\ 0 \end{pmatrix}, \zeta_2 = \begin{pmatrix} 1 \\ 0 \\ 1 \\ 0 \end{pmatrix}, \zeta_3 = \begin{pmatrix} -1 \\ 0 \\ 0 \\ 1 \end{pmatrix} ζ1= 1100 ,ζ2= 1010 ,ζ3= −1001
将 ζ 1 , ζ 2 , ζ 3 \zeta_1,\zeta_2,\zeta_3 ζ1,ζ2,ζ3 正交化 : 取 η 1 = ζ 1 \eta_1 = \zeta_1 η1=ζ1，
η 2 = ζ 2 − [ η 1 , ζ 2 ] ∣ ∣ η 1 ∣ ∣ η 1 = ( 1 0 1 0 ) − 1 2 ( 1 1 0 0 ) = 1 2 ( 1 − 1 2 0 ) η 3 = ζ 3 − [ η 3 , ζ 1 ] ∣ ∣ η 1 ∣ ∣ η 1 − [ η 3 , ζ 2 ] ∣ ∣ η 2 ∣ ∣ η 2 = ( − 1 0 0 1 ) + 1 2 ( 1 1 0 0 ) + 1 6 ( 1 − 1 2 0 ) = 1 3 ( − 1 1 1 3 ) \begin{aligned} & \eta_2 = \zeta_2 - \frac{[\eta_1,\zeta_2]}{||\eta_1||}\eta_1 = \begin{pmatrix} 1 \\ 0 \\ 1 \\ 0 \end{pmatrix}- \frac{1}{2} \begin{pmatrix} 1 \\ 1 \\ 0 \\ 0 \end{pmatrix}= \frac{1}{2} \begin{pmatrix} 1 \\ -1 \\ 2 \\ 0 \end{pmatrix} \\ & \eta_3 = \zeta_3 - \frac{[\eta_3,\zeta_1]}{||\eta_1||}\eta_1- \frac{[\eta_3,\zeta_2]}{||\eta_2||}\eta_2 = \begin{pmatrix} -1 \\ 0 \\ 0 \\ 1 \end{pmatrix} + \frac{1}{2} \begin{pmatrix} 1 \\ 1 \\ 0 \\ 0 \end{pmatrix} + \frac{1}{6} \begin{pmatrix} 1 \\ -1 \\ 2 \\ 0 \end{pmatrix} = \frac{1}{3} \begin{pmatrix} -1 \\ 1 \\ 1 \\ 3 \end{pmatrix} \end{aligned} η2=ζ2−∣∣η1∣∣[η1,ζ2]η1= 1010 −21 1100 =21 1−120 η3=ζ3−∣∣η1∣∣[η3,ζ1]η1−∣∣η2∣∣[η3,ζ2]η2= −1001 +21 1100 +61 1−120 =31 −1113
将 η 1 , η 2 , η 3 \eta_1,\eta_2,\eta_3 η1,η2,η3 单位化求得 p 1 , p 2 , p 3 p_1,p_2,p_3 p1,p2,p3
p 1 = 1 2 ( 1 , 1 , 0 , 0 ) T p 2 = 1 6 ( 1 , − 1 , 2 , 0 ) T p 3 = 1 12 ( − 1 , 1 , 1 , 3 ) T \begin{aligned} & p_1 = \frac{1}{\sqrt2} (1,1,0,0)^T \\ & p_2 = \frac{1}{\sqrt6} (1,-1,2,0)^T \\ & p_3 = \frac{1}{\sqrt{12}} (-1,1,1,3)^T \end{aligned} p1=21(1,1,0,0)Tp2=61(1,−1,2,0)Tp3=121(−1,1,1,3)T
把 λ 4 = − 3 \lambda_4=-3 λ4=−3 带入齐次方程组，得
A + 3 E = ( 3 1 1 − 1 1 3 − 1 1 1 − 1 3 1 − 1 1 1 3 ) = ( 1 0 0 − 1 0 1 0 1 0 0 1 1 0 0 0 0 ) A + 3E = \begin{pmatrix} 3 & 1 & 1 & -1 \\ 1 & 3 & -1 & 1 \\ 1 & -1 & 3 & 1 \\-1 & 1 & 1 & 3 \\ \end{pmatrix}= \begin{pmatrix} 1 & 0 & 0 & -1 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 0 \\ \end{pmatrix} A+3E= 311−113−111−131−1113 = 100001000010−1110 { x 1 = x 4 x 2 = − x 4 x 3 = − x 4 x 4 = x 4 = > x 4 ( 1 − 1 − 1 1 ) \begin{cases} x_1 = x_4 \\ x_2 = -x_4 \\ x_3 = -x_4 \\ x_4 = x_4 \\ \end{cases} => x_4 \begin{pmatrix} 1 \\ -1 \\ -1 \\ 1 \end{pmatrix} ⎩ ⎨ ⎧x1=x4x2=−x4x3=−x4x4=x4=>x4 1−1−11
得出基础解系 ζ 4 \zeta_4 ζ4，
ζ 4 = ( 1 − 1 − 1 1 ) \zeta_4 = \begin{pmatrix} 1 \\ -1 \\ -1 \\ 1 \end{pmatrix} ζ4= 1−1−11
将 ζ 4 \zeta_4 ζ4 单位化，得 p 4 p_4 p4，
p 4 = 1 2 ( 1 , − 1 , − 1 , 1 ) T p_4 = \frac{1}{2} (1,-1,-1,1)^T p4=21(1,−1,−1,1)T
将 p 1 , p 2 , p 3 , p 4 p_1,p_2,p_3,p_4 p1,p2,p3,p4 构成正交矩阵 P P P
P = ( p 1 , p 2 , p 3 , p 4 ) = ( 1 2 1 6 − 1 12 1 2 1 2 − 1 6 1 12 − 1 2 0 2 6 1 12 − 1 2 0 0 3 12 1 2 ) P = (p_1,p_2,p_3,p_4) = \begin{pmatrix} \frac{1}{\sqrt2} & \frac{1}{\sqrt6} & -\frac{1}{\sqrt{12}} & \frac{1}{2} \\ \frac{1}{\sqrt2} & -\frac{1}{\sqrt6} & \frac{1}{\sqrt{12}} & -\frac{1}{2} \\ 0 & \frac{2}{\sqrt6} & \frac{1}{\sqrt{12}} & -\frac{1}{2} \\ 0 & 0 & \frac{3}{\sqrt{12}} & \frac{1}{2} \\ \end{pmatrix} P=(p1,p2,p3,p4)= 21210061−61620−12112112112321−21−2121
有
P T A P = ( 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 − 3 ) P^{T}AP = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & -3 \\ \end{pmatrix} PTAP= 100001000010000−3
