10._施密特正交化例题

施密特正交化例题

在上一节解释了施密特正交化。下面举几个例子进行说明。很遗憾，施密特正交化即难理解，计算量又非常复杂，稍微不细心就会做错。但是，在期末考试或者考研里，必有一题和正交化相关试题，他是线性代数的核心内容。

例题

例 在欧氏空间 R3R^3 中，对于基 β1=(1,1,1)\beta_1=(1,1,1) ， β2=(1,1,0),β3=(1,0,0)\beta_2=(1,1,0), \quad \beta_3=(1,0,0) 施行正交化方法，求出 R3R^3 的一个标准正交基。

解 取 γ1=β1=(1,1,1)\gamma_1=\beta_1=(1,1,1) ，由施米特正交化方法：

γ2=β2−(β2,γ1)(γ1,γ1)γ1=(13,13,−23),γ3=β3−(β3,γ1)(γ1,γ1)γ1−(β3,γ2)(γ2,γ2)γ2=(12,−12,0),\begin{aligned} & \gamma_2=\beta_2-\frac{\left(\beta_2, \gamma_1\right)}{\left(\gamma_1, \gamma_1\right)} \gamma_1=\left(\frac{1}{3}, \frac{1}{3},-\frac{2}{3}\right), \\ & \gamma_3=\beta_3-\frac{\left(\beta_3, \gamma_1\right)}{\left(\gamma_1, \gamma_1\right)} \gamma_1-\frac{\left(\beta_3, \gamma_2\right)}{\left(\gamma_2, \gamma_2\right)} \gamma_2=\left(\frac{1}{2},-\frac{1}{2}, 0\right), \end{aligned}

所以 {γ1,γ2,γ3}\left\{\gamma_1, \gamma_2, \gamma_3\right\} 是 R3R^3 的一个正交基；

再令

α1=γ1∣γ1∣=(13,13,13),α2=γ2∣γ2∣=(16,16,−26),α3=γ3∣γ3∣=(12,−12,0),\begin{aligned} & \alpha_1=\frac{\gamma_1}{\left|\gamma_1\right|}=\left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right), \\ & \alpha_2=\frac{\gamma_2}{\left|\gamma_2\right|}=\left(\frac{1}{\sqrt{6}}, \frac{1}{\sqrt{6}},-\frac{2}{\sqrt{6}}\right), \\ & \alpha_3=\frac{\gamma_3}{\left|\gamma_3\right|}=\left(\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}, 0\right), \end{aligned}

则 {α1,α2,α3}\left\{\alpha_1, \alpha_2, \alpha_3\right\} 即为欧氏空间 R3R^3 的一个标准正交基。 此例可以看出欧氏空间中的标准正交基不唯一，对 R3R^3 而言 ε1=(1,0,0),ε2=(0,1,0),ε3=(0,0,1)\varepsilon_1=(1,0,0), \quad \varepsilon_2=(0,1,0), \quad \varepsilon_3=(0,0,1) 也是一个标准正交基。

例将线性无关向量组化为标准正交向量组

α1=(1,1,1,1)T,α2=(1,−2,−3,−4)T,α3=(1,2,2,3)T\alpha_1=( 1 , 1,1,1)^T, \alpha_2=(1,-2,-3,-4)^T, \alpha_3=(1,2,2,3)^T

解 (1) 正交化

β1=α1=(1,1,1,1)Tβ2=α2−(α2,β1)(β1,β1)β1=(3,0,−1,−2)T\begin{aligned} \beta_1 & =\alpha_1=( 1 , 1 , 1,1)^T \\ \beta_2 & =\alpha_2-\frac{\left(\alpha_2, \beta_1\right)}{\left(\beta_1, \beta_1\right)} \beta_1 \\ & =( 3 , 0 ,-1,-2)^T \end{aligned}
β3=α3−(α3,β1)(β1,β1)β1−(α3,β2)(β2,β2)β2=(114,0,−514,414)T\begin{aligned} \beta_3 & =\alpha_3-\frac{\left(\alpha_3, \beta_1\right)}{\left(\beta_1, \beta_1\right)} \beta_1-\frac{\left(\alpha_3, \beta_2\right)}{\left(\beta_2, \beta_2\right)} \beta_2 \\ & =\left(\frac{1}{14}, 0,-\frac{5}{14}, \frac{4}{14}\right)^T \end{aligned}

(2)单位化

γ1=1∣β1∣β1=112+12+12+12(1,1,1,1)T=(12,12,12,12)T,γ2=1∣β2∣β2=(314,0,−114,−214)T,γ3=1∣β3∣β3=(142,0,−542,442)T.\begin{aligned} & \gamma_1=\frac{1}{\left|\beta_1\right|} \beta_1=\frac{1}{\sqrt{1^2+1^2+1^2+1^2}}(1,1,1,1)^T=\left(\frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2}\right)^T, \\ & \gamma_2=\frac{1}{\left|\beta_2\right|} \beta_2=\left(\frac{3}{\sqrt{14}}, 0,-\frac{1}{\sqrt{14}},-\frac{2}{\sqrt{14}}\right)^T, \\ & \gamma_3=\frac{1}{\left|\beta_3\right|} \beta_3=\left(\frac{1}{\sqrt{42}}, 0,-\frac{5}{\sqrt{42}}, \frac{4}{\sqrt{42}}\right)^T . \end{aligned}

则 γ1,γ2,γ3\gamma_1, \gamma_2, \gamma_3 为所求.

例 设 A=[422242224]A =\left[\begin{array}{lll}4 & 2 & 2 \\ 2 & 4 & 2 \\ 2 & 2 & 4\end{array}\right] ，求正交阵 QQ ，使得 Q−1AQ=ΛQ ^{-1} A Q = \Lambda 为对角阵． 解 ∣λE−A∣=∣λ−4−2−2−2λ−4−2−2−2λ−4∣=(λ−8)(λ−2)2|\lambda E - A |=\left|\begin{array}{ccc}\lambda-4 & -2 & -2 \\ -2 & \lambda-4 & -2 \\ -2 & -2 & \lambda-4\end{array}\right|=(\lambda-8)(\lambda-2)^2 ，故 AA 的特征值为 λ1=8,λ2=λ3=2\lambda_1=8, \lambda_2=\lambda_3=2 ．

当 λ1=8\lambda_1=8 时，解齐次方程组

(8E−A)X=[4−2−2−24−2−2−24][x1x2x3]=[000](8 E - A ) X =\left[\begin{array}{ccc} 4 & -2 & -2 \\ -2 & 4 & -2 \\ -2 & -2 & 4 \end{array}\right]\left[\begin{array}{l} x_1 \\ x_2 \\ x_3 \end{array}\right]=\left[\begin{array}{l} 0 \\ 0 \\ 0 \end{array}\right]

得基础解系 ξ1=[1,1,1]T\xi _1=[1,1,1]^{ T } ．由于此基础解系只有一个解向量，故无需正交化，只需规范化，得

η1=1∥ξ1∥⋅ξ1=[13,13,13]T.\eta _1=\frac{1}{\left\| \xi _1\right\|} \cdot \xi _1=\left[\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right]^{T} .

当 λ2=λ3=2\lambda_2=\lambda_3=2 时，解齐次方程组

(2E−A)X=[−2−2−2−2−2−2−2−2−2]X=[000](2 E - A ) X =\left[\begin{array}{lll} -2 & -2 & -2 \\ -2 & -2 & -2 \\ -2 & -2 & -2 \end{array}\right] X =\left[\begin{array}{l} 0 \\ 0 \\ 0 \end{array}\right]

得基础解系 ξ2=[−1,1,0]T,ξ3=[−1,0,1]T\xi _2=[-1,1,0]^{ T }, \xi _3=[-1,0,1]^{ T } ，利用施密特正交化方法将其正交化得

β2=ξ2=[−1,1,0]T,β3=ξ3−(ξ3,β2)(β2,β2)⋅β2=[−101]−12[−110]=[−12−121].\begin{aligned} & \beta _2= \xi _2=[-1,1,0]^{T}, \\ & \beta _3= \xi _3-\frac{\left( \xi _3, \beta _2\right)}{\left( \beta _2, \beta _2\right)} \cdot \beta _2=\left[\begin{array}{c} -1 \\ 0 \\ 1 \end{array}\right]-\frac{1}{2}\left[\begin{array}{c} -1 \\ 1 \\ 0 \end{array}\right]=\left[\begin{array}{c} -\frac{1}{2} \\ -\frac{1}{2} \\ 1 \end{array}\right] . \end{aligned}

将 β2,β3\beta _2, \beta _3 单位化，得

η2=1∥β2∥⋅β2=[−12120],η3=1∥β3∥⋅β3=[−161626].\eta _2=\frac{1}{\left\| \beta _2\right\|} \cdot \beta _2=\left[\begin{array}{c} -\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} \\ 0 \end{array}\right], \quad \eta _3=\frac{1}{\left\| \beta _3\right\|} \cdot \beta _3=\left[\begin{array}{c} -\frac{1}{\sqrt{6}} \\ \frac{1}{\sqrt{6}} \\ \frac{2}{\sqrt{6}} \end{array}\right] .
 令 Q=[η1,η2,η3]=[13−12−161312−1613026], 则 Q−1AQ=Λ=[800020002]. \begin{aligned} &\text { 令 } Q =\left[ \eta _1, \eta _2, \eta _3\right]=\left[\begin{array}{ccc} \frac{1}{\sqrt{3}} & -\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{6}} \\ \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{6}} \\ \frac{1}{\sqrt{3}} & 0 & \frac{2}{\sqrt{6}} \end{array}\right] \text {, 则 } Q ^{-1} A Q = \Lambda =\left[\begin{array}{lll} 8 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{array}\right] \text {. }\\ \end{aligned}
上一章9._施密特正交化过程
下一章11._本章思维导图