Section02_标准化

配方法

  1. f(x1,x2,x3)=x12+2x1x2−2x2x3−5x32f(x_{1},x_{2},x_{3}) = x_{1}^{2} + 2x_{1}x_{2} - 2x_{2}x_{3} - 5x_{3}^{2} A=(11010−10−1−5),X=(x1x2x3)f=X⊺AXf(x1,x2,x3)=(x1+x2)2−(x2+x3)2−4x32令{y1=x1+x2y2=x2+x3y3=x3⇒{x1=y1−y2+y3x2=y2−y3x3=y3 即X=PYP=(1−1101−1001)可逆f=X⊺AX=X=PYY⊺(P⊺AP)Y=y12−y22−4y32 \begin{array}{ll} & \boldsymbol{A} = \left( \begin{smallmatrix} 1 & 1 & 0 \\ 1& 0 & -1 \\ 0 & -1 & -5 \end{smallmatrix} \right), \boldsymbol{X} = \left( \begin{smallmatrix} x_{1}\\x_{2}\\x_{3} \end{smallmatrix} \right)\\ & f = \boldsymbol{X}^{\intercal}\boldsymbol{AX} \\\\ & f(x_{1},x_{2},x_{3}) = (x_{1}+x_{2})^{2} - (x_{2}+x_{3})^{2} - 4x_{3}^{2}\\\\ & \text{令} \begin{cases} y_{1} = x_{1} + x_{2} \\ y_{2} = x_{2} +x_{3} \\ y_{3} = x_{3} \end{cases} \Rightarrow \begin{cases} x_{1} = y_{1} - y_{2} + y_{3} \\ x_{2} = y_{2} - y_{3} \\ x_{3} = y_{3} \end{cases}\text{ 即} \boldsymbol{X} = \boldsymbol{PY} \\ & \boldsymbol{P} = \left( \begin{smallmatrix} 1 & -1 & 1 \\ 0 & 1 & -1 \\ 0 & 0 & 1 \end{smallmatrix} \right)\text{可逆} \\\\ & f = \boldsymbol{X}^{\intercal}\boldsymbol{AX} \xlongequal{\boldsymbol{X} = \boldsymbol{PY}} \boldsymbol{Y}^{\intercal}(\boldsymbol{P}^{\intercal}\boldsymbol{A}\boldsymbol{P})\boldsymbol{Y} = y_{1}^{2} - y_{2}^{2} - 4y_{3}^{2} \end{array}
  2. f(x1,x2,x3)=x12+2x1x2−2x1x3−3x32f(x_{1},x_{2},x_{3}) = x_{1}^{2} + 2x_{1}x_{2} - 2x_{1}x_{3} -3x_{3}^{2} P=(11−1100−10−3),X=(x1x2x3)f=X⊺AXf(x1,x2,x3)=(x1+x2−x3)2+2x2x3−x22−4x32=(x1+x2−x3)2−(x2−x3)2−3x32令{y1=x1+x2−x3y2=x2−x3y3=x3⇒{x1=y1−y2x2=y2+y3x3=y3 即X=PYP=(1−10011001)可逆f=X⊺AX=X=PYY⊺(P⊺AP)Y=y12−y22−3y32 \begin{array}{ll} & \boldsymbol{P} = \left( \begin{smallmatrix} 1 & 1 & -1 \\ 1 & 0 & 0\\ -1 & 0 & -3 \end{smallmatrix} \right), \boldsymbol{X} = \left( \begin{smallmatrix} x_{1} \\x_{2} \\x_{3} \end{smallmatrix} \right) \\ & f = \boldsymbol{X}^{\intercal}\boldsymbol{AX} \\ \\ & f(x_{1},x_{2},x_{3}) = (x_{1} + x_{2} - x_{3})^{2} + 2x_{2}x_{3}-x_{2}^{2} -4x^{2}_{3} \\ & = (x_{1} + x_{2} -x_{3})^{2}-(x_{2}-x_{3})^{2} - 3x_{3}^{2} \\\\ & \text{令}\begin{cases} y_{1} = x_{1}+x_{2}-x_{3} \\ y_{2} = x_{2} - x_{3} \\ y_{3} = x_{3} \end{cases} \Rightarrow \begin{cases} x_{1} = y_{1} - y_{2} \\ x_{2} = y_{2} + y_{3} \\ x_{3} = y_{3} \end{cases}\text{ 即} \boldsymbol{X} = \boldsymbol{PY} \\ & \boldsymbol{P} = \left( \begin{smallmatrix} 1 & -1 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{smallmatrix} \right) \text{可逆} \\\\ & f = \boldsymbol{X}^{\intercal}\boldsymbol{AX} \xlongequal{\boldsymbol{X} = \boldsymbol{PY}} \boldsymbol{Y}^{\intercal}(\boldsymbol{P}^{\intercal}\boldsymbol{AP})\boldsymbol{Y} = y_{1}^{2} - y_{2}^{2} - 3y_{3}^{2} \end{array}

正交变化法

步骤

  1. f=X⊺AXf= \boldsymbol{X}^{\intercal}\boldsymbol{AX}，A⊺=A\boldsymbol{A}^{\intercal} = \boldsymbol{A}
  2. det⁡(λE−A)=0⇒λ1,⋯ ,λn\det(\lambda \boldsymbol{E} - \boldsymbol{A})= 0\Rightarrow \lambda_{1},\cdots,\lambda_{n}
  3. (λiE−A)X=0⇒α1,⋯ ,αn(\lambda_{i}\boldsymbol{E} - \boldsymbol{A})\boldsymbol{X} = \boldsymbol{0}\Rightarrow \boldsymbol{\alpha}_{1},\cdots, \boldsymbol{\alpha}_{n}
  4. α1,⋯ ,αn⇒标准化正交化γ1,⋯ ,γn\boldsymbol{\alpha}_{1}, \cdots,\boldsymbol{\alpha}_{n} \xRightarrow[\text{标准化}]{\text{正交化}} \boldsymbol{\gamma}_{1},\cdots, \boldsymbol{\gamma}_{n}
    • Aγ1=λ1γ1,⋯ ,Aγn=λnγn\boldsymbol{A\gamma}_{1} = \lambda_{1}\boldsymbol{\gamma}_{1}, \cdots, \boldsymbol{A\gamma}_{n}= \lambda_{n}\boldsymbol{\gamma}_{n}
    • 令 Q=(γ1⋯γn), Q⊺Q=E, Q⊺AQ=(λ1⋱λn)\boldsymbol{Q} = \left( \begin{matrix} \boldsymbol{\gamma}_{1} & \cdots & \boldsymbol{\gamma}_{n} \end{matrix} \right),\ \boldsymbol{Q}^{\intercal}\boldsymbol{Q} = \boldsymbol{E},\ \boldsymbol{Q}^{\intercal}\boldsymbol{AQ} = \left( \begin{matrix} \lambda_{1} \\ &\ddots \\ && \lambda_{n} \end{matrix} \right)
  5. f=X⊺AX=X=QYY⊺(Q⊺AQ)Y=λ1y12+⋯+λnyn2f=\boldsymbol{X}^{\intercal}\boldsymbol{AX} \xlongequal{\boldsymbol{X} = \boldsymbol{QY}} \boldsymbol{Y}^{\intercal}(\boldsymbol{Q}^{\intercal}\boldsymbol{AQ})\boldsymbol{Y} = \lambda_{1}y_{1}^{2} + \cdots + \lambda_{n}y_{n}^{2}

例题

  1. f(x1,x2,x3)=4x1x2+4x1x3+4x2x3f(x_{1},x_{2},x_{3}) = 4x_{1}x_{2} + 4x_{1}x_{3} + 4x_{2}x_{3} A=(022202220),X=(x1x2x3)⇒f=X⊺AXdet⁡(λE−A)=∣λ−2−2−2λ−2−2−2λ∣=(λ−4)∣1110λ+2000λ+2∣=(λ−4)(λ+2)2∴λ1=4,λ2=λ3=−24E−A=(4−2−2−24−2−2−24)→(10−101−1000)∴λ=4对应的无关特征向量α1=(111)2E+A→(111000000)∴λ=−2对应的无关特征向量α2=(−110),α3=(−101)∴β1=α1,β2=α2,β3=α3−(β2,α3)(β2,β2)β2=(−12−121)γ1=33β1,γ2=22β2,γ3=63β3令Q=(γ1γ2γ3)=(33−22−663322−6633063),Q⊺Q=Ef=X⊺AX=X=QYY⊺(Q⊺AQ)Y=4y12−2y22−2y32 \begin{array}{ll} & \boldsymbol{A} = \left( \begin{smallmatrix} 0 & 2 & 2 \\ 2 & 0 & 2 \\ 2 & 2 & 0 \\ \end{smallmatrix} \right), \boldsymbol{X} = \left( \begin{smallmatrix} x_{1} \\ x_{2} \\ x_{3} \end{smallmatrix} \right)\Rightarrow f= \boldsymbol{X}^{\intercal}\boldsymbol{AX} \\ & \det(\lambda\boldsymbol{E}- \boldsymbol{A}) = \left\vert \begin{smallmatrix} \lambda & -2 & -2 \\ -2 & \lambda & -2 \\ -2 & -2 & \lambda \end{smallmatrix} \right\vert = (\lambda -4) \left\vert \begin{smallmatrix} 1 & 1 & 1\\ 0 & \lambda +2 & 0 \\ 0 & 0 & \lambda +2 \end{smallmatrix} \right\vert = (\lambda-4)(\lambda+2)^{2} \\ \therefore & \lambda_{1} = 4, \lambda_{2} = \lambda_{3} = -2 \\ & 4 \boldsymbol{E} - \boldsymbol{A} = \left( \begin{smallmatrix} 4 & -2 & -2 \\ -2 & 4 & -2 \\ -2 & -2 & 4 \\ \end{smallmatrix} \right)\rightarrow \left( \begin{smallmatrix} 1 & 0 & -1 \\ 0 & 1 & -1 \\ 0 & 0 & 0 \end{smallmatrix} \right) \\ \therefore & \lambda=4 \text{对应的无关特征向量} \boldsymbol{\alpha}_{1} = \left( \begin{smallmatrix} 1 \\ 1 \\ 1 \end{smallmatrix} \right)\\ & 2 \boldsymbol{E} + \boldsymbol{A} \rightarrow \left( \begin{smallmatrix} 1 & 1 & 1 \\ 0 & 0 & 0\\ 0 & 0 & 0\\ \end{smallmatrix} \right) \\ \therefore & \lambda=-2 \text{对应的无关特征向量} \boldsymbol{\alpha}_{2} = \left( \begin{smallmatrix} -1 \\ 1 \\ 0 \end{smallmatrix} \right), \boldsymbol{\alpha}_{3} = \left( \begin{smallmatrix} -1 \\ 0 \\1 \end{smallmatrix} \right)\\ \therefore & \boldsymbol{\beta}_{1} = \boldsymbol{\alpha}_{1} , \boldsymbol{\beta}_{2} = \boldsymbol{\alpha}_{2}, \boldsymbol{\beta}_{3} = \boldsymbol{\alpha}_{3} - \frac{(\boldsymbol{\beta}_{2}, \boldsymbol{\alpha}_{3})}{(\boldsymbol{\beta}_{2},\boldsymbol{\beta}_{2})} \boldsymbol{\beta}_{2} = \left( \begin{smallmatrix} - \frac{1}{2} \\ -\frac{1}{2} \\ 1 \end{smallmatrix} \right)\\ & \boldsymbol{\gamma}_{1} = \frac{\sqrt{3}}{3}\boldsymbol{\beta}_{1}, \boldsymbol{\gamma}_{2} = \frac{\sqrt{2}}{2} \boldsymbol{\beta}_{2}, \boldsymbol{\gamma}_{3} = \frac{\sqrt{6}}{3}\boldsymbol{\beta}_{3} \\ & \text{令} \boldsymbol{Q} = \left( \begin{matrix} \boldsymbol{\gamma}_{1} & \boldsymbol{\gamma}_{2} & \boldsymbol{\gamma}_{3} \end{matrix} \right) = \left( \begin{smallmatrix} \frac{\sqrt{3}}{3} & -\frac{\sqrt{2}}{2} & -\frac{\sqrt{6}}{6} \\ \frac{\sqrt{3}}{3} & \frac{\sqrt{2}}{2} & -\frac{\sqrt{6}}{6} \\ \frac{\sqrt{3}}{3} & 0 & \frac{\sqrt{6}}{3} \\ \end{smallmatrix} \right), \boldsymbol{Q}^{\intercal}\boldsymbol{Q} = \boldsymbol{E} \\ & f = \boldsymbol{X}^{\intercal}\boldsymbol{AX} \xlongequal{\boldsymbol{X} = \boldsymbol{QY}} \boldsymbol{Y}^{\intercal}(\boldsymbol{Q}^{\intercal}\boldsymbol{AQ})\boldsymbol{Y} = 4y_{1}^{2} - 2y_{2}^{2}-2y_{3}^{2} \end{array}

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