Section03_对角化过程

对角化过程

  1. 求出特征值 det⁡(λE−A)=0⇒λ1,⋯ ,λn\det(\lambda \boldsymbol{E} - \boldsymbol{A}) = 0\Rightarrow \lambda_{1},\cdots, \lambda_{n}
  2. 求出线性无关特征向量 (λiE−A)X=0⇒α1,⋯ ,αm{线性无关m≤n(\lambda_{i}\boldsymbol{E} - \boldsymbol{A}) \boldsymbol{X} = \boldsymbol{0}\Rightarrow \boldsymbol{\alpha}_{1}, \cdots ,\boldsymbol{\alpha}_{m}\begin{cases} \text{线性无关} \\ m \le n \end{cases}
  3. 判别与构建
    1. m<nm<n 不可对角化
    2. m=nm = n 令P=(α1⋯αn)可逆∴AP=(Aα1⋯Aαn)=(λ1α1⋯λnαn)AP=P(λ1⋱λn)P−1AP=(λ1⋱λn) \begin{array}{ll} & \text{令} \boldsymbol{P} = \left( \begin{matrix} \boldsymbol{\alpha}_{1} & \cdots & \boldsymbol{\alpha}_{n} \end{matrix} \right) \text{可逆} \\ \therefore & \boldsymbol{AP} = \left( \begin{matrix} \boldsymbol{A\alpha}_{1} & \cdots & \boldsymbol{A\alpha}_{n} \end{matrix} \right) = \left( \begin{matrix} \lambda_{1}\boldsymbol{\alpha}_{1} & \cdots & \lambda_{n}\boldsymbol{\alpha}_{n} \end{matrix} \right) \\ & \boldsymbol{AP} = \boldsymbol{P} \left( \begin{matrix} \lambda_{1} \\ &\ddots \\ && \lambda_{n} \end{matrix} \right)\\ & \boldsymbol{P}^{-1}\boldsymbol{AP} = \left( \begin{matrix} \lambda_{1} \\ &\ddots \\ && \lambda_{n} \end{matrix} \right) \end{array}

附 正交

向量正交

  1. 概念 若 (α,β)=α⊺β=β⊺A=0(\boldsymbol{\alpha}, \boldsymbol{\beta}) = \boldsymbol{\alpha}^{\intercal}\boldsymbol{\beta} = \boldsymbol{\beta}^{\intercal}\boldsymbol{A} = 0，则称 A\boldsymbol{A} 与 B\boldsymbol{B} 正交，记为 A⊥B\boldsymbol{A}\perp \boldsymbol{B}
  2. 核心特质
    • α1,⋯ ,αn\boldsymbol{\alpha}_{1},\cdots, \boldsymbol{\alpha}_{n} 正交 ⇍⇒\nLeftarrow\Rightarrow α1,⋯ ,αn\boldsymbol{\alpha}_{1},\cdots, \boldsymbol{\alpha}_{n} 线性无关
  3. Schmidt 正交规范化 (已知 α1,α2,⋯ ,αn\boldsymbol{\alpha}_{1},\boldsymbol{\alpha}_{2}, \cdots, \boldsymbol{\alpha}_{n} 线性无关)
    1. 正交化 (β1,β2,⋯ ,βn\boldsymbol{\beta}_{1},\boldsymbol{\beta}_{2},\cdots, \boldsymbol{\beta}_{n} 两两正交)
      • β1=α1\boldsymbol{\beta}_{1} = \boldsymbol{\alpha}_{1}
      • β2=α2−(β1,α2)(β1,β1)β1\boldsymbol{\beta}_{2} = \boldsymbol{\alpha}_{2} - \frac{(\boldsymbol{\beta_{1},\alpha_{2}})}{(\boldsymbol{\beta}_{1},\boldsymbol{\beta}_{1})}\boldsymbol{\beta}_{1}
      • ⋯\cdots
      • βn=αn−(β1,αn)(β1,β1)β1−⋯−(βn−1,αn)(βn−1,βn−1)βn−1\boldsymbol{\beta}_{n} = \boldsymbol{\alpha}_{n} - \frac{(\boldsymbol{\beta_{1},\alpha_{n}})}{(\boldsymbol{\beta}_{1},\boldsymbol{\beta}_{1})}\boldsymbol{\beta}_{1} - \cdots - \frac{(\boldsymbol{\beta}_{n-1},\boldsymbol{\alpha}_{n})}{(\boldsymbol{\beta}_{n-1},\boldsymbol{\beta}_{n-1})}\boldsymbol{\beta}_{n-1}
    2. 规范化
      • γ1=1∣β1∣β1\boldsymbol{\gamma}_{1} = \frac{1}{\vert \boldsymbol{\beta}_{1} \vert}\boldsymbol{\beta}_{1}
      • γ2=1∣β2∣β2\boldsymbol{\gamma}_{2} = \frac{1}{\vert \boldsymbol{\beta}_{2} \vert}\boldsymbol{\beta}_{2}
      • ⋯\cdots
      • γn=1∣βn∣βn\boldsymbol{\gamma}_{n} = \frac{1}{\vert \boldsymbol{\beta}_{n} \vert}\boldsymbol{\beta}_{n}

        正交矩阵

        概念

  4. An×n\boldsymbol{A}_{n\times n}，若 A⊺A=E\boldsymbol{A}^{\intercal}\boldsymbol{A} = \boldsymbol{E} (或 AA⊺=E\boldsymbol{AA}^{\intercal}= \boldsymbol{E})，称 A\boldsymbol{A} 为正交矩阵

性质

  1. A⊺A=E⇒A−1=A⊺\boldsymbol{A}^{\intercal}\boldsymbol{A} = \boldsymbol{E} \Rightarrow \boldsymbol{A}^{-1} = \boldsymbol{A}^{\intercal}
  2. A⊺A=E⇒∣A∣=±1\boldsymbol{A}^{\intercal}\boldsymbol{A} = \boldsymbol{E} \Rightarrow \vert \boldsymbol{A} \vert = \pm 1

正交矩阵等价条件

  • Th An×n=(α1⋯αn)\boldsymbol{A}_{n\times n} = \left( \begin{matrix} \boldsymbol{\alpha}_{1} & \cdots & \boldsymbol{\alpha}_{n} \end{matrix} \right)，则 A⊺A=E⇔α1,⋯ ,αn\boldsymbol{A}^{\intercal}\boldsymbol{A} = \boldsymbol{E} \Leftrightarrow \boldsymbol{\alpha}_{1}, \cdots, \boldsymbol{\alpha}_{n} 两两正交且规范

    Proof A⊺A=(α1⊺⋮αn⊺)(α1⋯α1)=(α1⊺α1α1⊺α2⋯α1⊺αnα2⊺α1α2⊺α2⋯α2⊺αn⋮⋮⋱⋮αn⊺α1αn⊺α2⋯αn⊺αn)=(1⋱1)⇔{∣αi∣=αi⊺αi=1αi⊺αj≠0(i≠j), 即α1,⋯ ,αn两两正交且规范\begin{array}{ll} & \boldsymbol{A}^{\intercal}\boldsymbol{A} = \left( \begin{matrix} \boldsymbol{\alpha}_{1}^{\intercal} \\ \vdots \\ \boldsymbol{\alpha}_{n}^{\intercal} \\ \end{matrix} \right) \left( \begin{matrix} \boldsymbol{\alpha}_{1} & \cdots \boldsymbol{\alpha}_{1} \end{matrix} \right)\\ & = \left( \begin{matrix} \boldsymbol{\alpha}_{1}^{\intercal} \boldsymbol{\alpha}_{1} & \boldsymbol{\alpha}_{1}^{\intercal} \boldsymbol{\alpha}_{2} & \cdots & \boldsymbol{\alpha}_{1}^{\intercal} \boldsymbol{\alpha}_{n} \\ \boldsymbol{\alpha}_{2}^{\intercal} \boldsymbol{\alpha}_{1} & \boldsymbol{\alpha}_{2}^{\intercal} \boldsymbol{\alpha}_{2} & \cdots & \boldsymbol{\alpha}_{2}^{\intercal} \boldsymbol{\alpha}_{n} \\ \vdots & \vdots & \ddots & \vdots \\ \boldsymbol{\alpha}_{n}^{\intercal} \boldsymbol{\alpha}_{1} & \boldsymbol{\alpha}_{n}^{\intercal} \boldsymbol{\alpha}_{2} & \cdots & \boldsymbol{\alpha}_{n}^{\intercal} \boldsymbol{\alpha}_{n} \\ \end{matrix} \right) = \left( \begin{matrix} 1 \\ & \ddots \\ & & 1 \end{matrix} \right) \\ \Leftrightarrow & \begin{cases} \vert \boldsymbol{\alpha}_{i} \vert = \sqrt{\boldsymbol{\alpha}^{\intercal}_{i}\boldsymbol{\alpha}_{i}} = 1 \\ \boldsymbol{\alpha}_{i}^{\intercal}\boldsymbol{\alpha}_{j}\ne 0 \quad (i\ne j) \end{cases} \text{, 即}\boldsymbol{\alpha}_{1},\cdots, \boldsymbol{\alpha}_{n} \text{两两正交且规范} \end{array}

实对称矩阵的对角化

  1. 求出特征值 det⁡(λE−A)=0⇒λ1,⋯ ,λn\det(\lambda \boldsymbol{E} - \boldsymbol{A}) = 0\Rightarrow \lambda_{1},\cdots, \lambda_{n}
  2. 求出线性无关特征向量 (λiE−A)X=0⇒α1,⋯ ,αn{线性无关不同特征值对应的特征向量正交(\lambda_{i}\boldsymbol{E} - \boldsymbol{A}) \boldsymbol{X} = \boldsymbol{0}\Rightarrow \boldsymbol{\alpha}_{1}, \cdots ,\boldsymbol{\alpha}_{n}\begin{cases} \text{线性无关} \\ \text{不同特征值对应的特征向量正交} \end{cases}
  3. 两种情形
    • Cases1 找可逆阵 P\boldsymbol{P}: 令 P=(α1⋯αn)\boldsymbol{P} = \left( \begin{matrix} \boldsymbol{\alpha}_{1} & \cdots & \boldsymbol{\alpha}_{n} \end{matrix} \right)，则 P−1AP=(λ1⋱λn)\boldsymbol{P}^{-1}\boldsymbol{AP} = \left( \begin{matrix} \lambda_{1} \\ & \ddots \\ && \lambda_{n} \end{matrix} \right)
    • Cases2 找正交阵 Q⇔{Q⊺Q=EQ−1=Q⊺Q列向量两两正交且规范\boldsymbol{Q}\Leftrightarrow \begin{cases} \boldsymbol{Q}^{\intercal}\boldsymbol{Q} = \boldsymbol{E} \\ \boldsymbol{Q}^{-1} = \boldsymbol{Q}^{\intercal} \\ \boldsymbol{Q} \text{列向量两两正交且规范} \end{cases} α1,⋯ ,αn⇒规范化正交化γ1,⋯ ,γn\boldsymbol{\alpha}_{1},\cdots,\boldsymbol{\alpha}_{n} \xRightarrow[\text{规范化}]{\text{正交化}} \boldsymbol{\gamma}_{1},\cdots, \boldsymbol{\gamma}_{n}，令 Q=(γ1⋯γn)\boldsymbol{Q} = \left( \begin{matrix} \boldsymbol{\gamma}_{1} & \cdots & \boldsymbol{\gamma}_{n} \end{matrix} \right)，则 Q⊺AQ=(λ1⋱λn)\boldsymbol{Q}^{\intercal}\boldsymbol{AQ} = \left( \begin{matrix} \lambda_{1} \\ & \ddots \\ & & \lambda_{n} \end{matrix} \right)

例题

  1. A=(1−1−1−11−1−1−11)\boldsymbol{A} = \left( \begin{matrix} 1 & -1 & -1 \\ -1 & 1 & -1 \\ -1 & -1 & 1 \\ \end{matrix} \right) det⁡(λE−A)=∣λ−1111λ−1111λ−1∣=(λ+1)∣1110λ−2000λ−2∣=(λ+1)(λ−2)=0⇒λ1=−1,λ2=λ3=2λ1=−1⇒α1=(111)λ=2⇒α2=(−110),α3=(−101)Case 1: 找可逆阵PP=(α1α2α3)=(1−1−1110101)P−1AP=(−122)Case 2: 找正交阵Qβ1=α1,β2=α2,β3=α3−(α3,β2)(β2,β2)β2=(−12−121)γ1=13β1,γ2=12β2,γ3=63∴Q=(γ1γ2γ3)=(33−22−663322−6633063)Q⊺AQ=(−122) \begin{array}{ll} & \det(\lambda \boldsymbol{E} - \boldsymbol{A}) = \left\vert \begin{matrix} \lambda - 1 & 1 & 1 \\ 1 & \lambda - 1 & 1 \\ 1 & 1 & \lambda - 1 \\ \end{matrix} \right\vert = (\lambda + 1) \left\vert \begin{matrix} 1 & 1 & 1 \\ 0 & \lambda - 2 & 0 \\ 0 & 0 & \lambda - 2 \\ \end{matrix} \right\vert \\ & = (\lambda + 1)(\lambda -2) = 0 \Rightarrow \lambda_{1} = -1, \lambda_{2} = \lambda_{3} = 2\\ & \lambda_{1} = -1 \Rightarrow \boldsymbol{\alpha}_{1} = \left( \begin{matrix} 1 \\ 1 \\ 1 \end{matrix} \right) \\ & \lambda = 2 \Rightarrow \boldsymbol{\alpha}_{2} = \left( \begin{matrix} -1 \\ 1 \\ 0 \end{matrix} \right), \boldsymbol{\alpha}_{3} = \left( \begin{matrix} -1 \\ 0 \\ 1 \end{matrix} \right) \\\\ & \text{Case 1: 找可逆阵}\boldsymbol{P} \\ & \boldsymbol{P} = \left( \begin{matrix} \boldsymbol{\alpha}_{1} & \boldsymbol{\alpha}_{2} & \boldsymbol{\alpha}_{3} \end{matrix} \right) = \left( \begin{matrix} 1 & -1 & -1 \\ 1 & 1 & 0 \\ 1 & 0 & 1 \end{matrix} \right) \\ & \boldsymbol{P}^{-1} \boldsymbol{AP} = \left( \begin{matrix} -1 \\ & 2 \\ & & 2 \\ \end{matrix} \right) \\\\ & \text{Case 2: 找正交阵}\boldsymbol{Q} \\ & \boldsymbol{\beta}_{1} = \boldsymbol{\alpha}_{1}, \boldsymbol{\beta}_{2} = \boldsymbol{\alpha}_{2}, \boldsymbol{\beta}_{3} = \boldsymbol{\alpha}_{3} - \frac{(\boldsymbol{\alpha}_{3}, \boldsymbol{\beta}_{2})}{(\boldsymbol{\beta}_{2},\boldsymbol{\beta}_{2})} \boldsymbol{\beta}_{2} = \left( \begin{matrix} -\frac{1}{2} \\ -\frac{1}{2} \\ 1 \end{matrix} \right) \\ & \boldsymbol{\gamma}_{1} = \frac{1}{\sqrt{3}}\boldsymbol{\beta}_{1}, \boldsymbol{\gamma}_{2} = \frac{1}{\sqrt{2}}\boldsymbol{\beta}_{2}, \boldsymbol{\gamma}_{3} = \frac{\sqrt{6}}{3} \\ \therefore & \boldsymbol{Q} = \left( \begin{matrix} \boldsymbol{\gamma}_{1} & \boldsymbol{\gamma}_{2} & \boldsymbol{\gamma}_{3} \end{matrix} \right) = \left( \begin{matrix} \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{matrix} \right)\\ & \boldsymbol{Q}^{\intercal}\boldsymbol{AQ} = \left( \begin{matrix} -1 \\ & 2 \\ & & 2 \\ \end{matrix} \right) \\ \end{array}

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