3.1 向量的正交

  • α=[a1,a2,,an]T,β=[b1,b2,,bn]T\boldsymbol{\alpha}=\begin{bmatrix} a_1,a_2,\cdots,a_n \end{bmatrix}^T,\boldsymbol{\beta}=\begin{bmatrix} b_1,b_2,\cdots,b_n \end{bmatrix}^T

  • 內積(α,β)=αTβ=i=1naibi=a1b1+a2b2++anbn(\boldsymbol{\alpha},\boldsymbol{\beta}) = \boldsymbol{\alpha}^T\boldsymbol{\beta} = \sum\limits_{i=1}^{n}a_ib_i = a_1b_1+a_2b_2+\cdots+a_nb_n
  • α=αTα=i=1nai2=a12+a22++an2\begin{Vmatrix} \boldsymbol{\alpha} \end{Vmatrix} = \sqrt{\boldsymbol{\alpha}^T\boldsymbol{\alpha}} = \sqrt{\sum\limits_{i=1}^{n}a_i^2} = \sqrt{a_1^2+a_2^2+\cdots+a_n^2}

  • 正交(α,β)=αTβ=0(\boldsymbol{\alpha},\boldsymbol{\beta})=\boldsymbol{\alpha}^T\boldsymbol{\beta}=0
  • 單位向量α=αTα=1\begin{Vmatrix} \boldsymbol{\alpha} \end{Vmatrix} = \sqrt{\boldsymbol{\alpha}^T\boldsymbol{\alpha}} = 1

  • 標準正交向量組:若列向量 α1,α2,,αn\boldsymbol{\alpha}_1,\boldsymbol{\alpha}_2,\cdots,\boldsymbol{\alpha}_n 滿足 αiTαj={0,ij1,i=j\boldsymbol{\alpha}_i^T\boldsymbol{\alpha}_j=\begin{cases} 0,&i \ne j\\1,&i=j \end{cases}(均爲單位向量且任意兩不同向量正交),則稱 α1,α2,,αn\boldsymbol{\alpha}_1,\boldsymbol{\alpha}_2,\cdots,\boldsymbol{\alpha}_n 是標準(單位)正交向量組。
  • 線性無關向量組 α1,α2,,αn\boldsymbol{\alpha}_1,\boldsymbol{\alpha}_2,\cdots,\boldsymbol{\alpha}_n \rightarrow 標準正交向量組 η1,η2,,ηn\boldsymbol{\eta}_1,\boldsymbol{\eta}_2,\cdots,\boldsymbol{\eta}_n
    1. 正交化得正交向量組 β1,β2,,βn\boldsymbol{\beta}_1,\boldsymbol{\beta}_2,\cdots,\boldsymbol{\beta}_n
      β1=α1,β2=α2(α2,β1)(β1,β1)β1,βn=αn(αn,βn1)(βn1,βn1)βn1(αn,βn2)(βn2,βn2)βn2(αn,β1)(β1,β1)β1 \begin{aligned} & \boldsymbol{\beta}_1 = \boldsymbol{\alpha}_1,\\ & \boldsymbol{\beta}_2 = \boldsymbol{\alpha}_2 - \frac{(\boldsymbol{\alpha}_2,\boldsymbol{\beta}_1)}{(\boldsymbol{\beta}_1,\boldsymbol{\beta}_1)}\boldsymbol{\beta}_1,\\ & \cdots\\ & \boldsymbol{\beta}_n = \boldsymbol{\alpha}_n - \frac{(\boldsymbol{\alpha}_n,\boldsymbol{\beta}_{n-1})}{(\boldsymbol{\beta}_{n-1},\boldsymbol{\beta}_{n-1})}\boldsymbol{\beta}_{n-1} - \frac{(\boldsymbol{\alpha}_n,\boldsymbol{\beta}_{n-2})}{(\boldsymbol{\beta}_{n-2},\boldsymbol{\beta}_{n-2})}\boldsymbol{\beta}_{n-2} - \cdots - \frac{(\boldsymbol{\alpha}_n,\boldsymbol{\beta}_{1})}{(\boldsymbol{\beta}_{1},\boldsymbol{\beta}_{1})}\boldsymbol{\beta}_{1} \end{aligned}

    2. 單位化得標準正交向量組 η1,η2,,ηn\boldsymbol{\eta}_1,\boldsymbol{\eta}_2,\cdots,\boldsymbol{\eta}_n
      η1=β1β1,η2=β2β2,ηn=βnβn \begin{aligned} &\boldsymbol{\eta}_1 = \frac{\boldsymbol{\beta}_1}{\begin{Vmatrix} \boldsymbol{\beta}_1 \end{Vmatrix}},\\ &\boldsymbol{\eta}_2 = \frac{\boldsymbol{\beta}_2}{\begin{Vmatrix} \boldsymbol{\beta}_2 \end{Vmatrix}},\\ &\cdots\\ &\boldsymbol{\eta}_n = \frac{\boldsymbol{\beta}_n}{\begin{Vmatrix} \boldsymbol{\beta}_n \end{Vmatrix}} \end{aligned}

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