Section03_向量组等价,极大组,秩

  1. 向量组等价 I: α1,,αm; II: β1,β2,,βn\text{I: } \boldsymbol{\alpha}_{1}, \cdots, \boldsymbol{\alpha}_{m}\text{; II: } \boldsymbol{\beta}_{1},\boldsymbol{\beta}_{2},\cdots, \boldsymbol{\beta}_{n}
    • {α1=k11β1++k1nβnαm=km1β1++kmnβn(1)\begin{cases} \boldsymbol{\alpha}_{1} = k_{11}\boldsymbol{\beta}_{1} + \cdots + k_{1n}\boldsymbol{\beta}_{n} \\ \cdots \\ \boldsymbol{\alpha}_{m} = k_{m1}\boldsymbol{\beta}_{1} + \cdots + k_{mn}\boldsymbol{\beta}_{n} \\ \end{cases} \quad (1),称 I\text{I} 可由 II\text{II} 线性表示
    • {β1=11α1++1mαmβn=n1α1++nmαm(2)\begin{cases} \boldsymbol{\beta}_{1} = \ell_{11}\boldsymbol{\alpha}_{1} + \cdots + \ell_{1m}\boldsymbol{\alpha}_{m} \\ \cdots \\ \boldsymbol{\beta}_{n} = \ell_{n1}\boldsymbol{\alpha}_{1} + \cdots + \ell_{nm}\boldsymbol{\alpha}_{m} \\ \end{cases} \quad (2),称 II\text{II} 可由 I\text{I} 线性表示
    • (1),(2)(1),(2) 皆成立,则称 I, II\text{I, II} 等价
  2. 极大线性无关组(无关组) 若
    1.  r\exists\ r 个向量线性无关
    2.  r+1\forall\ r+1 个向量线性相关
    3. rr 个无关向量为最大线性无关组,rr 称为向量组的秩

      Notes

      1. 极大组不一定唯一
      2. 向量组与极大组等价
      3. 重点 对于 α1,,αn\boldsymbol{\alpha}_{1},\cdots, \boldsymbol{\alpha}_{n}
        • Case 1 α1,,αn\boldsymbol{\alpha}_{1},\cdots,\boldsymbol{\alpha}_{n} 无关 \Leftrightarrow α1,,αn\boldsymbol{\alpha}_{1},\cdots,\boldsymbol{\alpha}_{n} 的秩 =n=n
        • Case 2 α1,,αn\boldsymbol{\alpha}_{1},\cdots,\boldsymbol{\alpha}_{n} 相关 \Leftrightarrow α1,,αn\boldsymbol{\alpha}_{1},\cdots,\boldsymbol{\alpha}_{n} 的秩 <n<n
      4. I: α1,,αn, II: α1,,αn,β\text{I: } \boldsymbol{\alpha}_{1},\cdots, \boldsymbol{\alpha}_{n}\text{, II: } \boldsymbol{\alpha}_{1},\cdots,\boldsymbol{\alpha}_{n}, \boldsymbol{\beta}
        • Case 1 I\text{I} 的秩 == II\text{II} 的秩 \Leftrightarrow β\boldsymbol{\beta} 可由 α1,,αn\boldsymbol{\alpha}_{1},\cdots,\boldsymbol{\alpha}_{n} 线性表示 \Leftrightarrow ()(**) 有解
        • Case 2 I\text{I} 的秩 + 1 == II\text{II} 的秩 \Leftrightarrow β\boldsymbol{\beta} 不可由 α1,,αn\boldsymbol{\alpha}_{1},\cdots,\boldsymbol{\alpha}_{n} 线性表示 \Leftrightarrow ()(**) 无解
      5. Am×n=(α1αm)=(β1βn)\boldsymbol{A}_{m\times n} = \left( \begin{matrix} \boldsymbol{\alpha}_{1} \\ \vdots \\ \boldsymbol{\alpha}_{m} \end{matrix} \right) = \left( \begin{matrix} \boldsymbol{\beta}_{1} & \cdots & \boldsymbol{\beta}_{n} \end{matrix} \right)
        • α1,,αm\boldsymbol{\alpha}_{1},\cdots, \boldsymbol{\alpha}_{m} 称为 A\boldsymbol{A} 的行向量组,其的秩成为 A\boldsymbol{A}行秩
        • β1,,βn\boldsymbol{\beta}_{1},\cdots, \boldsymbol{\beta}_{n} 称为 A\boldsymbol{A} 的列向量组,其的秩成为 A\boldsymbol{A}列秩
      6. Am×n,Bn×s=(β1βs)\boldsymbol{A}_{m\times n}, \boldsymbol{B}_{n\times s} = \left( \begin{matrix} \boldsymbol{\beta}_{1} & \cdots & \boldsymbol{\beta}_{s} \end{matrix} \right) \Rightarrow AB=A(β1βs)=(Aβ1Aβs)\boldsymbol{AB} = \boldsymbol{A} \left( \begin{matrix} \boldsymbol{\beta}_{1} & \cdots & \boldsymbol{\beta}_{s} \end{matrix} \right) = \left( \begin{matrix} \boldsymbol{A\beta}_{1} & \cdots & \boldsymbol{A\beta}_{s} \end{matrix} \right)
      7. A(BC)=(ABAC)\boldsymbol{A} \left( \begin{array}{c:c} \boldsymbol{B} & \boldsymbol{C} \end{array}\right) = \left( \begin{array}{c:c} \boldsymbol{AB} & \boldsymbol{AC} \end{array}\right)
      8. α1,α2,α3\boldsymbol{\alpha}_{1},\boldsymbol{\alpha}_{2},\boldsymbol{\alpha}_{3}A=(α1α2α3)\boldsymbol{A} = \left( \begin{matrix} \boldsymbol{\alpha}_{1} & \boldsymbol{\alpha}_{2} & \boldsymbol{\alpha}_{3} \end{matrix} \right), B=(a1α1+a2α2+a3α3b1α1+b2α2+b3α3c1α1+c2α2+c3α3)=(a1b1c1a2b2c2a3b3c3)\begin{split} \boldsymbol{B} & = \left(\begin{smallmatrix} a_{1}\boldsymbol{\alpha}_{1} + a_{2}\boldsymbol{\alpha}_{2} + a_{3}\boldsymbol{\alpha}_{3} & b_{1}\boldsymbol{\alpha}_{1} + b_{2}\boldsymbol{\alpha}_{2} + b_{3}\boldsymbol{\alpha}_{3} & c_{1}\boldsymbol{\alpha}_{1} + c_{2}\boldsymbol{\alpha}_{2} + c_{3}\boldsymbol{\alpha}_{3} \end{smallmatrix}\right) \\ & = \boldsymbol{} \begin{pmatrix} a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \\ a_{3} & b_{3} & c_{3} \\ \end{pmatrix} \end{split}

性质

  1. 任何矩阵三秩相等
  2. I: α1,,αm, II: β1,,βn\text{I: }\boldsymbol{\alpha}_{1},\cdots, \boldsymbol{\alpha}_{m}\text{, II: } \boldsymbol{\beta}_{1},\cdots, \boldsymbol{\beta}_{n},若 I\text{I} 组可由 II\text{II} 组线性表示,则 I\text{I} 的秩 \le II\text{II} 的秩

    Proof A=(α1αm),B=(β1βn){α1=k11β1++k1nβnαm=km1β1++kmnβnA=B(k11k21km1k12k22km2k1nk2nkmn)=BKr(A)=r(BK)r(B)I的秩II的秩 \begin{array}{ll} & \boldsymbol{A} = \left( \begin{matrix} \boldsymbol{\alpha}_{1}& \cdots & \boldsymbol{\alpha}_{m} \end{matrix} \right) , \boldsymbol{B} = \left( \begin{matrix} \boldsymbol{\beta}_{1}& \cdots & \boldsymbol{\beta}_{n} \end{matrix} \right) \\ \because & \begin{cases} \boldsymbol{\alpha}_{1} = k_{11}\boldsymbol{\beta}_{1} + \cdots + k_{1n}\boldsymbol{\beta}_{n} \\ \cdots \\ \boldsymbol{\alpha}_{m} = k_{m1}\boldsymbol{\beta}_{1} + \cdots + k_{mn}\boldsymbol{\beta}_{n} \\ \end{cases} \\ \therefore & \boldsymbol{A} = \boldsymbol{B} \left( \begin{matrix} k_{11} & k_{21} & \cdots & k_{m1} \\ k_{12} & k_{22} & \cdots & k_{m2} \\ \vdots & \vdots & \ddots & \vdots \\ k_{1n} & k_{2n} & \cdots & k_{mn} \\ \end{matrix} \right) = \boldsymbol{BK}\\ \because & r(\boldsymbol{A}) = r(\boldsymbol{BK}) \le r(\boldsymbol{B}) \\ \therefore & \text{I的秩} \le \text{II的秩} \\ \end{array}

  3. 向量组 I\text{I} 与向量组 II\text{II} 等价 \nLeftarrow \Rightarrow I\text{I} 的秩 == II\text{II} 的秩

    Proof

    \Rightarrow I可由II线性表示I的秩II的秩II可由I线性表示II的秩I的秩I的秩=II的秩 \begin{array}{ll} \because & \text{I可由II线性表示} \Rightarrow \text{I的秩}\le \text{II的秩} \\ & \text{II可由I线性表示} \Rightarrow \text{II的秩}\le \text{I的秩} \\ \therefore & \text{I的秩} = \text{II的秩} \\ \end{array} \nLeftarrow 反例 I: α1=(1100),α2=(2100)II: α1=(0012),α2=(0013) \begin{array}{ll} & \text{I: } \boldsymbol{\alpha}_{1} = \left( \begin{smallmatrix} 1 \\ -1 \\ 0 \\ 0 \end{smallmatrix} \right), \boldsymbol{\alpha}_{2} = \left( \begin{smallmatrix} 2 \\ 1 \\ 0 \\ 0 \end{smallmatrix} \right) \\ & \text{II: } \boldsymbol{\alpha}_{1} = \left( \begin{smallmatrix} 0 \\ 0 \\ 1 \\ 2 \end{smallmatrix} \right), \boldsymbol{\alpha}_{2} = \left( \begin{smallmatrix} 0 \\ 0 \\ -1 \\ 3 \end{smallmatrix} \right) \end{array}

例题

  1. α1,α2\boldsymbol{\alpha}_{1}, \boldsymbol{\alpha}_{2} 线性无关,α3\boldsymbol{\alpha}_{3} 不可由 α1,α2\boldsymbol{\alpha}_{1},\boldsymbol{\alpha}_{2} 线性表示,α4\boldsymbol{\alpha}_{4} 可由 α1,α2\boldsymbol{\alpha}_{1},\boldsymbol{\alpha}_{2} 线性表示,证 α1,α2,α3+α4\boldsymbol{\alpha}_{1},\boldsymbol{\alpha}_{2}, \boldsymbol{\alpha}_{3} + \boldsymbol{\alpha}_{4} 线性无关 α1,α2线性无关, 且α3不可由α1,α2线性表示α1,α2,α3线性无关A=(α1α2α3),r(A)=3α4可由α1,α2线性表示α4=k1α1+k2α2B=(α1α2α3+α4)=A(10k101k2001)10k101k20010r(B)=r(A)=3α1,α2,α3+α4线性无关 \begin{array}{ll} \because & \boldsymbol{\alpha}_{1}, \boldsymbol{\alpha}_{2} \text{线性无关, 且} \boldsymbol{\alpha}_{3} \text{不可由} \boldsymbol{\alpha}_{1},\boldsymbol{\alpha}_{2}\text{线性表示} \\ \therefore & \boldsymbol{\alpha}_{1}, \boldsymbol{\alpha}_{2},\boldsymbol{\alpha}_{3}\text{线性无关} \\ & \boldsymbol{A} = \left( \begin{matrix} \boldsymbol{\alpha}_{1} &\boldsymbol{\alpha}_{2} &\boldsymbol{\alpha}_{3} \end{matrix} \right), r(\boldsymbol{A}) = 3 \\ \because & \boldsymbol{\alpha}_{4} \text{可由} \boldsymbol{\alpha}_{1},\boldsymbol{\alpha}_{2}\text{线性表示} \\ \therefore & \boldsymbol{\alpha}_{4} = k_{1}\boldsymbol{\alpha}_{1} + k_{2}\boldsymbol{\alpha}_{2} \\ \therefore & \boldsymbol{B} = \left( \begin{matrix} \boldsymbol{\alpha}_{1} & \boldsymbol{\alpha}_{2} & \boldsymbol{\alpha}_{3} + \boldsymbol{\alpha}_{4} \end{matrix} \right) = \boldsymbol{A} \left( \begin{matrix} 1 & 0 & k_{1} \\ 0 & 1 & k_{2} \\ 0 & 0 & 1 \\ \end{matrix} \right)\\ \because & \left\vert \begin{matrix} 1 & 0 & k_{1} \\ 0 & 1 & k_{2} \\ 0 & 0 & 1 \\ \end{matrix} \right\vert \ne 0\\ \therefore & r(\boldsymbol{B}) = r(\boldsymbol{A}) = 3\\ \therefore & \boldsymbol{\alpha}_{1},\boldsymbol{\alpha}_{2}, \boldsymbol{\alpha}_{3} + \boldsymbol{\alpha}_{4} \text{线性无关} \\ \end{array}
  2. I: α1,,αr, II: β1,,βs\text{I: } \boldsymbol{\alpha}_{1},\cdots, \boldsymbol{\alpha}_{r}\text{, II: } \boldsymbol{\beta}_{1},\cdots, \boldsymbol{\beta}_{s}I\text{I} 可由 II\text{II} 线性表示,且 r>sr>s,证明 α1,,αr\boldsymbol{\alpha}_{1},\cdots,\boldsymbol{\alpha}_{r} 线性相关 I可由II线性表示I的秩II的秩s<rα1,,αr线性相关 \begin{array}{ll} \because & \text{I可由II线性表示} \\ \therefore & \text{I的秩} \le \text{II的秩} \le s < r \\ \therefore & \boldsymbol{\alpha}_{1},\cdots,\boldsymbol{\alpha}_{r} \text{线性相关} \\ \end{array}

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