Section03_解的理论

Am×nAX=0AX=β \begin{array}{c} \boldsymbol{A}_{m\times n} \\ \boldsymbol{AX} = \boldsymbol{0} \\ \boldsymbol{AX} = \boldsymbol{\beta} \\ \end{array}

  • Th1 对 (∗)(*)
    1. (∗)(*) 仅有零解 ⇔\Leftrightarrow r(A)=nr(\boldsymbol{A}) = n
    2. (∗)(*) 仅有非零解 ⇔\Leftrightarrow r(A)<nr(\boldsymbol{A}) < n
  • Th 2 A=(α1⋯αn),Aˉ=(α1⋯αnβ)\boldsymbol{A} = \left( \begin{matrix} \boldsymbol{\alpha}_{1} & \cdots & \boldsymbol{\alpha}_{n} \end{matrix} \right), \bar{\boldsymbol{A}} = \left( \begin{array}{ccc:c} \boldsymbol{\alpha}_{1} & \cdots & \boldsymbol{\alpha}_{n} & \boldsymbol{\beta} \end{array} \right)，对 (∗∗)(**)
    1. (∗∗)(**) 有解 ⇔\Leftrightarrow r(A)=r(Aˉ){=n,唯一解<n,无数解r(\boldsymbol{A})=r(\bar{\boldsymbol{A}})\begin{cases} =n,& \text{唯一解} \\ <n,& \text{无数解} \end{cases}
    2. (∗∗)(**) 无解 ⇔\Leftrightarrow r(A)=r(Aˉ)r(Aˉ)=r(A)+1r(\boldsymbol{A}) = r(\bar{\boldsymbol{A}})\quad r(\bar{\boldsymbol{A}}) = r(\boldsymbol{A}) + 1

      Q1 AX=0\boldsymbol{AX} = \boldsymbol{0} 仅有零解与 AX=β\boldsymbol{AX} = \boldsymbol{\beta} 有唯一解的关系? AX=0仅有零解⇐⇏AX=β有唯一解⇕⇕r(A)=nr(A)=r(Aˉ)=n \begin{array}{ccc} \boldsymbol{AX} = \boldsymbol{0}\text{仅有零解} & \Leftarrow\nRightarrow & \boldsymbol{AX} = \boldsymbol{\beta} \text{有唯一解} \\ \Updownarrow & & \Updownarrow\\r(\boldsymbol{A}) = n & & r(\boldsymbol{A}) = r(\bar{\boldsymbol{A}}) = n \end{array} Q2 AX=0\boldsymbol{AX} = \boldsymbol{0} 有非零解与 AX=β\boldsymbol{AX} = \boldsymbol{\beta} 有无数解的关系? AX=0有非零解⇐⇏AX=β有无数解⇕⇕r(A)<nr(A)=r(Aˉ)<n \begin{array}{ccc} \boldsymbol{AX} = \boldsymbol{0}\text{有非零解} & \Leftarrow\nRightarrow & \boldsymbol{AX} = \boldsymbol{\beta} \text{有无数解} \\ \Updownarrow & & \Updownarrow\\r(\boldsymbol{A}) < n & & r(\boldsymbol{A}) = r(\bar{\boldsymbol{A}}) < n \end{array} Q3 若 r(A)=mr(\boldsymbol{A}) = m，问 AX=β\boldsymbol{AX} = \boldsymbol{\beta} 是否一定有解? Aˉ=(Aβ)∴r(Aˉ)≥r(A)=m∵r(Aˉ)≤m∴r(A)=r(Aˉ)=m∴AX=β一定有解 \begin{array}{ll} & \boldsymbol{\bar{A}} = \left( \begin{array}{c:c} \boldsymbol{A}& \boldsymbol{\beta} \end{array} \right) \\ \therefore & r(\bar{\boldsymbol{A}}) \ge r(\boldsymbol{A}) = m \\ \because & r(\bar{\boldsymbol{A}}) \le m \\ \therefore & r(\boldsymbol{A}) = r(\bar{\boldsymbol{A}}) = m \\ \therefore & \boldsymbol{AX} = \boldsymbol{\beta} \text{一定有解} \\ \end{array}

  • Th 3 Am×n,Bn×s=(β1⋯βs)\boldsymbol{A}_{m\times n}, \boldsymbol{B}_{n\times s} = \left( \begin{matrix} \boldsymbol{\beta}_{1} & \cdots & \boldsymbol{\beta}_{s} \end{matrix} \right)，若 AB=0\boldsymbol{AB} = \boldsymbol{0}，则 β1,⋯ ,βs\boldsymbol{\beta}_{1},\cdots ,\boldsymbol{\beta}_{s} 为 AX=0\boldsymbol{AX} = \boldsymbol{0} 的解

    Proof AB=A(β1⋯βs)=(Aβ1⋯Aβs)∵AB=0∴Aβ1=Aβ2=⋯=Aβs=0∴β1,β2,⋯ ,βs为AX=0的一组解 \begin{array}{ll} & \boldsymbol{AB} = \boldsymbol{A}\left( \begin{matrix} \boldsymbol{\beta}_{1} & \cdots & \boldsymbol{\beta}_{s} \end{matrix} \right) = \left( \begin{matrix} \boldsymbol{A}\boldsymbol{\beta}_{1} & \cdots & \boldsymbol{A}\boldsymbol{\beta}_{s} \end{matrix} \right) \\ \because & \boldsymbol{AB} = 0 \\ \therefore & \boldsymbol{A\beta}_{1} = \boldsymbol{A\beta}_{2}= \cdots = \boldsymbol{A\beta}_{s} = \boldsymbol{0} \\ \therefore & \boldsymbol{\beta}_{1}, \boldsymbol{\beta}_{2},\cdots, \boldsymbol{\beta}_{s} \text{为} \boldsymbol{AX} = \boldsymbol{0}\text{的一组解} \\ \end{array}

  • NOTICE 看到 AB=0⇒{r(A)+r(B)≤n(内标)B的列为AX=0的解\boldsymbol{AB}= \boldsymbol{0}\Rightarrow \begin{cases} r(\boldsymbol{A}) + r(\boldsymbol{B}) \le n\quad \text{(内标)} \\ \boldsymbol{B}\text{的列为}\boldsymbol{AX} = \boldsymbol{0}\text{的解} \end{cases}

results matching ""

    No results matching ""