Section02_线性相关与线性表示

背景

  1. 方程组的一般形式 {a11x1+⋯+a1nan=0⋯am1x1+⋯+amnan=0(∗){a11x1+⋯+a1nan=b1⋯am1x1+⋯+amnan=bm(∗∗) \begin{array}{ll} \begin{cases} a_{11}x_{1}+\cdots +a_{1n}a_{n} = 0 \\ \cdots \\ a_{m1}x_{1}+\cdots +a_{mn}a_{n} = 0 \\ \end{cases} & (*) \\\\ \begin{cases} a_{11}x_{1}+\cdots +a_{1n}a_{n} = b_{1} \\ \cdots \\ a_{m1}x_{1}+\cdots +a_{mn}a_{n} = b_{m} \\ \end{cases} & (**) \end{array}
  2. 方程组的矩阵形式 A=(a11a12⋯a1na21a22⋯a2n⋮⋮⋱⋮am1am2⋯amn)X=(x1x2⋮xn)β=(b1b2⋮bm)AX=0(∗)AX=β(∗∗) \begin{array}{c} \begin{array}{ll} \boldsymbol{A} = \left( \begin{matrix} a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{m1} & a_{m2} & \cdots & a_{mn} \\ \end{matrix} \right)\quad \boldsymbol{X} = \left( \begin{matrix} x_{1} \\ x_{2} \\ \vdots \\ x_{n} \\ \end{matrix} \right) \quad \boldsymbol{\beta} = \left( \begin{matrix} b_{1} \\ b_{2} \\ \vdots \\ b_{m} \\ \end{matrix} \right) \end{array} \\\\ \boldsymbol{AX} = \boldsymbol{0}\quad (*)\\ \boldsymbol{AX} = \boldsymbol{\beta}\quad (**)\\ \end{array}
  3. 方程组的向量形式 α1=(a11a21⋮am1), α2=(a12a22⋮am2),⋯ , αn=(a1na2n⋮amn)x1α1+x2α2+⋯+xnαn=0(∗)x1α1+x2α2+⋯+xnαn=β(∗∗) \begin{array}{c} \boldsymbol{\alpha}_{1} = \left( \begin{matrix} a_{11} \\ a_{21} \\ \vdots \\ a_{m1} \end{matrix} \right),\ \boldsymbol{\alpha}_{2} = \left( \begin{matrix} a_{12} \\ a_{22} \\ \vdots \\ a_{m2} \end{matrix} \right),\cdots,\ \boldsymbol{\alpha}_{n} = \left( \begin{matrix} a_{1n} \\ a_{2n} \\ \vdots \\ a_{mn} \end{matrix} \right) \\\\ x_{1}\boldsymbol{\alpha}_{1} + x_{2}\boldsymbol{\alpha}_{2} + \cdots + x_{n}\boldsymbol{\alpha}_{n} = \boldsymbol{0}\quad (*) \\ x_{1}\boldsymbol{\alpha}_{1} + x_{2}\boldsymbol{\alpha}_{2} + \cdots + x_{n}\boldsymbol{\alpha}_{n} = \boldsymbol{\beta} \quad (**)\\ \end{array}

(∗) (∗∗)(*)\ (**) 解的概况

  1. (∗) {仅有零解除零解外，有无数非零解(*)\ \begin{cases} \text{仅有零解} \\ \text{除零解外，有无数非零解} \end{cases}
  2. (∗∗) {有解{唯一解无数解无解(**)\ \begin{cases} \text{有解} \begin{cases} \text{唯一解} \\ \text{无数解} \end{cases} \\ \text{无解} \end{cases}

相关性与线性表示

概念

  1. 相关性 α1,α2,⋯ ,αn\boldsymbol{\alpha}_{1}, \boldsymbol{\alpha}_{2},\cdots,\boldsymbol{\alpha}_{n} x1α1+x2α2+⋯+xnαn=0(∗) x_{1}\boldsymbol{\alpha}_{1} + x_{2}\boldsymbol{\alpha}_{2}+\cdots + x_{n}\boldsymbol{\alpha}_{n} = \boldsymbol{0} \quad (*)
    • Cases 1 (∗)(*) 仅有零解，称 α1,α2,⋯ ,αn\boldsymbol{\alpha}_{1}, \boldsymbol{\alpha}_{2},\cdots,\boldsymbol{\alpha}_{n} 线性无关
    • Cases 2 (∗)(*) 有非零解，称 α1,α2,⋯ ,αn\boldsymbol{\alpha}_{1}, \boldsymbol{\alpha}_{2},\cdots,\boldsymbol{\alpha}_{n} 线性相关
  2. 线性表示 α1,α2,⋯ ,αn,β\boldsymbol{\alpha}_{1}, \boldsymbol{\alpha}_{2},\cdots,\boldsymbol{\alpha}_{n}, \boldsymbol{\beta} x1α1+x2α2+⋯+xnαn=β(∗∗) x_{1}\boldsymbol{\alpha}_{1} + x_{2}\boldsymbol{\alpha}_{2}+\cdots + x_{n}\boldsymbol{\alpha}_{n} = \boldsymbol{\beta} \quad (**)
    • Cases 1 (∗∗)(**) 有解，称 β\boldsymbol{\beta} 可由 α1,α2,⋯ ,αn\boldsymbol{\alpha}_{1}, \boldsymbol{\alpha}_{2},\cdots,\boldsymbol{\alpha}_{n} 线性表示
    • Cases 2 (∗∗)(**) 无解，称 β\boldsymbol{\beta} 不可由 α2,α2,⋯ ,αn\boldsymbol{\alpha}_{2}, \boldsymbol{\alpha}_{2},\cdots,\boldsymbol{\alpha}_{n} 线性表示

      性质

  3. α1,⋯ ,αn\boldsymbol{\alpha}_{1},\cdots,\boldsymbol{\alpha}_{n} 线性相关 ⇔\Leftrightarrow 至少一个向量可由其他向量线性表示

    Proof

    ⇒\Rightarrow ∃ 不全为0的k1,⋯kn, 使得k1α1+⋯+knαn=0设k1≠0⇒α1=−k2k1α2−⋯−knk1αn∴至少一个向量可由其他向量线性表示 \begin{array}{ll} & \exists\ \text{不全为} 0 \text{的}k_{1},\cdots k_{n}\text{, 使得} \\ & k_{1}\boldsymbol{\alpha}_{1} + \cdots + k_{n}\boldsymbol{\alpha}_{n} = 0 \\ & \text{设}k_{1} \ne 0 \Rightarrow \boldsymbol{\alpha}_{1} = -\frac{k_{2}}{k_{1}}\boldsymbol{\alpha}_{2}-\cdots-\frac{k_{n}}{k_{1}}\boldsymbol{\alpha}_{n} \\ \therefore & \text{至少一个向量可由其他向量线性表示} \\ \end{array} ⇐\Leftarrow 设αk=ℓ1α1+⋯+ℓnαn(右式中无αk)∴ℓ1α1+⋯+(−1)αk+⋯+ℓnαn=0x1α1+x2α2+⋯+xnαn=0 有非零解∴α1,⋯ ,αn线性相关 \begin{array}{ll} & \text{设}\boldsymbol{\alpha}_{k} = \ell_{1}\boldsymbol{\alpha}_{1} + \cdots + \ell_{n}\boldsymbol{\alpha}_{n} \quad(\text{右式中无}\boldsymbol{\alpha}_{k}) \\ \therefore & \ell_{1}\boldsymbol{\alpha}_{1} +\cdots + (-1)\boldsymbol{\alpha}_{k} +\cdots + \ell_{n}\boldsymbol{\alpha}_{n} = \boldsymbol{0} \\ & x_{1}\boldsymbol{\alpha}_{1} + x_{2}\boldsymbol{\alpha}_{2}+\cdots + x_{n}\boldsymbol{\alpha}_{n} = \boldsymbol{0} \text{ 有非零解} \\ \therefore & \boldsymbol{\alpha}_{1},\cdots, \boldsymbol{\alpha}_{n} \text{线性相关} \\ \end{array}

    Notes

    1. 含零向量的向量组成线性相关
    2. α,β\boldsymbol{\alpha}, \boldsymbol{\beta} 相关 ⇔\Leftrightarrow α,β\boldsymbol{\alpha},\boldsymbol{\beta} 成比例

      Proof ⇒\Rightarrow ∵α,β线性相关∴∃ 不全为零的k1,k2, 使得k1α+k2β=0设k1≠0⇒α=k2k1β∴α,β成比例 \begin{array}{ll} \because & \boldsymbol{\alpha},\boldsymbol{\beta} \text{线性相关}\\ \therefore & \exists\ \text{不全为零的} k_{1}, k_{2} \text{, 使得} \\ & k_{1}\boldsymbol{\alpha} + k_{2}\boldsymbol{\beta} = \boldsymbol{0} \\ & \text{设} k_{1}\ne 0 \Rightarrow \boldsymbol{\alpha} = \frac{k_{2}}{k_{1}}\boldsymbol{\beta} \\ \therefore & \boldsymbol{\alpha}, \boldsymbol{\beta} \text{成比例} \\ \end{array} ⇐\Leftarrow ∵α=kβ∴1×α−kα=0∴x1α+x2β=0有非零解∴α与β相关 \begin{array}{ll} \because & \boldsymbol{\alpha} = k\boldsymbol{\beta} \\ \therefore & 1\times\boldsymbol{\alpha} - k \boldsymbol{\alpha} = \boldsymbol{0} \\ \therefore & x_{1}\boldsymbol{\alpha} + x_{2}\boldsymbol{\beta} = \boldsymbol{0} \text{有非零解}\\ \therefore & \boldsymbol{\alpha} \text{与} \boldsymbol{\beta} \text{相关} \\ \end{array}

  4. α1,⋯ ,αn\boldsymbol{\alpha}_{1},\cdots, \boldsymbol{\alpha}_{n} 线性无关
    1. α1,⋯ ,αn,β\boldsymbol{\alpha}_{1},\cdots, \boldsymbol{\alpha}_{n}, \boldsymbol{\beta} 继续线性无关 ⇔\Leftrightarrow β\boldsymbol{\beta} 不可由 α1,⋯ ,αn\boldsymbol{\alpha}_{1},\cdots, \boldsymbol{\alpha}_{n} 线性表示
    2. α1,⋯ ,αn,β\boldsymbol{\alpha}_{1},\cdots, \boldsymbol{\alpha}_{n}, \boldsymbol{\beta} 线性相关 ⇔\Leftrightarrow β\boldsymbol{\beta} 可由 α1,⋯ ,αn\boldsymbol{\alpha}_{1},\cdots, \boldsymbol{\alpha}_{n} 线性表示
  5. 全组无关 ⇒\Rightarrow 部分组无关
  6. 部分组相关 ⇒\Rightarrow 全组相关
  7. ==重点== α1,α2,⋯ ,αn\boldsymbol{\alpha}_{1}, \boldsymbol{\alpha}_{2},\cdots, \boldsymbol{\alpha}_{n} 为 nn 个 nn 维向量
    1. α1,⋯ ,αn\boldsymbol{\alpha}_{1},\cdots, \boldsymbol{\alpha}_{n} 线性无关 ⇔\Leftrightarrow det⁡(α1,α2,⋯ ,αn)≠0\det(\boldsymbol{\alpha}_{1}, \boldsymbol{\alpha}_{2},\cdots, \boldsymbol{\alpha}_{n}) \ne 0
    2. α1,⋯ ,αn\boldsymbol{\alpha}_{1},\cdots, \boldsymbol{\alpha}_{n} 线性相关 ⇔\Leftrightarrow det⁡(α1,α2,⋯ ,αn)=0\det(\boldsymbol{\alpha}_{1}, \boldsymbol{\alpha}_{2},\cdots, \boldsymbol{\alpha}_{n}) = 0

      Proof

      令 A=(α1,⋯ ,αn)\boldsymbol{A} = (\boldsymbol{\alpha}_{1},\cdots ,\boldsymbol{\alpha}_{n})，则 x1α1+⋯+xnαn=0⇔AX=0x_{1}\boldsymbol{\alpha}_{1} + \cdots + x_{n}\boldsymbol{\alpha}_{n} = \boldsymbol{0} \Leftrightarrow \boldsymbol{AX} = \boldsymbol{0}

      1. α1,⋯ ,αn\boldsymbol{\alpha}_{1},\cdots, \boldsymbol{\alpha}_{n} 无关 ⇔\Leftrightarrow AX=0\boldsymbol{AX} = \boldsymbol{0} 仅有零解 ⇔\Leftrightarrow r(A)=nr(\boldsymbol{A}) = n ⇔\Leftrightarrow det⁡(A)≠0\det(\boldsymbol{A}) \ne 0
      2. α1,⋯ ,αn\boldsymbol{\alpha}_{1},\cdots, \boldsymbol{\alpha}_{n} 相关 ⇔\Leftrightarrow AX=0\boldsymbol{AX} = \boldsymbol{0} 有非零解 ⇔\Leftrightarrow r(A)<nr(\boldsymbol{A}) < n ⇔\Leftrightarrow det⁡(A)=0\det(\boldsymbol{A}) = 0
  8. α1,⋯ ,αn\boldsymbol{\alpha}_{1},\cdots, \boldsymbol{\alpha}_{n} 为 nn 个 mm 维向量，且 m<nm<n，则 α1,⋯ ,αn\boldsymbol{\alpha}_{1},\cdots, \boldsymbol{\alpha}_{n} 一定线性相关

    Proof Am×n=(α1⋯αn)x1α1+⋯+xnαn=0⇔AX=0∵r(A)≤min⁡{m,n}=m<n∴AX=0有非零解∴α1,⋯ ,αn线性相关\begin{array}{ll} & \boldsymbol{A}_{m\times n} = \left( \begin{matrix} \boldsymbol{\alpha}_{1} & \cdots & \boldsymbol{\alpha}_{n} \end{matrix} \\ \right) \\ & x_{1}\boldsymbol{\alpha}_{1} + \cdots + x_{n}\boldsymbol{\alpha}_{n} = \boldsymbol{0} \Leftrightarrow \boldsymbol{AX} = \boldsymbol{0} \\ \because & r(\boldsymbol{A})\le \min\{m,n\} = m < n \\ \therefore & \boldsymbol{AX} = 0 \text{有非零解} \\ \therefore & \boldsymbol{\alpha}_{1},\cdots, \boldsymbol{\alpha}_{n} \text{线性相关} \\ \end{array}

  9. 向某一向量组中
    1. 加向量，提高相关性
    2. 加维度，提高无关性
  10. α1,⋯ ,αn\boldsymbol{\alpha}_{1},\cdots, \boldsymbol{\alpha}_{n} 非零，且两两正交 ⇍⇒\nLeftarrow \Rightarrow α1,⋯ ,αn\boldsymbol{\alpha}_{1},\cdots, \boldsymbol{\alpha}_{n} 线性无关

    Proof

    ⇒\Rightarrow 设α1,⋯ ,αn非零两两正交令k1α1+⋯+knαn=0∵(α1,k1α1+⋯+knαn)=0∴k1(α1,α1)=0∵(α1,α1)=α1⊺α1=∣α1∣2>0∴k1=0∵(α2,k2α2+⋯+knαn)=0∴k2(α2,α2)=0∵(α2,α2)=α2⊺α2=∣α2∣2>0∴k2=0同理可得k1=k2=⋯=kn=0∴x1α1+⋯+xnαn=0仅有零解∴α1,⋯ ,αn线性无关 \begin{array}{ll} & \text{设} \boldsymbol{\alpha}_{1},\cdots, \boldsymbol{\alpha}_{n} \text{非零两两正交} \\ & \text{令} k_{1}\boldsymbol{\alpha}_{1} + \cdots + k_{n}\boldsymbol{\alpha}_{n} = \boldsymbol{0} \\ \because & (\boldsymbol{\alpha}_{1}, k_{1}\boldsymbol{\alpha}_{1}+\cdots + k_{n}\boldsymbol{\alpha}_{n}) = 0 \\ \therefore & k_{1}(\boldsymbol{\alpha}_{1},\boldsymbol{\alpha}_{1}) = 0 \\ \because & (\boldsymbol{\alpha}_{1},\boldsymbol{\alpha}_{1}) = \boldsymbol{\alpha}^{\intercal}_{1}\boldsymbol{\alpha}_{1} = \vert \boldsymbol{\alpha}_{1} \vert^{2} > 0 \\ \therefore & k_{1} = 0 \\ \because & (\boldsymbol{\alpha}_{2},k_{2}\boldsymbol{\alpha}_{2} + \cdots + k_{n}\boldsymbol{\alpha}_{n}) = 0 \\ \therefore & k_{2}(\boldsymbol{\alpha}_{2},\boldsymbol{\alpha}_{2}) = 0 \\ \because & (\boldsymbol{\alpha}_{2},\boldsymbol{\alpha}_{2}) = \boldsymbol{\alpha}^{\intercal}_{2}\boldsymbol{\alpha}_{2} = \vert \boldsymbol{\alpha}_{2} \vert^{2} > 0 \\ \therefore & k_{2} = 0 \\ & \text{同理可得} k_{1} = k_{2} = \cdots = k_{n} = 0 \\ \therefore & x_{1}\boldsymbol{\alpha}_{1} + \cdots + x_{n}\boldsymbol{\alpha}_{n} = \boldsymbol{0} \text{仅有零解}\\ \therefore & \boldsymbol{\alpha}_{1},\cdots, \boldsymbol{\alpha}_{n} \text{线性无关} \\ \end{array} ⇍\nLeftarrow 如 α1=(11),α2=(23)\boldsymbol{\alpha}_{1}= \left( \begin{matrix} 1 \\ 1 \end{matrix} \right), \boldsymbol{\alpha}_{2} = \left( \begin{matrix} 2 \\ 3 \end{matrix} \right)

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，问 α1,α2,α3+α4\boldsymbol{\alpha}_{1}, \boldsymbol{\alpha}_{2}, \boldsymbol{\alpha}_{3} + \boldsymbol{\alpha}_{4} 之间的关系? 设α1,α2,α3+α4四者线性相关∵α1,α2线性无关∴∃ k1,k2不全为0, 使得α3+α4=k1α1+k2α2∵α4可由α1,α2线性表示∴∃ ℓ1,ℓ2不全为0使得α4=ℓ1α1+ℓ2α2∴α3=(k1−ℓ1)α1+(k2−ℓ2)α2与题意矛盾∴α1,α2,α3+α4线性无关 \begin{array}{ll} & \text{设} \boldsymbol{\alpha}_{1}, \boldsymbol{\alpha}_{2}, \boldsymbol{\alpha}_{3} + \boldsymbol{\alpha}_{4} \text{四者线性相关} \\ \because & \boldsymbol{\alpha}_{1},\boldsymbol{\alpha}_{2} \text{线性无关} \\ \therefore & \exists\ k_{1},k_{2} \text{不全为}0 \text{, 使得} \\ & \boldsymbol{\alpha}_{3} + \boldsymbol{\alpha}_{4} = k_{1} \boldsymbol{\alpha}_{1} + k_{2}\boldsymbol{\alpha}_{2} \\ \because & \boldsymbol{\alpha}_{4} \text{可由} \boldsymbol{\alpha}_{1},\boldsymbol{\alpha}_{2} \text{线性表示} \\ \therefore & \exists\ \ell_{1},\ell_{2}\text{不全为}0 \text{使得} \\ & \boldsymbol{\alpha}_{4} = \ell_{1}\boldsymbol{\alpha}_{1} + \ell_{2}\boldsymbol{\alpha}_{2} \\ \therefore & \boldsymbol{\alpha}_{3} = (k_{1} - \ell_{1})\boldsymbol{\alpha}_{1} + (k_{2}-\ell_{2})\boldsymbol{\alpha}_{2} \\ & \text{与题意矛盾} \\ \therefore & \boldsymbol{\alpha}_{1}, \boldsymbol{\alpha}_{2}, \boldsymbol{\alpha}_{3} + \boldsymbol{\alpha}_{4} \text{线性无关} \\ \end{array}
  2. α1,α2,α3\boldsymbol{\alpha}_{1}, \boldsymbol{\alpha}_{2}, \boldsymbol{\alpha}_{3} 三者线性无关, α2,α3,α4\boldsymbol{\alpha}_{2}, \boldsymbol{\alpha}_{3}, \boldsymbol{\alpha}_{4} 线性相关，问 α4\boldsymbol{\alpha}_{4} 可否由 α1,α2,α3\boldsymbol{\alpha}_{1},\boldsymbol{\alpha}_{2}, \boldsymbol{\alpha}_{3} 线性表示？ ∵α2,α3,α4线性相关α2,α3线性无关∴∃ k1,k2不全为0，使得α4=k1α2+k2α3∴α4=0α1+k1α2+k2α3∴α4可由α1,α2,α3线性表示 \begin{array}{ll} \because & \boldsymbol{\alpha}_{2}, \boldsymbol{\alpha}_{3}, \boldsymbol{\alpha}_{4}\text{线性相关} \\ & \boldsymbol{\alpha}_{2}, \boldsymbol{\alpha}_{3} \text{线性无关} \\ \therefore & \exists\ k_{1},k_{2}\text{不全为}0 \text{，使得} \\ & \boldsymbol{\alpha}_{4} = k_{1}\boldsymbol{\alpha}_{2}+k_{2}\boldsymbol{\alpha}_{3}\\ \therefore & \boldsymbol{\alpha}_{4} = 0 \boldsymbol{\alpha}_{1} + k_{1}\boldsymbol{\alpha}_{2}+k_{2}\boldsymbol{\alpha}_{3} \\ \therefore & \boldsymbol{\alpha}_{4} \text{可由}\boldsymbol{\alpha}_{1}, \boldsymbol{\alpha}_{2}, \boldsymbol{\alpha}_{3}\text{线性表示} \\ \end{array}
  3. α1=(111)α2=(1−24)α3=(1aa2)\boldsymbol{\alpha}_{1} = \left( \begin{matrix} 1 \\ 1 \\ 1 \\ \end{matrix} \right) \boldsymbol{\alpha}_{2} = \left( \begin{matrix} 1 \\ -2 \\ 4 \\ \end{matrix} \right) \boldsymbol{\alpha}_{3} = \left( \begin{matrix} 1 \\ a \\ a^{2} \\ \end{matrix} \right) 线性相关，则 a=a=         ∵α1,α2,α3线性相关∴∣1111−2a14a2∣=0=(a+2)(a−1)(−2−1)∴a=−2或a=1 \begin{array}{ll} \because & \boldsymbol{\alpha}_{1}, \boldsymbol{\alpha}_{2}, \boldsymbol{\alpha}_{3}\text{线性相关} \\ \therefore & \left\vert \begin{matrix} 1 & 1 & 1 \\ 1 & -2 & a \\ 1 & 4 & a^{2} \\ \end{matrix} \right\vert = 0 = (a+2)(a-1)(-2-1)\\ \therefore & a= -2 \text{或} a=1 \\ \end{array}
  4. α1=(2−1100)α2=(13010)α3=(−12001)\boldsymbol{\alpha}_{1} = \left( \begin{matrix} 2 \\ -1 \\ 1 \\ 0 \\ 0 \end{matrix} \right) \boldsymbol{\alpha}_{2} = \left( \begin{matrix} 1 \\ 3 \\ 0 \\ 1 \\ 0\end{matrix} \right) \boldsymbol{\alpha}_{3} = \left( \begin{matrix} -1 \\ 2 \\ 0 \\ 0 \\ 1 \end{matrix} \right)，问三者相关性? ∵∣100010001∣=1≠0∴(100),(010),(001)三者线性无关∴α1=(2−1100)α2=(13010)α3=(−12001)线性无关 \begin{array}{ll} \because & \left\vert \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{matrix} \right\vert = 1 \ne 0\\ \therefore & \left( \begin{matrix} 1 \\ 0 \\ 0 \end{matrix} \right), \left( \begin{matrix} 0 \\ 1 \\ 0 \end{matrix} \right), \left( \begin{matrix} 0 \\ 0 \\ 1 \end{matrix} \right) \text{三者线性无关} \\ \therefore & \boldsymbol{\alpha}_{1} = \left( \begin{matrix} 2 \\ -1 \\ 1 \\ 0 \\ 0 \end{matrix} \right) \boldsymbol{\alpha}_{2} = \left( \begin{matrix} 1 \\ 3 \\ 0 \\ 1 \\ 0\end{matrix} \right) \boldsymbol{\alpha}_{3} = \left( \begin{matrix} -1 \\ 2 \\ 0 \\ 0 \\ 1 \end{matrix} \right) \text{线性无关} \\ \end{array}
  5. α1,α2,α3\boldsymbol{\alpha}_{1},\boldsymbol{\alpha}_{2},\boldsymbol{\alpha}_{3} 三者线性无关，β1=α1+α2,β2=α2+α3,β3=α3+α1\boldsymbol{\beta}_{1} = \boldsymbol{\alpha}_{1} + \boldsymbol{\alpha}_{2}, \boldsymbol{\beta}_{2} = \boldsymbol{\alpha}_{2} + \boldsymbol{\alpha}_{3}, \boldsymbol{\beta}_{3} =\boldsymbol{\alpha}_{3}+\boldsymbol{\alpha}_{1}，问 β1,β2,β3\boldsymbol{\beta}_{1},\boldsymbol{\beta}_{2}, \boldsymbol{\beta}_{3} 三者关系? A=(α1α2α3),B=(β1β2β3)∴B=A(101110011)∵∣101110011∣=1+1=2≠0α1,α2,α3线性无关∴r(B)=r(A)=3∴β1,β2,β3三者线性无关 \begin{array}{ll} & \boldsymbol{A} = \left( \begin{matrix} \boldsymbol{\alpha}_{1} & \boldsymbol{\alpha}_{2} &\boldsymbol{\alpha}_{3} \end{matrix} \right), \boldsymbol{B} = \left( \begin{matrix} \boldsymbol{\beta}_{1} & \boldsymbol{\beta}_{2} &\boldsymbol{\beta}_{3} \end{matrix} \right) \\ \therefore & \boldsymbol{B} = \boldsymbol{A} \left( \begin{matrix} 1 & 0 & 1\\ 1 & 1 & 0\\ 0 & 1 & 1 \end{matrix} \right)\\ \because & \left\vert \begin{matrix} 1 & 0 & 1\\ 1 & 1 & 0\\ 0 & 1 & 1 \end{matrix} \right\vert = 1 + 1 =2 \ne 0\\ & \boldsymbol{\alpha}_{1}, \boldsymbol{\alpha}_{2}, \boldsymbol{\alpha}_{3} \text{线性无关} \\ \therefore & r(\boldsymbol{B}) = r(\boldsymbol{A}) = 3 \\ \therefore & \boldsymbol{\beta}_{1}, \boldsymbol{\beta}_{2}, \boldsymbol{\beta}_{3}\text{三者线性无关} \\ \end{array}

  6. α1∼α4\boldsymbol{\alpha}_{1}\sim \boldsymbol{\alpha}_{4} 线性无关，β1=α1+α2,β2=α2+α3,β3=α3+α4,β4=α4+α1\boldsymbol{\beta}_{1} = \boldsymbol{\alpha}_{1}+ \boldsymbol{\alpha}_{2},\boldsymbol{\beta}_{2} = \boldsymbol{\alpha}_{2}+ \boldsymbol{\alpha}_{3}, \boldsymbol{\beta}_{3} = \boldsymbol{\alpha}_{3}+ \boldsymbol{\alpha}_{4},\boldsymbol{\beta}_{4} = \boldsymbol{\alpha}_{4}+ \boldsymbol{\alpha}_{1}，问 β1∼β4\boldsymbol{\beta}_{1}\sim\boldsymbol{\beta}_{4} 之间的关系?

    A=(α1α2α3α4),B=(β1β2β3β4)∴B=A(1001110001100011)∵∣1001110001100011∣=0∴det⁡(B)=0∴β1,β2,β3,β4线性相关 \begin{array}{ll} & \boldsymbol{A} = \left( \begin{matrix} \boldsymbol{\alpha}_{1} & \boldsymbol{\alpha}_{2} &\boldsymbol{\alpha}_{3} & \boldsymbol{\alpha}_{4} \end{matrix} \right), \boldsymbol{B} = \left( \begin{matrix} \boldsymbol{\beta}_{1} & \boldsymbol{\beta}_{2} &\boldsymbol{\beta}_{3} & \boldsymbol{\beta}_{4} \end{matrix} \right) \\ \therefore & \boldsymbol{B} = \boldsymbol{A} \left( \begin{matrix} 1 & 0 & 0 & 1 \\ 1 & 1 & 0 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 1 & 1 \\ \end{matrix} \right)\\ \because & \left\vert \begin{matrix} 1 & 0 & 0 & 1 \\ 1 & 1 & 0 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 1 & 1 \\ \end{matrix} \right\vert = 0\\ \therefore & \det(\boldsymbol{B}) = 0 \\ \therefore & \boldsymbol{\beta}_{1},\boldsymbol{\beta}_{2},\boldsymbol{\beta}_{3},\boldsymbol{\beta}_{4}\text{线性相关} \\ \end{array}

results matching ""

    No results matching ""