Notes 极大组不一定唯一 向量组与极大组等价 重点 对于 α1,⋯ ,αn\boldsymbol{\alpha}_{1},\cdots, \boldsymbol{\alpha}_{n}α1,⋯,αn Case 1 α1,⋯ ,αn\boldsymbol{\alpha}_{1},\cdots,\boldsymbol{\alpha}_{n}α1,⋯,αn 无关 ⇔\Leftrightarrow⇔ α1,⋯ ,αn\boldsymbol{\alpha}_{1},\cdots,\boldsymbol{\alpha}_{n}α1,⋯,αn 的秩 =n=n=n Case 2 α1,⋯ ,αn\boldsymbol{\alpha}_{1},\cdots,\boldsymbol{\alpha}_{n}α1,⋯,αn 相关 ⇔\Leftrightarrow⇔ α1,⋯ ,αn\boldsymbol{\alpha}_{1},\cdots,\boldsymbol{\alpha}_{n}α1,⋯,αn 的秩 <n<n<n 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}I: α1,⋯,αn, II: α1,⋯,αn,β Case 1 I\text{I}I 的秩 === II\text{II}II 的秩 ⇔\Leftrightarrow⇔ β\boldsymbol{\beta}β 可由 α1,⋯ ,αn\boldsymbol{\alpha}_{1},\cdots,\boldsymbol{\alpha}_{n}α1,⋯,αn 线性表示 ⇔\Leftrightarrow⇔ (∗∗)(**)(∗∗) 有解 Case 2 I\text{I}I 的秩 + 1 === II\text{II}II 的秩 ⇔\Leftrightarrow⇔ β\boldsymbol{\beta}β 不可由 α1,⋯ ,αn\boldsymbol{\alpha}_{1},\cdots,\boldsymbol{\alpha}_{n}α1,⋯,αn 线性表示 ⇔\Leftrightarrow⇔ (∗∗)(**)(∗∗) 无解 Am×n=(α1⋮αm)=(β1⋯βn)\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)Am×n=⎝⎛α1⋮αm⎠⎞=(β1⋯βn) α1,⋯ ,αm\boldsymbol{\alpha}_{1},\cdots, \boldsymbol{\alpha}_{m}α1,⋯,αm 称为 A\boldsymbol{A}A 的行向量组,其的秩成为 A\boldsymbol{A}A 的行秩 β1,⋯ ,βn\boldsymbol{\beta}_{1},\cdots, \boldsymbol{\beta}_{n}β1,⋯,βn 称为 A\boldsymbol{A}A 的列向量组,其的秩成为 A\boldsymbol{A}A 的列秩 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)Am×n,Bn×s=(β1⋯βs) ⇒\Rightarrow⇒ AB=A(β1⋯βs)=(Aβ1⋯Aβs)\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)AB=A(β1⋯βs)=(Aβ1⋯Aβs) 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)A(BC)=(ABAC) 如 α1,α2,α3\boldsymbol{\alpha}_{1},\boldsymbol{\alpha}_{2},\boldsymbol{\alpha}_{3}α1,α2,α3,A=(α1α2α3)\boldsymbol{A} = \left( \begin{matrix} \boldsymbol{\alpha}_{1} & \boldsymbol{\alpha}_{2} & \boldsymbol{\alpha}_{3} \end{matrix} \right)A=(α1α2α3), 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}B=(a1α1+a2α2+a3α3b1α1+b2α2+b3α3c1α1+c2α2+c3α3)=⎝⎛a1a2a3b1b2b3c1c2c3⎠⎞
Notes
极大组不一定唯一 向量组与极大组等价 重点 对于 α1,⋯ ,αn\boldsymbol{\alpha}_{1},\cdots, \boldsymbol{\alpha}_{n}α1,⋯,αn Case 1 α1,⋯ ,αn\boldsymbol{\alpha}_{1},\cdots,\boldsymbol{\alpha}_{n}α1,⋯,αn 无关 ⇔\Leftrightarrow⇔ α1,⋯ ,αn\boldsymbol{\alpha}_{1},\cdots,\boldsymbol{\alpha}_{n}α1,⋯,αn 的秩 =n=n=n Case 2 α1,⋯ ,αn\boldsymbol{\alpha}_{1},\cdots,\boldsymbol{\alpha}_{n}α1,⋯,αn 相关 ⇔\Leftrightarrow⇔ α1,⋯ ,αn\boldsymbol{\alpha}_{1},\cdots,\boldsymbol{\alpha}_{n}α1,⋯,αn 的秩 <n<n<n 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}I: α1,⋯,αn, II: α1,⋯,αn,β Case 1 I\text{I}I 的秩 === II\text{II}II 的秩 ⇔\Leftrightarrow⇔ β\boldsymbol{\beta}β 可由 α1,⋯ ,αn\boldsymbol{\alpha}_{1},\cdots,\boldsymbol{\alpha}_{n}α1,⋯,αn 线性表示 ⇔\Leftrightarrow⇔ (∗∗)(**)(∗∗) 有解 Case 2 I\text{I}I 的秩 + 1 === II\text{II}II 的秩 ⇔\Leftrightarrow⇔ β\boldsymbol{\beta}β 不可由 α1,⋯ ,αn\boldsymbol{\alpha}_{1},\cdots,\boldsymbol{\alpha}_{n}α1,⋯,αn 线性表示 ⇔\Leftrightarrow⇔ (∗∗)(**)(∗∗) 无解 Am×n=(α1⋮αm)=(β1⋯βn)\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)Am×n=⎝⎛α1⋮αm⎠⎞=(β1⋯βn) α1,⋯ ,αm\boldsymbol{\alpha}_{1},\cdots, \boldsymbol{\alpha}_{m}α1,⋯,αm 称为 A\boldsymbol{A}A 的行向量组,其的秩成为 A\boldsymbol{A}A 的行秩 β1,⋯ ,βn\boldsymbol{\beta}_{1},\cdots, \boldsymbol{\beta}_{n}β1,⋯,βn 称为 A\boldsymbol{A}A 的列向量组,其的秩成为 A\boldsymbol{A}A 的列秩 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)Am×n,Bn×s=(β1⋯βs) ⇒\Rightarrow⇒ AB=A(β1⋯βs)=(Aβ1⋯Aβs)\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)AB=A(β1⋯βs)=(Aβ1⋯Aβs) 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)A(BC)=(ABAC) 如 α1,α2,α3\boldsymbol{\alpha}_{1},\boldsymbol{\alpha}_{2},\boldsymbol{\alpha}_{3}α1,α2,α3,A=(α1α2α3)\boldsymbol{A} = \left( \begin{matrix} \boldsymbol{\alpha}_{1} & \boldsymbol{\alpha}_{2} & \boldsymbol{\alpha}_{3} \end{matrix} \right)A=(α1α2α3), 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}B=(a1α1+a2α2+a3α3b1α1+b2α2+b3α3c1α1+c2α2+c3α3)=⎝⎛a1a2a3b1b2b3c1c2c3⎠⎞
Case 1 α1,⋯ ,αn\boldsymbol{\alpha}_{1},\cdots,\boldsymbol{\alpha}_{n}α1,⋯,αn 无关 ⇔\Leftrightarrow⇔ α1,⋯ ,αn\boldsymbol{\alpha}_{1},\cdots,\boldsymbol{\alpha}_{n}α1,⋯,αn 的秩 =n=n=n Case 2 α1,⋯ ,αn\boldsymbol{\alpha}_{1},\cdots,\boldsymbol{\alpha}_{n}α1,⋯,αn 相关 ⇔\Leftrightarrow⇔ α1,⋯ ,αn\boldsymbol{\alpha}_{1},\cdots,\boldsymbol{\alpha}_{n}α1,⋯,αn 的秩 <n<n<n
Case 1 I\text{I}I 的秩 === II\text{II}II 的秩 ⇔\Leftrightarrow⇔ β\boldsymbol{\beta}β 可由 α1,⋯ ,αn\boldsymbol{\alpha}_{1},\cdots,\boldsymbol{\alpha}_{n}α1,⋯,αn 线性表示 ⇔\Leftrightarrow⇔ (∗∗)(**)(∗∗) 有解 Case 2 I\text{I}I 的秩 + 1 === II\text{II}II 的秩 ⇔\Leftrightarrow⇔ β\boldsymbol{\beta}β 不可由 α1,⋯ ,αn\boldsymbol{\alpha}_{1},\cdots,\boldsymbol{\alpha}_{n}α1,⋯,αn 线性表示 ⇔\Leftrightarrow⇔ (∗∗)(**)(∗∗) 无解
α1,⋯ ,αm\boldsymbol{\alpha}_{1},\cdots, \boldsymbol{\alpha}_{m}α1,⋯,αm 称为 A\boldsymbol{A}A 的行向量组,其的秩成为 A\boldsymbol{A}A 的行秩 β1,⋯ ,βn\boldsymbol{\beta}_{1},\cdots, \boldsymbol{\beta}_{n}β1,⋯,βn 称为 A\boldsymbol{A}A 的列向量组,其的秩成为 A\boldsymbol{A}A 的列秩
Proof A=(α1⋯αm),B=(β1⋯βn)∵{α1=k11β1+⋯+k1nβn⋯αm=km1β1+⋯+kmnβn∴A=B(k11k21⋯km1k12k22⋯km2⋮⋮⋱⋮k1nk2n⋯kmn)=BK∵r(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} ∵∴∵∴A=(α1⋯αm),B=(β1⋯βn)⎩⎨⎧α1=k11β1+⋯+k1nβn⋯αm=km1β1+⋯+kmnβnA=B⎝⎛k11k12⋮k1nk21k22⋮k2n⋯⋯⋱⋯km1km2⋮kmn⎠⎞=BKr(A)=r(BK)≤r(B)I的秩≤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} ∵∴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 ⇍\nLeftarrow⇍ 反例 I: α1=(1−100),α2=(2100)II: α1=(0012),α2=(00−13) \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} I: α1=(1−100),α2=(2100)II: α1=(0012),α2=(00−13)
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} ∵∴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 ⇍\nLeftarrow⇍ 反例 I: α1=(1−100),α2=(2100)II: α1=(0012),α2=(00−13) \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} I: α1=(1−100),α2=(2100)II: α1=(0012),α2=(00−13)