Section03_秩

定义

  • 对于 Am×n\boldsymbol{A}_{m\times n} 从 A\boldsymbol{A} 中任取 rr 行 rr 列而成的 rr 阶行列式，称为 A\boldsymbol{A} 的 rr 阶子式，若
    1. ∃ r阶子式不为0\exists\ r \text{阶子式不为} 0
    2. ∀ r+1阶子式(不一定有)皆为0\forall\ r+1 \text{阶子式(不一定有)皆为} 0
    3. 则称 rr 为 A\boldsymbol{A} 的秩，记为 r(A)=rr(\boldsymbol{A}) = r

Notes

  1. Am×n⇒{r(A)≤mr(A)≤n⇔r(A)≤min⁡{m,n}\boldsymbol{A}_{m\times n} \Rightarrow \begin{cases} r(\boldsymbol{A}) \le m \\ r(\boldsymbol{A}) \le n \\ \end{cases} \Leftrightarrow r(\boldsymbol{A}) \le \min\{m,n\}
  2. α=(a1a2⋮an)⇒r(α)≤1⇔r(α)={0,α=01,α≠0\boldsymbol{\alpha}= \left( \begin{matrix} a_{1} \\ a_{2} \\ \vdots \\ a_{n} \end{matrix} \right)\Rightarrow r(\boldsymbol{\alpha}) \le 1\Leftrightarrow r(\boldsymbol{\alpha}) = \begin{cases} 0, & \boldsymbol{\alpha} = \boldsymbol{0} \\ 1, & \boldsymbol{\alpha} \ne \boldsymbol{0} \end{cases}
  3. 对于An×n\boldsymbol{A}_{n\times n}
    • A可逆⇔∣A∣≠0⇔r(A)=n (A满秩)\boldsymbol{A}\text{可逆}\Leftrightarrow \vert \boldsymbol{A} \vert \ne 0 \Leftrightarrow r(\boldsymbol{A}) = n\ (\boldsymbol{A} \text{满秩})
    • A不可逆⇔∣A∣=0⇔r(A)<n (A降秩)\boldsymbol{A}\text{不可逆}\Leftrightarrow \vert \boldsymbol{A} \vert = 0 \Leftrightarrow r(\boldsymbol{A}) < n\ (\boldsymbol{A} \text{降秩})

秩求法

A→row(×⋯×⋯×⋯) \boldsymbol{A} \xrightarrow{\text{row}} \left( \begin{matrix} \times & \cdots \\ & \times & \cdots \\ & && \times & \cdots \\\\ \end{matrix} \right)

  • 计算行变化后的非 00 行数 rr, r(A)=rr(\boldsymbol{A}) = r
    • 如 A=(11−123145−1)\boldsymbol{A} = \left( \begin{matrix} 1 & 1 & -1 \\ 2 & 3 & 1 \\ 4 & 5 & -1 \end{matrix} \right) A→(11−1013013)→(11−1013000)∴r(A)=2 \begin{array}{ll} & \boldsymbol{A} \rightarrow \left( \begin{matrix} 1 & 1 & -1 \\ 0 & 1 & 3 \\ 0 & 1 & 3 \end{matrix} \right) \rightarrow \left( \begin{matrix} 1 & 1 & -1 \\ 0 & 1 & 3 \\ 0 & 0 & 0 \end{matrix} \right) \\ \therefore & r(\boldsymbol{A} ) = 2 \\ \end{array}

Notes

  1. r(A)=0⇔A=0r(\boldsymbol{A}) = 0 \Leftrightarrow \boldsymbol{A} = 0
  2. A≠0⇔r(A)≥1\boldsymbol{A} \ne \boldsymbol{0} \Leftrightarrow r(\boldsymbol{A}) \ge 1
  3. r(A)≥2⇔A至少两行不成比例r(\boldsymbol{A}) \ge 2 \Leftrightarrow \boldsymbol{A}\text{至少两行不成比例}

秩性质

  1. r(A)=r(A⊺)=r(A⊺A)=r(AA⊺)r(\boldsymbol{A}) = r(\boldsymbol{A}^{\intercal}) = r(\boldsymbol{A}^{\intercal}\boldsymbol{A}) = r(\boldsymbol{AA}^{\intercal})

    Note 见 A⊺A, AA⊺\boldsymbol{A}^{\intercal}\boldsymbol{A},\ \boldsymbol{AA}^{\intercal} 时使用

  2. r(A±B)≤r(A)+r(B)r(\boldsymbol{A}\pm \boldsymbol{B}) \le r(\boldsymbol{A}) + r(\boldsymbol{B})

    Note 见 {r(A+B)r(A−B)r(A)+r(B)\begin{cases} r(\boldsymbol{A}+\boldsymbol{B}) \\ r(\boldsymbol{A} - \boldsymbol{B}) \\ r(\boldsymbol{A}) + r(\boldsymbol{B}) \end{cases} 时使用

  3. 分块矩阵
    1. {r(AB)≥r(A)r(AB)≥r(B){r(AB)≥r(A)r(AB)≥r(B)\begin{cases} r\left( \begin{array}{c:c} \boldsymbol{A}& \boldsymbol{B} \end{array}\right)\ge r(\boldsymbol{A}) \\ r\left( \begin{array}{c:c} \boldsymbol{A}& \boldsymbol{B} \end{array}\right)\ge r(\boldsymbol{B}) \end{cases}\quad \begin{cases} r\left( \begin{array}{c} \boldsymbol{A}\\\hdashline \boldsymbol{B} \end{array}\right)\ge r(\boldsymbol{A}) \\ r\left( \begin{array}{c} \boldsymbol{A}\\\hdashline \boldsymbol{B} \end{array}\right)\ge r(\boldsymbol{B}) \end{cases}
    2. r(A00B)=r(A)+r(B)r \left( \begin{array}{c:c} \boldsymbol{A} & \boldsymbol{0} \\\hdashline \boldsymbol{0} & \boldsymbol{B} \\ \end{array}\right) = r(\boldsymbol{A}) + r(\boldsymbol{B})
  4. Am×n, Bn×s\boldsymbol{A}_{m\times n},\ \boldsymbol{B}_{n\times s}，则 {r(AB)≤r(A)r(AB)≤r(B)⇔r(AB)≤min⁡{r(A),r(B)}\begin{cases} r(\boldsymbol{AB})\le r(\boldsymbol{A}) \\ r(\boldsymbol{AB})\le r(\boldsymbol{B}) \end{cases}\Leftrightarrow r(\boldsymbol{AB})\le \min\{r(\boldsymbol{A}),r(\boldsymbol{B})\}
  5. (==重点==) Am×n, Bn×s\boldsymbol{A}_{m\times n},\ \boldsymbol{B}_{n\times s}，且 AB=0\boldsymbol{AB} = \boldsymbol{0}，则 r(A)+r(B)≤nr(\boldsymbol{A}) + r(\boldsymbol{B}) \le n

    Note 见 AB=0\boldsymbol{AB} = \boldsymbol{0} 时，使用

  6. Am×n,Pm×m,Qn×n\boldsymbol{A}_{m\times n}, \boldsymbol{P}_{m\times m}, \boldsymbol{Q}_{n\times n}，且 P,Q\boldsymbol{P},\boldsymbol{Q} 均可逆，则 r(A)=r(PA)=r(AQ)=r(PAQ)r(\boldsymbol{A}) = r(\boldsymbol{PA}) = r(\boldsymbol{AQ}) = r(\boldsymbol{PAQ})

    Proof

    令B=PAr(B)=r(PA)≤r(A)∵P可逆,A=P−1B∴r(A)=r(P−1B)≤r(B)=r(PA)∴r(A)=r(PA) \begin{array}{ll} & \text{令}\boldsymbol{B} = \boldsymbol{PA} \\ & r(\boldsymbol{B}) = r(\boldsymbol{PA}) \le r(\boldsymbol{A}) \\ \because & \boldsymbol{P} \text{可逆}, \boldsymbol{A} = \boldsymbol{P}^{-1}\boldsymbol{B} \\ \therefore & r(\boldsymbol{A}) = r(\boldsymbol{P}^{-1}\boldsymbol{B}) \le r(\boldsymbol{B}) = r(\boldsymbol{PA}) \\ \therefore & r(\boldsymbol{A}) = r(\boldsymbol{PA}) \\ \end{array}

  7. (==重点==) r(A∗)={n,r(A)=n1,r(A)=n−10,r(A)<n−1r(\boldsymbol{A}^{*}) = \begin{cases} n, & r(\boldsymbol{A}) = n \\ 1, & r(\boldsymbol{A}) = n-1 \\ 0, & r(\boldsymbol{A}) < n-1 \\ \end{cases}

    Proof

    r(A)=nr(\boldsymbol{A}) = n

    ∵AA∗=∣A∣E∴∣A∣∣A∗∣=∣A∣n∵∣A∣≠0∴∣A∗∣=∣A∣n−1≠0∴r(A∗)=n \begin{array}{ll} \because & \boldsymbol{AA}^{*} = \vert \boldsymbol{A} \vert \boldsymbol{E} \\ \therefore & \vert\boldsymbol{A}\vert \vert \boldsymbol{A}^{*} \vert = \vert \boldsymbol{A} \vert^{n} \\ \because & \vert \boldsymbol{A} \vert \ne 0 \\ \therefore & \vert \boldsymbol{A}^{*} \vert = \vert \boldsymbol{A} \vert^{n-1} \ne 0 \\ \therefore & r(\boldsymbol{A}^{*}) = n \\ \end{array}

    r(A)=n−1r(\boldsymbol{A}) = n-1

    ∵AA∗=∣A∣E=0∴r(A)+r(A∗)≤n∵r(A)=n−1∴r(A∗)≤1∵r(A)=n−1∴∃ Aij≠0,∴A∗≠0r(A∗)≥1∴r(A∗)=1 \begin{array}{ll} \because & \boldsymbol{AA}^{*} = \vert \boldsymbol{A} \vert \boldsymbol{E} = \boldsymbol{0} \\ \therefore & r(\boldsymbol{A}) + r(\boldsymbol{A}^{*}) \le n \\ \because & r(\boldsymbol{A}) = n-1 \\ \therefore & r(\boldsymbol{A}^{*}) \le 1 \\ \because & r(\boldsymbol{A}) = n-1 \\ \therefore & \exists\ A_{ij} \ne 0, \\ \therefore & \boldsymbol{A}^{*} \ne \boldsymbol{0} \\ & r(\boldsymbol{A}^{*}) \ge 1 \\ \therefore & r(\boldsymbol{A}^{*}) = 1 \\ \end{array}

    r(A)<n−1r(\boldsymbol{A}) < n-1

    ∵r(A)<n−1∴∀ Aij=0∴A∗=0∴r(A∗)=0 \begin{array}{ll} \because & r(\boldsymbol{A}) < n-1 \\ \therefore & \forall\ A_{ij} = 0 \\ \therefore & \boldsymbol{A}^{*} = \boldsymbol{0} \\ \therefore & r(\boldsymbol{A}^{*}) = 0 \\ \end{array}

==Notes==

  1. Am×n,Bn×s=(β1β2⋯βs)\boldsymbol{A}_{m\times n}, \boldsymbol{B}_{n\times s} = \left(\begin{matrix}\boldsymbol{\beta}_{1} & \boldsymbol{\beta}_{2}&\cdots & \boldsymbol{\beta}_{s}\end{matrix}\right)，则 AB=A(β1β2⋯βs)=(Aβ1Aβ2⋯Aβs)\color{#D0104C}\boldsymbol{AB} = \boldsymbol{A}\left(\begin{matrix}\boldsymbol{\beta}_{1} & \boldsymbol{\beta}_{2}&\cdots & \boldsymbol{\beta}_{s}\end{matrix}\right) = \left(\begin{matrix}\boldsymbol{A\beta}_{1} & \boldsymbol{A\beta}_{2}&\cdots & \boldsymbol{A\beta}_{s}\end{matrix}\right)
  2. A(BC)=(ABAC)\color{#D0104C} \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)
  3. A=(α1α2α3)\boldsymbol{A} = \left(\begin{matrix}\boldsymbol{\alpha}_{1} & \boldsymbol{\alpha}_{2}& \boldsymbol{\alpha}_{3}\end{matrix}\right)，则 B=(α1+α22α1−α3α2+2α3)=A(1201010−12)\color{#D0104C}\boldsymbol{B} = \left(\begin{matrix}\boldsymbol{\alpha}_{1}+\boldsymbol{\alpha}_{2} & 2\boldsymbol{\alpha}_{1}- \boldsymbol{\alpha}_{3}& \boldsymbol{\alpha}_{2} + 2 \boldsymbol{\alpha}_{3}\end{matrix}\right) = \boldsymbol{A} \left( \begin{matrix} 1 & 2 & 0 \\ 1 & 0 & 1 \\ 0 & -1 & 2 \end{matrix} \right)

例题

  1. Am×n\boldsymbol{A}_{m\times n}，A⊺A=0\boldsymbol{A}^{\intercal}\boldsymbol{A} = \boldsymbol{0}，证明 A=0\boldsymbol{A} = \boldsymbol{0} ∵r(A⊺A)=r(A), A⊺A=0∴r(A)=r(A⊺A)=0∴A=0 \begin{array}{ll} \because & r(\boldsymbol{A}^{\intercal}\boldsymbol{A}) = r(\boldsymbol{A}),\ \boldsymbol{A}^{\intercal}\boldsymbol{A} = \boldsymbol{0} \\ \therefore & r(\boldsymbol{A}) = r(\boldsymbol{A}^{\intercal}\boldsymbol{A}) = 0 \\ \therefore & \boldsymbol{A} = \boldsymbol{0} \\ \end{array}
  2. α=(a1a2⋮an), β=(b1b2⋮bn), A=αα⊺+ββ⊺\boldsymbol{\alpha} = \left( \begin{smallmatrix} a_{1} \\ a_{2} \\ \vdots \\ a_{n} \end{smallmatrix} \right),\ \boldsymbol{\beta} = \left( \begin{smallmatrix} b_{1} \\ b_{2} \\ \vdots \\ b_{n} \end{smallmatrix} \right),\ \boldsymbol{A} = \boldsymbol{\alpha\alpha}^{\intercal} + \boldsymbol{\beta\beta}^{\intercal}，证明 r(A)≤2r(\boldsymbol{A}) \le 2 ∵r(αα⊺)=r(α)=1r(ββ⊺)=r(β)=1∵r(A)≤r(αα⊺)+r(ββ⊺)∴r(A)≤2 \begin{array}{ll} \because & r(\boldsymbol{\alpha\alpha}^{\intercal}) = r(\boldsymbol{\alpha}) = 1 \\ & r(\boldsymbol{\beta\beta}^{\intercal}) = r(\boldsymbol{\beta}) = 1 \\ \because & r(\boldsymbol{A}) \le r(\boldsymbol{\alpha\alpha}^{\intercal}) + r(\boldsymbol{\beta\beta}^{\intercal}) \\ \therefore & r(\boldsymbol{A}) \le 2 \\ \end{array}
  3. 证明 r(AAB)=r(A)r\left( \begin{array}{c:c} \boldsymbol{A}& \boldsymbol{AB} \end{array}\right) = r(\boldsymbol{A}) (AAB)=A(EB)∴r(AAB)≤r(A)∵r(AAB)≥r(A)∴r(AAB)=r(A) \begin{array}{ll} & \left( \begin{array}{c:c} \boldsymbol{A} & \boldsymbol{AB} \end{array}\right) = \boldsymbol{A}\left( \begin{array}{c:c} \boldsymbol{E} & \boldsymbol{B} \end{array}\right) \\ \therefore & r\left( \begin{array}{c:c} \boldsymbol{A} & \boldsymbol{AB} \end{array}\right) \le r(\boldsymbol{A}) \\ \because & r\left( \begin{array}{c:c} \boldsymbol{A} & \boldsymbol{AB} \end{array}\right) \ge r(\boldsymbol{A}) \\ \therefore & r\left( \begin{array}{c:c} \boldsymbol{A} & \boldsymbol{AB} \end{array}\right) = r(\boldsymbol{A}) \\ \end{array}
  4. An×n\boldsymbol{A}_{n\times n} 可逆，证其逆矩阵唯一 设AB=E, AC=E∴AB−AC=A(B−C)=0r(A)+r(B−C)≤n∵A可逆∴r(A)=n∴r(B−C)=0∴B−C=0B=C∴A的逆矩阵唯一 \begin{array}{ll} & \text{设} \boldsymbol{AB} = \boldsymbol{E},\ \boldsymbol{AC} = \boldsymbol{E} \\ \therefore & \boldsymbol{AB} - \boldsymbol{AC} = \boldsymbol{A}(\boldsymbol{B} - \boldsymbol{C}) = \boldsymbol{0} \\ & r(\boldsymbol{A}) + r(\boldsymbol{B} - \boldsymbol{C}) \le n \\ \because & \boldsymbol{A}\text{可逆} \\ \therefore & r(\boldsymbol{A}) = n \\ \therefore & \boldsymbol{r}(\boldsymbol{B} - \boldsymbol{C}) = 0 \\ \therefore & \boldsymbol{B} - \boldsymbol{C} = \boldsymbol{0} \\ & \boldsymbol{B} = \boldsymbol{C} \\ \therefore & \boldsymbol{A}\text{的逆矩阵唯一} \\ \end{array}
  5. An×n, A2=4E\boldsymbol{A}_{n\times n},\ \boldsymbol{A}^{2} = 4 \boldsymbol{E}，证 r(2E−A)+r(2E+A)=nr(2\boldsymbol{E}-\boldsymbol{A})+r(2\boldsymbol{E} + \boldsymbol{A}) = n ∵A2=4E∴(2E−A)(2E+A)=0∴r[(2E−A)(2E+A)]=0∴r(2E−A)+r(2E+A)≤n∵r(2E−A)+r(2E+A)≥r(4E)=n∴r(2E−A)+r(2E+A)=n \begin{array}{ll} \because & \boldsymbol{A}^{2} = 4 \boldsymbol{E} \\ \therefore & (2 \boldsymbol{E} - \boldsymbol{A})(2 \boldsymbol{E} + \boldsymbol{A}) = \boldsymbol{0} \\ \therefore & r[(2 \boldsymbol{E} - \boldsymbol{A})(2 \boldsymbol{E} + \boldsymbol{A})] = 0 \\ \therefore & r(2\boldsymbol{E}-\boldsymbol{A})+r(2\boldsymbol{E} + \boldsymbol{A})\le n\\ \because & r(2\boldsymbol{E}-\boldsymbol{A})+r(2\boldsymbol{E} + \boldsymbol{A}) \ge r(4 \boldsymbol{E}) = n \\ \therefore & r(2\boldsymbol{E}-\boldsymbol{A})+r(2\boldsymbol{E} + \boldsymbol{A}) = n \\ \end{array}

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