矩阵秩(不)等式汇总与证明

矩阵秩(不)等式汇总与证明

Thm 1

$\text{Theorem1}$
$$\begin{array}{rl}0& ⩽\mathrm{rank}(A_{m,n})\stackrel{\text{def}}{}\underset{\det (A_k)\ne 0}{\max }\{k\}\\ & =\dim \mathrm{Im}(\mathcal{A}:V_{\mathbb{F}}^n\to U_{\mathbb{F}}^m)=\dim \mathrm{span}(A)\\ & ={\mathrm{rank}}_r(A)={\mathrm{rank}}_c(A)\\ & ⩽\min \{m,n\}\end{array}$$

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Thm 2

$\text{Theorem2}$
$$\begin{array}{rl}\mathrm{rank}(A_{mn})& \stackrel{c\in \mathbb{R}\backslash \{0\}}{}\mathrm{rank}(cA)=\mathrm{rank}(A^{\mathrm{T}})=\mathrm{rank}(A{A}^{\mathrm{T}})=\mathrm{rank}(A^{\mathrm{T}}A)\\ & \stackrel{\mathrm{rank}(P)=m}{}\mathrm{rank}(P_{sm}{A}_{mn})\stackrel{\mathrm{rank}(Q)=n}{}\mathrm{rank}(A_{mn}{Q}_{nt})\\ & =\mathrm{rank}(A,AB)=\mathrm{rank}\left(\begin{array}{c}A\\ CA\end{array}\right)\end{array}$$

$\text{Thm2.1 Proof:}$

由于
$$\begin{array}{r}Ax=0\Rightarrow A^{\mathrm{T}}Ax=0\Rightarrow x^{\mathrm{T}}{A}^{\mathrm{T}}Ax=(Ax{)}^{\mathrm{T}}(Ax)=0\Rightarrow Ax=0\end{array}$$
故齐次方程组 $Ax=0$ 与 $A^{\mathrm{T}}Ax=0$ 同解，于是 $A$ 与 $A^{\mathrm{T}}A$ 等秩；同理可证，$A^{\mathrm{T}}x=0$ 与 $A{A}^{\mathrm{T}}x=0$ 同解，故 $A^{\mathrm{T}}$ 与 $A{A}^{\mathrm{T}}$ 等秩；矩阵 $cA(c\ne 0),A^{\mathrm{T}}$ 与 $A$ 的最高阶子式相同，由秩定义可知三者等秩；综上所述，矩阵 $A,cA(c\ne 0),A^{\mathrm{T}},A{A}^{\mathrm{T}},A^{\mathrm{T}}A$ 皆等秩.

$\text{Thm2.3 Proof:}$

详见 Corollary7

$\text{Thm2.3 Proof:}$

设 $A=(\{{\alpha }_i{\}}_{i=1}^n{)}_{m\times n},B=(b_{ij}{)}_{n\times s}$，则
( A , A B ) = ( α 1 , ⋯   , α n ;   ( α 1 , ⋯   , α n ) ( b 11 ⋯ b 1 s ⋮ ⋱ ⋮ b n 1 ⋯ b n s ) ) = ( α 1 , ⋯   , α n ;   γ 1 , ⋯   , γ s ) \begin{align*} (A,AB)&=(\alpha_1,\cdots,\alpha_n;\ (\alpha_1,\cdots,\alpha_n)\left( \begin{matrix} b_{11}& \cdots& b_{1s}\\ \vdots& \ddots& \vdots\\ b_{n1}& \cdots& b_{ns}\\ \end{matrix} \right))\\&=(\alpha_1,\cdots,\alpha_n;\ \gamma_1,\cdots,\gamma_s) \end{align*} (A,AB)​=(α1​,⋯,αn​; (α1​,⋯,αn​) ​b11​⋮bn1​​⋯⋱⋯​b1s​⋮bns​​ ​)=(α1​,⋯,αn​; γ1​,⋯,γs​)​
所以向量组 $\{{\gamma }_j{\}}_{j=1}^s$ 可由 $\{{\alpha }_i{\}}_{i=1}^n$ 线性表示，于是
$$(A,AB)=({\alpha }_1,\cdots ,{\alpha }_n;{\gamma }_1,\cdots ,{\gamma }_s)\stackrel{c}{}({\alpha }_1,\cdots ,{\alpha }_n;0,\cdots ,0)$$
从而
$$\mathrm{rank}(A,AB)=\mathrm{rank}({\alpha }_1,\cdots ,{\alpha }_n;0,\cdots ,0)=\mathrm{rank}(A)$$
余下同理可证.

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Thm 3

$\text{Theorem3}$
$$\mathrm{rank}(A^{\ast })=\{\begin{array}{l}n,\mathrm{rank}(A)=n\\ 1,\mathrm{rank}(A)=n-1\\ 0,\mathrm{rank}(A)<n-1\end{array}$$

$\text{Thm3 Proof:}$

由秩定义

$$\{\begin{array}{l}\mathrm{rank}(A)=k\iff \exists |A_k|\ne 0\wedge \forall |A_{k+1}|=0\\ \mathrm{rank}(A)⩾k\iff \exists |A_k|\ne 0\\ \mathrm{rank}(A)<k\iff \forall |A_k|=0\end{array}$$

(1) 当 $\mathrm{rank}(A)=n$ 时，有 $|A|\ne 0$，故 $|A^{\ast }|=||A|A^{-1}|=|A{|}^{n-1}\ne 0$，于是有 $\mathrm{rank}(A^{\ast })=n$；

(2) 当 $\mathrm{rank}(A)=n-1$ 时，存在 $n-1$ 阶子式不为零，则 $A^{\ast }\ne O\Rightarrow \mathrm{rank}(A^{\ast })>0$，又 $A^{\ast }A=|A|I=O$，故由 Corollary7 有 $\mathrm{rank}(A^{\ast })+\mathrm{rank}(A)⩽n$，于是 $1⩽\mathrm{rank}(A^{\ast })⩽n-\mathrm{rank}(A)=n-(n-1)=1$，从而 $\mathrm{rank}(A^{\ast })=1$；

(3) 当 $\mathrm{rank}(A)<n-1$ 时，所有 $n-1$ 阶子式全为零，则 $A^{\ast }=O\Rightarrow \mathrm{rank}(A^{\ast })=0$.

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Thm 4

$\text{Theorem4}$

从一个矩阵中抽取某行、列交叉处元素构成新矩阵称为该原矩阵的子矩阵，其秩不超过原矩阵，即
$$\mathrm{rank}(\text{SubMatrix})⩽\mathrm{rank}(\text{Matrix})$$

$\text{Thm4 Proof:}$

设子矩阵 $\mathrm{rank}(M^{\prime })=r^{\prime }$，则其存在非零 $r^{\prime }$ 阶子式，即 $\exists \det (M_{r^{\prime }}^{\prime })\ne 0$，且 $r^{\prime }$ 阶子式对应矩阵 $M_{r^{\prime }}^{\prime }$ 亦为原矩阵 $M$ 的子矩阵，所以原矩阵 $M$ 中 $\exists r⩾r^{\prime },s.t.\det (M_r)\ne 0$，于是有 $\mathrm{rank}(\mathrm{M})=r⩾r^{\prime }=\mathrm{rank}({\mathrm{M}}^{\prime })$.

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Thm 5

$\text{Theorem5}$ 矩阵降阶公式

设分块矩阵 $M=\left(\begin{array}{cc}{A}_m& B\\ C& D_n\end{array}\right)$

若 $A$ 为非奇异矩阵，则
$$\mathrm{rank}(M)=\mathrm{rank}(A)+\mathrm{rank}(D-C{A}^{-1}B)$$

若 $D$ 为非奇异矩阵，则
$$\mathrm{rank}(M)=\mathrm{rank}(D)+\mathrm{rank}(A-B{D}^{-1}C)$$

若 $A,D$ 皆为非奇异矩阵，则
$$\mathrm{rank}(A)+\mathrm{rank}(D-C{A}^{-1}B)=\mathrm{rank}(D)+\mathrm{rank}(A-B{D}^{-1}C)$$

$\text{Thm5 Proof:}$

初等变换
$$\left(\begin{array}{cc}A& B\\ C& D\end{array}\right)\stackrel{r_2-C{A}^{-1}\cdot r_1}{}\left(\begin{array}{cc}A& B\\ O& D-C{A}^{-1}B\end{array}\right)$$
则
$$\mathrm{rank}(M)=\mathrm{rank}(A)+\mathrm{rank}(D-C{A}^{-1}B)$$
余下同理可证.

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Thm 6

$\text{Theorem6}$
$$\begin{array}{rl}\begin{array}{c}\max \{\mathrm{rank}(A),\mathrm{rank}(B)\}\\ \mathrm{rank}(A\pm B)\end{array}& ⩽\begin{array}{c}\mathrm{rank}(A,B)\\ \mathrm{rank}\left(\begin{array}{c}A\\ B\end{array}\right)\end{array}\\ & ⩽\mathrm{rank}(A)+\mathrm{rank}(B)\\ & =\mathrm{rank}\left(\begin{array}{cc}A& O\\ O& B\end{array}\right)\\ & ⩽\mathrm{rank}\left(\begin{array}{cc}A& C\\ O& B\end{array}\right)\\ & ⩽\mathrm{rank}(A)+\mathrm{rank}(B)+\mathrm{rank}(C)\end{array}$$

$\text{Corollary6}$

$$\mathrm{rank}(A)-\mathrm{rank}(B)⩽\mathrm{rank}(A-B)$$

$$\mathrm{rank}(A_n{B}_n-I_n)⩽\mathrm{rank}(A-I)+\mathrm{rank}(B-I)$$

$\text{Lemma6}$

设向量组 $\{{\alpha }_i{\}}_{i=1}^s$ 可由 $\{{\beta }_j{\}}_{j=1}^t$ 线性表示，

1. (1) 若 $s>t$，则 $\{{\alpha }_i{\}}_{i=1}^s$ 线性相关；(2) 若 $\{{\alpha }_i{\}}_{i=1}^s$ 线性无关，则 $s⩽t$；

2. $\mathrm{rank}(\{{\alpha }_i\})⩽\mathrm{rank}(\{{\beta }_j\})$.

$\text{Lemma6.1 Proof:}$

(1) $\{{\alpha }_i{\}}_{i=1}^s$ 可由 $\{{\beta }_j{\}}_{j=1}^t$线性表示，则有
r a n k ( α 1 , α 2 , ⋯   , α s ) = r a n k (   ( β 1 , ⋯   , β t ) [ b 11 ⋯ b 1 s ⋮ ⋱ ⋮ b t 1 ⋯ b t s ] ) ⩽ min ⁡ { t , s } ⩽ t < s \begin{aligned} \mathrm{rank}\left( \alpha _1,\alpha _2,\cdots ,\alpha _s \right) &=\mathrm{rank}(\ \left( \beta _1,\cdots ,\beta _t \right) \left[ \begin{matrix} b_{11}& \cdots& b_{1s}\\ \vdots& \ddots& \vdots\\ b_{t1}& \cdots& b_{ts}\\ \end{matrix} \right] ) \\&\leqslant \min \left\{ t,s \right\} \leqslant trank(α1​,α2​,⋯,αs​)​=rank( (β1​,⋯,βt​) ​b11​⋮bt1​​⋯⋱⋯​b1s​⋮bts​​ ​)⩽min{t,s}⩽t<s​
故向量组 $\{{\alpha }_i{\}}_{i=1}^s$ 线性相关.

(2) 将 (1) 取逆否命题.

$\text{Lemma6.2 Proof1:}$

分别取两组向量组的极大无关组 $$\{{\alpha }_{i_{{}_1}},\cdots ,{\alpha }_{i_{{}_n}}\}\subseteq \{{\alpha }_i{\}}_{i=1}^s,\{{\beta }_{j_{{}_1}},\cdots ,{\beta }_{j_{{}_m}}\}\subseteq \{{\beta }_j{\}}_{j=1}^t$$
则前者可由后者线性表示，由 Lemma6.1 可知
$$n⩽m\Rightarrow \mathrm{rank}(\{{\alpha }_i{\}}_{i=1}^s)⩽\mathrm{rank}(\{{\beta }_j{\}}_{j=1}^t)$$

$\text{Lemma6.2 Proof2:}$

记 $(\{{\alpha }_i{\}}_{i=1}^s)=A_{ns},(\{{\beta }_j{\}}_{j=1}^t)=B_{nt}$，则
$$\exists C\in M_{t,s}(\mathbb{F}),s.t.A_{ns}=B_{nt}{C}_{ts}$$
故由 Theorem7 有
$$\mathrm{rank}(A)=\mathrm{rank}(BC)⩽\min \{\mathrm{rank}(B),\mathrm{rank}(C)\}⩽\mathrm{rank}(B)$$

$\text{Thm6.1.1 Proof:}$

矩阵 $A,B$ 为 $(A,B),\left(\begin{array}{c}A\\ B\end{array}\right)$ 的子阵，故由 Theorem4 可知
$$\max \{\mathrm{rank}(A),\mathrm{rank}(B)\}⩽\begin{array}{c}\mathrm{rank}(A,B)\\ \mathrm{rank}\left(\begin{array}{c}A\\ B\end{array}\right)\end{array}$$

$\text{Thm6.1.2 Proof:}$

列向量组 $A\pm B=({\alpha }_1\pm {\beta }_1,\cdots ,{\alpha }_n\pm {\beta }_n)$ 可由 $(A,B)=({\alpha }_1,\cdots ,{\alpha }_n;{\beta }_1,\cdots ,{\beta }_n)$ 线性表示，由 Lemma6 可知 $$\mathrm{rank}(A\pm B)⩽\mathrm{rank}(A,B)$$
同理，由行向量组可证
$$\mathrm{rank}(A\pm B)⩽\mathrm{rank}\left(\begin{array}{c}A\\ B\end{array}\right)$$

$\text{Thm6.2 Proof1:}$

记矩阵 $(A_{mn},B_{ms})$ 极大无关向量组所构成的矩阵为 $C_{m,t}(0<t⩽\min \{m,n+s\})$，
若 $(A,B)$ 的极大无关组线性无关，则 $t=\mathrm{rank}(A)+\mathrm{rank}(B)$；若线性相关，则 $t<\mathrm{rank}(A)+\mathrm{rank}(B)$. 综上 $t⩽\mathrm{rank}(A)+\mathrm{rank}(B)$，于是
$$\begin{array}{rl}\mathrm{rank}(A,B)& =\mathrm{rank}(C_{m,t})\\ & ⩽\min \{m,t\}\\ & ⩽\min \{m,\mathrm{rank}(A)+\mathrm{rank}(B)\}\\ & ⩽\mathrm{rank}(A)+\mathrm{rank}(B)\end{array}$$
同理可证
$$\mathrm{rank}\left(\begin{array}{c}A\\ B\end{array}\right)⩽\mathrm{rank}(A)+\mathrm{rank}(B)$$

$\text{Thm6.2 Proof2:}$

$$\begin{array}{rl}\mathrm{rank}(A,B)& =\mathrm{rank}\left((I_m,I_m)\left(\begin{array}{cc}{A}_{m,n}& O\\ O& B_{m,s}\end{array}\right)\right)\\ & ⩽\min \{\mathrm{rank}(I,I),\mathrm{rank}\left(\begin{array}{cc}A& O\\ O& B\end{array}\right)\}\\ & ⩽\mathrm{rank}\left(\begin{array}{cc}A& O\\ O& B\end{array}\right)\\ & =\mathrm{rank}(A)+\mathrm{rank}(B)\end{array}$$
同理可证
$$\mathrm{rank}\left(\begin{array}{c}A\\ B\end{array}\right)⩽\mathrm{rank}(A)+\mathrm{rank}(B)$$

$\text{Thm6.3 Proof:}$

对矩阵进行初等行变换化至阶梯型
$$A_{mn}\stackrel{r}{}\left(\begin{array}{c}{A}_{\mathrm{rank}\left(A\right),n}\\ O\end{array}\right),B_{st}\stackrel{r}{}\left(\begin{array}{c}{B}_{\mathrm{rank}\left(B\right),t}\\ O\end{array}\right)$$
则
( A O O B ) → r ( A r a n k ( A ) , n O O O O B r a n k ( B ) , t O O ) → ( A r a n k ( A ) , n O O B r a n k ( B ) , t O O O O ) \left( \begin{matrix} A& O\\ O& B\\ \end{matrix} \right) \xrightarrow{r} \left( \begin{matrix} A_{\mathrm{rank}\left( A \right) ,n}& O\\ O& O\\ O& B_{\mathrm{rank}\left( B \right) ,t}\\ O& O\\ \end{matrix} \right) \rightarrow \left( \begin{matrix} A_{\mathrm{rank}\left( A \right) ,n}& O\\ O& B_{\mathrm{rank}\left( B \right) ,t}\\ O& O\\ O& O\\ \end{matrix} \right) (AO​OB​)r ​ ​Arank(A),n​OOO​OOBrank(B),t​O​ ​→ ​Arank(A),n​OOO​OBrank(B),t​OO​ ​
有限次初等变换不改变矩阵的秩，故有
$$\begin{array}{rl}\mathrm{rank}\left(\begin{array}{cc}A& O\\ O& B\end{array}\right)& =\mathrm{rank}\left(A_{\mathrm{rank}\left(A\right),n}\right)+\mathrm{rank}\left(B_{\mathrm{rank}\left(B\right),t}\right)\\ & =\mathrm{rank}(A)+\mathrm{rank}(B)\end{array}$$

$\text{Thm6.4 Proof:}$

记 $|A_{\mathrm{rank}(A)}|,|B_{\mathrm{rank}(B)}|$ 分别为矩阵 $A,B$ 的最高阶非零子式，则
∣ A r a n k ( A ) C ′ O B r a n k ( B ) ∣ = ∣ A r a n k ( A ) O O B r a n k ( B ) ∣ = ∣ A r a n k ( A ) ∣ ∣ B r a n k ( B ) ∣ ≠ 0 \left| \begin{matrix} A_{\mathrm{rank}\left( A \right)}& C'\\ O& B_{\mathrm{rank}\left( B \right)}\\ \end{matrix} \right| = \left|\begin{matrix} A_{\mathrm{rank}\left( A \right)}& O\\ O& B_{\mathrm{rank}\left( B \right)}\\ \end{matrix} \right|=|A_{\mathrm{rank}\left( A \right)}||B_{\mathrm{rank}\left( B \right)}|\ne 0 ​Arank(A)​O​C′Brank(B)​​ ​= ​Arank(A)​O​OBrank(B)​​ ​=∣Arank(A)​∣∣Brank(B)​∣=0
所以上述两个方阵皆为满秩，于是
$$\begin{array}{rl}\mathrm{rank}\left(\begin{array}{cc}A& C\\ O& B\end{array}\right)& ⩾\mathrm{rank}\left(\begin{array}{cc}{A}_{\mathrm{rank}\left(A\right)}& C^{\prime }\\ O& B_{\mathrm{rank}\left(B\right)}\end{array}\right)\\ & =\mathrm{rank}\left(\begin{array}{cc}{A}_{\mathrm{rank}\left(A\right)}& O\\ O& B_{\mathrm{rank}\left(B\right)}\end{array}\right)\\ & =\mathrm{rank}\left(A\right)+\mathrm{rank}\left(B\right)\\ & =\mathrm{rank}\left(\begin{array}{cc}A& O\\ O& B\end{array}\right)\end{array}$$

$\text{Thm6.5 Proof:}$

$$\mathrm{rank}\left(\begin{array}{cc}A& C\\ O& B\end{array}\right)⩽\mathrm{rank}(A,C)+\mathrm{rank}(O,B)⩽\mathrm{rank}(A)+\mathrm{rank}(C)+\mathrm{rank}(B)$$

$\text{Cor6.1 Proof:}$

$$\mathrm{rank}(A)=\mathrm{rank}((A-B)+B)⩽\mathrm{rank}(A-B)+\mathrm{rank}(B)$$

$\text{Cor6.2 Proof:}$

$$\begin{array}{rl}\mathrm{rank}(AB-I)& =\mathrm{rank}(A(B-I)+A-I)\\ & ⩽\mathrm{rank}(A(B-I))+\mathrm{rank}(A-I)\\ & ⩽\mathrm{rank}(B-I)+\mathrm{rank}(A-I)\end{array}$$

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Thm 7

$\text{Theorem7}$

若 $A\in M_{m,n}(\mathbb{F}),B\in M_{n,s}(\mathbb{F})$，则
$$\mathrm{rank}(A)+\mathrm{rank}(B)-n⩽r(AB)⩽\min \{\mathrm{rank}(A),\mathrm{rank}(B)\}$$
其中，左侧不等式称 $\text{Sylvester}$ 秩不等式.


$\text{Corollary7}$
$$\begin{array}{rl}& \text{LHS}\stackrel{AB=O}{}\mathrm{rank}(A)+\mathrm{rank}(B)⩽n\\ & \\ & \text{RHS}\iff \{\begin{array}{l}\mathrm{rank}(AB)⩽\mathrm{rank}(A)\stackrel{\mathrm{rank}(B)=n}{}\mathrm{rank}(AB)=\mathrm{rank}(A)\\ \\ \mathrm{rank}(AB)⩽\mathrm{rank}(B)\stackrel{\mathrm{rank}(A)=n}{}\mathrm{rank}(AB)=\mathrm{rank}(B)\end{array}\end{array}$$

$\text{Thm7 LHS Proof1:}$

构造矩阵，进行初等变换
$$\begin{array}{rl}\left(\begin{array}{cc}{I}_n& O\\ O& AB\end{array}\right)& \stackrel{r_2+A\cdot r_1}{}\left(\begin{array}{cc}{I}_n& O\\ A& AB\end{array}\right)\stackrel{c_2+c_1\cdot \left(-B\right)}{}\left(\begin{array}{cc}{I}_n& -B\\ A& O\end{array}\right)\\ & \stackrel{c_1\leftrightarrow -c_2}{}\left(\begin{array}{cc}B& I_n\\ O& A\end{array}\right)\end{array}$$
则
$$\begin{array}{rl}n+\mathrm{rank}\left(AB\right)& =\mathrm{rank}\left(I_n\right)+\mathrm{rank}\left(AB\right)\\ & =\mathrm{rank}\left(\begin{array}{cc}{I}_n& O\\ O& AB\end{array}\right)\\ & =\mathrm{rank}\left(\begin{array}{cc}B& I_n\\ O& A\end{array}\right)\\ & ⩾\mathrm{rank}\left(\begin{array}{cc}B& O\\ O& A\end{array}\right)\\ & =\mathrm{rank}\left(A\right)+\mathrm{rank}\left(B\right)\end{array}$$

$\text{Thm7 LHS Proof2:}$

设 $U,V,W$ 为有限维线性空间，$\mathcal{B}\in \mathrm{Hom}(U^s,V^n),\mathcal{A}\in \mathrm{Hom}(V^n,W^m)$，则 $\mathcal{AB}\in \mathrm{Hom}(U^s,W^m)$，

对 $\mathcal{A}$ 作限制映射 $\mathcal{A}{|}_{\mathrm{Im}\mathcal{B}}:\mathrm{Im}\mathcal{B}\to \mathcal{A}(\mathrm{Im}\mathcal{B})=\mathrm{Im}(\mathcal{AB})$，故有
$$\begin{array}{rl}\dim \mathrm{Im}(\mathcal{B})& =\dim \mathrm{Im}(\mathcal{A}{|}_{\mathrm{Im}\mathcal{B}})+\dim \mathrm{Ker}(\mathcal{A}{|}_{\mathrm{Im}\mathcal{B}})\\ & =\dim \mathrm{Im}(\mathcal{AB})+\dim \mathrm{Ker}(\mathcal{A}{|}_{\mathrm{Im}\mathcal{B}})\\ & =\dim \mathrm{Im}(\mathcal{AB})+\dim (\mathrm{Ker}\mathcal{A}\cap \mathrm{Im}\mathcal{B})\\ & ⩽\dim \mathrm{Im}(\mathcal{AB})+\dim \mathrm{Ker}(\mathcal{A})\\ & =\dim \mathrm{Im}(\mathcal{AB})+\dim (V^n)-\dim \mathrm{Im}(\mathcal{A})\\ & =\dim \mathrm{Im}(\mathcal{AB})+n-\dim \mathrm{Im}(\mathcal{A})\end{array}$$
得证.

$\text{Thm7 RHS Proof:}$

设 $A=(\{{\alpha }_i{\}}_{i=1}^n{)}_{m\times n},B=(b_{ij}{)}_{n\times s},AB=(\{{\gamma }_i{\}}_{i=1}^n{)}_{m\times s}$，则
( α 1 , ⋯   , α n ) [ b 11 ⋯ b 1 s ⋮ ⋱ ⋮ b n 1 ⋯ b n s ] = ( ∑ i = 1 n b i 1 α i , ⋯   , ∑ i = 1 n b i s α i ) : = ( γ 1 , ⋯   , γ s ) \left( \alpha _1,\cdots ,\alpha _n \right) \left[ \begin{matrix} b_{11}& \cdots& b_{1s}\\ \vdots& \ddots& \vdots\\ b_{n1}& \cdots& b_{ns}\\ \end{matrix} \right] =\left( \sum_{i=1}^n{b_{i1}\alpha _i},\cdots ,\sum_{i=1}^n{b_{is}\alpha _i} \right) :=\left( \gamma _1,\cdots ,\gamma _s \right) (α1​,⋯,αn​) ​b11​⋮bn1​​⋯⋱⋯​b1s​⋮bns​​ ​=(i=1∑n​bi1​αi​,⋯,i=1∑n​bis​αi​):=(γ1​,⋯,γs​)
$AB$ 任一列向量均可表示为 $A$ 列向量的线性组合，即 $AB$ 列向量可由 $A$ 线性表出，所以 $$\mathrm{rank}(AB)⩽\mathrm{rank}(A)$$

同理可得，$AB$ 行向量可由 $B$ 行向量的线性表出，故
$$\mathrm{rank}(AB)⩽\mathrm{rank}(B)$$

或者由上述不等式结论
$$\mathrm{rank}(AB)=\mathrm{rank}((AB{)}^{\mathrm{T}})=\mathrm{rank}(B^{\mathrm{T}}{A}^{\mathrm{T}})⩽\mathrm{rank}(B^{\mathrm{T}})=\mathrm{rank}(B)$$

$\text{Cor7 LHS Proof1:}$

$AB=O\iff \mathrm{rank}(AB)=0$，带入 $\mathrm{Sylvester}$ 不等式可知结论成立.

$\text{Cor7 LHS Proof2:}$

将矩阵 $B$ 写作形式行向量
$$AB=A(b_1,b_2,\cdots ,b_s)=(A{b}_1,A{b}_2,\cdots ,A{b}_s)=(0,0,\cdots ,0)$$
则齐次线性方程组 $Ax=0$ 存在非零解集，解空间维数高于子空间维数，即 $$n-\mathrm{rank}(A)⩾\mathrm{rank}(b_1,b_2,\cdots ,b_s)=\mathrm{rank}(B)$$

$\text{Cor7 RHS Proof1:}$

由于 $B\in M_{n,s}(\mathbb{F}),\mathrm{rank}(B)=n$，所以存在可逆矩阵 $Q_s$，使得 $$B_{n,s}{Q}_n=(I_n,O_{n,s-n})$$
故 $$A_{m,n}{B}_{n,s}{Q}_s=A_{m,n}(I_n,O_{n,s-n})=(A_{m,n},O_{m,s-n})$$
从而 $$\mathrm{rank}(AB)=\mathrm{rank}(ABQ)=\mathrm{rank}(A,O)=\mathrm{rank}(A)$$
即右乘行满秩矩阵秩不变.

$\text{Cor7 RHS Proof2:}$

由于
$$n=\mathrm{rank}(A_{mn})=\mathrm{rank}(A_{nm}^T{A}_{mn})$$
故 $A^{T}A$ 存在逆矩阵，于是
$$\begin{array}{rl}\mathrm{rank}(B)& =\mathrm{rank}((A^{T}A{)}^{-1}(A^{T}A)B)\\ & =\mathrm{rank}([(A^{T}A{)}^{-1}{A}^T](AB))\\ & ⩽\min \{\mathrm{rank}((A^{T}A{)}^{-1}{A}^T),\mathrm{rank}(AB)\}\\ & ⩽\mathrm{rank}(AB)\\ & ⩽\min \{\mathrm{rank}(A),\mathrm{rank}(B)\}\\ & ⩽\mathrm{rank}(B)\end{array}$$
从而
$$\mathrm{rank}(A_{mn}{B}_{ns})\stackrel{\mathrm{rank}(A)=n}{}\mathrm{rank}(B_{ns})$$
即左乘列满秩矩阵秩不变.

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Thm 8

$\text{Theorem8}$ $\text{Frobenius}$ 秩不等式

若 $A\in M_{m,n}(\mathbb{F}),B\in M_{n,s}(\mathbb{F}),C\in M_{s,t}(\mathbb{F})$，则
$$\mathrm{rank}(ABC)⩾\mathrm{rank}(AB)+\mathrm{rank}(BC)-\mathrm{rank}(B)$$
取 $B=I_n,C\in M_{n,t}(\mathbb{F})$，即为 $\text{Sylvester}$ 不等式，$\text{Frobenius}$ 不等式为其推广形式.

$\text{Thm8 Proof1:}$

$$\begin{array}{rl}\left(\begin{array}{cc}B& O\\ O& ABC\end{array}\right)& \stackrel{r_2+A\cdot r_1}{}\left(\begin{array}{cc}B& O\\ AB& ABC\end{array}\right)\stackrel{c_2-c_1\cdot C}{}\left(\begin{array}{cc}B& -BC\\ AB& O\end{array}\right)\\ & \stackrel{c_1\leftrightarrow -c_2}{}\left(\begin{array}{cc}BC& B\\ O& AB\end{array}\right)\end{array}$$
则
$$\mathrm{rank}\left(\begin{array}{cc}B& O\\ O& ABC\end{array}\right)=\mathrm{rank}\left(\begin{array}{cc}BC& B\\ O& AB\end{array}\right)⩾\mathrm{rank}\left(\begin{array}{cc}BC& O\\ O& AB\end{array}\right)$$
即
$$\mathrm{rank}(ABC)+\mathrm{rank}(B)⩾\mathrm{rank}(AB)+\mathrm{rank}(BC)$$

$\text{Thm8 Proof2:}$

设 $T,U,V,W$ 为有限维线性空间，$\mathcal{C}\in \mathrm{Hom}(T^t,U^s),\mathcal{B}\in \mathrm{Hom}(U^s,V^n),\mathcal{A}\in \mathrm{Hom}(V^n,W^m),\mathcal{BC}\in \mathrm{Hom}(T^t,V^n)$，

对 $\mathcal{A}$ 作限制映射 $\mathcal{A}{|}_{\mathrm{Im}\mathcal{B}},\mathcal{A}{|}_{\mathrm{Im}\mathcal{BC}}$，则有
$$\begin{array}{rl}\dim \mathrm{Im}(\mathcal{B})& =\dim \mathrm{Im}(\mathcal{AB})+\dim (\mathrm{Ker}\mathcal{A}\cap \mathrm{Im}\mathcal{B})\\ \dim \mathrm{Im}(\mathcal{BC})& =\dim \mathrm{Im}(\mathcal{ABC})+\dim (\mathrm{Ker}\mathcal{A}\cap \mathrm{Im}\mathcal{BC})\end{array}$$
且 $$\mathrm{Ker}\mathcal{A}\cap \mathrm{Im}\mathcal{BC}\subseteq \mathrm{Ker}\mathcal{A}\cap \mathrm{Im}\mathcal{B}\Rightarrow \dim (\mathrm{Ker}\mathcal{A}\cap \mathrm{Im}\mathcal{BC})⩽\dim (\mathrm{Ker}\mathcal{A}\cap \mathrm{Im}\mathcal{B})$$
故
$$\dim \mathrm{Im}(\mathcal{B})-\dim \mathrm{Im}(\mathcal{AB})⩽\dim \mathrm{Im}(\mathcal{BC})-\dim \mathrm{Im}(\mathcal{ABC})$$

---

Thm 9

$\text{Theorem9}$
若 $(A_n+c_{{}_1}{I}_n)(A_n+c_{{}_2}{I}_n)=O$，其中 $c_1,c_2\in \mathbb{R}(c_1\ne c_2)$，则
$$\mathrm{rank}(A+c_{{}_1}I)+\mathrm{rank}(A+c_{{}_2}I)=n$$

$\text{Corollary9.1}$
方阵 $A_n$ 为幂等矩阵 (即 $A^2=A$) 的充要条件为 $\mathrm{rank}(A)+\mathrm{rank}(A-I)=n$.

$\text{Corollary9.2}$
方阵 $A_n$ 为对合矩阵 (即 $A^2=I$ ) 的充要条件为 $\mathrm{rank}(A+I)+\mathrm{rank}(A-I)=n$.

$\text{Thm9 Proof:}$

$$\begin{array}{rl}n& =\mathrm{rank}(I)\\ & =\mathrm{rank}((c_1-c_2)I)\\ & =\mathrm{rank}((A+c_{1}I)-(A+c_{2}I))\\ & ⩽\mathrm{rank}(A+c_{1}I)+\mathrm{rank}(A+c_{2}I)\\ & ⩽n\end{array}$$
故 $$\mathrm{rank}(A+c_{{}_1}I)+\mathrm{rank}(A+c_{{}_2}I)=n$$

$\text{Cor9 Proof1:}$

分别取 $(c_1,c_2)=(0,1),(1,-1)$，由 Theorem 8 可知结论成立.

$\text{Cor9.1 Proof2:}$

构造矩阵作初等变换

$$\begin{array}{rl}\left(\begin{array}{cc}A& O\\ O& I-A\end{array}\right)& \stackrel{r_2+r_1}{}\left(\begin{array}{cc}A& O\\ A& I-A\end{array}\right)\stackrel{c_2+c_1}{}\left(\begin{array}{cc}A& A\\ A& I\end{array}\right)\\ & \stackrel{r_1+\left(-A\right)r_2}{}\left(\begin{array}{cc}A-A^2& O\\ O& I\end{array}\right)\end{array}$$
则
$$\mathrm{rank}\left(\begin{array}{cc}A& O\\ O& I-A\end{array}\right)=\mathrm{rank}\left(\begin{array}{cc}A-A^2& O\\ O& I\end{array}\right)$$
从而
$$\begin{array}{rl}\mathrm{rank}\left(A\right)+\mathrm{rank}\left(I-A\right)& =\mathrm{rank}\left(A-A^2\right)+\mathrm{rank}\left(I\right)\\ & \stackrel{{}_{A^2-A=O}}{}0+n=n\end{array}$$

$\text{Cor9.2 Proof2:}$

构造矩阵作初等变换

$$\left(\begin{array}{cc}A+I& O\\ O& A-I\end{array}\right)\to \left(\begin{array}{cc}2I& O\\ -\left(A+I\right)& -\frac{\left(A+I\right)\left(A-I\right)}{2}\end{array}\right)\to \left(\begin{array}{cc}2I& O\\ O& -\frac{\left(A^2-I\right)}{2}\end{array}\right)$$
则
$$\begin{array}{rl}\mathrm{rank}(A+I)+\mathrm{rank}(A-I)& =\mathrm{rank}(2I)+\mathrm{rank}\left(\frac{I-A^2}{2}\right)\\ & \stackrel{{}_{A^2-I=O}}{}n+0=n\end{array}$$