Group 1
Theorem 1:
0 ⩽ r a n k ( A m , n ) = max ∣ A k ∣ ≠ 0 { k } ⇒ { r a n k ( A ) = k ⇔ ∃ ∣ A k ∣ ≠ 0 ∧ ∀ ∣ A k + 1 ∣ = 0 r a n k ( A ) ≥ k ⇔ ∃ ∣ A k ∣ ≠ 0 r a n k ( A ) < k ⇔ ∀ ∣ A k ∣ = 0 = dim I m ( A : V F n → U F m ) = dim s p a n ( A ) = r a n k r ( A ) = r a n k c ( A ) = P , Q 非奇异 r a n k ( P m A m , n ) = r a n k ( A m , n Q n ) = r a n k ( P m A m , n Q n ) = r a n k ( c A ) c ∈ R \ { 0 } = r a n k ( A T ) = r a n k ( A A T ) = r a n k ( A T A ) ⩽ min { m , n } { \begin{aligned} 0&\leqslant \mathrm{rank}(A_{m,n}) \\&=\underset{|A_{k}|\ne 0 }{\max}\{k\} \Rightarrow \left\{ \begin{array}{l} \mathrm{rank}(A) = k \Leftrightarrow \exist |A_k|\ne 0\land \forall |A_{k+1}|=0\\ \mathrm{rank}(A) \ge k \Leftrightarrow \exist |A_k|\ne 0\\ \mathrm{rank}(A) < k \Leftrightarrow \forall |A_k|=0\\ \end{array} \right. \\&=\dim\mathrm{Im}(\mathscr{A}:V_{\mathbb{F}}^n \to U_{\mathbb{F}}^m) =\dim\mathrm{span}(A)=\mathrm{rank}_r(A)=\mathrm{rank}_c(A) \\&\xlongequal{P,Q\ \text{非奇异}}\mathrm{rank}(P_mA_{m,n})=\mathrm{rank}(A_{m,n}Q_n)=\mathrm{rank}(P_mA_{m,n}Q_n) \\&=\underset{c\ \in\ \mathbb{R}\backslash\{0\}}{\mathrm{rank}(cA)}=\mathrm{rank}(A^\mathrm{T})=\mathrm{rank}(AA^\mathrm{T})=\mathrm{rank}(A^\mathrm{T}A) \\&\leqslant \min\{m,n\} \end{aligned} }0⩽rank(Am,n)=∣Ak∣=0max{k}⇒⎩⎨⎧rank(A)=k⇔∃∣Ak∣=0∧∀∣Ak+1∣=0rank(A)≥k⇔∃∣Ak∣=0rank(A)<k⇔∀∣Ak∣=0=dimIm(A:VFn→UFm)=dimspan(A)=rankr(A)=rankc(A)P,Q 非奇异rank(PmAm,n)=rank(Am,nQn)=rank(PmAm,nQn)=c ∈ R\{0}rank(cA)=rank(AT)=rank(AAT)=rank(ATA)⩽min{m,n}
Proof:
1） 矩阵c A ( c ≠ 0 ) , A T cA(c\ne 0),A^{\mathrm{T}}cA(c=0),AT 与 A AA 的最高阶子式相同，所以三者等秩；
2）A x = 0 ⇒ A T A x = 0 ⇒ x T A T A x = ( A x ) T ( A x ) = 0 ⇒ A x = 0 Ax=0 \Rightarrow A^{\mathrm{T}}Ax=0 \Rightarrow x^{\mathrm{T}}A^{\mathrm{T}}Ax=(Ax)^{\mathrm{T}}(Ax)=0\Rightarrow Ax=0Ax=0⇒ATAx=0⇒xTATAx=(Ax)T(Ax)=0⇒Ax=0
齐次方程组 A x = 0 Ax=0Ax=0 与 A T A x = 0 A^{\mathrm{T}}Ax=0ATAx=0 同解，所以 A AA 与 A T A A^{\mathrm{T}}AATA 等秩；
3）同理可证，齐次方程组 A T x = 0 A^{\mathrm{T}}x=0ATx=0 与 A A T x = 0 AA^{\mathrm{T}}x=0AATx=0 同解，所以 A T A^{\mathrm{T}}AT 与 A A T AA^{\mathrm{T}}AAT 等秩；
综上所述，矩阵A , c A ( c ≠ 0 ) , A T , A T A , A A T A,cA(c\ne 0),A^{\mathrm{T}},A^{\mathrm{T}}A,AA^{\mathrm{T}}A,cA(c=0),AT,ATA,AAT 皆等秩.
Theorem 2:
r a n k ( A ∗ ) = { n , r a n k ( A ) = n 1 , r a n k ( A ) = n − 1 0 , r a n k ( A ) < n − 1 { \mathrm{rank}(A^*)= \left\{ \begin{array}{l} n \ , \ \mathrm{rank}(A) = n\\ 1 \ , \ \mathrm{rank}(A) = n-1\\ 0 \ , \ \mathrm{rank}(A) < n-1\\ \end{array} \right. }rank(A∗)=⎩⎨⎧n , rank(A)=n1 , rank(A)=n−10 , rank(A)<n−1
Proof:
1）当 r a n k ( A ) = n \mathrm{rank}(A)=nrank(A)=n 时，∣ A ∣ = 0 |A|=0∣A∣=0，则∣ A ∗ ∣ = ∣ ∣ A ∣ A − 1 ∣ = ∣ A ∣ n − 1 ≠ 0 |A^*|=||A|A^{-1}|=|A|^{n-1}\ne 0∣A∗∣=∣∣A∣A−1∣=∣A∣n−1=0，所以 r a n k ( A ∗ ) = n \mathrm{rank}(A^*)=nrank(A∗)=n；
2）当 r a n k ( A ) = n − 1 \mathrm{rank}(A)=n-1rank(A)=n−1 时，则存在 n − 1 n-1n−1 阶子式不为零，则 A ∗ ≠ O , r a n k ( A ∗ ) > 0 A^*\ne O,\mathrm{rank}(A^*)> 0A∗=O,rank(A∗)>0，又 A ∗ A = ∣ A ∣ I = O A^*A=|A|I=OA∗A=∣A∣I=O，所以 r a n k ( A ∗ ) + r a n k ( A ) ⩽ n \mathrm{rank}(A^*)+\mathrm{rank}(A)\leqslant nrank(A∗)+rank(A)⩽n，于是有 1 ⩽ r a n k ( A ∗ ) ⩽ n − r a n k ( A ) = n − ( n − 1 ) = 1 1\leqslant \mathrm{rank}(A^*)\leqslant n-\mathrm{rank}(A)=n-(n-1)=11⩽rank(A∗)⩽n−rank(A)=n−(n−1)=1，即r a n k ( A ∗ ) = 1 \mathrm{rank}(A^*)=1rank(A∗)=1；
3）当 r a n k ( A ) < n − 1 \mathrm{rank}(A)<n-1rank(A)<n−1 时，所有 n − 1 n-1n−1 阶子式全为零，则 A ∗ = O , r a n k ( A ∗ ) = 0 A^*= O,\mathrm{rank}(A^*)= 0A∗=O,rank(A∗)=0.
Group 2
Theorem 3:
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 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
r a n k ( 子矩阵 ) ⩽ r a n k ( 原矩阵 ) \mathrm{rank}(\text{子矩阵}) \leqslant \mathrm{rank}(\text{原矩阵})rank(子矩阵)⩽rank(原矩阵)
Proof:
设子矩阵 r a n k ( M ′ ) = r ′ \mathrm{rank}(M')=r'rank(M′)=r′，则 ∃ det ( M r ′ ′ ) ≠ 0 \exists \det(M'_{r'})\ne 0∃det(Mr′′)=0，且 M r ′ ′ M'_{r'}Mr′′ 亦为原矩阵 M MM 的子矩阵，所以原矩阵 M MM 中 ∃ r ⩾ r ′ , s . t . det ( M r ) ≠ 0 \exists r\geqslant r', s.t. \det(M_{r})\ne 0∃r⩾r′,s.t.det(Mr)=0，于是有 r a n k ( M ) ⩾ r a n k ( M ′ ) \mathrm{rank(M)}\geqslant \mathrm{rank(M')}rank(M)⩾rank(M′).
Theorem 4: (矩阵降阶公式)
\text{ }
设分块矩阵 M = ( A m , m B C D n , n ) 设分块矩阵 \ M=\left(\begin{matrix} A_{m,m}& B\\ C& D_{n,n}\\ \end{matrix}\right)设分块矩阵 M=(Am,mCBDn,n)若 A 为非奇异矩阵，则 若\ A \ 为非奇异矩阵，则若 A 为非奇异矩阵，则
r a n k ( M ) = r a n k ( A ) + r a n k ( D − C A − 1 B ) \mathrm{rank}(M)=\mathrm{rank}(A)+\mathrm{rank}(D-CA^{-1}B)rank(M)=rank(A)+rank(D−CA−1B)若 D 为非奇异矩阵，则 若\ D \ 为非奇异矩阵，则若 D 为非奇异矩阵，则
r a n k ( M ) = r a n k ( D ) + r a n k ( A − B D − 1 C ) \mathrm{rank}(M)=\mathrm{rank}(D)+\mathrm{rank}(A-BD^{-1}C)rank(M)=rank(D)+rank(A−BD−1C)若 A , D 皆为非奇异矩阵，则 若\ A,D\ 皆为非奇异矩阵，则若 A,D 皆为非奇异矩阵，则
r a n k ( A ) + r a n k ( D − C A − 1 B ) = r a n k ( D ) + r a n k ( A − B D − 1 C ) \mathrm{rank}(A)+\mathrm{rank}(D-CA^{-1}B)=\mathrm{rank}(D)+\mathrm{rank}(A-BD^{-1}C)rank(A)+rank(D−CA−1B)=rank(D)+rank(A−BD−1C)
Proof: 初等变换即可
Theorem 5:
max { r a n k ( A ) , r a n k ( B ) } r a n k ( A ± B ) ⩽ r a n k ( A , B ) r a n k ( A B ) ⩽ r a n k ( A ) + r a n k ( B ) = r a n k ( A O O B ) ⩽ r a n k ( A C O B ) ( A X + Y B = C 有解时取等 ) { \begin{aligned} &\begin{array}{c} \max\{\mathrm{rank}(A),\mathrm{rank}(B)\}\\ \\ \mathrm{rank}(A\pm B)\\ \end{array} \leqslant \begin{array}{c} \mathrm{rank}(A,B)\\ \ \ \mathrm{rank}\left(\begin{array}{c} A\\ B\\ \end{array} \right)\\ \end{array} \\&\leqslant \mathrm{rank}(A)+\mathrm{rank}(B) =\mathrm{rank}\left( \begin{matrix} A& O\\ O& B\\ \end{matrix} \right) \\&\leqslant \mathrm{rank}\left( \begin{matrix} A& C\\ O& B\\ \end{matrix} \right) \ \ ( { AX+YB=C} 有解时取等) \end{aligned} }max{rank(A),rank(B)}rank(A±B)⩽rank(A,B) rank(AB)⩽rank(A)+rank(B)=rank(AOOB)⩽rank(AOCB) (AX+YB=C有解时取等)
Lemma:
1 ) 若向量组 { α i } i = 1 s 可由 { β j } j = 1 t 线性表示 , 且 s > t , 则 { α i } i = 1 s 线性相关 1)\ 若向量组 \{\alpha_i\}_{i=1}^{s} 可由 \{\beta_j\}_{j=1}^{t} 线性表示,且 s>t,则 \{\alpha_i\}_{i=1}^{s} 线性相关1) 若向量组{αi}i=1s可由{βj}j=1t线性表示,且s>t,则{αi}i=1s线性相关.2 ) 若向量组 { α i } i = 1 s 可由 { β j } j = 1 t 线性表示 , 且 { α i } i = 1 s 线性无关 , 则 s ⩽ t 2)\ 若向量组 \{\alpha_i\}_{i=1}^{s} 可由 \{\beta_j\}_{j=1}^{t} 线性表示,且 \{\alpha_i\}_{i=1}^{s}线性无关,则 s\leqslant t2) 若向量组{αi}i=1s可由{βj}j=1t线性表示,且{αi}i=1s线性无关,则s⩽t.
3 ) 若向量组 { α i } i = 1 s 可由 { β j } j = 1 t 线性表示 , 则 r a n k ( { α i } i = 1 s ) ⩽ r a n k ( { β j } j = 1 t ) 3)\ 若向量组 \{\alpha_i\}_{i=1}^{s} 可由 \{\beta_j\}_{j=1}^{t} 线性表示,则 \mathrm{rank}( \{\alpha_i\}_{i=1}^{s} ) \leqslant \mathrm{rank}( \{\beta_j\}_{j=1}^{t} )3) 若向量组{αi}i=1s可由{βj}j=1t线性表示,则rank({αi}i=1s)⩽rank({βj}j=1t).
Proof:
1） { α i } i = 1 s \{\alpha_i\}_{i=1}^{s}{αi}i=1s 可由 { β j } j = 1 t \{\beta_j\}_{j=1}^{t}{βj}j=1t线性表示，则
r a n k ( α 1 , α 2 , ⋯ , α s ) = r a n k ( ( β 1 , β 2 , ⋯ , β t ) [ l b 11 b 12 ⋯ b 1 s b 21 b 22 ⋯ b 2 s ⋮ ⋮ ⋱ ⋮ b t 1 b t 2 ⋯ b t s ] ) ⩽ min { t , s } ⩽ t < s \begin{aligned} \mathrm{rank}\left( \alpha _1,\alpha _2,\cdots ,\alpha _s \right) &=\mathrm{rank}\left( \left( \beta _1,\beta _2,\cdots ,\beta _t \right) \left[ \begin{matrix}{l} b_{11}& b_{12}& \cdots& b_{1s}\\ b_{21}& b_{22}& \cdots& b_{2s}\\ \vdots& \vdots& \ddots& \vdots\\ b_{t1}& b_{t2}& \cdots& b_{ts}\\ \end{matrix} \right] \right) \\&\leqslant \min \left\{ t,s \right\} \leqslant t<s \end{aligned}rank(α1,α2,⋯,αs)=rank(β1,β2,⋯,βt)lb11b21⋮bt1b12b22⋮bt2⋯⋯⋱⋯b1sb2s⋮bts⩽min{t,s}⩽t<s
故 { α i } i = 1 s \{\alpha_i\}_{i=1}^{s}{αi}i=1s 线性相关.2）（1）取逆否命题.
3）证法一：
分别取两组向量组的极大无关组 { α s 1 , ⋯ , α s n } , { β t 1 , ⋯ , β t m } \{\alpha_{s_{_1}},\cdots,\alpha_{s_{_n}}\},\{\beta_{t_{_1}},\cdots,\beta_{t_{_m}}\}{αs1,⋯,αsn},{βt1,⋯,βtm} ，则前者可由后者线性表示，由（2）可知，n ⩽ m n\leqslant mn⩽m，即 r a n k ( { α i } i = 1 s ) ⩽ r a n k ( { β j } j = 1 t ) \mathrm{rank}( \{\alpha_i\}_{i=1}^{s} ) \leqslant \mathrm{rank}( \{\beta_j\}_{j=1}^{t} )rank({αi}i=1s)⩽rank({βj}j=1t).
证法二：
记 ( { α i } i = 1 s ) = A n , s , ( { β j } j = 1 t ) = B n , t \ (\{\alpha_i\}_{i=1}^{s})=A_{n,s} \ , \ (\{\beta_j\}_{j=1}^{t})=B_{n,t} ({αi}i=1s)=An,s , ({βj}j=1t)=Bn,t，则存在矩阵 C t s C_{ts}Cts，使得 A n , s = B n , t C t , s \ A_{n,s}=B_{n,t}C_{t,s} An,s=Bn,tCt,s，
所以有 r a n k ( A ) = r a n k ( B C ) ⩽ min { r a n k ( B ) , r a n k ( C ) } ⩽ r a n k ( B ) \mathrm{rank}(A)=\mathrm{rank}(BC)\leqslant \min\{\mathrm{rank}(B),\mathrm{rank}(C)\}\leqslant \mathrm{rank}(B)rank(A)=rank(BC)⩽min{rank(B),rank(C)}⩽rank(B).
Proof 1-1:
矩阵 A , B A,BA,B 为 ( A , B ) , ( A B ) (A,B),\left(\begin{array}{c} A\\B\\ \end{array} \right)(A,B),(AB) 的子阵，所以
max { r a n k ( A ) , r a n k ( B ) } ⩽ r a n k ( A , B ) r a n k ( A B ) \max\{\mathrm{rank}(A),\mathrm{rank}(B)\}\leqslant \begin{array}{c} \mathrm{rank}(A,B)\\ \ \ \mathrm{rank}\left(\begin{array}{c} A\\B\\\end{array} \right)\\\end{array}max{rank(A),rank(B)}⩽rank(A,B) rank(AB)
Proof 1-2:
列向量组 A ± B = ( α 1 ± β 1 , ⋯ , α n ± β n ) A\pm B=(\alpha_1\pm\beta_1,\cdots,\alpha_n\pm\beta_n)A±B=(α1±β1,⋯,αn±βn)可由( A , B ) = ( α 1 , ⋯ , α n ; β 1 , ⋯ , β n ) (A,B)=(\alpha_1,\cdots,\alpha_n;\beta_1,\cdots,\beta_n)(A,B)=(α1,⋯,αn;β1,⋯,βn)线性表示，由「 Lemma^」可知 r a n k ( A ± B ) ⩽ r a n k ( A , B ) \mathrm{rank}(A\pm B)\leqslant \mathrm{rank}(A,B)rank(A±B)⩽rank(A,B)；
同理，由行向量组可证 r a n k ( A ± B ) ⩽ r a n k ( A B ) \mathrm{rank}(A\pm B)\leqslant \mathrm{rank}\left(\begin{array}{c} A\\B\\ \end{array} \right)rank(A±B)⩽rank(AB)
Proof 2:
记矩阵 ( A m , n , B m , s ) (A_{m,n},B_{m,s})(Am,n,Bm,s) 极大无关列向量组构成的矩阵为C m , t ( 0 < t ⩽ min { m , n + s } ) C_{m,t} (0<t\leqslant \min\{m,n+s\})Cm,t(0<t⩽min{m,n+s})，
若矩阵 A AA 与 B BB 的极大无关组线性无关,则 t = r a n k ( A ) + r a n k ( B ) t=\mathrm{rank}(A)+\mathrm{rank}(B)t=rank(A)+rank(B)；
若线性相关，则 t < r a n k ( A ) + r a n k ( B ) t<\mathrm{rank}(A)+\mathrm{rank}(B)t<rank(A)+rank(B). 于是有
r a n k ( A , B ) = r a n k ( C m , t ) ⩽ min { m , t } ⩽ min { m , r a n k ( A ) + r a n k ( B ) } ⩽ r a n k ( A ) + r a n k ( B ) \begin{aligned} \mathrm{rank}(A,B) &=\mathrm{rank}(C_{m,t})\leqslant \min\{m,t\} \\&\leqslant \min\{m,\mathrm{rank}(A)+\mathrm{rank}(B)\} \\&\leqslant \mathrm{rank}(A)+\mathrm{rank}(B) \end{aligned}rank(A,B)=rank(Cm,t)⩽min{m,t}⩽min{m,rank(A)+rank(B)}⩽rank(A)+rank(B)
或者
r a n k ( A , B ) = r a n k ( ( I m , I m ) ( A m , n O O B m , s ) ) ⩽ min { r a n k ( I , I ) , r a n k ( A O O B ) } ⩽ r a n k ( A O O B ) = r a n k ( A ) + r a n k ( B ) \begin{aligned} \mathrm{rank}(A,B)&=\mathrm{rank}\left((I_m,I_m)\left( \begin{matrix}A_{m,n}&O\\O&B_{m,s}\\\end{matrix} \right) \right) \\&\leqslant \min\{\mathrm{rank}(I,I),\mathrm{rank}\left( \begin{matrix}A&O\\O&B\\\end{matrix} \right)\} \\&\leqslant\mathrm{rank}\left( \begin{matrix}A&O\\O&B\\\end{matrix} \right) =\mathrm{rank}(A)+\mathrm{rank}(B) \end{aligned}rank(A,B)=rank((Im,Im)(Am,nOOBm,s))⩽min{rank(I,I),rank(AOOB)}⩽rank(AOOB)=rank(A)+rank(B)
同理可证 r a n k ( A B ) ⩽ r a n k ( A ) + r a n k ( B ) \mathrm{rank}\left(\begin{array}{c}A\\B\\\end{array}\right) \leqslant \mathrm{rank}(A)+\mathrm{rank}(B)rank(AB)⩽rank(A)+rank(B)
Proof 3:
对矩阵进行初等行变换化至阶梯型
A m n → r ( A r a n k ( A ) , n O ) , B s t → r ( B r a n k ( B ) , t O ) A_{mn}\xrightarrow{r} \left( \begin{array}{c} A_{\mathrm{rank}\left( A \right) ,n}\\ O\\ \end{array} \right) ,\ B_{st}\xrightarrow{r} \left( \begin{array}{c} B_{\mathrm{rank}\left( B \right) ,t}\\ O\\ \end{array} \right)Amnr(Arank(A),nO), Bstr(Brank(B),tO)
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)(AOOB)rArank(A),nOOOOOBrank(B),tO→Arank(A),nOOOOBrank(B),tOO
有限次初等变换不改变矩阵的秩，故有
r a n k ( A O O B ) = r a n k ( A r a n k ( A ) , n ) + r a n k ( B r a n k ( B ) , t ) = r a n k ( A ) + r a n k ( B ) \begin{aligned} \mathrm{rank}\left( \begin{matrix} A& O\\ O& B\\ \end{matrix} \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{aligned}rank(AOOB)=rank(Arank(A),n)+rank(Brank(B),t)=rank(A)+rank(B)
Proof 4:
设 ∣ A r a n k ( A ) ∣ |A_{\mathrm{rank}(A)}|∣Arank(A)∣, ∣ B r a n k ( B ) ∣ |B_{\mathrm{rank}(B)}|∣Brank(B)∣ 分别为矩阵 A AA,B BB 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 0Arank(A)OC′Brank(B)=Arank(A)OOBrank(B)=∣Arank(A)∣∣Brank(B)∣=0
所以上述两个方阵皆为满秩，于是
r a n k ( A C O B ) ⩾ r a n k ( A r a n k ( A ) C ′ O B r a n k ( B ) ) = r a n k ( A r a n k ( A ) O O B r a n k ( B ) ) = r a n k ( A ) + r a n k ( B ) = r a n k ( A O O B ) \begin{aligned} \mathrm{rank}\left( \begin{matrix} A& C\\ O& B\\ \end{matrix} \right) &\geqslant \mathrm{rank}\left( \begin{matrix} A_{\mathrm{rank}\left( A \right)}& C'\\ O& B_{\mathrm{rank}\left( B \right)}\\ \end{matrix} \right) \\&= \mathrm{rank}\left( \begin{matrix} A_{\mathrm{rank}\left( A \right)}& O\\ O& B_{\mathrm{rank}\left( B \right)}\\ \end{matrix} \right) \\&=\mathrm{rank}\left( A \right) + \mathrm{rank}\left( B \right) \\&= \mathrm{rank}\left( \begin{matrix} A& O\\ O& B\\ \end{matrix} \right) \end{aligned}rank(AOCB)⩾rank(Arank(A)OC′Brank(B))=rank(Arank(A)OOBrank(B))=rank(A)+rank(B)=rank(AOOB)
Corollary:
1 ) r a n k ( A ) − r a n k ( B ) ⩽ r a n k ( A − B ) 2 ) r a n k ( A n n B n n − I n ) ⩽ r a n k ( A − I ) + r a n k ( B − I ) \begin{aligned} &1) \ {\mathrm{rank}(A)-\mathrm{rank}(B)\leqslant \mathrm{rank}(A-B)} \\&2) \ \mathrm{rank}(A_{nn}B_{nn}-I_n)\leqslant \mathrm{rank}(A-I)+\mathrm{rank}(B-I) \end{aligned}1) rank(A)−rank(B)⩽rank(A−B)2) rank(AnnBnn−In)⩽rank(A−I)+rank(B−I)
Proof:
1)
r a n k ( A ) = r a n k ( ( A − B ) + B ) ⩽ r a n k ( A − B ) + r a n k ( B ) \mathrm{rank}(A)=\mathrm{rank}((A-B)+B)\leqslant \mathrm{rank}(A-B)+\mathrm{rank}(B)rank(A)=rank((A−B)+B)⩽rank(A−B)+rank(B)
2)
r a n k ( A B − I ) = r a n k ( A ( B − I ) + A − I ) ⩽ r a n k ( A ( B − I ) ) + r a n k ( A − I ) ⩽ r a n k ( B − I ) + r a n k ( A − I ) \begin{aligned}\mathrm{rank}(AB-I)&=\mathrm{rank}(A(B-I)+A-I)\\&\leqslant \mathrm{rank}(A(B-I))+ \mathrm{rank}(A-I)\\&\leqslant \mathrm{rank}(B-I)+\mathrm{rank}(A-I)\end{aligned}rank(AB−I)=rank(A(B−I)+A−I)⩽rank(A(B−I))+rank(A−I)⩽rank(B−I)+rank(A−I)
Group 3
Theorem 6:
A ∈ M m , n ( F ) , B ∈ M n , s ( F ) r a n k ( A ) + r a n k ( B ) − n ⩽ r ( A B ) ⏟ S y l v e s t e r 秩不等式 ⩽ min { r a n k ( A ) , r a n k ( B ) } \begin{aligned} &A\in M_{m,n}(\mathbb{F}),B\in M_{n,s}(\mathbb{F}) \\&\underset{\mathrm{Sylvester \ \text{秩不等式}}}{\underbrace{\mathrm{rank}(A)+\mathrm{rank}(B)-n \leqslant r(AB)}} \leqslant \min\{\mathrm{rank}(A),\mathrm{rank}(B)\} \end{aligned}A∈Mm,n(F),B∈Mn,s(F)Sylvester 秩不等式rank(A)+rank(B)−n⩽r(AB)⩽min{rank(A),rank(B)}
Corollary:
L H S ⇒ A B = O r a n k ( A ) + r a n k ( B ) ⩽ n R H S ⇔ { r a n k ( A B ) ⩽ r a n k ( A ) ⇒ r a n k ( B ) = n r a n k ( A B ) = r a n k ( A ) r a n k ( A B ) ⩽ r a n k ( B ) ⇒ r a n k ( A ) = n r a n k ( A B ) = r a n k ( B ) \begin{aligned} &LHS\xRightarrow{AB=O \ }\mathrm{rank}(A)+\mathrm{rank}(B) \leqslant n \\&RHS\Leftrightarrow \left\{ \begin{array}{l} \mathrm{rank}(AB)\leqslant \mathrm{rank}(A) \xRightarrow{\mathrm{rank}(B)=n \ }\mathrm{rank}(AB)=\mathrm{rank}(A) \\\mathrm{rank}(AB)\leqslant\mathrm{rank}(B) \xRightarrow{\mathrm{rank}(A)=n \ }\mathrm{rank}(AB)=\mathrm{rank}(B) \end{array} \right. \end{aligned}LHSAB=O rank(A)+rank(B)⩽nRHS⇔{rank(AB)⩽rank(A)rank(B)=n rank(AB)=rank(A)rank(AB)⩽rank(B)rank(A)=n rank(AB)=rank(B)
Proof LHS:
证法一：
构造矩阵，进行初等变换
( I n O O A B ) → r 2 + A ⋅ r 1 ( I n O A A B ) → c 2 + c 1 ⋅ ( − B ) ( I n − B A O ) → c 1 ↔ − c 2 ( B I n O A ) \begin{aligned} \left( \begin{matrix} I_n& O\\ O& AB\\ \end{matrix} \right) &\xrightarrow{r_2+A\cdot r_1}\left( \begin{matrix} I_n& O\\ A& AB\\ \end{matrix} \right) \xrightarrow{c_2+c_1\cdot \left( -B \right)}\left( \begin{matrix} I_n& -B\\ A& O\\ \end{matrix} \right) \\&\xrightarrow{c_1\leftrightarrow -c_2}\left( \begin{matrix} B& I_n\\ O& A\\ \end{matrix} \right) \end{aligned}(InOOAB)r2+A⋅r1(InAOAB)c2+c1⋅(−B)(InA−BO)c1↔−c2(BOInA)
则
n + r a n k ( A B ) = r a n k ( I n ) + r a n k ( A B ) = r a n k ( I n O O A B ) = r a n k ( B I n O A ) ⩾ r a n k ( B O O A ) = r a n k ( A ) + r a n k ( B ) \begin{aligned} n+\mathrm{rank}\left( AB \right) &=\mathrm{rank}\left( I_n \right) +\mathrm{rank}\left( AB \right) \\& =\mathrm{rank}\left( \begin{matrix} I_n& O\\ O& AB\\ \end{matrix} \right) \\&=\mathrm{rank}\left( \begin{matrix} B& I_n\\ O& A\\ \end{matrix} \right) \\&\geqslant\mathrm{rank}\left( \begin{matrix} B& O\\ O& A\\ \end{matrix} \right) \\&= \mathrm{rank}\left( A \right) +\mathrm{rank}\left( B \right) \end{aligned}n+rank(AB)=rank(In)+rank(AB)=rank(InOOAB)=rank(BOInA)⩾rank(BOOA)=rank(A)+rank(B)
证法二：
设 U , V , W U,V,WU,V,W 为有限维线性空间，B ∈ H o m ( U s , V n ) , A ∈ H o m ( V n , W m ) \mathscr{B}\in \mathrm{Hom}(U^s,V^n) ,\ \mathscr{A}\in \mathrm{Hom}(V^n,W^m)B∈Hom(Us,Vn), A∈Hom(Vn,Wm)，则 A B ∈ H o m ( U s , W m ) \mathscr{AB}\in \mathrm{Hom}(U^s,W^m)AB∈Hom(Us,Wm)，
对 A \mathscr{A}A 作限制映射 A ∣ I m B : I m B → A ( I m B ) = I m ( A B ) \mathscr{A}|_{\mathrm{Im}\mathscr{B}}:\mathrm{Im}\mathscr{B}\to \mathscr{A}(\mathrm{Im}\mathscr{B})=\mathrm{Im}(\mathscr{AB})A∣ImB:ImB→A(ImB)=Im(AB)
故有
dim I m ( B ) = dim I m ( A ∣ I m B ) + dim K e r ( A ∣ I m B ) = dim I m ( A B ) + dim K e r ( A ∣ I m B ) = dim I m ( A B ) + dim ( K e r A ∩ I m B ) ⩽ dim I m ( A B ) + dim K e r ( A ) = dim I m ( A B ) + dim ( V n ) − dim I m ( A ) = dim I m ( A B ) + n − dim I m ( A ) \begin{aligned} \dim\mathrm{Im}(\mathscr{B}) &=\dim\mathrm{Im}(\mathscr{A}|_{\mathrm{Im}\mathscr{B}})+\dim\mathrm{Ker}(\mathscr{A}|_{\mathrm{Im}\mathscr{B}}) \\&=\dim\mathrm{Im}(\mathscr{AB})+\dim\mathrm{Ker}(\mathscr{A}|_{\mathrm{Im}\mathscr{B}}) \\&=\dim\mathrm{Im}(\mathscr{AB})+\dim(\mathrm{Ker}\mathscr{A}\cap\mathrm{Im}\mathscr{B}) \\&\leqslant \dim\mathrm{Im}(\mathscr{AB})+\dim\mathrm{Ker}(\mathscr{A}) \\&=\dim\mathrm{Im}(\mathscr{AB})+\dim(V^n)-\dim\mathrm{Im}(\mathscr{A}) \\&=\dim\mathrm{Im}(\mathscr{AB})+n-\dim\mathrm{Im}(\mathscr{A}) \end{aligned}dimIm(B)=dimIm(A∣ImB)+dimKer(A∣ImB)=dimIm(AB)+dimKer(A∣ImB)=dimIm(AB)+dim(KerA∩ImB)⩽dimIm(AB)+dimKer(A)=dimIm(AB)+dim(Vn)−dimIm(A)=dimIm(AB)+n−dimIm(A)
得证
Proof RHS:
1）设 A = ( { α i } i = 1 n ) , B = ( b i j ) n s , A B = ( { γ i } i = 1 n ) A=(\{\alpha _i\}_{i=1}^{n}),B=(b_{ij})_{ns},AB=(\{\gamma _i\}_{i=1}^{n})A=({αi}i=1n),B=(bij)ns,AB=({γi}i=1n)，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∑nbi1αi,⋯,i=1∑nbisαi)=(γ1,⋯,γs)
A B ABAB 任一列向量均可表示为 A AA 列向量的线性组合，即 A B ABAB 列向量可由 A AA 线性表出，所以 r a n k ( A B ) ⩽ r a n k ( A ) \mathrm{rank}(AB)\leqslant \mathrm{rank}(A)rank(AB)⩽rank(A);
2）同理可得，A B ABAB 行向量可由 B BB 行向量的线性表出，故 r a n k ( A B ) ⩽ r a n k ( B ) \mathrm{rank}(AB)\leqslant \mathrm{rank}(B)rank(AB)⩽rank(B).
或者由第一问结论 r a n k ( A B ) = r a n k ( ( A B ) T ) = r a n k ( B T A T ) ⩽ r a n k ( B T ) = r a n k ( B ) \mathrm{rank}(AB)=\mathrm{rank}((AB)^{\mathrm{T}})=\mathrm{rank}(B^{\mathrm{T}}A^{\mathrm{T}})\leqslant \mathrm{rank}(B^{\mathrm{T}})=\mathrm{rank}(B)rank(AB)=rank((AB)T)=rank(BTAT)⩽rank(BT)=rank(B)
Proof Corollary LHS:
证法一：
A B = O ⇔ r a n k ( A B ) = 0 AB=O\Leftrightarrow \mathrm{rank}(AB)=0AB=O⇔rank(AB)=0，带入 S y l v e s t e r \mathrm{Sylvester}Sylvester 不等式可知结论成立.
证法二：
将矩阵 B BB 写作形式行向量，有 A B = A ( b 1 , b 2 , ⋯ , b s ) = ( A b 1 , A b 2 , ⋯ , A b s ) = ( 0 , 0 , ⋯ , 0 ) AB=A(b_1,b_2,\cdots,b_s)=(Ab_1,Ab_2,\cdots,Ab_s)=(0,0,\cdots,0)AB=A(b1,b2,⋯,bs)=(Ab1,Ab2,⋯,Abs)=(0,0,⋯,0)，则齐次线性方程组 A x = 0 Ax=0Ax=0 存在非零解集，解空间维数高于子空间维数，即 n − r a n k ( A ) ⩾ r a n k ( b 1 , b 2 , ⋯ , b s ) = r a n k ( B ) n-\mathrm{rank}(A) \geqslant \mathrm{rank}(b_1,b_2,\cdots,b_s)=\mathrm{rank}(B)n−rank(A)⩾rank(b1,b2,⋯,bs)=rank(B).
Proof Corollary RHS:
由于 B ∈ M n , s ( F ) , r a n k ( B ) = n B\in M_{n,s}(\mathbb{F}),\mathrm{rank}(B)=nB∈Mn,s(F),rank(B)=n，故存在非奇异阵 Q s Q_sQs，使得 B Q = ( I n , O ) BQ=(I_n,O)BQ=(In,O)，于是 A B Q = A ( I n , O ) = ( A , O ) ABQ=A(I_n,O)=(A,O)ABQ=A(In,O)=(A,O)，从而 r a n k ( A B ) = r a n k ( A B Q ) = r a n k ( A , O ) = r a n k ( A ) \mathrm{rank}(AB)=\mathrm{rank}(ABQ)=\mathrm{rank}(A,O)=\mathrm{rank}(A)rank(AB)=rank(ABQ)=rank(A,O)=rank(A).
同理可证另一个等式.
Theorem 7: (Frobenius 秩不等式)
A ∈ M m , n ( F ) , B ∈ M n , s ( F ) , C ∈ M s , t ( F ) r a n k ( A B C ) ⩾ r a n k ( A B ) + r a n k ( B C ) − r a n k ( B ) \begin{aligned} &A\in M_{m,n}(\mathbb{F}),B\in M_{n,s}(\mathbb{F}),C\in M_{s,t}(\mathbb{F}) \\&\mathrm{rank}(ABC)\geqslant \mathrm{rank}(AB)+\mathrm{rank}(BC)-\mathrm{rank}(B) \end{aligned}A∈Mm,n(F),B∈Mn,s(F),C∈Ms,t(F)rank(ABC)⩾rank(AB)+rank(BC)−rank(B)
Proof :
证法一：
( B O O A B C ) → r 2 + A ⋅ r 1 ( B O A B A B C ) → c 2 − c 1 ⋅ C ( B − B C A B O ) → c 1 ↔ − c 2 ( B C B O A B ) \begin{aligned} \left( \begin{matrix} B& O\\ O& ABC\\ \end{matrix} \right) &\xrightarrow{r_2+A\cdot r_1}\left( \begin{matrix} B& O\\ AB& ABC\\ \end{matrix} \right) \xrightarrow{c_2-c_1\cdot C }\left( \begin{matrix} B& -BC\\ AB& O\\ \end{matrix} \right) \\ &\xrightarrow{c_1\leftrightarrow -c_2}\left( \begin{matrix} BC& B\\ O& AB\\ \end{matrix} \right) \end{aligned}(BOOABC)r2+A⋅r1(BABOABC)c2−c1⋅C(BAB−BCO)c1↔−c2(BCOBAB)
则
r a n k ( B O O A B C ) = r a n k ( B C B O A B ) ⩾ r a n k ( B C O O A B ) \mathrm{rank}\left( \begin{matrix} B& O\\ O& ABC\\ \end{matrix} \right) =\mathrm{rank}\left( \begin{matrix} BC& B\\ O& AB\\ \end{matrix} \right) \geqslant \mathrm{rank}\left( \begin{matrix} BC& O\\ O& AB\\ \end{matrix} \right)rank(BOOABC)=rank(BCOBAB)⩾rank(BCOOAB)
即
r a n k ( A B C ) + r a n k ( B ) ⩾ r a n k ( A B ) + r a n k ( B C ) \mathrm{rank}(ABC)+\mathrm{rank}(B)\geqslant \mathrm{rank}(AB)+\mathrm{rank}(BC)rank(ABC)+rank(B)⩾rank(AB)+rank(BC)
证法二：
设 T , U , V , W T,U,V,WT,U,V,W 为有限维线性空间，C ∈ H o m ( T t , U s ) , B ∈ H o m ( U s , V n ) , A ∈ H o m ( V n , W m ) , B C ∈ H o m ( T t , V n ) \mathscr{C}\in\mathrm{Hom}(T^t,U^s),\mathscr{B}\in\mathrm{Hom}(U^s,V^n),\mathscr{A}\in\mathrm{Hom}(V^n,W^m),\mathscr{BC}\in\mathrm{Hom}(T^t,V^n)C∈Hom(Tt,Us),B∈Hom(Us,Vn),A∈Hom(Vn,Wm),BC∈Hom(Tt,Vn)，
对 A \mathscr{A}A 作限制映射 A ∣ I m B , A ∣ I m B C \mathscr{A}|_{\mathrm{Im}\mathscr{B}} \ ,\ \mathscr{A}|_{\mathrm{Im}\mathscr{BC}}A∣ImB , A∣ImBC，则有
dim I m ( B ) = dim I m ( A B ) + dim ( K e r A ∩ I m B ) dim I m ( B C ) = dim I m ( A B C ) + dim ( K e r A ∩ I m B C ) \dim\mathrm{Im}(\mathscr{B}) =\dim\mathrm{Im}(\mathscr{AB})+\dim(\mathrm{Ker}\mathscr{A}\cap\mathrm{Im}\mathscr{B}) \\ \dim\mathrm{Im}(\mathscr{BC}) =\dim\mathrm{Im}(\mathscr{ABC})+\dim(\mathrm{Ker}\mathscr{A}\cap\mathrm{Im}\mathscr{BC})dimIm(B)=dimIm(AB)+dim(KerA∩ImB)dimIm(BC)=dimIm(ABC)+dim(KerA∩ImBC)
且 K e r A ∩ I m B C ⊆ K e r A ∩ I m B ⇒ dim ( K e r A ∩ I m B C ) ⩽ dim ( K e r A ∩ I m B ) \mathrm{Ker}\mathscr{A}\cap\mathrm{Im}\mathscr{BC}\subseteq\mathrm{Ker}\mathscr{A}\cap\mathrm{Im}\mathscr{B}\Rightarrow \dim(\mathrm{Ker}\mathscr{A}\cap\mathrm{Im}\mathscr{BC})\leqslant \dim(\mathrm{Ker}\mathscr{A}\cap\mathrm{Im}\mathscr{B})KerA∩ImBC⊆KerA∩ImB⇒dim(KerA∩ImBC)⩽dim(KerA∩ImB)
故有
dim I m ( B ) − dim I m ( A B ) ⩽ dim I m ( B C ) − dim I m ( A B C ) \dim\mathrm{Im}(\mathscr{B}) -\dim\mathrm{Im}(\mathscr{AB})\leqslant \dim\mathrm{Im}(\mathscr{BC}) -\dim\mathrm{Im}(\mathscr{ABC})dimIm(B)−dimIm(AB)⩽dimIm(BC)−dimIm(ABC)
remark :
取 B = I n , C ∈ M n , t ( F ) B=I_n\ ,\ C\in M_{n,t}(\mathbb{F})B=In , C∈Mn,t(F)，Frobenius \text{Frobenius}Frobenius 不等式即为 Sylvester \text{Sylvester}Sylvester 不等式.
Group 4
Theorem 8:
若 ( A n n + c 1 I n ) ( A n n + c 2 I n ) = O , 且 c 1 , 2 ∈ R , c 1 ≠ c 2 ， 则 r a n k ( A + c 1 I ) + r a n k ( A + c 2 I ) = n . \begin{aligned} &\qquad 若 \ (A_{nn}+c_{_1}I_n)(A_{nn}+c_{_2}I_n)=O,且 \ c_{_{1,2}} \in\R,c_1\ne c_2， \\&\qquad则\ \mathrm{rank}(A+c_{_1}I)+\mathrm{rank}(A+c_{_2}I)=n. \end{aligned}若 (Ann+c1In)(Ann+c2In)=O,且 c1,2∈R,c1=c2，则 rank(A+c1I)+rank(A+c2I)=n.
Proof:
n = r a n k ( I ) = r a n k ( ( c 1 − c 2 ) I ) = r a n k ( ( A + c 1 I ) − ( A + c 2 I ) ) ⩽ r a n k ( A + c 1 I ) + r a n k ( A + c 2 I ) ⩽ n \begin{aligned} n&=\mathrm{rank}(I)=\mathrm{rank}((c_1-c_2)I) \\&=\mathrm{rank}((A+c_1I)-(A+c_2I)) \\&\leqslant \mathrm{rank}(A+c_1I)+\mathrm{rank}(A+c_2I) \leqslant n \end{aligned}n=rank(I)=rank((c1−c2)I)=rank((A+c1I)−(A+c2I))⩽rank(A+c1I)+rank(A+c2I)⩽n
Corollary:
A n n 为幂等矩阵 ( 即 A 2 = A ) 的充要条件为 r a n k ( A ) + r a n k ( A − I ) = n . A n n 为对合矩阵 ( 即 A 2 = I ) 的充要条件为 r a n k ( A + I ) + r a n k ( A − I ) = n . \begin{aligned} &\quad A_{nn} \ 为幂等矩阵(即A^2=A )的充要条件为 \ \mathrm{rank}(A)+\mathrm{rank}(A-I)=n. \\&\quad A_{nn} \ 为对合矩阵(即A^2=I )的充要条件为 \ \mathrm{rank}(A+I)+\mathrm{rank}(A-I)=n. \end{aligned}Ann 为幂等矩阵(即A2=A)的充要条件为 rank(A)+rank(A−I)=n.Ann 为对合矩阵(即A2=I)的充要条件为 rank(A+I)+rank(A−I)=n.
分别取 ( c 1 , c 2 ) = ( 0 , 1 ) , ( 1 , − 1 ) (c_1,c_2)=(0,1),(1,-1)(c1,c2)=(0,1),(1,−1)，由 Theorem 8 可知结论成立，下面给出其他证法.
Proof 1:
( A O O I − A ) → r 2 + r 1 ( A O A I − A ) → c 2 + c 1 ( A A A I ) → r 1 + ( − A ) r 2 ( A − A 2 O O I ) \begin{aligned} \left( \begin{matrix} A& O\\ O& I-A\\ \end{matrix} \right) &\xrightarrow{r_2+r_1}\left( \begin{matrix} A& O\\ A& I-A\\ \end{matrix} \right) \xrightarrow{c_2+c_1}\left( \begin{matrix} A& A\\ A& I\\ \end{matrix} \right) \\&\xrightarrow{r_1+\left( -A \right) r_2}\left( \begin{matrix} A-A^2& O\\ O& I\\ \end{matrix} \right) \end{aligned}(AOOI−A)r2+r1(AAOI−A)c2+c1(AAAI)r1+(−A)r2(A−A2OOI)
则
r a n k ( A O O I − A ) = r a n k ( A − A 2 O O I ) \mathrm{rank}\left( \begin{matrix} A& O\\ O& I-A\\ \end{matrix} \right) =\mathrm{rank}\left( \begin{matrix} A-A^2& O\\ O& I\\ \end{matrix} \right)rank(AOOI−A)=rank(A−A2OOI)
从而
r a n k ( A ) + r a n k ( I − A ) = r a n k ( A − A 2 ) + r a n k ( I ) = A 2 − A = O 0 + n = n \begin{aligned} \mathrm{rank}\left( A \right) +\mathrm{rank}\left( I-A \right) &=\mathrm{rank}\left( A-A^2 \right) +\mathrm{rank}\left( I \right) \\&\xlongequal{ _{\ A^2-A=O \ }} 0+n=n \end{aligned}rank(A)+rank(I−A)=rank(A−A2)+rank(I) A2−A=O 0+n=n
Proof 2:
( A + I O O A − I ) → ( 2 I O − ( A + I ) − ( A + I ) ( A − I ) 2 ) → ( 2 I O O − ( A 2 − I ) 2 ) \quad \left( \begin{matrix} A+I& O\\ O& A-I\\ \end{matrix} \right) \rightarrow \left( \begin{matrix} 2I& O\\ -\left( A+I \right)& -\frac{\left( A+I \right) \left( A-I \right)}{2}\\ \end{matrix} \right) \rightarrow \left( \begin{matrix} 2I& O\\ O& -\frac{\left( A^2-I \right)}{2}\\ \end{matrix} \right)(A+IOOA−I)→(2I−(A+I)O−2(A+I)(A−I))→(2IOO−2(A2−I))
则
r a n k ( A + I ) + r a n k ( A − I ) = r a n k ( 2 I ) + r a n k ( I − A 2 2 ) = A 2 − I = O n + 0 = n \begin{aligned} \mathrm{rank}(A+I)+\mathrm{rank}(A-I)&=\mathrm{rank}(2I)+\mathrm{rank}(\frac{I-A^2}{2}) \\&\xlongequal{_{\ A^2-I=O\ }} n+0 = n \end{aligned}rank(A+I)+rank(A−I)=rank(2I)+rank(2I−A2) A2−I=O n+0=n