证明:设 A = ( a i j ) m × s , B = ( b i j ) s × n A=\left(a_{i j}\right)_{m \times s}, B=\left(b_{i j}\right)_{s \times n} A=(aij)m×s,B=(bij)s×n,记 A B = C = ( c i j ) m × n , B T A T = D = ( d i j ) n × m A B=C=\left(c_{i j}\right)_{m \times n}, B^{T} A^{T}=D=\left(d_{i j}\right)_{n \times m} AB=C=(cij)m×n,BTAT=D=(dij)n×m,于是根据矩阵乘法公式,有
c j i = ∑ k = 1 s a j k b k i c_{j i}=\sum_{k=1}^{s} a_{j k} b_{k i} cji=k=1∑sajkbki
而 B T \boldsymbol{B}^{\mathbf{T}} BT的第 i i i行为 ( b 1 i , ⋯ , b s i ) , A T \left(b_{1 i}, \cdots, b_{s i}\right), \boldsymbol{A}^{\mathrm{T}} (b1i,⋯,bsi),AT的第 j j j列为 ( a j 1 , ⋯ , a j s ) T \left(a_{j 1}, \cdots, a_{j s}\right)^{T} (aj1,⋯,ajs)T,因此
d i j = ∑ k = 1 s b k i a j k = ∑ k = 1 s a j k b k i d_{i j}=\sum_{k=1}^{s} b_{k i} a_{j k}=\sum_{k=1}^{s} a_{j k} b_{k i} dij=k=1∑sbkiajk=k=1∑sajkbki
所以
d i j = c j i ( i = 1 , 2 , ⋯ , n ; j = 1 , 2 , ⋯ , m ) d_{i j}=c_{j i} \quad(i=1,2, \cdots, n ; j=1,2, \cdots, m) dij=cji(i=1,2,⋯,n;j=1,2,⋯,m)
即 D = C T D=C^{\mathrm{T}} D=CT
这个由定义就很容易理解,不再证明
分析:对于向量组 α 1 , α 2 , ⋯ , α s \alpha_{1}, \alpha_{2}, \cdots, \alpha_{s} α1,α2,⋯,αs及 β 1 , β 2 , ⋯ , β t \beta_{1}, \beta_{2}, \cdots, \beta_{t} β1,β2,⋯,βt,若任一 β i ( i = 1 , 2 , ⋯ , t ) \boldsymbol{\beta}_{i}(i=1,2, \cdots, t) βi(i=1,2,⋯,t)均可由 α 1 , α 2 , ⋯ , α s \alpha_{1}, \alpha_{2}, \cdots, \alpha_{s} α1,α2,⋯,αs线性表出,则 r ( β 1 , β 2 , ⋯ , β t ) ⩽ r ( α 1 , α 2 , ⋯ , α s ) r\left(\boldsymbol{\beta}_{1}, \boldsymbol{\beta}_{2}, \cdots, \boldsymbol{\beta}_{t}\right) \leqslant r\left(\boldsymbol{\alpha}_{1}, \boldsymbol{\alpha}_{2}, \cdots, \boldsymbol{\alpha}_{s}\right) r(β1,β2,⋯,βt)⩽r(α1,α2,⋯,αs),这是本题证明的关键依据
**证明:**不妨设 A m × n , B n × s A_{m \times n}, B_{n \times s} Am×n,Bn×s,将 B , A B B, A B B,AB按行分块为 B = [ β 1 β 2 ⋮ β n ] , A B = C = [ γ 1 γ 2 ⋮ γ m ] B=\left[\begin{array}{c}\beta_{1} \\ \beta_{2} \\ \vdots \\ \beta_{n}\end{array}\right], A B=C=\left[\begin{array}{c}\gamma_{1} \\ \gamma_{2} \\ \vdots \\ \gamma_{m}\end{array}\right] B=⎣⎢⎢⎢⎡β1β2⋮βn⎦⎥⎥⎥⎤,AB=C=⎣⎢⎢⎢⎡γ1γ2⋮γm⎦⎥⎥⎥⎤,于是
A B = [ a 11 a 12 ⋯ a 1 n a 21 a 22 ⋯ a 2 n ⋮ ⋮ ⋮ a m 1 a m 2 ⋯ a n n ] [ β 1 β 2 ⋮ β n ] = [ a 11 β 1 + a 12 β 2 + ⋯ + a 1 n β n a 21 β 1 + a 22 β 2 + ⋯ + a 2 n β n ⋮ a m 1 β 1 + a m 2 β 2 + ⋯ + a m β n ] = [ γ 1 γ 2 ⋮ γ m ] A B=\left[\begin{array}{cccc} a_{11} & a_{12} & \cdots & a_{1 n} \\ a_{21} & a_{22} & \cdots & a_{2 n} \\ \vdots & \vdots & & \vdots \\ a_{m 1} & a_{m 2} & \cdots & a_{n n} \end{array}\right]\left[\begin{array}{c} \beta_{1} \\ \beta_{2} \\ \vdots \\ \beta_{n} \end{array}\right]=\left[\begin{array}{c} a_{11} \beta_{1}+a_{12} \beta_{2}+\cdots+a_{1 n} \beta_{n} \\ a_{21} \beta_{1}+a_{22} \beta_{2}+\cdots+a_{2 n} \beta_{n} \\ \vdots \\ a_{m 1} \beta_{1}+a_{m 2} \beta_{2}+\cdots+a_{m} \beta_{n} \end{array}\right]=\left[\begin{array}{c} \gamma_{1} \\ \gamma_{2} \\ \vdots \\ \gamma_{m} \end{array}\right] AB=⎣⎢⎢⎢⎡a11a21⋮am1a12a22⋮am2⋯⋯⋯a1na2n⋮ann⎦⎥⎥⎥⎤⎣⎢⎢⎢⎡β1β2⋮βn⎦⎥⎥⎥⎤=⎣⎢⎢⎢⎡a11β1+a12β2+⋯+a1nβna21β1+a22β2+⋯+a2nβn⋮am1β1+am2β2+⋯+amβn⎦⎥⎥⎥⎤=⎣⎢⎢⎢⎡γ1γ2⋮γm⎦⎥⎥⎥⎤
所以 A B AB AB的行向量 γ i ( i = 1 , 2 , ⋯ , m ) \gamma_{i}(i=1,2, \cdots, m) γi(i=1,2,⋯,m)均可由 B B B的行向量线性表出,故
r ( A B ) ⩽ r ( B ) r(A B) \leqslant r(B) r(AB)⩽r(B)
同理可证 r ( A B ) ⩽ r ( A ) r(A B) \leqslant r(A) r(AB)⩽r(A),故有 r ( A B ) ⩽ min { r ( A ) , r ( B ) } r(A B) \leqslant \min \{r(A), r(B)\} r(AB)⩽min{r(A),r(B)}
证明:设 r ( A ) = p , r ( B ) = q r(A)=p, r(\boldsymbol{B})=q r(A)=p,r(B)=q,将 A , B A,B A,B按列分块为
A = [ α 1 , α 2 , ⋯ , α s ] , B = [ β 1 , β 2 , ⋯ , β s ] A=\left[\alpha_{1}, \alpha_{2}, \cdots, \alpha_{s}\right], B=\left[\beta_{1}, \beta_{2}, \cdots, \beta_{s}\right] A=[α1,α2,⋯,αs],B=[β1,β2,⋯,βs]
于是
[ A , B ] = [ α 1 , α 2 , ⋯ , α s , β 1 , β 2 , ⋯ , β s ] A + B = [ α 1 + β 1 , α 2 + β 2 , ⋯ , α s + β s ] \begin{array}{c} {[A, B]=\left[\alpha_{1}, \alpha_{2}, \cdots, \alpha_{s}, \beta_{1}, \beta_{2}, \cdots, \beta_{s}\right]} \\ A+B=\left[\alpha_{1}+\beta_{1}, \alpha_{2}+\beta_{2}, \cdots, \alpha_{s}+\beta_{s}\right] \end{array} [A,B]=[α1,α2,⋯,αs,β1,β2,⋯,βs]A+B=[α1+β1,α2+β2,⋯,αs+βs]
因 α i + β i ( i = 1 , 2 , ⋯ , s ) \boldsymbol{\alpha}_{i}+\boldsymbol{\beta}_{i}(i=1,2, \cdots, s) αi+βi(i=1,2,⋯,s)均可由向量组 α 1 , α 2 , ⋯ , α s , β 1 , β 2 , ⋯ , β s \alpha_{1}, \alpha_{2}, \cdots, \alpha_{s}, \beta_{1}, \beta_{2}, \cdots, \beta_{s} α1,α2,⋯,αs,β1,β2,⋯,βs线性表出,故
r ( A + B ) ⩽ r ( [ A , B ] ) r(A+B) \leqslant r([A, B]) r(A+B)⩽r([A,B])
又设 A , B A,B A,B的列向量的极大线性无关组分别为 α 1 , α 2 , ⋯ , α p \alpha_{1}, \alpha_{2}, \cdots, \alpha_{p} α1,α2,⋯,αp和 β 1 , β 2 , ⋯ , β q \boldsymbol{\beta}_{1}, \boldsymbol{\beta}_{2}, \cdots, \boldsymbol{\beta}_{q} β1,β2,⋯,βq。将 A A A的极大线性无关组 α 1 , α 2 , ⋯ , α p \alpha_{1}, \alpha_{2}, \cdots, \alpha_{p} α1,α2,⋯,αp扩充成 [ A , B ] [A,B] [A,B]的极大线性无关组,设为 α 1 , α 2 , ⋯ , α p , β 1 , β 2 , ⋯ , β w \alpha_{1}, \alpha_{2}, \cdots, \alpha_{p}, \beta_{1}, \beta_{2}, \cdots, \beta_{w} α1,α2,⋯,αp,β1,β2,⋯,βw,显然 w ⩽ q w \leqslant q w⩽q,故有
r ( [ A , B ] ) = p + w ⩽ p + q = r ( A ) + r ( B ) r([A, B])=p+w \leqslant p+q=r(A)+r(B) r([A,B])=p+w⩽p+q=r(A)+r(B)
故 r ( A + B ) ⩽ r ( [ A , B ] ) ⩽ r ( A ) + r ( B ) r(A+B) \leqslant r([A, B]) \leqslant r(A)+r(B) r(A+B)⩽r([A,B])⩽r(A)+r(B)
r ( [ A O O B ] ) = r ( A ) + r ( B ) r\left(\left[\begin{array}{ll}A & O \\ O & B\end{array}\right]\right)=r(A)+r(B) r([AOOB])=r(A)+r(B)
r ( A ) + r ( B ) ⩽ r ( [ A O C B ] ) ⩽ r ( A ) + r ( B ) + r ( C ) r(\boldsymbol{A})+r(\boldsymbol{B}) \leqslant r\left(\left[\begin{array}{cc}\boldsymbol{A} & \boldsymbol{O} \\ \boldsymbol{C} & \boldsymbol{B}\end{array}\right]\right) \leqslant r(\boldsymbol{A})+r(\boldsymbol{B})+r(\boldsymbol{C}) r(A)+r(B)⩽r([ACOB])⩽r(A)+r(B)+r(C)
证明:
r ( A ) + r ( B ) = r ( [ A , O ] ) + r ( [ O , B ] ) = r ( [ A O O B ] ) ⩽ r ( [ A O C B ] ) ⩽ r ( [ A , O ] ) + r ( [ C , B ] ) ⩽ r ( A ) + r ( B ) + r ( C ) \begin{aligned} r(A)+r(B) &=r([A, O])+r([O, B]) \\ &=r\left(\left[\begin{array}{cc} A & O \\ O & B \end{array}\right]\right) \leqslant r\left(\left[\begin{array}{cc} A & O \\ C & B \end{array}\right]\right) \\ & \leqslant r([A, O])+r([C, B]) \\ & \leqslant r(A)+r(B)+r(C) \end{aligned} r(A)+r(B)=r([A,O])+r([O,B])=r([AOOB])⩽r([ACOB])⩽r([A,O])+r([C,B])⩽r(A)+r(B)+r(C)
(因 r ( B ) ⩽ r ( [ C , B ] ) ⩽ r ( B ) + r ( C ) r(\boldsymbol{B}) \leqslant r([\boldsymbol{C}, \boldsymbol{B}]) \leqslant r(\boldsymbol{B})+r(\boldsymbol{C}) r(B)⩽r([C,B])⩽r(B)+r(C))
注解:第五题也为本题的一个特例
证明::对 [ A O E n B ] \left[\begin{array}{ll}\mathbf{A} & \boldsymbol{O} \\ \boldsymbol{E}_{n} & \boldsymbol{B}\end{array}\right] [AEnOB]做初等变换,得:
[ A O E n B ] → [ O − A B E n B ] → [ O − A B E n O ] → [ E n O O A B ] \left[\begin{array}{cc} A & O \\ E_{n} & B \end{array}\right] \rightarrow\left[\begin{array}{cc} O & -A B \\ E_{n} & B \end{array}\right] \rightarrow\left[\begin{array}{cc} O & -A B \\ E_{n} & O \end{array}\right] \rightarrow\left[\begin{array}{cc} E_{n} & O \\ O & A B \end{array}\right] [AEnOB]→[OEn−ABB]→[OEn−ABO]→[EnOOAB]
又显然 r ( A ) + r ( B ) = r ( [ A O O B ] ) ⩽ r ( [ A O E n B ] ) r(A)+r(B)=r\left(\left[\begin{array}{ll}A & O \\ O & B\end{array}\right]\right) \leqslant r\left(\left[\begin{array}{ll}A & O \\ E_{n} & B\end{array}\right]\right) r(A)+r(B)=r([AOOB])⩽r([AEnOB]),则
r ( A ) + r ( B ) ⩽ r ( [ E n O O A B ] ) = r ( E n ) + r ( A B ) = n + r ( A B ) r(\boldsymbol{A})+r(\boldsymbol{B}) \leqslant r\left(\left[\begin{array}{cc} \boldsymbol{E}_{n} & \boldsymbol{O} \\ \boldsymbol{O} & \boldsymbol{A} \boldsymbol{B} \end{array}\right]\right)=r\left(\boldsymbol{E}_{n}\right)+r(\boldsymbol{A} \boldsymbol{B})=n+r(\boldsymbol{A} \boldsymbol{B}) r(A)+r(B)⩽r([EnOOAB])=r(En)+r(AB)=n+r(AB)
所以 r ( A B ) ⩾ r ( A ) + r ( B ) − n r(A B) \geqslant r(A)+r(B)-n r(AB)⩾r(A)+r(B)−n
r ( A ) = r ( A T ) = r ( A A T ) = r ( A T A ) r(A)=r\left(A^{\mathrm{T}}\right)=r\left(A A^{\mathrm{T}}\right)=r\left(A^{\mathrm{T}} A\right) r(A)=r(AT)=r(AAT)=r(ATA)
r ( A ∗ ) = { n , r ( A ) = n 1 , r ( A ) = n − 1 0 , r ( A ) < n − 1 r\left(A^{*}\right)=\left\{\begin{array}{ll}n, & r(A)=n \\ 1, & r(A)=n-1 \\ 0, & r(A)
证明::当 r ( A ) = n r(\boldsymbol{A})=n r(A)=n时, A A A可逆, ∣ A ∣ ≠ 0 |A| \neq 0 ∣A∣=0,由 A A ∗ = ∣ A ∣ E A A^{*}=|A| E AA∗=∣A∣E可知, A A A和 A ∗ A^{*} A∗均是可逆矩阵,故 r ( A ∗ ) = n r\left(A^{*}\right)=n r(A∗)=n
当 r ( A ) = n − 1 r(\boldsymbol{A})=n-1 r(A)=n−1时,由矩阵的秩的定义知, ∣ A ∣ |A| ∣A∣中存在 n − 1 n-1 n−1阶子式不等于零,而 A ∗ 由 A^{*}由 A∗由 ∣ A ∣ |A| ∣A∣的元素的 a i j a_{i j} aij的代数余子式 A i j A_{i j} Aij组成,故 r ( A ∗ ) ⩾ 1 r\left(\boldsymbol{A}^{*}\right) \geqslant 1 r(A∗)⩾1,又 r ( A ) = n − 1 , ∣ A ∣ = 0 , A A ∗ = ∣ A ∣ E = O r(\boldsymbol{A})=n-1,|\boldsymbol{A}|=0, \boldsymbol{A} \boldsymbol{A}^{*}=|\boldsymbol{A}| \boldsymbol{E}=\boldsymbol{O} r(A)=n−1,∣A∣=0,AA∗=∣A∣E=O,得 r ( A ) + r ( A ∗ ) ⩽ n r(A)+r\left(A^{*}\right) \leqslant n r(A)+r(A∗)⩽n,而 r ( A ) = n − 1 r(A)=n-1 r(A)=n−1,故 r ( A ∗ ) ⩽ 1 r\left(A^{*}\right) \leqslant 1 r(A∗)⩽1(或 A ∗ A^{*} A∗的每一列均是 A x = 0 Ax=0 Ax=0的解向量,故 r ( A ∗ ) ⩽ 1 r\left(\boldsymbol{A}^{*}\right) \leqslant 1 r(A∗)⩽1),所以 r ( A ∗ ) = 1 r\left(\boldsymbol{A}^{*}\right)=1 r(A∗)=1
当 r ( A ) < n − 1 r(\boldsymbol{A})
- 注意伴随矩阵、可逆矩阵一定是方阵
证明:由 A 2 = A A^{2}=A A2=A,得 A ( A − E ) = O \boldsymbol{A}(\boldsymbol{A}-\boldsymbol{E})=\boldsymbol{O} A(A−E)=O,故 r ( A ) + r ( A − E ) ⩽ n r(\boldsymbol{A})+r(\boldsymbol{A}-\boldsymbol{E}) \leqslant n r(A)+r(A−E)⩽n
又 r ( A ) + r ( A − E ) = r ( A ) + r ( E − A ) ⩾ r ( A + E − A ) = r ( E ) = n r(A)+r(A-E)=r(A)+r(E-A) \geqslant r(A+E-A)=r(E)=n r(A)+r(A−E)=r(A)+r(E−A)⩾r(A+E−A)=r(E)=n,得证 r ( A ) + r ( A − E ) = n r(\boldsymbol{A})+r(\boldsymbol{A}-\boldsymbol{E})=n r(A)+r(A−E)=n
解析:由题设 A 2 = E A^{2}=E A2=E知
( A + E ) ( A − E ) = A 2 − E = O (A+E)(A-E)=A^{2}-E=O (A+E)(A−E)=A2−E=O
于是有
r ( A + E ) + r ( A − E ) ⩽ n r(A+E)+r(A-E) \leqslant n r(A+E)+r(A−E)⩽n
注意到 r ( A − E ) = r ( − A + E ) r(\boldsymbol{A}-\boldsymbol{E})=r(-\boldsymbol{A}+\boldsymbol{E}) r(A−E)=r(−A+E),则
r ( A + E ) + r ( A − E ) = r ( A + E ) + r ( − A + E ) ⩾ r ( A + E − A + E ) = r ( 2 E ) = n \begin{aligned} r(A+E)+r(A-E) &=r(A+E)+r(-A+E) \\ & \geqslant r(A+E-A+E) \\ &=r(2 E)=n \end{aligned} r(A+E)+r(A−E)=r(A+E)+r(−A+E)⩾r(A+E−A+E)=r(2E)=n
综上所述 r ( A + E ) + r ( A − E ) = n r(A+E)+r(A-E)=n r(A+E)+r(A−E)=n得证