矩阵论(四)——矩阵的广义逆

矩阵论(四)——矩阵的广义逆

  • 1. 广义逆矩阵
  • 2. 减号广义逆
  • 3. 极小范数广义逆
  • 4. 最小二乘广义逆
  • 5. 加号广义逆
  • 6.方程通解与最小二乘解
    • 6.1 相容方程的通解
    • 6.2 矛盾方程的最小二乘解

1. 广义逆矩阵

∀ A ∈ R m × n , ∃ G ∈ R n × m \forall A \in R^{m \times n},\exists G \in R^{n \times m} ARm×nGRn×m,满足下列的一个、多个或者全部,则称G为A的广义逆矩阵
( 1 ) A G A = A (1)\quad AGA = A (1)AGA=A
( 2 ) G A G = G (2)\quad GAG = G (2)GAG=G
( 3 ) ( A G ) H = A G (3)\quad (AG)^H = AG (3)(AG)H=AG
( 4 ) ( G A ) H = G A (4)\quad (GA)^H = GA (4)(GA)H=GA
若满足第i个条件,记为 G = A ( i ) , A { i } = { G ∣ G = A ( i ) } G = A^{(i)},A\{i\} = \{G | G = A^{(i)}\} G=A(i)A{i}={GG=A(i)}
若满足第i, j个条件,记为 G = A ( i , j ) , A { i , j } = { G ∣ G = A ( i , j ) } G = A^{(i, j)},A\{i, j\} = \{G | G = A^{(i, j)}\} G=A(i,j)A{i,j}={GG=A(i,j)}

减号逆( A − A^{-} A): A − = A ( 1 ) A^{-} = A^{(1)} A=A(1),满足 A G A = A AGA = A AGA=A

极小范数广义逆( A m − A^{-}_m Am): A ( 1 , 4 ) A^{(1, 4)} A(1,4),满足 A G A = A ,   ( G A ) H = G A AGA = A,\ (GA)^H = GA AGA=A, (GA)H=GA

最小二乘广义逆( A l − A^{-}_l Al): A ( 1 , 3 ) A^{(1, 3)} A(1,3),满足 A G A = A ,   ( A G ) H = A G AGA = A,\ (AG)^H = AG AGA=A, (AG)H=AG

加号广义逆( A + A^+ A+): A ( 1 , 2 , 3 , 4 ) A^{(1, 2, 3, 4)} A(1,2,3,4),满足 A G A = A ,   G A G = A ,   ( A G ) H = A G ,   ( G A ) H = G A AGA = A,\ GAG = A,\ (AG)^H = AG,\ (GA)^H = GA AGA=A, GAG=A, (AG)H=AG, (GA)H=GA

2. 减号广义逆

A ∈ C m × n A \in C^{m \times n} ACm×n,若存在 G ∈ C n × m G \in C^{n \times m} GCn×m,使 A G A = A AGA=A AGA=A,则称G为A的一个减号广义逆,记为 A − , A { 1 } = { G ∣ A G A = A } A^{-},A\{1\} = \{G | AGA = A\} AA{1}={GAGA=A}

A − A^- A存在的条件及求法: ∀ A ∈ C m × n \forall A \in C^{m \times n} ACm×n
(1) 若秩(A) = 0,即 A = 0 A = 0 A=0,则 A − A^- A存在,即 ∀ G ∈ C n × m , 均 有 A G A = A \forall G \in C^{n \times m},均有AGA=A GCn×mAGA=A
(2) 若秩(A) = n(m = n),即 ∣ A ∣ ≠ 0 |A| \neq 0 A=0,则 A − A^- A不唯一,且 A − = A − 1 , A G A = A A^- = A^{-1},AGA = A A=A1AGA=A
(3) 若秩(A) = r,有可逆P,Q,使 P A Q = ( I r 0 ) m × n PAQ = \begin{pmatrix}I_r & \\ & 0\end{pmatrix}_{m \times n} PAQ=(Ir0)m×n,即 A = P − 1 ( I r 0 ) Q − 1 A = P^{-1} \begin{pmatrix}I_r & \\ & 0\end{pmatrix}Q^{-1} A=P1(Ir0)Q1
G ∈ A { 1 }    ⟺    G = Q ( I r U V W ) P G \in A\{1\} \iff G = Q \begin{pmatrix}I_r & U \\ V & W\end{pmatrix} P GA{1}G=Q(IrVUW)P,其中U、V、W为相应的任意矩阵

A ∈ C m × n , λ ∈ C A \in C^{m \times n},\lambda \in C ACm×nλC,则 A − A^{-} A满足一下性质:
( 1 ) r a n k ( A ) ≤ r a n k ( A − ) (1)\quad rank(A) \leq rank(A^-) (1)rank(A)rank(A)
( 2 ) A A − 与 A − A (2)\quad AA^-与A^-A (2)AAAA都是幂等矩阵,且 r a n k ( A ) = r a n k ( A A − ) = r a n k ( A − A ) rank(A) = rank(AA^{-}) = rank(A^{-}A) rank(A)=rank(AA)=rank(AA)
( 3 ) R ( A A − ) = R ( A ) , N ( A − A ) = N ( A ) (3)\quad R(AA^-) = R(A),N(A^-A) = N(A) (3)R(AA)=R(A)N(AA)=N(A)

例如:
矩阵论(四)——矩阵的广义逆_第1张图片
说明:可以先对 ( A I 3 ) \begin{pmatrix}A & I_3\end{pmatrix} (AI3)做行变换,将变换后的结果记为 ( A ′ P ) \begin{pmatrix}A' & P \end{pmatrix} (AP),然后对 ( A ′ I 4 ) \begin{pmatrix} A' \\ I_4 \end{pmatrix} (AI4)做列变换,即可得到最终的P和Q

3. 极小范数广义逆

极小范数广义逆: A ∈ C m × n A \in C^{m \times n} ACm×n,若存在 G ∈ C n × m G \in C^{n \times m} GCn×m,使 ( 1 ) A G A = A , ( 4 ) ( G A ) H = G A (1) AGA=A,\quad (4)(GA)^H = GA (1)AGA=A(4)(GA)H=GA,则称G为A的一个极小范数广义逆
记为 A m − , A { 1 , 4 } = { G ∣ A G A = A , ( G A ) H = G A } A^{-}_m,A\{1,4\} = \{G | AGA = A,(GA)^H = GA\} AmA{14}={GAGA=A(GA)H=GA}

A m − A^-_m Am存在的条件及求法: A ∈ C m × n A \in C^{m \times n} ACm×n,A的奇异分解: A = U ( Δ r 0 0 0 ) V H A = U \begin{pmatrix}\Delta_r & 0 \\ 0 & 0\end{pmatrix} V^H A=U(Δr000)VH,则
G ∈ A { 1 , 4 }    ⟺    G = V ( Δ r − 1 K 0 M ) U H G \in A\{1,4\} \iff G = V \begin{pmatrix}\Delta_r^{-1} & K \\ 0 & M\end{pmatrix} U^H GA{14}G=V(Δr10KM)UH,K和M为相应任意矩阵
奇异分解求解过程可参考https://blog.csdn.net/u011609063/article/details/102680734#32_SVD_250

相容方程: 方程组Ax=b有解
矛盾(不相容)方程: 方程组Ax=b无解

极小范数解: 方程 A x = b Ax = b Ax=b有无数解,在所有的解中最小的解就是极小范数解。即 x 0 = m i n ( x ) x_0 = min(x) x0=min(x) x 0 x_0 x0 A x = b Ax = b Ax=b的极小范数解

G ∈ A { 1 , 4 } , 则 G b = m i n ( x ) G \in A\{1,4\},则Gb = min(x) GA{14}Gb=min(x),即 x 0 = G b x_0 = Gb x0=Gb是相容方程组Ax=b的极小范数解
A ∈ C m × n , G ∈ A { 1 , 4 }    ⟺    x 0 = G b A \in C^{m \times n},G \in A\{1,4\} \iff x_0 = Gb ACm×nGA{14}x0=Gb是相容方程Ax=b的极小范数解
∀ G 1 , G 2 ∈ A { 1 , 4 } , 必 有 G 1 b = G 2 b \forall G_1,G_2 \in A\{1,4\},必有G_1 b = G_2 b G1G2A{14}G1b=G2b,即相容方程的极小范数解是唯一的。

例如:

4. 最小二乘广义逆

最小二乘广义逆: A ∈ C m × n A \in C^{m \times n} ACm×n,若存在 G ∈ C n × m G \in C^{n \times m} GCn×m,使 ( 1 ) A G A = A , ( 3 ) ( A G ) H = A G (1) AGA=A, (3)(AG)^H = AG (1)AGA=A(3)(AG)H=AG,则称G为A的一个最小二乘广义逆,记为 A l − , A { 1 , 3 } = { G ∣ A G A = A , ( A G ) H = A G } A^{-}_l,A\{1,3\} = \{G | AGA = A,(AG)^H = AG\} AlA{13}={GAGA=A(AG)H=AG}

A l − A^-_l Al存在的条件及求法: A ∈ C m × n A \in C^{m \times n} ACm×n,A的奇异分解: A = U ( Δ r 0 0 0 ) V H A = U \begin{pmatrix}\Delta_r & 0 \\ 0 & 0\end{pmatrix} V^H A=U(Δr000)VH,则
G ∈ A { 1 , 3 }    ⟺    G = V ( Δ r − 1 0 L M ) U H G \in A\{1,3\} \iff G = V \begin{pmatrix}\Delta_r^{-1} & 0 \\ L & M\end{pmatrix} U^H GA{13}G=V(Δr1L0M)UH,L和M为相应任意矩阵
奇异分解求解过程可参考https://blog.csdn.net/u011609063/article/details/102680734#32_SVD_250

最小二乘解: A x − b Ax-b Axb取最小值时x的解,即矛盾方程 A x = b , ∃ u ∈ C n , ∣ ∣ A u − b ∣ ∣ 2 = m i n ∣ ∣ A x − b ∣ ∣ 2 ( ∀ x ∈ C n ) Ax = b,\exists u \in C^n,||Au - b||_2 = min||Ax -b||_2(\forall x \in C^n) Ax=buCnAub2=minAxb2(xCn)时u的值

G ∈ A { 1 ,   3 } , 则 A H A G = A H G \in A\{1,\ 3\},则A^HAG = A^H GA{1, 3}AHAG=AH

G ∈ A { 1 ,   3 } , 则 x = G b G \in A\{1,\ 3\},则x = Gb GA{1, 3}x=Gb是矛盾方程Ax=b的最小二乘解,即 A G b − b = m i n ( A x − b ) AGb - b = min(Ax - b) AGbb=min(Axb)

例如:
矩阵论(四)——矩阵的广义逆_第2张图片

5. 加号广义逆

加号广义逆: A ∈ C m × n ,   ∃ G ∈ C n × m A \in C^{m \times n},\ \exists G \in C^{n \times m} ACm×n, GCn×m,满足 ( 1 ) A G A = A ( 2 ) G A G = G ( 3 ) ( A G ) H = A G ( 4 ) ( G A ) H = G A (1)AGA = A \quad (2)GAG = G \quad (3)(AG)^H = AG \quad (4)(GA)^H = GA (1)AGA=A(2)GAG=G(3)(AG)H=AG(4)(GA)H=GA
则称G为A的Moore-Penrose广义逆加号广义逆,简称A的M-P逆,记为 A + , A { 1 ,   2 ,   3 ,   4 } A^+,A\{1,\ 2,\ 3,\ 4\} A+A{1, 2, 3, 4}

A + A^+ A+存在的条件及求法: A ∈ C m × n A \in C^{m \times n} ACm×n,则 A + A^+ A+存在且唯一
A的奇异分解: A = U ( Σ r 0 0 0 ) V H A = U \begin{pmatrix}\Sigma_r & 0 \\ 0 & 0\end{pmatrix} V^H A=U(Σr000)VH,则 A + = V ( Σ r − 1 0 0 0 ) U H A^+ = V \begin{pmatrix}\Sigma^{-1}_r & 0 \\ 0 & 0\end{pmatrix} U^H A+=V(Σr1000)UH
奇异分解求解过程可参考https://blog.csdn.net/u011609063/article/details/102680734#32_SVD_250

设rank(A) = r,A的一个满值分解为 A = B C , B ∈ C m × r , C ∈ C r × n , r a n k ( B ) = r a n k ( C ) = r A = BC,B \in C^{m \times r},C \in C^{r \times n},rank(B) = rank(C) = r A=BCBCm×rCCr×nrank(B)=rank(C)=r
A + = C H ( C C H ) − 1 ( B H B ) − 1 B H A^+ = C^H(CC^H)^{-1}(B^HB)^{-1}B^H A+=CH(CCH)1(BHB)1BH

A ∈ C m × n , λ ∈ C m × n , 则 A + A \in C ^{m \times n},\lambda \in C^{m \times n},则A^+ ACm×nλCm×nA+满足以下性质:
( 1 ) ( A + ) + = A (1) \quad (A^+)^+ = A (1)(A+)+=A
( 2 ) ( A + ) H = ( A H ) + (2) \quad (A^+)^H = (A^H)^+ (2)(A+)H=(AH)+
( 3 ) ( λ A ) + = λ + A + (3) \quad (\lambda A)^+ = \lambda^+ A^+ (3)(λA)+=λ+A+,其中 λ + = { 1 λ λ ≠ 0 0 λ = 0 \lambda^+ = \begin{cases} \frac{1}{\lambda} & \lambda \neq 0 \\ 0 & \lambda = 0\end{cases} λ+={λ10λ=0λ=0
( 4 ) A + = { ( A H A ) − 1 A H A 为 列 满 秩 A H ( A A H ) − 1 A 为 行 满 秩 (4) \quad A^+ = \begin{cases}(A^HA)^{-1}A^H & A为列满秩 \\ A^H(AA^H)^{-1} & A为行满秩\end{cases} (4)A+={(AHA)1AHAH(AAH)1AA

例如:
矩阵论(四)——矩阵的广义逆_第3张图片

6.方程通解与最小二乘解

6.1 相容方程的通解

相容方程: A x = b , A ∈ C m × n Ax = b,A \in C^{m \times n} Ax=bACm×n
x = A − b + ( I n − A − A ) z , z ∈ C n x = A^- b + (I_n - A^- A) z,z \in C^n x=Ab+(InAA)zzCn

6.2 矛盾方程的最小二乘解

矛盾方程: A x = b , A ∈ C m × n Ax = b,A \in C^{m \times n} Ax=bACm×n
x = A l − b + ( I n − A l − A ) z , ∀ z ∈ C n x = A^-_lb + (I_n - A^-_l A)z,\forall z \in C^n x=Alb+(InAlA)zzCn

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