Moore-Penrose 广义逆

对于任意一个矩阵 Am×n∈Cm×n,Am×n∈Cm×n,A_{m \times n} \in \mathbb C ^{m \times n}, 若矩阵 Xn×m∈Cn×mXn×m∈Cn×mX_{n \times m} \in \mathbb C ^{n \times m} 满足：
AXA=A(1)(1)AXA=AAXA = A \tag {1}
XAX=X(2)(2)XAX=XXAX = X \tag {2}
(AX)H=AX(3)(3)(AX)H=AX{(AX)} ^H = AX \tag {3}
(XA)H=XA(4)(4)(XA)H=XA{(XA)} ^H = XA \tag {4}
则称 XXX 为 A" role="presentation">AAA 的 Moore-Penrose 广义逆，记为 A+A+A^+ 。

定理

对于任意一个矩阵 Am×n∈Cm×n,Am×n∈Cm×n,A_{m \times n} \in \mathbb C ^{m \times n}, A+A+A^+ 存在并唯一。

证明

1. 存在性

对 AAA 做奇异值分解：
A=UBVH" role="presentation">A=UBVHA=UBVHA = U B V^H
其中 Um×m,Vn×nUm×m,Vn×nU_{m \times m}, V_{n \times n} 均为酉矩阵，
B=(Λ0(m−r)×(n−r))m×nB=(Λ0(m−r)×(n−r))m×nB = \begin{pmatrix} \Lambda & \\ }_{(m - r) \times (n - r)} \end{pmatrix} _{m \times n}
Λ=⎛⎝⎜⎜λ1−−√⋱λr−−√⎞⎠⎟⎟r×r,Λ=(λ1⋱λr)r×r,\Lambda = \begin{pmatrix} \sqrt{\lambda_1} & & \\ & \ddots & \\ & & \sqrt{\lambda_r} \end{pmatrix} _{r \times r}, 其中 {λi,i∈N,1≤i≤r}{λi,i∈N,1≤i≤r}\{\lambda_i , i \in \mathbb N, 1 \le i \le r\} 为 AHAAHAA^H A 的 rrr 个正特征值。
记 C=(Λ−10(m−r)×(n−r))m×n,D=(I0(m−r)×(n−r))m×n" role="presentation">C=(Λ−10(m−r)×(n−r))m×n,D=(I0(m−r)×(n−r))m×nC=(Λ−10(m−r)×(n−r))m×n,D=(I0(m−r)×(n−r))m×nC = \begin{pmatrix} \Lambda ^{-1} & \\ }_{(m - r) \times (n - r)} \end{pmatrix} _{m \times n}, D = \begin{pmatrix} I & \\ }_{(m - r) \times (n - r)} \end{pmatrix} _{m \times n}
取 X=VCUH,X=VCUH,X = VCU^H, 则：
AX=(UBVH)(VCUH)=UBCUH=UDUHAX=(UBVH)(VCUH)=UBCUH=UDUHAX = (UBV^H) (VCU^H) = UBCU^H = UDU^H
XA=(VCUH)(UBVH)=VCBVH=VDVHXA=(VCUH)(UBVH)=VCBVH=VDVHXA = (VCU^H) (UBV^H) = VCBV^H = VDV^H
因此：
(AX)H=(UDUH)H=UDUH=AX(AX)H=(UDUH)H=UDUH=AX{(AX)}^H = {(UDU^H)}^H = UDU^H = AX
(XA)H=(VDVH)H=VDVH=XA(XA)H=(VDVH)H=VDVH=XA{(XA)}^H = {(VDV^H)}^H = VDV^H = XA
XAX=(VDVH)(VCUH)=VDCUH=VCUH=XXAX=(VDVH)(VCUH)=VDCUH=VCUH=XXAX = (VDV^H) (VCU^H) = VDCU^H = VCU^H = X
AXA=(UBVH)(VDVH)=UBDVH=UBVH=AAXA=(UBVH)(VDVH)=UBDVH=UBVH=AAXA = (UBV^H) (VDV^H) = UBDV^H = UBV^H = A

2. 唯一性

设存在两个矩阵 X,YX,YX, Y 均满足条件，则
AX=(AYA)X=(AY)(AX)=(AY)H(AX)H=(AXAY)H=(AY)H=AYAX=(AYA)X=(AY)(AX)=(AY)H(AX)H=(AXAY)H=(AY)H=AYAX = (AYA) X = (AY) (AX) = {(AY)}^H {(AX)}^H = {(AXAY)} ^H = {(AY)}^H = AY
XA=X(AYA)=(XA)(YA)=(XA)H(YA)H=(YAXA)H=(YA)H=YAXA=X(AYA)=(XA)(YA)=(XA)H(YA)H=(YAXA)H=(YA)H=YAXA = X(AYA) = (XA) (YA) = {(XA)}^H {(YA)}^H = {(YAXA)} ^H = {(YA)}^H = YA
因此：
X=XAX=(XA)X=(YA)X=Y(AX)=Y(AY)=YAY=YX=XAX=(XA)X=(YA)X=Y(AX)=Y(AY)=YAY=YX = XAX = (XA) X = (YA) X = Y(AX) = Y(AY) = YAY = Y

性质

r(A+)=r(A)r(A+)=r(A)r(A^+) = r(A)
证明：由条件1，2可得。
rangeA+=rangeAH,kerA+=kerAHrange⁡A+=range⁡AH,ker⁡A+=ker⁡AH\operatorname {range} A^+ = \operatorname {range} A^H, \ker A^+ = \ker A^H
证明：
AHα=β⇒β=AHα=(AXA)Hα=(XA)HAHα=(XA)AHα=X(AAHα)⇒rangeAH⊆rangeA+AHα=β⇒β=AHα=(AXA)Hα=(XA)HAHα=(XA)AHα=X(AAHα)⇒range⁡AH⊆range⁡A+A^H\alpha = \beta \Rightarrow \beta = A^H\alpha = (AXA)^H \alpha = (XA)^H A^H\alpha = (XA) A^H\alpha = X(AA^H\alpha) \Rightarrow \operatorname {range} A^H \subseteq \operatorname {range} A^+
Xα=β⇒β=Xα=(XAX)α=(XA)Xα=(XA)HXα=(AHXH)Xα=AH(XHXα)⇒rangeA+⊆rangeAHXα=β⇒β=Xα=(XAX)α=(XA)Xα=(XA)HXα=(AHXH)Xα=AH(XHXα)⇒range⁡A+⊆range⁡AHX\alpha = \beta \Rightarrow \beta = X\alpha = (XAX) \alpha = (XA) X\alpha = (XA)^H X\alpha = (A^HX^H) X\alpha = A^H ( X^H X\alpha ) \Rightarrow \operatorname {range} A^+ \subseteq \operatorname {range} A^H
AHα=0⃗ ⇒Xα=(XAX)α=X(AX)Hα=X(XHAH)α=(XXH)(AHα)=0⃗ ⇒kerAH⊆kerA+AHα=0→⇒Xα=(XAX)α=X(AX)Hα=X(XHAH)α=(XXH)(AHα)=0→⇒ker⁡AH⊆ker⁡A+A^H\alpha = \vec 0 \Rightarrow X \alpha = (XAX) \alpha = X(AX)^H \alpha = X(X^HA^H) \alpha = (XX^H) (A^H \alpha) = \vec 0 \Rightarrow \ker A^H \subseteq \ker A^+
Xα=0⃗ ⇒AHα=(AXA)Hα=AH(AX)Hα=AH(AX)α=(AHA)(Xα)=0⃗ ⇒kerA+⊆kerAHXα=0→⇒AHα=(AXA)Hα=AH(AX)Hα=AH(AX)α=(AHA)(Xα)=0→⇒ker⁡A+⊆ker⁡AHX \alpha = \vec 0 \Rightarrow A^H\alpha = (AXA)^H \alpha = A^H (AX)^H \alpha = A^H (AX) \alpha = (A^H A) (X \alpha) = \vec 0 \Rightarrow \ker A^+ \subseteq \ker A^H