广义逆阵

广义逆阵$A^+$

设$A=A_{n \times n}$，具有如下四个性质：

(1)$AXA=A$

(2)$XAX=X$

(3)$(AX)^{H}=AX$

(4)$(XA)^{H}=XA$

称$X$为$A$的广义逆阵，记为$X=A^+$

 

常见的$A^+$：

(1)$0_{m \times n}^+=0_{n \times n}$

(2)可逆方阵$A=A_{n \times n}$，$A^+=A^{-1}$

(3)一阶复数阵a：$a=0 \Rightarrow a^+=0$，$a \ne 0 \Rightarrow a^+=\frac{1}{a}$

 

$A^+$的唯一性：给定A后，只有唯一的$A^+$。

 

求解$B^+$：

(1)(高阵)若$B=B_{m \times r}$，$rank(B)=r$，即B为高阵，则$B^+=B_L=(B^HB)^{-1}B^H$，且$B^+B=I$

(1)(低阵)若$B=B_{r \times n}$，$rank(B)=r$，即B为低阵，则$B^+=B_L=B^H(BB^H)^{-1}$，且$BB^+=I$

(3)高低分解公式：$A=A_{m \times n}=BC$为高低分解，则有$A^+=C^+B^+$且$B^+B=CC^+=I$，其中$B^+=(B^HB)^{-1}B^H, C^+=C^H(CC^H)^{-1}$。

(4)

\[\begin{array}{l}
B = \left[ {\begin{array}{*{20}{c}}
A&0
\end{array}} \right] \Rightarrow {B^ + } = \left[ {\begin{array}{*{20}{c}}
{{A^ + }}\\
{{0^ + }}
\end{array}} \right]\\
B = \left[ {\begin{array}{*{20}{c}}
A\\
0
\end{array}} \right] \Rightarrow {B^ + } = \left[ {\begin{array}{*{20}{c}}
{{A^ + }}&{{0^ + }}
\end{array}} \right]
\end{array}\]

(5)(正SVD分解)若$A=P \Delta Q^H$，为正SVD，P和Q为列半酉阵，$P^HP=Q^HQ=I$，则$A^+=Q\Delta^{-1}P^H$

证明：

1、\[A{A^ + }A = (P\Delta {Q^H})(Q{\Delta ^{ - 1}}{P^H})(P\Delta {Q^H}) = P\Delta {Q^H} = A\]

2、\[{A^ + }A{A^ + } = (Q{\Delta ^{ - 1}}{P^H})(P\Delta {Q^H})(Q{\Delta ^{ - 1}}{P^H}) = Q{\Delta ^{ - 1}}{P^H} = {A^ + }\]

3、\[A{A^ + } = P{P^H}(Hermite)\]

4、\[{A^ + }A = {Q^H}Q(Hermite)\]

(6)(秩1分解)若$A=A_{m \times n}$且$rank(A)=1$，则$A = \frac{{{A^H}}}{{\sum {{{\left| {{a_{ij}}} \right|}^2}} }}$

(7)(QR分解)若$A=QR$，$Q^HQ=I$，则$A^+=R^+Q^+=R^{-1}Q^+$

(8)(谱分解)若A为正规阵，且有谱分解$A=\lambda_1G_1+\lambda_2G_2+...+\lambda_kG_k$，则$A^+=\lambda_1^+G_1+\lambda_2G_2^++...+\lambda_kG_k^+$

(9)

(i)若P为列半酉阵($Q^HQ=I$)，则$P^+=P^H$

(ii)若P为列半酉阵($QQ^H=I$)，则$P^+=P^H$

(iii)$(A^H)^+=(A^+)^H$

(10)(分块矩阵)

\[A = \left[ {\begin{array}{*{20}{c}}
B&0\\
0&D
\end{array}} \right] \Rightarrow {A^ + } = \left[ {\begin{array}{*{20}{c}}
{{B^ + }}&0\\
0&{{D^ + }}
\end{array}} \right]\]

证明：

证明条件(1)

\[A{A^ + }A = \left[ {\begin{array}{*{20}{c}}
B&0\\
0&D
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
{{B^ + }}&0\\
0&{{D^ + }}
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
B&0\\
0&D
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
{B{B^ + }B}&0\\
0&{D{D^ + }D}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
B&0\\
0&D
\end{array}} \right]\]

证明条件(2)

\[{A^ + }A{A^ + } = \left[ {\begin{array}{*{20}{c}}
{{B^ + }}&0\\
0&{{D^ + }}
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
B&0\\
0&D
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
{{B^ + }}&0\\
0&{{D^ + }}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
{{B^{\rm{ + }}}B{B^{\rm{ + }}}}&0\\
0&{{D^ + }D{D^ + }}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
{{B^ + }}&0\\
0&{{D^ + }}
\end{array}} \right]\]

证明条件(3)：

\[{(A{A^{\rm{ + }}})^H} = {\left[ {\begin{array}{*{20}{c}}
{{B^{\rm{ + }}}B}&0\\
0&{{D^{\rm{ + }}}D}
\end{array}} \right]^H} = \left[ {\begin{array}{*{20}{c}}
{{{({B^{\rm{ + }}}B)}^H}}&0\\
0&{{{({D^{\rm{ + }}}D)}^H}}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
{{B^{\rm{ + }}}B}&0\\
0&{{D^{\rm{ + }}}D}
\end{array}} \right] = A{A^ + }\]

证明条件(4)：

\[{({A^ + }A)^H} = {\left[ {\begin{array}{*{20}{c}}
{B{B^ + }}&0\\
0&{D{D^ + }}
\end{array}} \right]^H} = \left[ {\begin{array}{*{20}{c}}
{{{(B{B^ + })}^H}}&0\\
0&{{{(D{D^ + })}^H}}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
{B{B^ + }}&0\\
0&{D{D^ + }}
\end{array}} \right] = {A^ + }A\]