广义逆阵$A^+$
设$A=A_{n imes n}$，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)$0_{m imes n}^+=0_{n imes n}$
(2)可逆方阵$A=A_{n imes n}$，$A^+=A^{-1}$
(3)一阶复数阵a：$a=0 Rightarrow a^+=0$，$a e 0 Rightarrow a^+=frac{1}{a}$
$A^+$的唯一性：给定A后，只有唯一的$A^+$。
求解$B^+$：
(1)(高阵)若$B=B_{m imes r}$，$rank(B)=r$，即B为高阵，则$B^+=B_L=(B^HB)^{-1}B^H$，且$B^+B=I$
(1)(低阵)若$B=B_{r imes n}$，$rank(B)=r$，即B为低阵，则$B^+=B_L=B^H(BB^H)^{-1}$，且$BB^+=I$
(3)高低分解公式：$A=A_{m imes n}=BC$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)
[egin{array}{l}
B = left[ {egin{array}{*{20}{c}}
A&0
end{array}}
ight] Rightarrow {B^ + } = left[ {egin{array}{*{20}{c}}
{{A^ + }}\
{{0^ + }}
end{array}}
ight]\
B = left[ {egin{array}{*{20}{c}}
A\
0
end{array}}
ight] Rightarrow {B^ + } = left[ {egin{array}{*{20}{c}}
{{A^ + }}&{{0^ + }}
end{array}}
ight]
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$A^+=QDelta^{-1}P^H$
证明：
1、[A{A^ + }A = (PDelta {Q^H})(Q{Delta ^{ - 1}}{P^H})(PDelta {Q^H}) = PDelta {Q^H} = A]
2、[{A^ + }A{A^ + } = (Q{Delta ^{ - 1}}{P^H})(PDelta {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$A=A_{m imes n}$且$rank(A)=1$，则$A = frac{{{A^H}}}{{sum {{{left| {{a_{ij}}} ight|}^2}} }}$
(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[ {egin{array}{*{20}{c}}
B&0\
0&D
end{array}}
ight] Rightarrow {A^ + } = left[ {egin{array}{*{20}{c}}
{{B^ + }}&0\
0&{{D^ + }}
end{array}}
ight]]
证明：
证明条件(1)
[A{A^ + }A = left[ {egin{array}{*{20}{c}}
B&0\
0&D
end{array}}
ight]left[ {egin{array}{*{20}{c}}
{{B^ + }}&0\
0&{{D^ + }}
end{array}}
ight]left[ {egin{array}{*{20}{c}}
B&0\
0&D
end{array}}
ight] = left[ {egin{array}{*{20}{c}}
{B{B^ + }B}&0\
0&{D{D^ + }D}
end{array}}
ight] = left[ {egin{array}{*{20}{c}}
B&0\
0&D
end{array}}
ight]]
证明条件(2)
[{A^ + }A{A^ + } = left[ {egin{array}{*{20}{c}}
{{B^ + }}&0\
0&{{D^ + }}
end{array}}
ight]left[ {egin{array}{*{20}{c}}
B&0\
0&D
end{array}}
ight]left[ {egin{array}{*{20}{c}}
{{B^ + }}&0\
0&{{D^ + }}
end{array}}
ight] = left[ {egin{array}{*{20}{c}}
{{B^{
m{ + }}}B{B^{
m{ + }}}}&0\
0&{{D^ + }D{D^ + }}
end{array}}
ight] = left[ {egin{array}{*{20}{c}}
{{B^ + }}&0\
0&{{D^ + }}
end{array}}
ight]]
证明条件(3)：
[{(A{A^{
m{ + }}})^H} = {left[ {egin{array}{*{20}{c}}
{{B^{
m{ + }}}B}&0\
0&{{D^{
m{ + }}}D}
end{array}}
ight]^H} = left[ {egin{array}{*{20}{c}}
{{{({B^{
m{ + }}}B)}^H}}&0\
0&{{{({D^{
m{ + }}}D)}^H}}
end{array}}
ight] = left[ {egin{array}{*{20}{c}}
{{B^{
m{ + }}}B}&0\
0&{{D^{
m{ + }}}D}
end{array}}
ight] = A{A^ + }]
证明条件(4)：
[{({A^ + }A)^H} = {left[ {egin{array}{*{20}{c}}
{B{B^ + }}&0\
0&{D{D^ + }}
end{array}}
ight]^H} = left[ {egin{array}{*{20}{c}}
{{{(B{B^ + })}^H}}&0\
0&{{{(D{D^ + })}^H}}
end{array}}
ight] = left[ {egin{array}{*{20}{c}}
{B{B^ + }}&0\
0&{D{D^ + }}
end{array}}
ight] = {A^ + }A]