從穆爾-彭羅斯條件出發，彭羅斯推導出了穆爾-彭羅斯廣義逆的一些性質[ 3]
(
A
H
)
†
=
(
A
†
)
H
{\displaystyle ({\boldsymbol {A}}^{H})^{\dagger }=({\boldsymbol {A}}^{\dagger })^{H}}
A
†
A
A
H
=
A
H
A
A
†
=
A
H
{\displaystyle {\boldsymbol {A}}^{\dagger }{\boldsymbol {A}}{\boldsymbol {A}}^{H}={\boldsymbol {A}}^{H}{\boldsymbol {A}}{\boldsymbol {A}}^{\dagger }={\boldsymbol {A}}^{H}}
A
A
H
(
A
H
)
†
=
(
A
H
)
†
A
H
A
=
A
{\displaystyle {\boldsymbol {A}}{\boldsymbol {A}}^{H}({\boldsymbol {A}}^{H})^{\dagger }=({\boldsymbol {A}}^{H})^{\dagger }{\boldsymbol {A}}^{H}{\boldsymbol {A}}={\boldsymbol {A}}}
A
†
A
{\displaystyle {\boldsymbol {A}}^{\dagger }{\boldsymbol {A}}}
A
A
†
{\displaystyle {\boldsymbol {A}}{\boldsymbol {A}}^{\dagger }}
(
I
−
A
†
A
)
{\displaystyle ({\boldsymbol {I}}-{\boldsymbol {A}}^{\dagger }{\boldsymbol {A}})}
(
I
−
A
†
A
)
{\displaystyle ({\boldsymbol {I}}-{\boldsymbol {A}}^{\dagger }{\boldsymbol {A}})}
偽逆存在且唯一：對於任何矩陣
A
{\displaystyle A}
A
†
{\displaystyle A^{\dagger }}
[ 4]
滿足該定義的第一個條件的矩陣被稱為廣義逆。如果該矩陣也滿足第二個定義，它就被稱為廣義反身逆陣 （generalized reflexive inverse）。廣義逆矩陣總存在，但一般不唯一。唯一性是最後兩個條件的結果。
這些性質的證明可以在維基教科書中找到。
如果
A
{\displaystyle A}
A
†
{\displaystyle A^{\dagger }}
如果
A
{\displaystyle A}
A
†
=
A
−
1
{\displaystyle A^{\dagger }=A^{-1}}
[ 5] :243
零矩陣 的偽逆是它的轉置。矩陣偽逆的偽逆是原矩陣，即：
(
A
†
)
†
=
A
{\displaystyle \left(A^{\dagger }\right)^{\dagger }=A}
[ 5] :245
偽轉置與轉置、復共軛和共軛轉置可以交換：[ 5] :245
(
A
T
)
†
=
(
A
†
)
T
{\displaystyle \left(A^{\textsf {T}}\right)^{\dagger }=\left(A^{\dagger }\right)^{\textsf {T}}}
(
A
¯
)
†
=
A
†
¯
{\displaystyle \left({\overline {A}}\right)^{\dagger }={\overline {A^{\dagger }}}}
(
A
∗
)
†
=
(
A
†
)
∗
{\displaystyle \left(A^{*}\right)^{\dagger }=\left(A^{\dagger }\right)^{*}}
矩陣
A
{\displaystyle A}
A
†
{\displaystyle A^{\dagger }}
(
α
A
)
†
=
α
−
1
A
†
{\displaystyle \left(\alpha A\right)^{\dagger }=\alpha ^{-1}A^{\dagger }}
α
≠
0
{\displaystyle \alpha \neq 0}
下面的恆等式可以用來判定部分涉及偽逆的子表達式的正確性：
A
=
A
A
∗
A
†
∗
=
A
†
∗
A
∗
A
{\displaystyle A={}A{}A^{*}{}A^{\dagger *}{}={}A^{\dagger *}{}A^{*}{}A}
A
†
{\displaystyle A^{\dagger }}
A
{\displaystyle A}
A
†
=
A
†
A
†
∗
A
∗
=
A
∗
A
†
∗
A
†
{\displaystyle A^{\dagger }={}A^{\dagger }{}A^{\dagger *}{}A^{*}{}={}A^{*}{}A^{\dagger *}{}A^{\dagger }}
A
∗
{\displaystyle A^{*}}
A
{\displaystyle A}
A
∗
=
A
∗
A
A
+
=
A
+
A
A
∗
.
{\displaystyle A^{*}={}A^{*}{}A{}A^{+}{}={}A^{+}{}A{}A^{*}.}
偽逆的計算可以簡化為其在埃爾米特 情況下的構造，這可以通過等價關係實現：
A
+
=
(
A
∗
A
)
+
A
∗
,
{\displaystyle A^{+}=\left(A^{*}A\right)^{+}A^{*},}
A
+
=
A
∗
(
A
A
∗
)
+
,
{\displaystyle A^{+}=A^{*}\left(AA^{*}\right)^{+},}
A
∗
A
{\displaystyle A^{*}A}
A
A
∗
{\displaystyle AA^{*}}
令
A
∈
k
m
×
n
,
B
∈
k
n
×
p
{\displaystyle A\in \mathbb {k} ^{m\times n},\ B\in \mathbb {k} ^{n\times p}}
[ 6]
(
A
B
)
†
=
B
†
A
†
{\displaystyle (AB)^{\dagger }=B^{\dagger }A^{\dagger }}
A
†
A
B
B
∗
A
∗
=
B
B
∗
A
∗
,
B
B
†
A
∗
A
B
=
A
∗
A
B
.
{\textstyle {\begin{aligned}A^{\dagger }ABB^{*}A^{*}&=BB^{*}A^{*},\\BB^{\dagger }A^{*}AB&=A^{*}AB.\end{aligned}}}
(
A
†
A
B
B
∗
)
∗
=
A
†
A
B
B
∗
,
(
A
∗
A
B
B
†
)
∗
=
A
∗
A
B
B
†
.
{\displaystyle {\begin{aligned}\left(A^{\dagger }ABB^{*}\right)^{*}&=A^{\dagger }ABB^{*},\\\left(A^{*}ABB^{\dagger }\right)^{*}&=A^{*}ABB^{\dagger }.\end{aligned}}}
A
†
A
B
B
∗
A
∗
A
B
B
†
=
B
B
∗
A
∗
A
{\displaystyle A^{\dagger }ABB^{*}A^{*}ABB^{\dagger }=BB^{*}A^{*}A}
A
†
A
B
=
B
(
A
B
)
†
A
B
,
B
B
†
A
∗
=
A
∗
A
B
(
A
B
)
†
.
{\displaystyle {\begin{aligned}A^{\dagger }AB&=B(AB)^{\dagger }AB,\\BB^{\dagger }A^{*}&=A^{*}AB(AB)^{\dagger }.\end{aligned}}}
下方列出了
(
A
B
)
+
=
B
+
A
+
{\displaystyle (AB)^{+}=B^{+}A^{+}}
A
{\displaystyle A}
A
∗
A
=
A
†
A
=
I
n
{\displaystyle A^{*}A=A^{\dagger }A=I_{n}}
B
{\displaystyle B}
B
B
∗
=
B
B
†
=
I
n
{\displaystyle BB^{*}=BB^{\dagger }=I_{n}}
A
{\displaystyle A}
A
†
A
=
I
{\displaystyle A^{\dagger }A=I}
B
{\displaystyle B}
B
B
†
=
I
{\displaystyle BB^{\dagger }=I}
B
=
A
∗
{\displaystyle B=A^{*}}
B
=
A
†
{\displaystyle B=A^{\dagger }}
下方列出了
(
A
B
)
†
=
B
†
A
†
{\displaystyle (AB)^{\dagger }=B^{\dagger }A^{\dagger }}
(
A
†
A
)
(
B
B
†
)
=
(
B
B
†
)
(
A
†
A
)
{\displaystyle (A^{\dagger }A)(BB^{\dagger })=(BB^{\dagger })(A^{\dagger }A)}
由最後一個充分條件得出等式：
(
A
A
∗
)
+
=
A
+
∗
A
+
,
(
A
∗
A
)
+
=
A
+
A
+
∗
.
{\displaystyle {\begin{aligned}\left(AA^{*}\right)^{+}&=A^{+*}A^{+},\\\left(A^{*}A\right)^{+}&=A^{+}A^{+*}.\end{aligned}}}
(
A
B
)
†
=
B
†
A
†
{\displaystyle (AB)^{\dagger }=B^{\dagger }A^{\dagger }}
(
(
1
1
0
0
)
(
0
0
1
1
)
)
+
=
(
1
1
0
0
)
+
=
(
1
2
0
1
2
0
)
≠
(
1
4
0
1
4
0
)
=
(
0
1
2
0
1
2
)
(
1
2
0
1
2
0
)
=
(
0
0
1
1
)
+
(
1
1
0
0
)
+
{\displaystyle {\Biggl (}{\begin{pmatrix}1&1\\0&0\end{pmatrix}}{\begin{pmatrix}0&0\\1&1\end{pmatrix}}{\Biggr )}^{+}={\begin{pmatrix}1&1\\0&0\end{pmatrix}}^{+}={\begin{pmatrix}{\tfrac {1}{2}}&0\\{\tfrac {1}{2}}&0\end{pmatrix}}\quad \neq \quad {\begin{pmatrix}{\tfrac {1}{4}}&0\\{\tfrac {1}{4}}&0\end{pmatrix}}={\begin{pmatrix}0&{\tfrac {1}{2}}\\0&{\tfrac {1}{2}}\end{pmatrix}}{\begin{pmatrix}{\tfrac {1}{2}}&0\\{\tfrac {1}{2}}&0\end{pmatrix}}={\begin{pmatrix}0&0\\1&1\end{pmatrix}}^{+}{\begin{pmatrix}1&1\\0&0\end{pmatrix}}^{+}}
P
=
A
A
†
{\displaystyle P=AA^{\dagger }}
Q
=
A
†
A
{\displaystyle Q=A^{\dagger }A}
P
=
P
∗
{\displaystyle P=P^{*}}
Q
=
Q
∗
{\displaystyle Q=Q^{*}}
P
2
=
P
{\displaystyle P^{2}=P}
Q
2
=
Q
{\displaystyle Q^{2}=Q}
P
A
=
A
Q
=
A
{\displaystyle PA=AQ=A}
A
†
P
=
Q
A
†
=
A
†
{\displaystyle A^{\dagger }P=QA^{\dagger }=A^{\dagger }}
P
{\displaystyle P}
A
{\displaystyle A}
A
∗
{\displaystyle A^{*}}
核 的正交補空間）。
Q
{\displaystyle Q}
A
∗
{\displaystyle A^{*}}
A
{\displaystyle A}
(
I
−
Q
)
=
(
I
−
A
†
A
)
{\displaystyle (I-Q)=\left(I-A^{\dagger }A\right)}
A
{\displaystyle A}
(
I
−
P
)
=
(
I
−
A
A
†
)
{\displaystyle (I-P)=\left(I-AA^{\dagger }\right)}
A
∗
{\displaystyle A^{*}}
[ 4] 最後兩條性質隱含了下列等式：
A
(
I
−
A
†
A
)
=
(
I
−
A
A
†
)
A
=
0
{\displaystyle A\,\ \left(I-A^{\dagger }A\right)=\left(I-AA^{\dagger }\right)A\ \ =0}
A
∗
(
I
−
A
A
†
)
=
(
I
−
A
†
A
)
A
∗
=
0
{\displaystyle A^{*}\left(I-AA^{\dagger }\right)=\left(I-A^{\dagger }A\right)A^{*}=0}
如果
A
∈
k
n
×
n
{\displaystyle A\in \mathbb {k} ^{n\times n}}
B
∈
k
m
×
n
{\displaystyle B\in \mathbb {k} ^{m\times n}}
[ 7]
A
(
B
A
)
†
=
(
B
A
)
†
{\displaystyle A(BA)^{\dagger }=(BA)^{\dagger }}
C
=
B
A
{\displaystyle C=BA}
D
=
A
(
B
A
)
†
{\displaystyle D=A(BA)^{\dagger }}
A
{\displaystyle A}
D
{\displaystyle D}
C
{\displaystyle C}
A
∈
k
n
×
n
{\displaystyle A\in \mathbb {k} ^{n\times n}}
B
∈
k
n
×
m
{\displaystyle B\in \mathbb {k} ^{n\times m}}
(
A
B
)
†
A
=
(
A
B
)
†
{\displaystyle (AB)^{\dagger }A=(AB)^{\dagger }}
當
A
{\displaystyle A}
A
†
=
A
{\displaystyle A^{\dagger }=A}
如果我們把矩陣看作是一個在數域
k
{\displaystyle \mathbb {k} }
A
:
k
n
→
k
m
{\displaystyle A:\mathbb {k} ^{n}\to \mathbb {k} ^{m}}
A
†
:
k
m
→
k
n
{\displaystyle A^{\dagger }:\mathbb {k} ^{m}\to \mathbb {k} ^{n}}
⊕
{\displaystyle \oplus }
⊥
{\displaystyle \perp }
ker
{\displaystyle \ker }
ran
{\displaystyle \operatorname {ran} }
k
n
=
(
ker
A
)
⊥
⊕
ker
A
{\displaystyle \mathbb {k} ^{n}=\left(\ker A\right)^{\perp }\oplus \ker A}
k
m
=
ran
A
⊕
(
ran
A
)
⊥
{\displaystyle \mathbb {k} ^{m}=\operatorname {ran} A\oplus \left(\operatorname {ran} A\right)^{\perp }}
A
:
(
ker
A
)
⊥
→
ran
A
{\displaystyle A:\left(\ker A\right)^{\perp }\to \operatorname {ran} A}
A
†
{\displaystyle A^{\dagger }}
ran
A
{\displaystyle \operatorname {ran} A}
(
ran
A
)
⊥
{\displaystyle \left(\operatorname {ran} A\right)^{\perp }}
換而言之，對於給定的
b
∈
k
m
{\displaystyle b\in \mathbb {k} ^{m}}
A
†
b
{\displaystyle A^{\dagger }b}
b
{\displaystyle b}
A
{\displaystyle A}
p
(
b
)
{\displaystyle p(b)}
A
−
1
(
{
p
(
b
)
}
)
{\displaystyle A^{-1}(\{p(b)\})}
k
n
{\displaystyle \mathbb {k} ^{n}}
A
{\displaystyle A}
p
(
b
)
{\displaystyle p(b)}
k
n
{\displaystyle \mathbb {k} ^{n}}
A
{\displaystyle A}
A
+
b
{\displaystyle A^{+}b}
A
−
1
(
{
p
(
b
)
}
)
{\displaystyle A^{-1}(\{p(b)\})}
A
{\displaystyle A}
以上描述與線性系統的最小範數解密切相關。
ker
(
A
+
)
=
ker
(
A
∗
)
ran
(
A
+
)
=
ran
(
A
∗
)
{\displaystyle {\begin{aligned}\ker \left(A^{+}\right)&=\ker \left(A^{*}\right)\\\operatorname {ran} \left(A^{+}\right)&=\operatorname {ran} \left(A^{*}\right)\end{aligned}}}
偽逆可以由極限定義：
A
†
=
lim
δ
↘
0
(
A
∗
A
+
δ
I
)
−
1
A
∗
=
lim
δ
↘
0
A
∗
(
A
A
∗
+
δ
I
)
−
1
{\displaystyle A^{\dagger }=\lim _{\delta \searrow 0}\left(A^{*}A+\delta I\right)^{-1}A^{*}=\lim _{\delta \searrow 0}A^{*}\left(AA^{*}+\delta I\right)^{-1}}
吉洪諾夫正則化 ）。當
(
A
A
∗
)
−
1
{\displaystyle \left(AA^{*}\right)^{-1}}
(
A
∗
A
)
−
1
{\displaystyle \left(A^{*}A\right)^{-1}}
[ 4] :263
與一般的矩陣求逆不同，求偽逆的過程並不連續：如果序列
(
A
n
)
{\displaystyle \left(A_{n}\right)}
A
{\displaystyle A}
(
A
n
)
†
{\displaystyle (A_{n})^{\dagger }}
A
†
{\displaystyle A^{\dagger }}
A
n
{\displaystyle A_{n}}
A
{\displaystyle A}
(
A
n
)
†
{\displaystyle (A_{n})^{\dagger }}
A
†
{\displaystyle A^{\dagger }}
[ 8]
實值偽逆矩陣的導數，該矩陣在某點
x
{\displaystyle x}
[ 9]
d
d
x
A
†
(
x
)
=
−
A
†
(
d
d
x
A
)
A
†
+
A
†
A
†
T
(
d
d
x
A
T
)
(
I
−
A
A
†
)
+
(
I
−
A
†
A
)
(
d
d
x
A
T
)
A
†
T
A
†
{\displaystyle {\frac {\mathrm {d} }{\mathrm {d} x}}A^{\dagger }(x)=-A^{\dagger }\left({\frac {\mathrm {d} }{\mathrm {d} x}}A\right)A^{\dagger }~+~A^{\dagger }A^{\dagger {\textsf {T}}}\left({\frac {\mathrm {d} }{\mathrm {d} x}}A^{\textsf {T}}\right)\left(I-AA^{\dagger }\right)~+~\left(I-A^{\dagger }A\right)\left({\frac {\text{d}}{{\text{d}}x}}A^{\textsf {T}}\right)A^{\dagger {\textsf {T}}}A^{\dagger }}
3rd. Baltimore: Johns Hopkins. 1996: 257–258. ISBN 978-0-8018-5414-9.
.
