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]
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{\displaystyle ({\boldsymbol {A}}^{H})^{\dagger }=({\boldsymbol {A}}^{\dagger })^{H}}
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{\displaystyle {\boldsymbol {A}}^{\dagger }{\boldsymbol {A}}{\boldsymbol {A}}^{H}={\boldsymbol {A}}^{H}{\boldsymbol {A}}{\boldsymbol {A}}^{\dagger }={\boldsymbol {A}}^{H}}
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{\displaystyle {\boldsymbol {A}}{\boldsymbol {A}}^{H}({\boldsymbol {A}}^{H})^{\dagger }=({\boldsymbol {A}}^{H})^{\dagger }{\boldsymbol {A}}^{H}{\boldsymbol {A}}={\boldsymbol {A}}}
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{\displaystyle {\boldsymbol {A}}^{\dagger }{\boldsymbol {A}}}
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{\displaystyle {\boldsymbol {A}}{\boldsymbol {A}}^{\dagger }}
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{\displaystyle ({\boldsymbol {I}}-{\boldsymbol {A}}^{\dagger }{\boldsymbol {A}})}
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{\displaystyle ({\boldsymbol {I}}-{\boldsymbol {A}}^{\dagger }{\boldsymbol {A}})}
偽逆存在且唯一：對於任何矩陣
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{\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。
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{\displaystyle A^{\dagger }}
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{\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：
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{\displaystyle \left(A^{\dagger }\right)^{\dagger }=A}
[ 5] :245
偽轉置與轉置、複共軛和共軛轉置可以交換：[ 5] :245
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{\displaystyle \left(A^{\textsf {T}}\right)^{\dagger }=\left(A^{\dagger }\right)^{\textsf {T}}}
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{\displaystyle \left({\overline {A}}\right)^{\dagger }={\overline {A^{\dagger }}}}
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{\displaystyle \left(A^{*}\right)^{\dagger }=\left(A^{\dagger }\right)^{*}}
矩陣
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{\displaystyle A}
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{\displaystyle A^{\dagger }}
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{\displaystyle \left(\alpha A\right)^{\dagger }=\alpha ^{-1}A^{\dagger }}
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{\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：
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{\displaystyle A={}A{}A^{*}{}A^{\dagger *}{}={}A^{\dagger *}{}A^{*}{}A}
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{\displaystyle A^{\dagger }}
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{\displaystyle A}
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{\displaystyle A^{\dagger }={}A^{\dagger }{}A^{\dagger *}{}A^{*}{}={}A^{*}{}A^{\dagger *}{}A^{\dagger }}
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{\displaystyle A^{*}}
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{\displaystyle A}
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{\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：
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{\displaystyle A^{+}=\left(A^{*}A\right)^{+}A^{*},}
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{\displaystyle A^{+}=A^{*}\left(AA^{*}\right)^{+},}
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{\displaystyle A^{*}A}
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{\displaystyle AA^{*}}
令
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{\displaystyle A\in \mathbb {k} ^{m\times n},\ B\in \mathbb {k} ^{n\times p}}
[ 6]
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{\displaystyle (AB)^{\dagger }=B^{\dagger }A^{\dagger }}
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{\textstyle {\begin{aligned}A^{\dagger }ABB^{*}A^{*}&=BB^{*}A^{*},\\BB^{\dagger }A^{*}AB&=A^{*}AB.\end{aligned}}}
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{\displaystyle {\begin{aligned}\left(A^{\dagger }ABB^{*}\right)^{*}&=A^{\dagger }ABB^{*},\\\left(A^{*}ABB^{\dagger }\right)^{*}&=A^{*}ABB^{\dagger }.\end{aligned}}}
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{\displaystyle A^{\dagger }ABB^{*}A^{*}ABB^{\dagger }=BB^{*}A^{*}A}
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{\displaystyle {\begin{aligned}A^{\dagger }AB&=B(AB)^{\dagger }AB,\\BB^{\dagger }A^{*}&=A^{*}AB(AB)^{\dagger }.\end{aligned}}}
下方列出了
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{\displaystyle (AB)^{+}=B^{+}A^{+}}
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{\displaystyle A}
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{\displaystyle A^{*}A=A^{\dagger }A=I_{n}}
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{\displaystyle B}
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{\displaystyle BB^{*}=BB^{\dagger }=I_{n}}
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{\displaystyle A}
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{\displaystyle A^{\dagger }A=I}
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{\displaystyle BB^{\dagger }=I}
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{\displaystyle B=A^{*}}
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{\displaystyle B=A^{\dagger }}
下方列出了
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{\displaystyle (AB)^{\dagger }=B^{\dagger }A^{\dagger }}
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{\displaystyle (A^{\dagger }A)(BB^{\dagger })=(BB^{\dagger })(A^{\dagger }A)}
由最後一個充分條件得出等式：
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{\displaystyle {\begin{aligned}\left(AA^{*}\right)^{+}&=A^{+*}A^{+},\\\left(A^{*}A\right)^{+}&=A^{+}A^{+*}.\end{aligned}}}
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{\displaystyle (AB)^{\dagger }=B^{\dagger }A^{\dagger }}
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{\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}}^{+}}
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{\displaystyle P=AA^{\dagger }}
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{\displaystyle Q=A^{\dagger }A}
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{\displaystyle P=P^{*}}
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{\displaystyle Q=Q^{*}}
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{\displaystyle P^{2}=P}
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{\displaystyle Q^{2}=Q}
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{\displaystyle PA=AQ=A}
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{\displaystyle A^{\dagger }P=QA^{\dagger }=A^{\dagger }}
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{\displaystyle P}
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{\displaystyle A}
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{\displaystyle A^{*}}
核 的正交補餘空間）。
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{\displaystyle A^{*}}
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{\displaystyle (I-Q)=\left(I-A^{\dagger }A\right)}
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{\displaystyle (I-P)=\left(I-AA^{\dagger }\right)}
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{\displaystyle A^{*}}
[ 4] 最後兩條性質隱含了下列等式：
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{\displaystyle A\,\ \left(I-A^{\dagger }A\right)=\left(I-AA^{\dagger }\right)A\ \ =0}
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{\displaystyle A^{*}\left(I-AA^{\dagger }\right)=\left(I-A^{\dagger }A\right)A^{*}=0}
如果
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{\displaystyle A\in \mathbb {k} ^{n\times n}}
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{\displaystyle B\in \mathbb {k} ^{m\times n}}
[ 7]
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{\displaystyle A(BA)^{\dagger }=(BA)^{\dagger }}
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{\displaystyle C=BA}
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{\displaystyle D=A(BA)^{\dagger }}
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{\displaystyle A\in \mathbb {k} ^{n\times n}}
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{\displaystyle B\in \mathbb {k} ^{n\times m}}
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{\displaystyle (AB)^{\dagger }A=(AB)^{\dagger }}
當
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{\displaystyle A}
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{\displaystyle A^{\dagger }=A}
如果我們把矩陣看作是一個在數體
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{\displaystyle \mathbb {k} }
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{\displaystyle A:\mathbb {k} ^{n}\to \mathbb {k} ^{m}}
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{\displaystyle A^{\dagger }:\mathbb {k} ^{m}\to \mathbb {k} ^{n}}
⊕
{\displaystyle \oplus }
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{\displaystyle \perp }
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{\displaystyle \ker }
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{\displaystyle \operatorname {ran} }
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{\displaystyle \mathbb {k} ^{n}=\left(\ker A\right)^{\perp }\oplus \ker A}
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{\displaystyle \mathbb {k} ^{m}=\operatorname {ran} A\oplus \left(\operatorname {ran} A\right)^{\perp }}
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{\displaystyle A:\left(\ker A\right)^{\perp }\to \operatorname {ran} A}
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{\displaystyle A^{\dagger }}
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{\displaystyle \operatorname {ran} A}
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{\displaystyle \left(\operatorname {ran} A\right)^{\perp }}
換而言之，對於給定的
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{\displaystyle b\in \mathbb {k} ^{m}}
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{\displaystyle A^{\dagger }b}
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{\displaystyle b}
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{\displaystyle A}
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{\displaystyle p(b)}
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{\displaystyle A^{-1}(\{p(b)\})}
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{\displaystyle \mathbb {k} ^{n}}
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{\displaystyle A}
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{\displaystyle p(b)}
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{\displaystyle \mathbb {k} ^{n}}
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{\displaystyle A}
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{\displaystyle A^{+}b}
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{\displaystyle A^{-1}(\{p(b)\})}
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{\displaystyle A}
以上描述與線性系統的最小範數解密切相關。
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{\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：
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{\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
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{\displaystyle \left(AA^{*}\right)^{-1}}
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{\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
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{\displaystyle \left(A_{n}\right)}
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{\displaystyle A}
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{\displaystyle (A_{n})^{\dagger }}
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{\displaystyle A^{\dagger }}
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{\displaystyle A_{n}}
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{\displaystyle A}
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{\displaystyle (A_{n})^{\dagger }}
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{\displaystyle A^{\dagger }}
[ 8]
實值偽逆矩陣的導數，該矩陣在某點
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{\displaystyle x}
[ 9]
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{\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.
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