Abstract
In this paper, we introduce a generalized core inverse (called the C-S inverse) and give some properties and characterizations of the inverse. By applying the C-S inverse, we introduce a binary relation (denoted “ \({\mathop \le \limits ^{\textcircled {S}}}\) ”) and a partial order (called the C-S partial order and denoted “ \({\mathop \le \limits ^{\textcircled {{c\!s}}}}\) ”), study their properties and consider the relationship between them and some classical partial orders. At last, by applying the C-S inverse, we give some characterizations of the EP matrix and i-EP matrix.
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1 Introduction
In this paper, \({\mathbb {C}}_{m, n}\) denotes the set of \(m\times n\) matrices with complex entries; \(A^{*}\), \({{\mathcal {R}}}\left( {A} \right) \), \(\text {rk}\left( A \right) \) and \(A^{\dagger }\) denote the conjugate transpose, range space, rank and Moore–Penrose of A, respectively. It is widely known that \(X= A^{\dagger }\) is the unique solution of \( AXA = A\), \(XAX=X\), \((AX)^{*}=AX \) and \( (XA)^{*}=XA \), [4, 24]. And denote
The smallest positive integer k satisfying \(\text {rk}\left( A^{k+1} \right) =\text { rk}\left( A^{k} \right) \) is called the index of \(A\in {\mathbb {C}}_{n,n}\) and is denoted by \({\hbox { Ind}}(A)\). The Drazin inverse of \(A\in {\mathbb {C}}_{n,n}\) with \({\hbox {Ind}}(A)=k\) is denoted as \(A^{D}\), which is the unique solution of \( AXA^{k}=A^{k}\), \(XAX =X\) and \(AX=XA\) [4, 24]. In general, when \(k=1\), it is called the group inverse of A, and is denoted as \(X=A^{\#}\). Therefore, we can also consider the Drazin inverse as one generalization of the group inverse.
Let \(A\in {\mathbb {C}}_{n, n}\) with \({\hbox { Ind}}(A)=k\), and denote \(A^{N}=A-AA^{D}A\). Greville [9] proves that
is the unique solution of
in which \(l\ge k\) and \(A^{N}=A-AA^{D}A\). In [1], the solution is denoted by \(A^\mathrm{(S)}\). We call it the G-S inverse of A. Obviously, when \(k=1\), \(A^{\mathrm{{(S)}}}=A^\#\). Therefore, the G-S inverse is also a generalization of the group inverse.
In [1], we see that the G-S inverse is a kind of S-inverse. The S-inverse is one special generalized inverse: “ for every \(\lambda \) and every vector x, x is a \(\lambda \)-vector of A of grade p if and only if it is a \(\lambda ^\dag \)-vector of X of grade p,” [1]. And A is one S-inverse of X, too. When \(k=1\), since \(A^\#=A^{\text { {(s)}}}\), we see that group inverse is also an S-inverse. When the index of A is 1, \(A^\#\) is the unique S-inverse of A in \(A\{1\} \cup A\{2\}\), in which \(A \{ 1 \} = \{ X: AXA=A \}\) and \(A\{2\}=\{X: XAX=X\}\), [1]. It is worth noting that Drazin inverse is a generalized group inverse, but it is not the S-inverse of A. It is one \(S'\)-inverse of A, [1]. There are also some interesting results of the G-S inverse:
-
(1)
we call it the inverse, but the G-S inverse is neither \(\{1\}\)- nor \(\{2\}\)- inverse;
-
(2)
\((A^{\text { {(S)}}})^{\text {{(S)}}}=A\);
-
(3)
\(\text {rk} (A) =\text {rk} (A^{\text { {(S)}}})\);
-
(4)
\({\text {Ind}}(A)={\text {Ind}}(A^{\text { {(S)}}})\).
More properties of the G-S inverse can be found in [1, 9].
In [2], Baksalary and Trenkler define the core inverse of a square matrix: let \(A\in {\mathbb {C}}^{n,n}\) with \({\text {Ind}}(A)=1\), then the solution of
is unique, and is denoted as . Subsequently, the definition of core inverse is extended to a larger class of matrices. For example, let \(A\in {\mathbb {C}}_{n,n}\) with \({\text {Ind}}(A)=k\), Prasad and Mohana introduce the core-EP inverse \({A^{\textcircled {\dag }}}\) of A, and prove \({A^{\textcircled {\dag }}} =A^k\left( {\left( {A^*} \right) ^kA^{k + 1}} \right) ^ \dag \left( {A^*} \right) ^k\), [21]; Baksalary and Trenkler introduce the BT inverse \(A^{\diamond }\) of A and prove \(A^{\diamond }=(AP_{A})^{\dag }\), [3]; Malik and Thome introduce the DMP inverse \(A^{D,{\dagger }}\), and prove \(A^{D,{\dagger }}=A^D AA^\dag \), [15]; Mehdipour and Salemi introduce the CMP inverse \(A^{C,{\dagger }}\) of A, and prove \(A^{C,{\dagger }}=A^\dagger AA^D AA^\dagger \), [16]. It is easy to check that, when \(k=1\), \({A^{\textcircled {\#}} =A^{\textcircled {\dag }} =A^{\diamond } =A^{{D,\dagger }} =A^{C,{\dagger }}}\), that is, the above four generalized inverse are all generalized core inverses. Furthermore, Wang and Chen [26] introduce a generalized group inverse: the solution of \(AX^{2} = X\) and \({AX=A^{\textcircled {\dag }}}A\) is unique, which is called the WG inverse of A and is denoted as \({X=A^{\textcircled {W}}}\).
A variety of generalized inverses have been established successively, arousing the interest of scholars in the study of generalized inverses including their properties, algorithms, applications, etc. For example, generalized inverses are one of the tools for characterizing special matrices. let \(A \in {\mathbb {C}}_{n, n}\), then it is EP if and only if \(AA^\dag =A^\dag A\), [24]; it is i-EP if and only if \(A^k(A^k)^\dag =(A^k)^\dag A^k\), [14, 23]; it is a WG matrix if and only if \({AA^{\textcircled {W}}=A^{\textcircled {W}} A}\), [27], where k is the index of A. Generalized inverses are also one of the main tools for constructing and characterizing partial orders. For example,
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(1)
\(A\mathop \le \limits ^{-}B\): \(A, B\in {\mathbb {C}}_{m, n}\), \(\text {rk}(B-A)=\text {rk}(B)-\text {rk}(A)\), [10];
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(2)
\(A\mathop \le \limits ^{*}B\): \(A, B\in {\mathbb {C}}_{m, n}\), \(AA^{*}=BA^{*}\) and \(A^{*}A=A^{*}B\), [7].
More results about generalized inverses and their applications can be seen in [1, 4, 12, 13, 18, 19, 22, 24, 28].
In this paper, matrix decompositions are mainly used to study the properties and characterizations of the new inverse (called it the C-S inverse). A new binary relation is introduced by C-S inverse, and its relationship with the star partial order is discussed. Based on the binary relation, a new partial order (C-S partial order) is introduced. The characterizations of the partial order and the star partial order are given by using the core-EP decomposition. And it is proved that the C-S partial order is a special kind of star partial order. Finally, the C-S partial order is applied to give characterizations of special matrices such as EP matrix.
2 Preliminaries
In this section, we present some preliminary results.
Theorem 1
([25], Core-EP decomposition) Let \(A\in {\mathbb {C}}_{n, n}\), \({\text {Ind}}(A)=k\) and \({rk}\left( A^k\right) =t\). Then there exist \( {{A}}_{1}\) and \({{A}}_{2}\), which satisfy
where \( {{A}}_{1}\in {\mathbb {C}}^{{\texttt {CM}}}_{n}\), \({ {{A}}_{2}^{k}}=0\) and \({ {{A}}_{1}^{*}} {{A}}_{2}= {{A}}_{2} {{A}}_{1}=0\). Furthermore, there exists a unitary matrix U satisfying
where \(T\in {\mathbb {C}}_{t, t}\) is nonsingular and N is nilpotent.
By applying the above decomposition, Ferreyra, Levis and Thome [8] get that
where
Of course, the core-EP decomposition can also be used to characterize generalized inverses. For example, characterizations of \(A^{\dag }\), \(A^{D}\) and \(A^{\textcircled {\dag }}\) are given in [8, 25]:
where \(\triangle =\left( TT^{*}+S\left( I_{n-t}-N^\dag N\right) S^{*}\right) ^{-1}\). Furthermore, by applying (1) and (7), we get a characterization of the G-S inverse:
Theorem 2
([8, 25]) Let \(A\in {\mathbb {C}}_{n, n}\) and \({\text {Ind}}(A)=k\), then
Theorem 3
([28], EP-nilpotent decomposition) Let \(A\in {\mathbb {C}}_{n, n}\), \({\text {Ind}}(A)=k\) and \({rk}\left( A^k\right) =t\). Then, there exist \({\widehat{A}}_{1}\) and \({{\widehat{A}}}_{2}\), which satisfy
where \({\widehat{A}}_{1}\in {\mathbb {C}}^{{\texttt {EP}}}_{n}\), \({{\widehat{A}}^{k+1}}_{2}=0\) and \({\widehat{A}}_{2}{\widehat{A}}_{1}=0\). Furthermore, there exists a unitary matrix U satisfying
where \(T\in {\mathbb {C}}_{t, t}\) is nonsingular, and N is nilpotent.
Theorem 4
( [20]) Let \(A\in {\mathbb {C}}_{n, n}\) and \(\text {rk}({A})=r\). Then A is EP if and only if there exists a unitary matrix U satisfying
where \(T\in {\mathbb {C}}_{r, r}\) is invertible.
Theorem 5
( [27]) Let \(A\in {\mathbb {C}}_{n,n}\), \(\text {Ind}(A)=k\) and \({rk}\left( A^k\right) =t\). Then, A is i-EP if and only if there exists a unitary matrix U, such that
where \(T\in {\mathbb {C}}_{t, t}\) is nonsingular, and N is nilpotent with \(\text {Ind}(N)=k\).
3 The C-S Inverse
In this section, we will give the definition of the C-S inverse, study its properties, its differences from and its connections with several generalized inverses.
3.1 Definition of the C-S Inverse
Theorem 6
Let \(A\in {\mathbb {C}}_{n, n}\) and \({\text {Ind}}(A)=k\). Then, the solution of
is unique. Furthermore, there exists a unitary matrix U such that
where \(T\in {\mathbb {C}}_{\text {rk}\left( A^k\right) , {rk}\left( A^k\right) }\) is nonsingular, N is nilpotent, and \(A_2=A-AA^{\textcircled {\dag }}A\).
Proof
Let the core-EP decomposition of A be as shown in (2). Substituting (3) and (15) in (14), we have
Therefore, we get that (14) is consistent, and (15) is one of the solutions.
Next, we apply the EP-nilpotent decomposition to study uniqueness of the solution of (14). Let the EP-nilpotent decomposition of A be as in (11). Write
Since \({\widehat{A}}_{2}{\widehat{A}}_{1}=0\) and \({\widehat{A}}_{2}^{k+1}=0\), by applying (11) it is easy to check that
Since \(XA^{k+1}=A^k\), by applying (21), we get
By applying (21) and (22) we get
Since \( A-X={A}^{k}{X}^{k}(A-X)\), by applying (22), we get
It follows that
Therefore, by applying (23), (24) and (25) to (18), we get
Furthermore, denote
in which \(X_{1}={{\widehat{A}}}_{1}^{\dagger }\), \(X_{21}=P_{{\widehat{A}}_1}XF_{{\widehat{A}}_1}\) and \(X_{22}=F_{{\widehat{A}}_1}AF_{{\widehat{A}}_1} \).
Since \({\widehat{A}}_{1}\) is EP, \(F_{{\widehat{A}}_1}{\widehat{A}}_1^\dag =0\), \(X_{2}X_{1}=0\) and
By applying \(X_{22}^k=X_{22}X_{21}=0\), we get
Since \(A^{k}X^{k}\) and \({{\widehat{A}}}_{1}^{k}({{\widehat{A}}}_{1}^{\dagger })^{k}\) are Hermitian, we get that \({\widehat{A}}_{1}^{k}{\sum \limits _{i =0}^{k-1} {X_1^{i}X_{21}X_{22}^{k-1-i}} }\) is Hermitian. For \({\widehat{A}}_{1}^{k}{\sum \limits _{i =0}^{k-1} {X_1^{i}X_{21}X_{22}^{k-1-i}} } \subseteq {\mathcal {R}}\left( {\widehat{A}}_1 \right) \), \(\left( {\widehat{A}}_{1}^{k}{\sum \limits _{i =0}^{k-1} {X_1^{i}X_{21}X_{22}^{k-1-i}} }\right) ^*\subseteq {\mathcal {R}}\left( F_{{\widehat{A}}_1} \right) \) and \({\mathcal {R}}\left( {\widehat{A}}_1 \right) \cap {\mathcal {R}}\left( F_{{\widehat{A}}_1} \right) =0\), we get
Since \({\widehat{A}}_1\) is group invertible, by applying \(X_{22}^k=0\) and \( {\mathcal {R}}\left( X_{21}\right) \subseteq {\mathcal {R}}\left( {\widehat{A}}_1 \right) \), we get \(X_{21}=0\), that is,
Since \({\widehat{A}}_1\) and \({\widehat{A}}_2\) are unique, by applying (23), (24), (25) and (29) to (18), we get that solution of (14) is unique. \(\square \)
Definition 1
Let \(A\in {\mathbb {C}}_{n, n}\) with \({\text {Ind}}(A)=k\). The C-S inverse of A is defined as the solution of (14) and is denoted as \(A^{\textcircled {S}}\).
Remark 1
From (15), we see that \(A^{\textcircled {S}}=A^{\textcircled {\#}}\), when \(k=1\). Thus, the C-S inverse is one more generalization of the core inverse.
3.2 Proprieties and Characterizations of C-S Inverse
Let \(A\in {\mathbb {C}}_{n, n}\) with \({\text {Ind}}(A)=k\), the core-EP decomposition of A be as in Theorem 1, and the C-S inverse of A be as in (15). It is obvious that \({rk}({A})= \text {rk}({T}) +\text {rk}({N})\) and \({rk}\left( {A^{\textcircled {S}}}\right) = \textrm{rk}\left( {T^{-1}}\right) +\text {rk}({N})\). Therefore, we have Theorem 7.
Theorem 7
Let \(A\in {\mathbb {C}}_{n, n}\) with \({\text {Ind}}(A)=k\), then
The following example shows that \(A^{\textcircled {S}}\) is not the same as \(A^{\dagger }\), \(A^{D}\), \(A^{\textcircled {\dag }}\), \(A^{\diamond } \), \(A^{D,{\dagger }}\), \(A^{C,{\dagger }}\), \(A^{{\dagger },D}\), \(A^{\textcircled {W}}\) and \(A^{\mathrm{{(S)}}}\).
Example 1
Let \(A ={ \left[ \begin{matrix} 1 &{}0 &{}1 &{}0 \\ 0 &{}1 &{}0 &{}1 \\ 0 &{}0 &{}0 &{}1 \\ 0 &{}0 &{}0 &{}0 \end{matrix} \right] }\). Then,
and
Theorem 8
Let \(A\in {\mathbb {C}}_{n, n}\) and \({\text {Ind}}(A)=k\). Then, the following conditions are equivalent:
-
(1)
\(X =A^{\textcircled {S}}\);
-
(2)
\(A^{k} X^{k} = AA^{\textcircled {\dag }}\), \(A^{k} X^{k+1} = A^{\textcircled {\dag }}\), \(A-X=A^{k} X^{k}(A-X)\);
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(3)
\(A^{k} X^{k} = P_{A^{k}}\), \( A^{k} X^{k+1} = A^{D}P_{A^{k}}\), \( A-X=A^{k} X^{k}(A-X)\);
-
(4)
\( P_{A^{k}} X = A^{\textcircled {\dag }}\), \( F_{A^{k}} (A-X)=0\);
-
(5)
\(F_{A^{k}} A +2 P_{A^{k}} X=A^{\textcircled {\dag }}+X\).
Proof
Let A be of the form (2), then C-S inverse of A is of the form (15).
“(1)\(\Rightarrow \)(2)” By applying (8) and (16), we get \(A^{k} X^{k} = AA^{\textcircled {\dag }}\). By applying (15) and (16), we get \(A^{k} X^{k+1} = U \left[ {\begin{matrix} T^{-1} &{}0 \\ 0 &{}0 \end{matrix}} \right] U^{*}\). It follows from (8) that we get \(A^{k} X^{k+1} = A^{\textcircled {\dag }}\). Furthermore, by applying (15), we get \( A-A^{\textcircled {S}} =U \left[ {\begin{matrix} T-T^{-1} &{}0 \\ 0 &{}0 \end{matrix}} \right] U^{*}\). It follows from applying (16) that we get \(A-X=A^{k} X^{k}(A-X)\).
“(2)\(\Rightarrow \)(1)” Write
in which \(X_{1}\in {\mathbb {C}}_{t, t}\). By applying \(A^{k} X^{k} = AA^{\textcircled {\dag }}\), \(A^{k} X^{k+1} = A^{\textcircled {\dag }}\) and \(A-X=A^{k} X^{k}(A-X)\), we have
It follows that \(X_{1} =T^{-1} \), \(X_{2}=0\), \(X_{3}=0\) and \(X_{4}=N\). Therefore, \(X=A^{\textcircled {S}}\).
“(2)\(\Leftrightarrow \)(3)” By applying (10), we get that (2) and (3) are equivalent.
“(1)\(\Rightarrow \)(4)” By using (13) and (15), we have
“(4)\(\Rightarrow \)(5)” For \( P_{A^{k}} X = A^{\textcircled {\dag }}\) and \( F_{A^{k}} (A-X)=0\), we have \( P_{A^{k}} X + F_{A^{k}} (A-X)= A^{\textcircled {\dag }}\), that is, \(F_{A^{k}} A +2 P_{A^{k}} X=A^{\textcircled {\dag }}+X\).
“(5)\(\Rightarrow \)(1)” Let X be of the form (30). Then,
From \(F_{A^{k}} A +2 P_{A^{k}} X=A^{\textcircled {\dag }}+X\), it follows that \(X_{1} =T^{-1} \), \(X_{2}=0\), \(X_{3}=0\) and \(X_{4}=N\). Therefore, we get \(X=A^{\textcircled {S}}\). \(\square \)
Theorem 9
Let \(A\in {\mathbb {C}}_{n, n}\) and \({\text {Ind}}(A)=k\). Then,
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(1)
\(A^{\textcircled {S}}=0\) \(\Leftrightarrow \) \(A=0\);
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(2)
\(A^{\textcircled {S}}=A\) \(\Leftrightarrow \) \(T^{-1}=T\), \(S=0\);
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(3)
\(A^{\textcircled {S}}=A^{*}\) \(\Leftrightarrow \) T is unitary, \(S=0\), \(N=0\);
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(4)
\(A^{\textcircled {S}}=P_{A}\) \(\Leftrightarrow \) \(T=I_{\textrm{rk}(A^k)}\), \(N=0\).
Proof
Let the core-EP decomposition of A be of the form (2). Then,
(1) Since \(A^{\textcircled {S}}=0\), \(N=0\) and T is null. Therefore, \(A=0\).
(2) Since \(A^{\textcircled {S}}=A\), \( U\left[ \begin{matrix} T^{ - 1} &{}0 \\ 0 &{}N \end{matrix} \right] U^{*} = U\left[ \begin{matrix} T &{}S \\ 0 &{}N \end{matrix} \right] U^{*} \). Therefore, \(T^{-1}=T\) and \(S=0\).
(3) Since \(A^{\textcircled {S}}=A^{*}\), \(U\left[ \begin{matrix} T^{ - 1} &{}0 \\ 0 &{}N \end{matrix} \right] U^{*} = U\left[ \begin{matrix} T^{*} &{}0 \\ S^{*} &{}N^{*} \end{matrix} \right] U^{*}\), that is, \(T^{-1}=T^{*}\), \(S=0\) and \(N=N^{*}\). Therefore, T is unitary, \(S=0\) and N is Hermitian.
Since N is Hermitian, if k is greater than 1, then by applying \(N^k=0\) gives \(N^{k-1} N^*=0\), \(N^{k-1} N^\dag =0\), \(N^{k-1} N^\dag N=0\) and \(N^{k-1}=0\). Therefore, the index of N is not greater than \(k-1\). This contradicts with \({\textrm{Ind}}(A)=k\). So, \(N=0\).
(4) Since \(A^{\textcircled {S}}=P_{A}\), \(A^{\textcircled {S}}=AA^{\dag }\). Furthermore, we get \( U\left[ \begin{matrix} T^{ - 1} &{}0 \\ 0 &{}N \end{matrix} \right] U^{*} = U\left[ \begin{matrix} I_{\text {rk}(A^k)} &{}0 \\ 0 &{}NN^{\dag } \end{matrix} \right] U^{*}\). It follows that \(T=I_{\text {rk}(A^k)}\) and \(N=0\). \(\square \)
Theorem 10
Let \(A\in {\mathbb {C}}_{n, n}\). Then, the following conditions are equivalent:
-
(1)
\(A^{\textcircled {S}}=A^{\dag }\);
-
(2)
\(A^{\textcircled {S}}=A^D\);
-
(3)
A is EP.
Proof
Let \(A\in {\mathbb {C}}_{n, n}\), \({\text {Ind}}(A)=k\) and the core-EP decomposition of A be as shown in (2).
“(1)\(\Rightarrow \)(3)” Applying (6) and (15), we get that \(A^{\textcircled {S}}=A^{\dag }\) is equivalent to \( T^{*}\triangle =T^{-1}\), \(-T^{*}\triangle SN^{\dagger }=0\), \(\left( I_{n-t}-N^\dag N\right) S^{*}\triangle =0\) and \(N^{\dagger }-\left( I_{n-t}-N^\dag N\right) S^{*}\triangle SN^{\dagger }=N\).
Since T and \(\triangle \) are nonsingular, from \(-T^{*}\triangle SN^{\dagger }=0\) and \(\left( I_{n-t}-N^\dag N\right) S^{*}\triangle =0\) we get \(SN^{\dagger }=0\) and \(S\left( I_{n-t}-N^\dag N\right) =0\). Therefore, \(S=0\). Substituting \(N=0\) in \(N^{\dagger }-\left( I_{n-t}-N^\dag N\right) S^{*}\triangle SN^{\dagger }=N\), we get \(N^{\dagger }=N\). Then \(N=0\). Since \(S=0\) and \(N=0\), by Theorem 5 it follows that A is EP.
“(2)\(\Rightarrow \)(3)” Since \( \text {rk} \left( A^D\right) = \textrm{rk} \left( T\right) \), from Theorem 7 we get that \( \text {rk}\left( {A^{\textcircled {S}}}\right) =\text {rk} \left( A^D\right) \) is equivalent to \(N=0\). Substituting \(N=0\) in (7), we get
If \(A^{\textcircled {S}}=A^D\), applying (15) and (31) we get \(S=0\). By Theorem 5, it follows that A is EP.
“(3)\(\Rightarrow \)(1) and (2)” When A is EP, the two equalities in (1) and (2) can trivially be derived applying (12). \(\square \)
4 Applications of the C-S Inverse
In this section, we discuss several applications of the C-S inverse.
4.1 A Binary Relation Based on the C-S Inverse
In this subsection, based on the C-S inverse, we introduce a binary relation:
Theorem 11
Let \(A, B\in {\mathbb {C}}_{n, n}\). Then, \(A\mathop \le \limits ^{\textcircled {S}}B\) if and only if there exists a unitary matrix U such that
where T is invertible, N is nilpotent, and \(N\mathop \le \limits ^{*}B_{4}\).
Proof
“\(\Rightarrow \)” Let the core-EP decomposition of A be as shown in (2), and \(A^{\textcircled {S}}\) be of the form (15). And write
Applying \(A\left( A^{\textcircled {S}}\right) ^{*}=B\left( A^{\textcircled {S}}\right) ^{*}\),
gives
From (36), we have
Since \((A^{\textcircled {S}})^{*}A=(A^{\textcircled {S}})^{*}B\), by applying (37) we get
It follows that
Since T is nonsingular, by applying (39), we have
By applying (37) and (40) to (34) we get
Therefore, by applying (2) and (41), we ge (33). Furthermore, by applying (35) and (38), we get \(N\mathop \le \limits ^{*}B_{4}\).
“\(\Leftarrow \)” Let A and B be as in (33), then
It follows from \(N\mathop \le \limits ^{*}B_{4}\) that
Therefore, \(A\mathop \le \limits ^{\textcircled {S}}B\). \(\square \)
Theorem 12
The binary relation “ \(\mathop \le \limits ^{\textcircled {S}}\) ” is anti-symmetric.
Proof
Let A and B be as in (33) and \(A\mathop \le \limits ^{\textcircled {S}}B\). Then,
Since \(B\mathop \le \limits ^{\textcircled {S}}A\), we have
It follows from (42) that \(B_{4}\mathop \le \limits ^{*}N\). And for \(N\mathop \le \limits ^{*}B_{4}\), we have \(N=B_{4}\). Therefore, by applying \(A\mathop \le \limits ^{\textcircled {S}}B\) and \(B\mathop \le \limits ^{\textcircled {S}}A\), we get \(A=B\), that is, the binary relation “ \( \mathop \le \limits ^{\textcircled {S}} \) ” is anti-symmetric. \(\square \)
Remark 2
The binary relation “ \( \mathop \le \limits ^{\textcircled {S}}\) ” is not transitive as the following example shows.
Example 2
Let
It is easy to get
and
Therefore, we have \(A\mathop \le \limits ^{\textcircled {S}}B\) and \(B\mathop \le \limits ^{\textcircled {S}}C\). Furthermore, since
we get that \(A\mathop \le \limits ^{\textcircled {S}}C\) is not true.
Corollary 1
Let \(A\mathop \le \limits ^{\textcircled {S}}B\). Then, \(AA^{\textcircled {\dag }}A=AA^{\textcircled {\dag }}B\).
Proof
Let A and B be as in (33). Then,
Therefore, \(AA^{\textcircled {\dag }}A=AA^{\textcircled {\dag }}B\). \(\square \)
Corollary 2
Let \(A\mathop \le \limits ^{\textcircled {S}}B\). Then, \(\text {rk}\left( {A} \right) \le \text {rk}\left( {B} \right) \).
Proof
Let A and B be as in (33). Then,
For \(A\mathop \le \limits ^{\textcircled {S}}B\), by applying Theorem 11, we have \(N\mathop \le \limits ^{*}B_{4}\). Therefore, \(\text {rk}\left( {N} \right) \le \text {rk}\left( {B}_{4} \right) \). It follows from (43) that \(\text {rk}\left( {A} \right) \le \text {rk}\left( {B} \right) \). \(\square \)
Theorem 13
Let \(A\mathop \le \limits ^{*} B\), and A be as in Theorem 1. Then,
where \( N^{*}B_3=0\), \(N\mathop \le \limits ^{*} B_{4}\) and \(NS^{*}= B_3T^{*}+B_{4}S^{*}\).
Proof
Let the core-EP decomposition of A be as in Theorem 1, and B be of the form (34). Then,
From \(A\mathop \le \limits ^{*} B\), we have \(AA^{*}=BA^{*}\) and \(A^{*}A=A^{*}B\). From \(A^{*}A=A^{*}B\), we have
By using (45) and \(AA^{*}=BA^{*}\), we get
Therefore,
By applying (45) and (46), we get
\(\square \)
Next, we discuss the relationship between the above binary relation and star partial order on \({\mathbb {C}}_{n, n}\).
Remark 3
Comparing Theorems 11 and 13, we get
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(1)
Applying \(A\mathop \le \limits ^{\textcircled {S}}B\) gives \(B_3=0\). But we can’t get \( NS^{*}= B_{4}S^{*}\). Therefore, \(A\mathop \le \limits ^{\textcircled {S}}B\) cannot deduce \(A\mathop \le \limits ^{ *}B\). (see Example 3)
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(2)
Since only \(N^{*}B_3=0\) is required in \(A\mathop \le \limits ^{ *}B\), \(B_3\) may not be equal to 0. Therefore, \(A\mathop \le \limits ^{ *}B\) cannot deduce \(A\mathop \le \limits ^{\textcircled {S}}B\). (see Example 4)
Example 3
Let
in which \(U=I_3\), \(T=1\), \(S=\left[ \begin{matrix}1&1\end{matrix} \right] \), \(N=\left[ \begin{matrix} 0 &{}0\\ 0 &{}0 \end{matrix} \right] \), \(B_3=\left[ \begin{matrix} 1 \\ 0 \end{matrix} \right] \) and \(B_4=\left[ \begin{matrix}0&{}-1 \\ 0 &{}0 \end{matrix} \right] \). Then,
It is easy to check that \(N\mathop \le \limits ^{*} B_{4}\), \(N^{*}B_3=0\) and \(NS^{*}= B_3T^{*}+B_{4}S^{*}\), that is,
And since
we get \(A(A^{\textcircled {S}})^{*} \ne B(A^{\textcircled {S}})^{*}\). It follows that \(A \mathop \le \limits ^{\textcircled {S}} B\) is not true.
Example 4
Let
in which \(U=I_3\), \(T=1\), \(S=\left[ \begin{matrix}0&1\end{matrix} \right] \), \(N=\left[ \begin{matrix} 0 &{}0\\ 0 &{}0 \end{matrix} \right] \), \(B_3=\left[ \begin{matrix} 0 \\ 0 \end{matrix} \right] \) and \(B_4=\left[ \begin{matrix}0&{}1 \\ 0 &{}0 \end{matrix} \right] \). Then,
It is easy to check that \(NS^{*}\ne B_3T^{*}+B_{4}S^{*}\), that is, A is not below B under the partial order “ \(\mathop \le \limits ^{*}\).”
And by applying
we get \(A \mathop \le \limits ^{\textcircled {S}} B\).
4.2 C-S Partial Order
Based on Sect. 4.1, we will introduce a new partial order on \({\mathbb {C}}_{n, n}\).
Lemma 1
Let \(N_5\) be a nilpotent matrix of order q, \(N_{5}\mathop \le \limits ^{*}C_{4}\) and \(C_{4}{S}_{3}^{*}=N_{5}{S}_{3}^{*}\). Then,
in which \(T_1\) is an invertible matrix of appropriate order.
Proof
Since \(N_{5}\mathop \le \limits ^{*}C_{4}\), we have \(N_{5}^*N_5 =N_5^*C_{4}\) and \(C_{4}{S}_{3}^{*}=N_{5}{S}_{3}^{*}\). From
it follows that
Therefore,
Similarly, by applying (48), \(C_{4}{S}_{3}^{*}=N_{5}{S}_{3}^{*}\) and
we have
that is,
By applying (49) and (50), we get (48). \(\square \)
Based on (32), we introduce a new binary relation “ \( \mathop \le \limits ^{\textcircled {c\!s}}\) ” on \(C\in {\mathbb {C}}_{n, n}\):
Definition 2
Let \( A, B\in {\mathbb {C}}_{n, n}\). The binary relation “ \(\mathop \le \limits ^{\textcircled {c\!s}}\) ” is defined as
Theorem 14
The binary relation “ \(\mathop \le \limits ^{\textcircled {c\!s}}\) ” is transitive on \(C\in {\mathbb {C}}_{n, n}\).
Proof
Let A, B and \(C\in {\mathbb {C}}_{n, n}\), \(A\mathop \le \limits ^{\textcircled {c\!s}}B\), \({\textrm{Ind}}(A)=k\), \(\text {rk}\left( A^k\right) =t\), \({\text {Ind}}(B)=l\), \(\text {rk}\left( B^l\right) =t+p\) and \(B\mathop \le \limits ^{\textcircled {c\!s}}C\). Write \(q=n-\textrm{rk}\left( B^l\right) \). Next, we prove \(A\mathop \le \limits ^{\textcircled {c\!s}}C\), that is, \(A\mathop \le \limits ^{\textcircled {S}}C\) and \( CA^{*}AA^{\textcircled {\dag }}=AA^{*}AA^{\textcircled {\dag }} \).
Since \(A\mathop \le \limits ^{\textcircled {S}}B\) and \(B\mathop \le \limits ^{\textcircled {S}}C\), by applying the core-EP decomposition and Theorem 13, we have
where \(\left[ \begin{matrix} N_{1} &{}N_{2} \\ N_{3} &{}N_{4} \end{matrix} \right] \mathop \le \limits ^{*} \left[ \begin{matrix} T _{1} &{}S_{3} \\ 0 &{}N_{5} \end{matrix} \right] \), \(T_1\in {\mathbb {C}}_{p, p}\) is nonsingular, \(N_{5}\) is nilpotent, and \(N_{5}\mathop \le \limits ^{*}C_{4}\).
By applying (52), we have
Since \(B\mathop \le \limits ^{\textcircled {S}}C\), \(BB^{*}BB^{\textcircled {\dag }}=CB^{*}BB^{\textcircled {\dag }}\). Therefore,
For \(N_{5}\mathop \le \limits ^{*}C_{4}\) and \(N_{5}{S}_{3}^{*}=C_{4}{S}_{3}^{*}\), by applying Lemma 1, we get \(\left[ {\begin{matrix} T_{1} &{}S_{3} \\ 0 &{}N_{5} \end{matrix}} \right] \mathop \le \limits ^{*} \left[ {\begin{matrix} T_{1} &{}S_{3} \\ 0 &{}C_{4} \end{matrix}} \right] \). Since the star partial order is transitive, it follows from \(\left[ {\begin{matrix} N_{1} &{}N_{2} \\ N_{3} &{}N_{4} \end{matrix}} \right] \mathop \le \limits ^{*} \left[ {\begin{matrix} T _{1} &{}S_{3} \\ 0 &{}N_{5} \end{matrix}} \right] \) that
By applying (52), it is easy to check that
Then, from (53), \(AA^{*}AA^{\textcircled {\dag }}=BA^{*}AA^{\textcircled {\dag }}\) and \(BB^{*}BB^{\textcircled {\dag }}=CB^{*}BB^{\textcircled {\dag }}\), we get \(N_{1}{S}_{1}^{*}+N_{2}{S}_{2}^{*} =T{S}_{1}^{*}+S_{3}{S}_{2}^{*}\), \(N_{3}{S}_{1}^{*}+N_{4}{S}_{2}^{*} =N_{5}{S}_{2}^{*}\) and \(N_{5}{S}_{2}^{*}=C_{4}{S}_{2}^{*}\). It follows from (55) and (56) that
From (52), (54) and (57), we get \(A\mathop \le \limits ^{\textcircled {c\!s}}C\). Therefore, the binary relation “ \(\mathop \le \limits ^{\textcircled {c\!s}}\) ” is transitive. \(\square \)
Theorem 15
The binary relation “ \(\mathop \le \limits ^{\textcircled {{c\!s}} }\) ” is a partial order on \({\mathbb {C}}_{n, n}\). We call it C-S partial order.
Proof
Reflexivity is trivial and transitivity follows by Theorem 14. For transitivity, let A, \(B \in {\mathbb {C}}_{n, n}\), \(A\mathop \le \limits ^{\textcircled {c\!s}}B\) and \(A\mathop \le \limits ^{\textcircled {c\!s}}B\). Then \(A\mathop \le \limits ^{\textcircled {S}}B\) and \(B\mathop \le \limits ^{\textcircled {S}}A\). It follows from applying Theorem 12 that we get \(A=B\). Therefore, the binary relation “ \( \mathop \le \limits ^{\textcircled {c\!s}} \) ” is anti-symmetric. \(\square \)
Theorem 16
Let \(A\mathop \le \limits ^{\textcircled {{c\!s}}}B\). Then, \(A\mathop \le \limits ^{*}B\).
Proof
Let A, \(B \in {\mathbb {C}}_{n, n}\), and A be below B under the C-S partial order. By applying (51), we have \(A\mathop \le \limits ^{\textcircled {S}}B\) and \(BA^{*}AA^{\textcircled {\dag }}=AA^{*}AA^{\textcircled {\dag }}\). Since \(A\mathop \le \limits ^{\textcircled {S}}B\), A and B have the forms as shown in (33). Then,
and
From (58), \(BA^{*}AA^{\textcircled {\dag }}=AA^{*}AA^{\textcircled {\dag }}\) and \(N\mathop \le \limits ^{*}B_{4}\), we have
Therefore, by applying (60) to (59) we get \(AA^{*}=BA^{*}\) and \(A^{*}A=A^{*}B\), that is, \(A\mathop \le \limits ^{*}B\). \(\square \)
Remark 4
In Theorem 16, we see that if \(A\mathop \le \limits ^{\textcircled {c\!s}}B\) then \(A\mathop \le \limits ^{*}B\). However, the converse does not hold (see Example 5)
Example 5
Let
From
it follows that \(A\mathop \le \limits ^{*}B\). For
we get
Therefore, A is not below B under the C-S partial order.
4.3 Characterizations of Special Matrices
Generalized inverses are one of the tools for characterizing special matrices. In this subsection, we try to use C-S inverse to characterize special matrices such as EP matrix and i-EP matrix.
Theorem 17
Let \(A \in {\mathbb {C}}_{n, n}\). Then, the following conditions are equivalent:
-
(1)
A is i-EP;
-
(2)
\(AA^{\textcircled {S}} = A^{\textcircled {S}}A\).
Proof
“(1)\(\Leftarrow \)(2)” Let \(A \in {\mathbb {C}}_{n, n}\). Write \({\text {Ind}}(A)=k\) and \(\text {rk}(A^k)=t\). By applying the core-EP decomposition of A, we have
Applying (61) and \(AA^{\textcircled {S}}= A^{\textcircled {S}}A\) gives
Post-multiplying \(N^{k-1}\) on \(S-TSN=0\) gives \(SN^{k-1}-TSN^{k}=0\). It follows from \(N^k=0\) that \(SN^{k-1}=0\). Post-multiplying \(N^{k-2}\) on \(S-TSN=0\) gives \(SN^{k-2}-TSN^{k-1}=0\), that is, \(SN^{k-2}=0\). And so on, we can get \(SN=0\). It follows from \(S-TSN=0\) that \(S=0\). Therefore, we get that A is i-EP.
(1)\( \Rightarrow \)(2) Obviously. \(\square \)
We see that \(\left( A^{\text { {(S)}}}\right) ^{\mathrm{{(S)}}}=A\) for any square matrix A. But in general, it is easy to see that \(\left( A^{\textcircled {S}}\right) ^{\textcircled {S}}\ne A \). So under what conditions are they equal? In other words, what kind of special matrix is A, when \(\left( A^{\textcircled {S}}\right) ^{\textcircled {S}}=A \)?
Theorem 18
Let \(A \in {\mathbb {C}}_{n, n}\). Then, the following conditions are equivalent:
-
(1)
A is i-EP;
-
(2)
\(\left( A^{\textcircled {S}}\right) ^{\textcircled {S}}= A \).
Proof
Let the decomposition of A be as in (2). Then,
If \((A^{\textcircled {S}})^{\textcircled {S}}=A\), we get \(S=0\). By applying Theorem 5, we get that A is i-EP.
Conversely, let A is i-EP, by applying Theorem 5, it is easy to check that \((A^{\textcircled {S}})^{\textcircled {S}}=A\). \(\square \)
In Example 1, we can see that in general, C-S inverse is not equal to Moore–Penrose inverse and G-S inverse. Next, we consider the case when the C-S inverse is equal to G-S inverse and Moore–Penrose inverse, respectively.
Theorem 19
Let \(A \in {\mathbb {C}}_{n, n}\). Then, the following conditions are equivalent:
-
(1)
A is i-EP;
-
(2)
\( A^{\textcircled {S}} = A^{\text {{(S)}}} \).
Proof
Let the decomposition of A be as in (2). If A is i-EP, then \(S=0\). By applying (9), it is easy to check that
Conversely, let \( A^{\textcircled {S}} = A^{\mathrm{{(S)}}}\). By applying (9) and (15), we get
that is, \(T^{-k-1}SN^{k-1} +T^{-k}SN^{k-2}+\cdots +T^{-3}SN +T^{-2}S +S - T^{-k+1}SN^{k-1}-T^{-k+2}SN^{k-2}-\cdots -T^{-1}SN -S =0\). After simplifying the equation, we get
Post-multiplying \(N^{k-1}\) on (62), we get
Since T is nonsingular, by applying \(N^k=0\), we get \(SN^{k-1}=0\). Therefore,
In the same way, by applying \(SN^{k-1}=0\), we get \(SN^{k-2}=0\); by applying \(SN^{k-2}=0\), we get \(SN^{k-3}=0\); \(\ldots \); by applying \(SN^2=0\), we get \(SN =0\). Therefore, from (62), we get \(T^{-2}S=0\). Since T is nonsingular, \(S=0\). It follows from Theorem 5 that A is i-EP. \(\square \)
Theorem 20
Let \(A \in {\mathbb {C}}_{n, n}\). Then, the following conditions are equivalent:
-
(1)
A is EP;
-
(2)
\( A^{\textcircled {S}} = A^\dag \).
Proof
“(1)\(\Leftarrow \)(2)” Let the decomposition of A and the C-S inverse of A be as in Theorem 6. If \( A^{\textcircled {S}} =A^\dag \), then
Therefore, \(N^3=N\). Since N is a nilpotent matrix, we get \(N=0\). By applying \(N=0\) and \(\left( A^{\textcircled {S}}A\right) ^*=A^{\textcircled {S}}A\), we get
Since T is invertible, it follows that \(S=0\).
For \(S=0\) and \(N=0\), by applying Theorem 4, we get that A is EP.
“(1)\(\Rightarrow \)(2)” If A is EP, by applying Theorem 4, it is easy to check that \( A^{\textcircled {S}} = A^\dag \). \(\square \)
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Acknowledgements
The authors wish to extend their sincere gratitude to the referees for their precious comments and suggestions, which helped to greatly improve this paper.
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The authors were supported by Xiangsihu Young Scholars Innovative Research Team of Guangxi University for Nationalities [No. 2019RSCXSHQN03], Thousands of Young and Middle-aged Key Teachers Training Programme in Guangxi Colleges and Universities [No. GUIJIAOSHIFAN2019-81HAO] and Special Fund for Science and Technological Bases and Talents of Guangxi [No. GUIKE AD19245148].
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Wang, H., Liu, N. The C-S Inverse and Its Applications. Bull. Malays. Math. Sci. Soc. 46, 90 (2023). https://doi.org/10.1007/s40840-023-01478-2
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DOI: https://doi.org/10.1007/s40840-023-01478-2