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Theorem 12.2([4, Theorem 4.2]) Let \( M \) be as in (12.53) and let \( s = d - b{a}^{-1}c \) . Assume that \( s\{ 1,4\} \neq \varnothing \) and let \( {s}^{\left( 1,4\right) } \in s\{ 1,4\} \) and \( f = {1}_{l} - {s}^{\left( 1,4\right) }s \) . Then \( M\{ 1,4\} \neq \varnothing \) if and only if \( v = {1}_{m} + {a}^{... | Proof We use the factorization (12.54). Choose \( {A}^{\left( 1,4\right) } = \left\lbrack \begin{matrix} {a}^{-1} & 0 \\ 0 & {s}^{\left( 1,4\right) } \end{matrix}\right\rbrack \), which\ngives \( I - {A}^{\left( 1,4\right) }A = \left\lbrack \begin{array}{ll} 0 & 0 \\ 0 & f \end{array}\right\rbrack \), where \( f = {1}_... | Yes |
Theorem 12.3([4, Theorem 4.3]) Let \( M \) be as in (12.53) and let \( s = d - b{a}^{-1}c \) . Assume that \( {s}^{ \dagger } \) exists and set \( e = {1}_{k} - s{s}^{ \dagger } \) and \( f = {1}_{l} - {s}^{ \dagger }s \) . Then \( {M}^{ \dagger } \) exists if and only if both \( u = {1}_{m} + {\left( b{a}^{-1}\right) ... | Proof In view of the factorization (12.54), we have that \( M \) is Moore-Penrose invertible if and only if \( {PA} \) has a \( \{ 1,3\} \) -inverse and \( {AQ} \) has a \( \{ 1,4\} \) -inverse. Now, since \( {s}^{ \dagger } \) exists we can consider \( {s}^{\left( 1,3\right) } = {d}^{ \dagger } \) in Theorem 12.1 and ... | Yes |
Theorem 12.5( \( {}^{\left( 4,\text{ Theorem 4.5 }\right) } \) Let \( e, f \), and \( c \) be as in (12.64) and \( T \) be as in (12.65). Assume that \( c\{ 1,4\} \neq \varnothing \) and let \( {c}^{\left( 1,4\right) } \in c\{ 1,4\} \) . Then \( T\{ 1,4\} \neq \varnothing \) if and only if \( v = {1}_{l} + {d}^{\left( ... | \[ {T}^{\left( 1,4\right) } = \left\lbrack \begin{matrix} {a}^{\left( 1,4\right) } - {hf}{\left( {d}^{\left( 1,2,4\right) }b\right) }^{ * }\mu & {hf}{\left( {d}^{\left( 1,2,4\right) }b\right) }^{ * }\rho + {c}^{\left( 1,4\right) }e \\ - \mu & \rho \end{matrix}\right\rbrack \left\lbrack \begin{matrix} {1}_{m} & 0 \\ - b... | Yes |
Proposition 12.6( \( {}^{\left( \left\lbrack 4,\text{Proposition 4.6}\right\rbrack \right) } \) Let \( e, f \) and \( c \) be as in (12.67). If any of the following conditions hold, then \( {c}^{ \dagger } \) exists.\n\n(1) \( w = c{c}^{ * } + d{d}^{ * } \) is invertible.\n\n(2) \( z = {c}^{ * }c + {a}^{ * }a \) is inv... | Proof (1) First, we prove that if \( w \) is invertible then \( c \) is regular. Let \( x = {c}^{ * }{w}^{-1} \) . Since \( {ew} = c{c}^{ * } \) we also have \( e = c{c}^{ * }{w}^{-1} = {cx} \) . Using this, we get \( {cxc} = {ec} = c \) . Hence, \( x \) is a \( \{ 1\} \) -inverse of \( c \) .\n\nNow, choose \( {c}^{\l... | Yes |
Theorem 12.7([4, Theorem 4.7]) Let \( T \) be as in (12.60) and let \( e, f \), and \( c \) be as in (12.67). If \( {c}^{ \dagger } \) exists, then \( {T}^{ \dagger } \) exists if and only if both \( u = {1}_{m} + {\left( b{a}^{ \dagger }\right) }^{ * }{egb}{a}^{ \dagger } \) and \( v = {1}_{l} + {d}^{ \dagger }{bhf}{\... | Proof We can apply Theorem 12.4 with \( {a}^{\left( 1,2,3\right) } = {a}^{ \dagger },{d}^{\left( 1,3\right) } = {d}^{ \dagger } \), and \( {c}^{\left( 1,3\right) } = {c}^{ \dagger } \) to obtain that \( {T}^{\left( 1,3\right) } \) exists if and only if \( u = {1}_{m} + {\left( b{a}^{ \dagger }\right) }^{ * }{egb}{a}^{ ... | Yes |
Lemma 14. \( {\mathbf{2}}^{\left( \left\lbrack {48},\text{ Lemma }2\right\rbrack \right) } \) Suppose that \( \varphi : X \rightarrow Y \) is a morphism of \( \mathfrak{C} \) with a \( \{ 1,2, i\} \) -inverse \( {\varphi }^{\left( 1,2, i\right) } : Y \rightarrow X \) (with respect to the involution \( * \) ) and that \... | Proof Follows from direct calculation and Jacobson's Lemma. | No |
Proposition 14.7( \( {}^{\left( \left\lbrack {48},\text{ Proposition }4\right\rbrack \right) } \) Suppose that \( \varphi : X \rightarrow Y \) is a morphism of \( \mathfrak{C} \) with Moore-Penrose inverse \( {\varphi }^{ \dagger } : Y \rightarrow X \), and that \( \eta : X \rightarrow Y \) is a morphism of \( \mathfra... | Proof The equivalence of conditions (1)-(3) can be obtained from Propositions 14.3 and 14.5.\n\nBy Proposition 14.3, \( {\left( {1}_{X} - \lambda \right) }^{-1}{\left( {1}_{X} + {\varphi }^{ \dagger }\eta \right) }^{-1}{\varphi }^{ \dagger } \in f\{ 1,2,4\} \).\n\nBy Proposition 14.5, \( {\varphi }^{ \dagger }{\left( {... | Yes |
Lemma 2.5 If \( \sup \left\{ {{G}_{C} - {\operatorname{pd}}_{R}\left( M\right) \mid M}\right. \) is an \( R \) -module \( \} < \infty \), then for any \( n \geq 0 \) , the following are equivalent:\n\n(1) \( \sup \left\{ {{G}_{C} - {\operatorname{pd}}_{R}\left( M\right) \mid M}\right. \) is an \( R \) -module \( \} \le... | Proof Use [8, Proposition 2.12] and [6, Theorem 9.8]. | No |
Lemma 3.3 Assume that \( \mathcal{W} \) is a cogenerator for \( \mathcal{X} \) and \( M \in \operatorname{Mod}R \) with \( \mathcal{X} \) - \( {\operatorname{pd}}_{R}\left( M\right) = n < \infty \) . If \( {\operatorname{id}}_{\mathcal{W}}\left( M\right) = 0 \) and \( {\operatorname{id}}_{\mathcal{X}}\left( M\right) < ... | Proof By assumption and \( \left\lbrack {4\text{, Corollary 4.5}}\right\rbrack \), there exists an exact sequence of \( R \) -modules\n\n\[ 0 \rightarrow M \rightarrow N \rightarrow X \rightarrow 0 \]\n\n\( \left( *\right) \)\n\nsuch that \( \mathcal{W} - {\operatorname{pd}}_{R}\left( N\right) \leq n \) and \( X \in \m... | Yes |
Corollary 3.4 Let \( M \) be an \( R \) -module with \( {\operatorname{id}}_{R}\left( M\right) < \infty \) and \( {G}_{C} - {\operatorname{pd}}_{R}\left( M\right) < \infty \) . Then \( {\mathcal{P}}_{C} - {\operatorname{pd}}_{R}\left( M\right) = {G}_{C} - {\operatorname{pd}}_{R}\left( M\right) < \infty \) . | Proof It is clear that \( {\operatorname{id}}_{\mathcal{X}}\left( M\right) \leq {\operatorname{id}}_{R}\left( M\right) < \infty \) for any subclass \( \mathcal{X} \) of Mod \( R \) . The finiteness of \( {\operatorname{id}}_{R}\left( M\right) \) implies \( {\operatorname{Ext}}_{R}^{i \geq 1}\left( {C, M}\right) = 0 \) ... | Yes |
Theorem 1.3 \( {}^{\left\lbrack 1,\text{ Lemma }2\right\rbrack }\; \) Let \( A \in {\mathbb{C}}^{n \times n} \) with rank \( \left( A\right) = r \) and have representation (1.1). Then the necessary and sufficient condition for the existence of \( {A}^{ \circledast } \) is that \( K \) is nonsingular. Furthermore, when ... | Proof First, observe that \[ {P}_{A} = U\left\lbrack \begin{matrix} {I}_{r} & 0 \\ 0 & 0 \end{matrix}\right\rbrack {U}^{ * } \] Suppose now that \( B \in {\mathbb{C}}^{n \times n} \), partitioned as \[ B = U\left\lbrack \begin{array}{ll} W & X \\ Y & Z \end{array}\right\rbrack {U}^{ * } \] where \( W \in {\mathbb{C}}^{... | Yes |
Theorem 1.5 \( {}^{\left\lbrack {32},\text{ Theorem }{2.4}\right\rbrack } \) Let \( A \in {\mathbb{C}}^{n \times n} \) . If \( A = {GH} \) is a full-rank decomposition, then \( {A}^{ \circledast } \) exists if and only if \( {HG} \) is invertible. In this case, | Proof The proof is based on Theorem 1.4 and [3, Theorem 1.7]. Moreover, by \( \left\lbrack 1\right\rbrack ,{A}^{ \circledast } = {A}^{\# }A{A}^{ \dagger } \) . Hence, \( {A}^{ \circledast } = G{\left( HG\right) }^{-1}{\left( {G}^{ * }G\right) }^{-1}{G}^{ * } \) . | Yes |
Theorem 2.3 \( {}^{\left\lbrack {29},\text{ Theorem }{2.14}\right\rbrack }\; \) Let \( a \in R \) . The following assertions are equivalent:\n\n(1) \( a \) is core invertible.\n\n(2) There exists \( x \in R \) such that \( \mathrm{{axa}} = a,{}^{ \circ }x = {}^{ \circ }a \) and \( {x}^{ \circ } = {\left( {a}^{ * }\righ... | Proof (1) \( \Rightarrow \) (2). Suppose that \( a \) is core invertible and let \( x = {a}^{ \circledast } \) . By definition, \( {axa} = a,{xR} = {aR} \) and \( {Rx} = R{a}^{ * } \) . By [3, Lemma 2.21], it follows that \( {}^{ \circ }x{ = }^{ \circ }a \) and \( {x}^{ \circ } = {\left( {a}^{ * }\right) }^{ \circ }.\n... | Yes |
Theorem 2.6 \( {}^{\left\lbrack {33},\text{ Theorem 2.6 }\right\rbrack } \) Let \( a \in R \) . Then \( a \in {R}^{ \circledast } \) if and only if \( a \in \) \( {R}^{\# } \cap {R}^{\{ 1,3\} } \) . In this case, \( {a}^{ \circledast } = {a}^{\# }a{a}^{\left( 1,3\right) } \) . | Proof Suppose \( a \in {R}^{ \circledast } \) . Then \( {a}^{\# } = {\left( {a}^{ \circledast }\right) }^{2}a \) by Lemma 2.5 and \( a \in {R}^{\{ 1,3\} } \) . Conversely, suppose \( a \in {R}^{\# } \cap {R}^{\{ 1,3\} }. \) Then \( a{a}^{(1,3)}a = a \) and \( (a{a}^{(1,3)}{)}^{ \ast } = a{a}^{(1,3)} \), we have\n\n\[ a... | Yes |
Corollary 2. \( {7}^{\left\lbrack {33},\text{ Corollary }{2.7}\right\rbrack } \) Let \( a \in R \) . Then the following conditions are equivalent:\n\n(1) \( a \in {R}^{ \circledast } \) .\n\n(2) \( a \in {R}^{\# } \) and there exists \( x \in R \) such that \( {\left( ax\right) }^{ * } = {ax} \) and \( x{a}^{2} = a \) ... | Proof (1) \( \Rightarrow \) (2). It is clear by Theorem 2.3 and Lemma 2.5.\n\n\( \left( 2\right) \Rightarrow \left( 3\right) . \) It is sufficient to prove that \( x{a}^{2} = a \) implies \( {xa} = {a}^{\# }a \) . It is easy to see that by \( {xa} = x{a}^{2}{a}^{\# } = a{a}^{\# } = {a}^{\# }a \) .\n\n(3) \( \Rightarrow... | Yes |
Theorem 2.8 \( {}^{\left\lbrack {33},\text{ Theorem }{3.3}\right\rbrack } \) Let \( a, b \in R \) . If \( {Ra} = R{a}^{2} \), then the following are equivalent:\n\n(1) \( a \in {R}^{ \circledast } \) with core inverse \( b \) .\n\n(2) \( {aba} = a,{\left( ab\right) }^{ * } = {ab} \) and \( a{b}^{2} = b \) . | Proof \( \;\left( 1\right) \Rightarrow \left( 2\right) \) It is trivial by Theorem 2.3.\n\n\( \left( 2\right) \Rightarrow \left( 1\right) \) Suppose that \( {aba} = a, a{b}^{2} = b,{\left( ab\right) }^{ * } = {ab}. \) Then \( a = {aba} = {a}^{2}{b}^{2}a \in {a}^{2}R. \) Thus \( a \in {R}^{\# } \) by the hypothesis \( {... | Yes |
Theorem 2.10 \( {}^{\left\lbrack {33},\text{ Theorem }{3.5}\right\rbrack } \) Let \( a, b \in R \) . If \( {aR} = {a}^{2}R \), then the following are equivalent:\n\n(1) \( a \in {R}^{ \circledast } \) with core inverse \( {a}^{ \circledast } = b \) .\n\n(2) \( {bab} = b,{\left( ab\right) }^{ * } = {ab} \) and \( b{a}^{... | Proof \( \left( 1\right) \Rightarrow \left( 2\right) \) . It is trivial by Theorem 2.3.\n\n(2) \( \Rightarrow \) (1). Suppose \( {bab} = b,{\left( ab\right) }^{ * } = {ab} \) and \( b{a}^{2} = a \) . Then \( a = b{a}^{2} \in R{a}^{2} \) , \( a \in {R}^{\# } \) by the hypothesis \( {aR} = {a}^{2}R \) and [14, Propositio... | Yes |
Theorem 2.13 \( {}^{\left\lbrack {33},\text{ Theorem }{3.9}\right\rbrack } \) The following conditions are equivalent:\n\n(1) \( R \) is a Dedekind-finite ring.\n\n(2) Let \( a, b \in R \) . Then \( {aba} = a,{\left( ab\right) }^{ * } = {ab}, a{b}^{2} = b \) if and only if \( a \in {R}^{ \circledast } \) with \( {a}^{ ... | Proof \( \;\left( 1\right) \Rightarrow \left( 2\right) \) . Suppose \( a \in {R}^{ \circledast } \) with \( {a}^{ \circledast } = b \) . Then \( {aba} = a,{\left( ab\right) }^{ * } = {ab} \) , \( a{b}^{2} = b \) by Theorem 2.3. Conversely, suppose that \( R \) is a Dedekind-finite ring and \( {aba} = a,{\left( ab\right... | Yes |
Theorem 2.14 \( {}^{\left\lbrack {33},\text{ Theorem }{3.10}\right\rbrack } \) Consider the following conditions:\n\n(1) \( R \) is a Dedekind-finite ring.\n\n(2) Let \( a, b \in R \) . If \( {bab} = b,{\left( ab\right) }^{ * } = {ab} \) and \( b{a}^{2} = a \), then \( a \in {R}^{ \oplus } \) with \( {a}^{ \oplus } = b... | Proof \( \;\left( 1\right) \Rightarrow \left( 2\right) \) . Suppose \( R \) is a Dedekind-finite ring. By \( {bab} = b \) and \( b{a}^{2} = a \) , we have\n\n\[ b = {bab} = b\left( {b{a}^{2}}\right) b = {b}^{2}{a}^{2}b \in {b}^{2}R. \]\n\nSince \( b = {bab} \) gives \( a \in b\{ 1\} \), by [4, Theorem 3.1], \( b + 1 - ... | Yes |
Proposition 2.15 \( {}^{\left\lbrack {33},\text{ Proposition }{3.11}\right\rbrack }\; \) Let \( a, b \in R \) . If \( {aR} = {a}^{2}R \), then the following conditions are equivalent:\n\n(1) \( a \in {R}^{ \circledast } \) with \( {a}^{ \circledast } = b \) .\n\n(2) \( b{a}^{2} = a,{\left( ab\right) }^{ * } = {ab},{bR}... | Proof \( \;\left( 1\right) \Rightarrow \left( 2\right) \) Suppose \( a \in {R}^{ \circledast } \) with \( {a}^{ \circledast } = b \) . Then \( b{a}^{2} = a,{\left( ab\right) }^{ * } = {ab} \) and \( b = a{b}^{2} \) by Theorem 2.3, thus \( {bR} \subseteq {aR} \) .\n\n(1) \( \Rightarrow \) (3) Suppose \( a \in {R}^{ \cir... | Yes |
Proposition 2.16 \( {}^{\left\lbrack {33},\text{ Proposition }{2.11}\right\rbrack } \) Let \( a \in R \) . Then the following conditions are equivalent:\n\n(1) \( a \in {R}^{ \circledast } \) .\n\n(2) \( R = {aR} \oplus {\left( {a}^{ * }\right) }^{ \circ } = {aR} \oplus {a}^{ \circ } \) .\n\n(3) \( R = {aR} + {\left( {... | Proof By Theorem 2.6, we have that \( a \in {R}^{ \circledast } \) if and only if \( a \in {R}^{\# } \cap {R}^{\{ 1,3\} } \) and\n\n\[ {a}^{ \circledast } = {a}^{\# }a{a}^{\left( 1,3\right) }.\]\n\n(2.7)\n\nThus it is easy to see (1)-(9) are equivalent by [3, Theorem 8.11] and [4, Proposition 2.3]. Suppose \( 1 = a{x}_... | Yes |
Proposition 2.17 \( {}^{\left\lbrack {33},\text{Proposition 2.12}\right\rbrack }\; \) Let \( a \in R \) . Then the following conditions are equivalent:\n\n(1) \( a \in {R}_{ \circledast } \) .\n\n(2) \( R = {a}^{ * }R \oplus {a}^{ \circ } = {aR} \oplus {a}^{ \circ } \) .\n\n(3) \( R = {a}^{ * }R + {a}^{ \circ } = {aR} ... | In this case,\n\n\[ {a}_{ \circledast } = {x}_{1}^{ * }a{y}_{1} = {x}_{1}^{ * }a{y}_{2}^{2}a = {x}_{2}a{y}_{1} = {x}_{2}a{y}_{2}^{2}a, \]\n\nwhere \( 1 = {a}^{ * }{x}_{1} + {u}_{1} = {x}_{2}a + {u}_{2} = a{y}_{1} + {v}_{1} = {y}_{2}a + {v}_{2} \) for some \( {x}_{1},{x}_{2},{y}_{1},{y}_{2} \in R \) , \( {u}_{2} \in {}^... | Yes |
Proposition 2.18 \( {}^{\left\lbrack {19},\text{ Proposition }{2.9}\right\rbrack } \) Let \( a \in R, n \geq 1 \) . We have the following results:\n\n(I) the following conditions are equivalent:\n\n(1) \( a \in R{\left( {a}^{ * }\right) }^{n}a \) .\n\n(2) \( a \in R{a}^{ * }a \cap {a}^{n}R \) .\n\n(3) \( R = {}^{ \circ... | Proof \( \left( I\right) \left( 1\right) \Rightarrow \left( 2\right) . \) It is clear to see that \( a \in R{a}^{ * }a \) follows from \( a \in R{\left( {a}^{ * }\right) }^{n}a, \) and there exists \( r \in R \) such that\n\n\[ a = r{\left( {a}^{ * }\right) }^{n}a = r{\left( {a}^{ * }\right) }^{n - 1}{a}^{ * }a, \]\n\n... | Yes |
Theorem 2.19 \( {}^{\left\lbrack {19},\text{ Theorem }{2.10}\right\rbrack }\; \) Let \( a \in R, n \geq 2. \) We have the following results:\n\n(1) \( a \in {R}^{ \circledast } \) if and only if \( a \in R{\left( {a}^{ * }\right) }^{n}a \cap R{a}^{n} \) . In this case, \( {a}^{ \circledast } = {a}^{n - 1}{s}^{ * } \) f... | Proof (1). Since \( a \in {R}^{\# } \) if and only if \( a \in {a}^{2}R \cap R{a}^{2} \) by [4, Theorem 2.2], there exist \( x, y \in R \) such that \( a = {a}^{2}x = y{a}^{2} \) . Further, we have\n\n\[ a = {a}^{2}x = a\left( {{a}^{2}x}\right) x = {a}^{3}{x}^{2} = \cdots = {a}^{n}{x}^{n - 1} \in {a}^{n}R \]\n\nand\n\n... | Yes |
Theorem 2.20 \( {}^{\left\lbrack {19},\text{ Theorem 2.11 }\right\rbrack } \) Let \( a \in R, n \geq 2 \) . The following conditions are equivalent:\n\n(1) \( a \in {R}^{ \dagger } \cap {R}^{\# } \) .\n\n(2) \( a \in {R}^{ \circledast } \cap {R}_{ \circledast } \) .\n\n(3) \( a \in a{\left( {a}^{ * }\right) }^{n}R \cap... | Proof The equivalences of seven conditions above and the representations of \( {a}^{ \circledast } \) and \( {a}_{ \circledast } \) can be easily obtained by Proposition 2.18 and Theorem 2.19. We will give the representations of \( {a}^{ \dagger } \) and \( {a}^{\# } \) in the following.\n\nSuppose \( a = s{\left( {a}^... | Yes |
Theorem 3.2 \( {}^{\left\lbrack {19},\text{ Theorem }{3.4}\right\rbrack }\; \) Let \( a \in R. \) The following conditions are equivalent:\n\n(1) \( a \in {R}^{ \circledast } \) .\n\n(2) There exists a unique projection \( p \) such that \( {pa} = 0, u = a + p \in {R}^{-1} \) .\n\n(3) There exists an Hermitian element ... | Proof \( \;\left( 1\right) \Rightarrow \left( 2\right) \) . Let \( p = 1 - a{a}^{ \circledast } \) . Then \( p \) is a projection satisfying \( {pa} = 0 \), and the proof of the uniqueness of \( p \) is similar to Theorem 3.1. It is easy to verify\n\n\[ \n\left( {a + 1 - a{a}^{ \circledast }}\right) \left( {{a}^{ \circ... | Yes |
Theorem 3.5 \( {}^{\left\lbrack {19},\text{ Theorem }{3.7}\right\rbrack } \) Let \( R \) be a Dedekind-finite ring and \( a \in R \) . The following conditions are equivalent:\n\n(1) \( a \in {R}^{ \circledast } \) .\n\n(2) There exists a unique projection \( p \) such that \( {pa} = 0,{a}^{ * }a + p \in {R}^{-1} \) .\... | Proof Since \( {a}^{ * }a + p \) is Hermitian. Thus \( {a}^{ * }a + p \) is one-sided invertible if and only if it is invertible, hence the conditions (2), (3) and (4) are equivalent. Next, we mainly show the equivalence between the conditions (1) and (2).\n\n(1) \( \Rightarrow \) (2). Assume that \( a \in {R}^{ \circl... | Yes |
Theorem 3.6 \( {}^{\left\lbrack {19},\text{ Theorem }{3.8}\right\rbrack } \) Let \( a \in R, n \geq 1 \) . The following conditions are equivalent:\n\n(1) \( a \) is EP.\n\n(2) There exists a unique projection \( p \) such that \( {pa} = {ap} = 0,{a}^{n} + p \in {R}^{-1} \).\n\n(3) There exists an Hermitian element \( ... | Proof (1) \( \Rightarrow \) (2). Suppose \( a \) is EP, so \( a \in {R}^{ \dagger } \cap {R}^{\# } \) and \( {a}^{ \dagger } = {a}^{\# } \) . Let \( p = \) \( 1 - {a}^{\# }a = 1 - {a}^{ \dagger }a \), it is easily seen that \( p \) is a projection satisfying \( {pa} = {ap} = 0 \) . Since\n\n\[ \left( {{a}^{n} + 1 - {a}... | Yes |
Theorem 3.7 \( {}^{\left\lbrack 7,\text{ Theorem 4.2 }\right\rbrack } \) Let \( a \in R \) be regular. Then the following conditions are equivalent:\n\n(1) \( a \in {R}^{ \circledast } \) .\n\n(2) \( a + 1 - a{a}^{ - } \) and \( {a}^{ * } + 1 - a{a}^{ - } \) are invertible for some \( {a}^{ - } \in a\{ 1\} \) .\n\n(3) ... | Proof \( \;\left( 1\right) \; \Rightarrow \left( 2\right) .\; \) Since \( a \in {\mathbb{R}}^{ \circledast } \) then \( a \in {\mathbb{R}}^{\# } \cap {\mathbb{R}}^{\left( 1,3\right) } \) by Theorem 2.6. Let \( {a}^{ - } \in a\{ 1,3\} \) . Then \( a + 1 - a{a}^{ - } \) is invertible by [4, Theorem 2.7] and hence \( {a}^... | Yes |
Proposition 3.9 \( {}^{\left\lbrack 7,\text{ Proposition }{4.4}\right\rbrack }\; \) Let \( k \geq 1 \) be an integer and suppose that \( a \in R \) is regular. If \( {\left( {a}^{ * }\right) }^{k} + 1 - a{a}^{ - } \in {R}^{-1} \) for any \( {a}^{ - } \in a\{ 1\} \), then \( a \in {R}^{ \oplus } \) . | Proof Let \( u = {\left( {a}^{ * }\right) }^{k} + 1 - a{a}^{ - } \) . As \( u \) is invertible, we have \( a = {u}^{-1}{\left( {a}^{ * }\right) }^{k}a \in R{a}^{ * }a \) . Hence \( a \) is \( \{ 1,3\} \) -invertible by [3, Lemma 8.3].\n\n\[ \text{As}{\left( {\left( {a}^{ * }\right) }^{k} + 1 - a{a}^{\left( 1,3\right) }... | Yes |
Let \( R \) be the ring as Remark 3.10. Given \( a = \left\lbrack \begin{matrix} 1 & - 2 \\ 1 & - 2 \end{matrix}\right\rbrack \in R \), then \( {a}^{2} = - a \) and hence \( {a}^{\# } \) exists. So, \( {a}^{\# } \) exists. | Taking \( {a}^{ - } = \left\lbrack \begin{matrix} \frac{2}{3} & \frac{1}{3} \\ 0 & 0 \end{matrix}\right\rbrack \) , then \( {a}^{ * } + 1 - a{a}^{ - } = \frac{1}{3}\left\lbrack \begin{matrix} 4 & 2 \\ - 8 & - 4 \end{matrix}\right\rbrack ,{\left( {a}^{ * }\right) }^{2} + 1 - a{a}^{ - } = \frac{1}{3}\left\lbrack \begin{m... | Yes |
Theorem 3.12 \( {}^{\left\lbrack 7,\text{ Theorem }{4.7}\right\rbrack }\; \) Let \( k \geq 1 \) be an integer and suppose \( a \in {R}^{ \oplus } \) . Then the following conditions are equivalent for any \( {a}^{ - } \in a\{ 1\} \) :\n\n(1) \( {\left( {a}^{ * }\right) }^{k} + 1 - a{a}^{ - } \in {R}^{-1} \) .\n\n(2) \( ... | Proof As \( a \in {R}^{ \circledast } \), we get \( a \in {R}^{\# } \) by Theorem 2.6. Hence, \( a + 1 - a{a}^{ \circledast } \in {R}^{-1} \) from \( \left\lbrack {4\text{, Theorem 2.7}}\right\rbrack \) . Note that \( a{a}^{ \circledast } = a{a}^{\left( 1,3\right) } \) and \( {a}^{ * }a{a}^{ \circledast } = {a}^{ * } \... | Yes |
Proposition 3.14 \( {}^{\left\lbrack 7,\text{ Proposition 4.9}\right\rbrack } \) Let \( k \geq 1 \) be an integer and suppose \( a \in R \) . Then the following conditions are equivalent:\n\n(1) \( a \in {R}^{ \circledast } \) .\n\n(2) \( a \in {R}^{\left( 1,3\right) } \) and \( {\left( {a}^{ * }\right) }^{k} + 1 - a{a... | Proof \( \left( 1\right) \Rightarrow \left( 2\right) . \) It follows from Theorem 2.6 that \( a \in {\mathbb{R}}^{ \circledast } \) implies \( a \in {\mathbb{R}}^{\# } \cap {\mathbb{R}}^{\left( 1,3\right) }. \) Hence, \( {a}^{k} + 1 - a{a}^{\left( 1,3\right) } \in {R}^{-1} \) from [4, Theorem 2.7].\n\n\[ \text{So,}{\le... | Yes |
Proposition 3.18 \( {}^{\left\lbrack 7,\text{ Proposition }{4.13}\right\rbrack }\; \) Let \( a \in {R}^{ \circledast } \) and suppose \( u = {a}^{ * } + 1 - a{a}^{ - } \in \) \( {R}^{-1} \) for some \( {a}^{ - } \in a\{ 1\} \) . Then \( {a}^{ \circledast } = {\left( {u}^{-1}\right) }^{ * }a{\left( {u}^{-1}\right) }^{ *... | Proof \ | No |
Corollary 3.19 \( {}^{\left\lbrack 7,\text{ Corollary }{4.14}\right\rbrack } \) Let \( R \) be a Dedekind-finite ring. Then the following conditions are equivalent:\n\n(1) \( a \in {R}^{ \circledast } \) .\n\n(2) \( a \in {R}^{\left( 1,3\right) } \) and \( {a}^{ * }a + 1 - a{a}^{\left( 1,3\right) } \) is invertible for... | Proof Let \( u = {a}^{ * } + 1 - a{a}^{\left( 1,3\right) } \) and \( v = {a}^{ * }a + 1 - a{a}^{\left( 1,3\right) } \) . Then \( v = u{u}^{ * } \) . Since \( R \) is a Dedekind-finite ring, we have \( v \in {R}^{-1} \) if and only if \( u \in {R}^{-1} \) . By Corollary 3.16, \( {a}^{ \circledast } = {\left( {u}^{-1}\ri... | Yes |
Theorem 3.23 \( {}^{\left\lbrack 7,\text{ Theorem }{5.6}\right\rbrack }\; \) Let \( a \in R \) be regular with an inner inverse \( {a}^{ - } \) . Then the following conditions are equivalent:\n\n(1) \( a \in {R}^{\# } \cap {R}^{ \dagger } \) .\n\n(2) \( a \in {R}^{ \circledast } \cap {R}_{ \circledast } \) .\n\n(3) \( ... | Proof (1) \( \Leftrightarrow \) (2). By Theorem 2.6.\n\n(3) \( \Leftrightarrow \) (5) and (4) \( \Leftrightarrow \) (6). By Jacobson’s lemma.\n\n\( \left( 1\right) \Rightarrow \left( 3\right) . \) From \( \left\lbrack {4,\text{ Theorem }{2.7}}\right\rbrack \) and \( \left\lbrack {3,\text{ Theorem }{8.14}}\right\rbrack ... | Yes |
Corollary 3.24 \( {}^{\left\lbrack 7,\text{ Corollary }{5.7}\right\rbrack } \) Let \( a \in {R}^{ \dagger } \) . Then the following conditions are equivalent:\n\n(1) \( a \in {R}^{ \oplus } \) .\n\n(2) \( a \in {R}_{ \oplus } \) .\n\n(3) \( u = a{a}^{ * }a + 1 - a{a}^{ \dagger } \in {R}^{-1} \) .\n\n(4) \( v = a{a}^{ *... | Proof As \( a \in {R}^{ \dagger } \), it follows that \( a \in {R}^{ \circledast } \) if and only if \( a \in {R}^{\# } \) if and only if \( a \in {R}_{ \circledast } \) by Theorem 2.6. So (1)-(6) are equivalent by Theorem 3.23. Moreover, \( {a}^{ \circledast } = {u}^{-1}a{a}^{ * } \) and \( {a}_{ \circledast } = {a}^{... | Yes |
Proposition 3.25 \( {}^{\left\lbrack 7,\text{ Proposition }{5.8}\right\rbrack }\; \) Let \( a \in {R}^{ \dagger } \) . Then the following conditions are equivalent:\n\n(1) \( a \in {R}^{ \circledast } \) .\n\n(2) \( a \in {R}^{\# } \) .\n\n(3) \( {a}^{ * } + 1 - a{a}^{ \dagger } \in {R}^{-1} \) .\n\nIn this case, \( {a... | Proof \( \left( 1\right) \Leftrightarrow \left( 2\right) \) . It follows by Theorem 3.23 \( \left( 1\right) \Leftrightarrow \left( 2\right) \) .\n\n(2) \( \Leftrightarrow \) (3). Note that \( {a}^{ * } + 1 - a{a}^{ \dagger } = {\left( a + 1 - a{a}^{ \dagger }\right) }^{ * } \) . It follows from [4, Theorem 2.7] that \(... | Yes |
Proposition 3.26 \( {}^{\left\lbrack 7,\text{ Proposition }{5.9}\right\rbrack }\; \) Let \( a \in {R}^{\# } \) . Then the following conditions are equivalent:\n\n(1) \( a \in {R}^{ \circledast } \cap {R}_{ \circledast } \) .\n\n(2) \( a \in {R}^{ \dagger } \) .\n\n(3) \( {a}^{ * } + 1 - a{a}^{\# } \in {R}^{-1} \) .\n\n... | Proof (1) \( \Leftrightarrow \) (2). It follows by Theorem 3.23 (1) \( \Leftrightarrow \) (2).\n\n\( \left( 2\right) \Rightarrow \left( 3\right) . \) Note that \( a \in {R}^{ \dagger } \) implies \( {a}^{ * }a + 1 - {a}^{\# }a \in {R}^{-1} \) by [3, Theorem 8.14]. As \( a \in {R}^{\# } \), then \( a + 1 - a{a}^{ \dagge... | Yes |
Lemma 3.32 \( \left\lbrack {{26},\text{ Theorem }{3.2}}\right\rbrack ,\left\lbrack {{25},\text{ Theorem }{1.3}}\right\rbrack \; \) Let \( d \in R \) be regular with \( {d}^{ - } \in d\{ 1\} . \) The following conditions are equivalent:\n\n(1) \( a \) is invertible along \( d \) .\n\n(2) \( u = {da} + 1 - d{d}^{ - } \) ... | In [26, Theorem 3.2], Mary and Patrício proved that \( m \) is invertible along \( d \) if and only if \( d \in {dmdR} \cap {Rdmd} \) . Comparing Theorem 3.29 and Lemma 3.32, we have the following corollary. | No |
Proposition 3.35 \( {}^{\left\lbrack {19},\text{ Proposition 4.4 }\right\rbrack } \) Let \( a \in R, n \geq 1 \), we have the following results:\n\n(1) If \( a \in {Ra}{\left( {a}^{ * }\right) }^{n}a \), then \( a \in {a}^{n}{a}^{ * }{a}^{n}R \) . | Proof (1). Suppose \( a \in {Ra}{\left( {a}^{ * }\right) }^{n}a \), there exists \( x \in R \) such that\n\n\[ a = {xa}{\left( {a}^{ * }\right) }^{n}a. \]\n\n(3.17)\n\nTaking involution on (3.17), we get\n\n\[ {a}^{ * } = {a}^{ * }{a}^{n}{a}^{ * }{x}^{ * }. \]\n\n(3.18)\n\nAgain by (3.17), we obtain\n\n\[ {\left\lbrack... | Yes |
Theorem 3.38 \( {}^{\left\lbrack {19},\text{ Theorem 4.6 }\right\rbrack } \) Let \( a \in R, n \geq 2 \) . Consider the following conditions:\n\n(1) \( R \) is a Dedekind-finite ring.\n\n(2) \( a \in a{\left( {a}^{ * }\right) }^{n}{aR} \) if and only if \( a \in {Ra}{\left( {a}^{ * }\right) }^{n}a \) .\n\n(3) \( a{a}^{... | Proof \( \;\left( 1\right) \Rightarrow \left( 2\right) . \) Since \( \;R\; \) is a Dedekind-finite ring, it follows that \( \;{\left( {a}^{ * }\right) }^{n}a + 1{-a}^{ - }a \) is right invertible if and only if \( {\left( {a}^{ * }\right) }^{n}a + 1 - {a}^{ - }a \) is left invertible. Hence, \( a \in a{\left( {a}^{ * }... | Yes |
Theorem 4. \( {\mathbf{6}}^{\left\lbrack {4.6},\text{ Theorem }{4.8}\right\rbrack } \) Let \( a, b \in {R}_{ \oplus } \) with dual core inverses \( {a}_{ \oplus } \) and \( {b}_{ \oplus }, \) respectively. If \( {ab} = 0 \) and \( a{b}^{ * } = 0 \), then \( a + b \in {R}_{ \oplus } \) . Moreover, | \[ {\left( a + b\right) }_{ \circledast } = {a}_{ \circledast } + {b}_{ \circledast }\left( {1 - a{a}_{ \circledast }}\right) . \] | No |
Theorem 4.11 \( {}^{\left\lbrack {39},\text{ Theorem 4.4 }\right\rbrack } \) Let \( a, b \in {R}^{ \oplus } \) . If \( b{a}^{ \oplus }a = a = a{a}^{ \oplus }b \), then \( a - b \in \) \( {R}^{ \circledast } \) and \[ {\left( a - b\right) }^{ \circledast } = a{a}^{ \circledast }{b}^{ \circledast } - {b}^{ \circledast }.... | Proof \( \; \) Since \( a = a{a}^{ \circledast }b,\;a{b}^{ \circledast } = a{a}^{ \circledast }b{b}^{ \circledast } \) . According to the proof of Theorem 4.10, we have the following equalities: \[ a{a}^{ \circledast } = a{a}^{ \circledast }b{b}^{ \circledast },\;{b}^{ \circledast }{ba} = a = a{a}^{ \circledast }{b}^{ ... | Yes |
Theorem 5. \( {1}^{\left\lbrack {17},\text{ Theorem }{2.6}\right\rbrack } \) Let \( a, p, q \in R \) be such that \( a\{ 1,4\} \neq \varnothing \) . Suppose that there exist \( {p}^{\prime },{q}^{\prime } \in R \) such that \( {p}^{\prime }{pa} = a = {aq}{q}^{\prime } \) . Then the following are equivalent:\n\n(1) \( {... | Proof By Corollary 3.6, \( {paq} \in {R}^{ \circledast } \) if and only if \( \left( {paq}\right) \{ 1,3\} \neq \varnothing \) and \( v = {paq} + 1 - {paq}{\left( paq\right) }^{\left( 1,3\right) } \) is a unit, and in this case \( {\left( paq\right) }^{ \circledast } = {v}^{-1}{paq}{\left( paq\right) }^{\left( 1,3\righ... | Yes |
Theorem 5. \( {2}^{\left\lbrack {17},\text{ Theorem }{2.7}\right\rbrack } \) Let \( a, p, q \in R \) be such that \( a\{ 1,3\} \neq \varnothing \) . Suppose that there exist \( {p}^{\prime },{q}^{\prime } \in R \) such that \( {p}^{\prime }{pa} = a = {aq}{q}^{\prime } \) . Then the following are equivalent:\n\n(1) \( {... | Proof In view of [4, Corollary 3.3], \( {\left( paq\right) }^{\# } \) exists if and only if \( u = {aqpa}{a}^{\left( 1,3\right) } + 1 - \) \( a{a}^{\left( 1,3\right) } \) is a unit if and only if \( v = {a}^{\left( 1,3\right) }{aqpa} + 1 - {a}^{\left( 1,3\right) }a \) is a unit when \( {p}^{\prime }{pa} = a = {aq}{q}^{... | Yes |
Let \( a, p, q \in R \) be such that \( a\{ 1,4\} \neq \varnothing \) and \( p, q \) be invertible. Then \( {\left( paq\right) }_{ \oplus } \) exists if and only if \( s = {aqp} + 1 - a{a}^{\left( 1,4\right) } \) is a unit and \( \left( {paq}\right) \{ 1,4\} \neq \varnothing \), or, equivalently, \( t = {qpa} + 1 - {a}... | In this case, \[ {\left( paq\right) }_{ \circledast } = {\left( paq\right) }^{\left( 1,4\right) }{paqp}{s}^{-2}{aq} = {\left( paq\right) }^{\left( 1,4\right) }{paqpa}{t}^{-2}q. \] | Yes |
Theorem 5. \( {7}^{\left\lbrack {17},\text{ Theorem }{2.12}\right\rbrack } \) Let \( a, p, q \in R \) be such that \( a \in {R}^{ \dagger } \) . Suppose that there exist \( {p}^{\prime },{q}^{\prime } \in R \) such that \( {p}^{\prime }{pa} = a = {aq}{q}^{\prime } \) . Then the following are equivalent:\n\n(1) \( {\lef... | In this case,\n\n\[ {\left( paq\right) }^{ \circledast } = {w}^{-2}{paqpa}{u}^{-1}{\left( pa\right) }^{ * } = {paq}{v}^{-2}{pa}{u}^{-1}{\left( pa\right) }^{ * },\]\n\n\[ {\left( paq\right) }_{ \circledast } = {\left( aq\right) }^{ * }{s}^{-1}{aq}{w}^{-2}{paq} = {\left( aq\right) }^{ * }{s}^{-1}{aqpaq}{v}^{-2}. \] | Yes |
Theorem 5. \( {8}^{\left\lbrack {17},\text{ Thoerme }{3.1}\right\rbrack }\; \) Suppose \( a \in R \) is regular with \( {a}^{ - } \in a\{ 1\} \) and \( p, q \in \) \( R \) ,\n\n\[ \nu = {aq}{\left( paq\right) }^{ * }{pa}{a}^{ - } + 1 - a{a}^{ - },\;v = {a}^{ - }{aq}{\left( paq\right) }^{ * }{pa} + 1 - {a}^{ - }a, \]\n\... | Proof From [4, Theorem 3.1], \( {sR} = R \) if and only if \( {aqpaR} = {aR} \) if and only if \( {tR} = R \) . By symmetry it follows that \( {Rs} = R \) is equivalent to \( R \) a \( {qpa} = {Ra} \), or equivalently, \( {Rt} = R \) . So \( s \) is a unit if and only if \( t \) is a unit if and only if \( {Ra} = {Raqp... | Yes |
Theorem 6.3. \( {}^{\left( \left\lbrack {13},\text{ Theorem }{2.1}\right\rbrack \right) } \) If \( L = \left\lbrack \begin{matrix} 0 & a \\ {I}_{n - 1} & \beta \end{matrix}\right\rbrack \) with \( \beta = {\left\lbrack {a}_{1},{a}_{2},\cdots ,{a}_{n - 1}\right\rbrack }^{\mathrm{T}} \) is a companion matrix over \( R \)... | Proof Consider the factorization of \( L \) :\n\n\[ \nL = \left\lbrack \begin{matrix} 0 & a \\ {I}_{n - 1} & \beta \end{matrix}\right\rbrack = \left\lbrack \begin{matrix} 0 & 1 \\ {I}_{n - 1} & 0 \end{matrix}\right\rbrack \left\lbrack \begin{matrix} {I}_{n - 1} & 0 \\ 0 & a \end{matrix}\right\rbrack \left\lbrack \begin... | Yes |
Theorem 7.1. \( {}^{\left\lbrack {21},\text{ Theorem }{2.1}\right\rbrack }\; \) Let \( \;\mathcal{C} \) be an additive category with an involution \( * \) . Suppose that \( \varphi : X \rightarrow X \) is a morphism of \( \mathcal{C} \) with core inverse \( {\varphi }^{ \circledast } \) and \( \eta : X \rightarrow X \)... | Proof By \( \left\lbrack {3,\text{ Lemma }{14.1}}\right\rbrack ,{\left( {1}_{X} + {\varphi }^{ \circledast }\eta \right) }^{-1}{\varphi }^{ \circledast } \in f\{ 1,2\} .\n\nLet \( {f}_{0} = {\left( {1}_{X} + {\varphi }^{ \circledast }\eta \right) }^{-1}{\varphi }^{ \circledast } = \alpha {\varphi }^{ \circledast } = {\... | Yes |
Theorem 7.4. \( {}^{\left\lbrack {21},\text{ Theorem }{3.1}\right\rbrack }\; \) If \( a \in {R}^{ \circledast } \) with core inverse \( {a}^{ \circledast } \) and \( j \in J\left( R\right) \), then\n\n\[ a + j \in {R}^{ \circledast }\text{ if and only if }\varepsilon = \left( {1 - a{a}^{ \circledast }}\right) j{\left( ... | Proof Remark first that, if \( j \in J\left( R\right) \), then \( 1 + {a}^{ \circledast }j \in {R}^{-1} \) and \( {j}^{ * } \in J\left( R\right) \).\n\nSet \( \phi = {\left( a + j\right) }^{ \circledast } \), then \( \phi \in \varepsilon \{ 1\} \) by [3, Lemma 14.1]. This shows that the element \( \varepsilon \in J\lef... | Yes |
Theorem 8.2. \( {}^{\left\lbrack {23},\text{ Theorem }{3.5}\right\rbrack }\; \) Let \( A \in {\mathbb{C}}^{n \times n} \) with ind \( \left( A\right) = k \) . Then the core-EP inverse of \( A \) always exists and it is unique. Further, \( {A}^{\circledR } = {A}^{k}{\left( {\left( {A}^{ * }\right) }^{k}{A}^{k + 1}\right... | Proof From the definition, we know that core-EP inverse of \( A \), is an outer inverse of \( A \) with its column and row spaces are identical, and equals \( \mathcal{R}\left( {A}^{k}\right) \) . By [22, Lemma 4.3], we get the core-EP inverse of \( A \) exists if and only if \( \operatorname{rank}\left( {{\left( {A}^{... | Yes |
Theorem 8.3. \( {}^{\left\lbrack {11},\text{ Theorem 2.3}\right\rbrack } \) Let \( A \in {\mathbb{C}}^{n \times n} \) with ind \( \left( A\right) = k \) . Then \( {A}^{\text{① }} = \) \( {A}^{D}{A}^{k}{\left( {A}^{k}\right) }^{ \dagger } \) . | Proof Let \( X = {A}^{D}{A}^{k}{\left( {A}^{k}\right) }^{ \dagger } \) . Then it is easy to check that \( X \) is an outer inverse of \( A \) and \( \mathcal{R}\left( X\right) = \mathcal{R}\left( {X}^{ * }\right) = \mathcal{R}\left( {A}^{k}\right) \) . | No |
Theorem 8.4. Let \( A \in {\mathbb{C}}^{n \times n} \) be of the form (1.1). Then\n\n\[ \n{A}^{ \oplus } = U\left\lbrack \begin{matrix} {\left( \sum K\right) }^{ \oplus } & 0 \\ 0 & 0 \end{matrix}\right\rbrack {U}^{ * }\n\] | Proof Let \( X = U\left\lbrack \begin{matrix} {\left( \sum K\right) }^{\left( 1\right) } & 0 \\ 0 & 0 \end{matrix}\right\rbrack {U}^{ * } \) . Then it is easy to show that there exits \( k \geq \operatorname{ind}\left( {\sum K}\right) \) such that \( {XAX} = X, X{A}^{k + 1} = {A}^{k} \) and \( {\left( AX\right) }^{ * }... | Yes |
Lemma 8.5. \( {}^{\left\lbrack 6\right\rbrack } \) Let \( {\lambda }_{1} \neq {\lambda }_{2}, a \in \mathbb{C} \) . Then \( \left\lbrack \begin{matrix} {\lambda }_{1} & a \\ 0 & {\lambda }_{2} \end{matrix}\right\rbrack \) is unitary similar to \( \left\lbrack \begin{matrix} {\lambda }_{2} & b \\ 0 & {\lambda }_{1} \end... | Proof Let \( k = \frac{a}{\lambda }, Q = \frac{1}{\sqrt{{k}^{ * }k + 1}}\left\lbrack \begin{matrix} k & 1 \\ 1 & - {k}^{ * } \end{matrix}\right\rbrack \) . Then \( Q \) is unitary matrix and \( {Q}^{ * }\left\lbrack \begin{matrix} {\lambda }_{1} & a \\ 0 & {\lambda }_{2} \end{matrix}\right\rbrack Q = \left\lbrack \begi... | Yes |
Theorem 8.6 \( {}^{\left\lbrack 6\right\rbrack } \) Let \( A \in {\mathbb{C}}^{n \times n} \) . Then \( A = U\left\lbrack \begin{matrix} D & L \\ 0 & N \end{matrix}\right\rbrack {U}^{ * } \), where \( U \in {\mathbb{C}}^{n \times n} \) is unitary, \( D \in {\mathbb{C}}^{t \times t} \) is nonsingular or zero, \( N \in {... | Proof From Schur’s lemma, there exists a unitary matrix \( U \) such that \( {U}^{ * }{AU} = \n\n\( T \) is a upper triangular matrix. Set \( T = \left\lbrack \begin{matrix} {\lambda }_{1} & {t}_{12} & \cdots & {t}_{1n} \\ & {\lambda }_{2} & \cdots & {t}_{2n} \\ & & \ddots & \vdots \\ & & & {\lambda }_{n} \end{matrix}\... | Yes |
Theorem 8.7. \( {}^{\left\lbrack {10},\text{ Theorem 2.7 }\right\rbrack } \) Let \( A \in {\mathbb{C}}^{n \times n} \) with \( \operatorname{ind}\left( A\right) = k \) . Then \( {A}^{\text{① }} = \) \( {\left( A{P}_{{A}^{k}}\right) }^{ \dagger } \) . | Proof Let \( A \in {\mathbb{C}}^{n \times n} \) . Then \( A = U\left\lbrack \begin{matrix} D & L \\ 0 & N \end{matrix}\right\rbrack {U}^{ * } \), and \( {A}^{\left( \oplus \right) } = U\left\lbrack \begin{matrix} {D}^{-1} & 0 \\ 0 & 0 \end{matrix}\right\rbrack {U}^{ * } \) . \n\nSince \( A{P}_{{A}^{k}} = U\left\lbrack ... | No |
Theorem 8.8. \( {}^{\left\lbrack {11},\text{ Theorem }{5.2}\right\rbrack } \) Let \( A \in {\mathbb{C}}^{n \times n} \) be of the form (8.69). Then\n\n\[ \n{A}^{ \oplus } = {Q}_{1}{D}^{-1}{\left( {Q}_{1}^{ * }{Q}_{1}\right) }^{-1}{Q}_{1}^{ * }\n\] | Proof From \( P = \left\lbrack \begin{array}{l} {P}_{1} \\ {P}_{2} \end{array}\right\rbrack ,{P}^{-1} = \left\lbrack {{Q}_{1},{Q}_{2}}\right\rbrack \) and \( A = {P}^{-1}\left\lbrack \begin{array}{ll} D & O \\ O & N \end{array}\right\rbrack P \), we have\n\n\[ \n{A}^{D} = {P}^{-1}\left\lbrack \begin{matrix} {D}^{-1} & ... | Yes |
Theorem 8.9. Let \( A \in {\mathbb{C}}^{n \times n} \) with \( \operatorname{ind}\left( A\right) = k \) . We perform a sequence of full-rank decompositions:\n\n\[ \nA = {B}_{1}{G}_{1},{G}_{i}{B}_{i} = {B}_{i + 1}{G}_{i + 1},\;i = 1,2,\ldots, k - 1.\n\]\n\nThen\n\[ \n{A}^{ \oplus } = \left\{ \begin{array}{rr} B{\left( {... | Proof Set \( B = {B}_{1}{B}_{2}\cdots {B}_{k} \) and \( G = {G}_{k}\cdots {G}_{2}{G}_{1} \) . Then\n\n\[ \n\begin{matrix} {A}^{\circledR } & = & {A}^{D}{A}^{k}{\left( {A}^{k}\right) }^{ \dagger } = B{\left( {G}_{k}{B}_{k}\right) }^{-k - 1}{GBG}{G}^{ * }{\left( G{G}^{ * }\right) }^{-1}{\left( {B}^{ * }B\right) }^{-1}{B}... | Yes |
Lemma 9.3. \( {}^{\left\lbrack {11},\text{ Lemma }{2.1}\right\rbrack }\; \) Let \( a \in R \) . If there exist \( x \in R \) and \( k \in {\mathbb{N}}^{ + } \) such that \( x{a}^{k + 1} = {a}^{k} \) and \( a{x}^{2} = x \), then\n\n(1) \( {ax} = {a}^{m}{x}^{m} \) for arbitrary positive integer \( m \) .\n\n(2) \( x \) a... | Proof (1). Since \( a{x}^{2} = x \), we have \( {ax} = a\left( {a{x}^{2}}\right) = {a}^{2}{x}^{2} = {a}^{2}\left( {a{x}^{2}}\right) x = {a}^{3}{x}^{3} = \) \( \cdots = {a}^{m}{x}^{m} \) for arbitrary positive integer \( m \) .\n\n(2). As \( {ax} = {a}^{k + 1}{x}^{k + 1} = {a}^{k}{x}^{k} \) follows from (1), we have\n\n... | Yes |
Theorem 9.4. \( {}^{\left\lbrack {11},\text{ Theorem }{2.2}\right\rbrack } \) Let \( a \in R \) . Then \( a \) has at most one pseudo core inverse in \( R \) . | Proof Suppose \( x \) and \( y \) satisfy condition (1)-(3) of Definition 9.1, with \( m, n \) as pseudo core index respectively. Let \( k = \max \{ m, n\} \) . Then by Lemma 9.3,\n\n\[ \n{x}^{k + 1}{a}^{k} = {x}^{m + 1}{a}^{m} = {a}^{D} = {y}^{n + 1}{a}^{n} = {y}^{k + 1}{a}^{k}, \n\]\n\nand\n\n\[ \n{a}^{k}{x}^{k}{a}^{... | Yes |
Theorem 9.5 \( {}^{\left\lbrack {11},\text{ Theorem 2.3}\right\rbrack }\; \) Let \( a \in R \) and \( m \in {\mathbb{N}}^{ + } \) with \( m \geq k. \) Then \( a \in {R}^{\bigodot } \) with \( I\left( a\right) = k \) if and only if \( a \in {R}^{D} \) with \( \operatorname{ind}\left( a\right) = k \) and \( {a}^{m} \in {... | Proof Suppose \( a \in {R}^{ \odot } \) with \( I\left( a\right) = k \) . By Lemma 9.3, we have \( a \in {R}^{D} \) with ind \( \left( a\right) \leq k \) and \( {a}^{m} \in {R}^{\{ 1,3\} } \) for arbitrary \( m \geq k \) . ind \( \left( a\right) < k \) would mean that \( x{a}^{k} = \) \( {a}^{k - 1} \), which is contra... | Yes |
Theorem 9.7. \( {}^{\left\lbrack {11},\text{ Theorem }{2.5}\right\rbrack }\; \) Let \( a \in R. \) Then \( a \in {R}^{\circledR } \) if and only if \( {a}^{m} \in {R}^{\circledR } \) for some positive integer \( m \) . In this case, \( {\left( {a}^{m}\right) }^{ \circledast } = {\left( {a}^{\circledR }\right) }^{m} \) ... | Proof Suppose \( {a}^{ \odot } = x \) with \( I\left( a\right) = m \) . Setting \( y = {x}^{m} \), by the definition of pseudo core inverse and by Lemma 9.3, we can check that\n\n\[ y{\left( {a}^{m}\right) }^{2} = {x}^{m}{\left( {a}^{m}\right) }^{2} = \left( {{x}^{m}{a}^{m}}\right) {a}^{m} = \left( {{x}^{m + 1}{a}^{m +... | Yes |
Theorem 9.9. \( {}^{\left\lbrack {11},\text{ Theorem }{2.7}\right\rbrack } \) Let \( a \in R \) . If \( a \in {R}^{\circledR } \), then \( {a}^{\circledR } \in {R}^{\circledR } \) . In fact \( {a}^{\circledR } \) is core invertible whenever it exists, and \( {\left( {a}^{\circledR }\right) }^{\circledR } = {\left( {a}^... | Proof To prove this, one has merely to verify that if \( x \) satisfies condition (1)-(3) of Definition 9.1, then \( y = {a}^{2}x \) satisfies\n\n\[ \ny{x}^{2} = x, x{y}^{2} = y,{\left( xy\right) }^{ * } = {xy}.\n\]\n\nHere we omit the details. | No |
Proposition 9.10. \( {}^{\left\lbrack {11},\text{ Proposition }{2.8}\right\rbrack } \) Let \( a \in {R}^{\circledR } \) . Then \( {\left( {\left( {a}^{\circledR }\right) }^{\circledR }\right) }^{\circledR } = {a}^{\circledR } \). | Proof Suppose \( a \in {R}^{ \odot } \) with \( I\left( a\right) = m \) . By Theorem 9.9, we have\n\n\[ \n{\left( {\left( {a}^{ \oplus }\right) }^{ \oplus }\right) }^{ \oplus } = {\left( {a}^{ \oplus }\right) }^{2}{\left( {a}^{ \oplus }\right) }^{ \oplus } = {\left( {a}^{ \oplus }\right) }^{2}{a}^{2}{a}^{ \oplus }.\n\]... | Yes |
Theorem 9.11. \( {}^{\left\lbrack {11},\text{ Theorem 2.9 }\right\rbrack } \) Let \( a \in R \) . Then \( a \in {R}^{\text{® }} \) with \( I\left( a\right) = m \) if and only if \( a \in {\mathbb{R}}^{D} \) with ind \( \left( a\right) = m \) and \( {c}_{a} \in {\mathbb{R}}^{ \circledast }. \) In this case, \( {a}^{\cir... | Proof Supposing \( a \in {R}^{ \odot } \) with \( I\left( a\right) = m \), we have \( a \in {R}^{D} \) with \( \operatorname{ind}\left( a\right) = m \) by Theorem 9.5 and\n\n\[ \n{a}^{\circledR }{c}_{a}^{2} = {a}^{\circledR }{\left( a{a}^{D}a\right) }^{2} = {a}^{\circledR }{a}^{3}{a}^{D} = \n\]\n\n\[ \n{a}^{\circledR }... | Yes |
Theorem 9.12. \( {}^{\left\lbrack {11},\text{ Theorem }{2.10}\right\rbrack } \) Let \( a, x \in R \) . Then the following conditions are equivalent:\n\n(1) \( {a}^{ \odot } = x \) .\n\n(2) \( {ax} = x \), and \( {xR} = {x}^{ * }R = {a}^{m}R \) for some positive integer \( m \) .\n\n(3) \( {xax} = x,{xR} = {a}^{m}R \) a... | Proof (1) \( \Rightarrow \) (2). Suppose \( {a}^{ \odot } = x \) with \( I\left( a\right) = m \) . Then by the definition of pseudo core inverse, we have \( {xax} = x \).\n\n\[ \n{xR} \subseteq {x}^{ * }R\text{ since }x = a{x}^{2} = \left( {ax}\right) x = {\left( ax\right) }^{ * }x = {x}^{ * }{a}^{ * }x \in {x}^{ * }R;... | Yes |
Theorem 9.13. \( {}^{\left\lbrack {11},\text{ Theorem 2.12 }\right\rbrack } \) Let \( a \in R \) . Then \( a \in {R}^{\text{(C) }} \) if and only if there exist \( u, v \in R \) and positive integer \( p, q \) such that\n\n\[ \n{a}^{p} = u{\left( {a}^{ * }\right) }^{p + 1}{a}^{p},{a}^{q} = v{a}^{q + 1}.\n\]\n | Proof Suppose \( {a}^{ \odot } = x \) with \( I\left( a\right) = m \), then\n\n\[ \nx{a}^{m + 1} = {a}^{m}, a{x}^{2} = x,{\left( ax\right) }^{ * } = {ax}.\n\]\n\nBy Lemma 9.3, we have \( {a}^{m}{x}^{m}{a}^{m} = {a}^{m},\;({a}^{m}{x}^{m}{)}^{ \ast } = {a}^{m}{x}^{m} \) which yields \( {a}^{m} = ({a}^{m}{x}^{m}{)}^{ \ast... | Yes |
Proposition 9.15. \( {}^{\left\lbrack {11},\text{ Proposition }{2.14}\right\rbrack } \) Let \( a, x \in R \) . Then \( \left( 1\right) \Leftrightarrow \left( 2\right) ,\left( 3\right) \Rightarrow \left( 1\right) \) , where\n\n(1) \( {a}^{ \circledast } = x \) .\n\n(2) \( x{a}^{m + 1} = {a}^{m}, a{x}^{2} = x,{\left( ax\... | Proof \( \left( 1\right) \Rightarrow \left( 2\right) \) . It is clear.\n\n\( \left( 2\right) \Rightarrow \left( 1\right) \) . By Theorem 9.3, we have\n\n\[ x = {xax} = x{a}^{m}{x}^{m} = x\left( {x{a}^{m + 1}}\right) {x}^{m} = {x}^{2}{a}^{m + 1}{x}^{m}, \]\n\nthen \( x = {x}^{m + 1}{a}^{2m}{x}^{m} \) by induction. So \(... | Yes |
Theorem 9.18. \( {}^{\left\lbrack {37},\text{ Theorem }{3.3}\right\rbrack } \) Let \( a \in R \) and \( k \in {\mathbb{N}}^{ + } \) . Then the following conditions are equivalent:\n\n(1) \( a \in {R}^{ \odot } \) with \( I\left( a\right) \leq k \) .\n\n(2) There exists a projection \( p \in R \) such that \( {pR} = {a}... | Proof \( \;\left( 1\right) \Rightarrow \left( 2\right) .\; \) Assume that \( a \in \mathbb{R} \) is pseudo core invertible and \( I\left( a\right) \leq k. \) Set \( p = a{a}^{\circledR } \) . By Lemma 9.3 (1), we have \( p = a{a}^{\circledR } = {a}^{k}{\left( {a}^{\circledR }\right) }^{k} = {a}^{k + 1}{\left( {a}^{\cir... | Yes |
Theorem 9.20. \( {}^{\left\lbrack {37},\text{ Theorem }{3.5}\right\rbrack } \) Let \( a \in {R}^{\circledR } \) with \( I\left( a\right) \leq k \) . Then \( {a}^{\circledR } \) is EP and \( {\left( {a}^{\circledR }\right) }^{ \dagger } = {\left( {a}^{\circledR }\right) }^{\# } = {a}^{k + 1}{\left( {a}^{k}\right) }^{\le... | Proof Suppose that \( x = {a}^{k + 1}{\left( {a}^{k}\right) }^{\left( 1,3\right) } \) . From Theorem 9.5, we know that \( {a}^{\circledR } = \) \( {a}^{D}{a}^{k}{\left( {a}^{k}\right) }^{\left( 1,3\right) } \) . Then we have\n\n\[ \n{a}^{\circledR }x{a}^{\circledR } = {a}^{\circledR }{a}^{k + 1}{\left( {a}^{k}\right) }... | Yes |
Theorem 9.24. \( {}^{\left\lbrack {37},\text{ Theorem }{3.9}\right\rbrack } \) Let \( a \in {R}^{\circledR } \) with \( I\left( a\right) = k \) . Then the following conditions are equivalent:\n\n(1) \( {a}^{k} \) is EP.\n\n(2) There exists a unit \( u \in R \) such that \( {a}^{\circledR } = u{a}^{k} \) .\n\n(3) There ... | Proof \( \;\left( 1\right) \Rightarrow \left( 2\right) . \) Suppose that \( {a}^{k} \) is EP. We know that \( a \in {\mathbb{R}}^{\bigodot } \) with \( I\left( a\right) = k. \) By Theorem 9.5 and [4, Lemma 11.6], we have\n\n\[ \n{a}^{\circledR } = {a}^{D}{a}^{k}{\left( {a}^{k}\right) }^{\left( 1,3\right) } = {a}^{D}{a}... | Yes |
Proposition 10.1. \( {}^{\left\lbrack {11},\text{ Proposition }{4.2}\right\rbrack } \) Let \( a, x \in R \) with \( {ax} = {xa},{a}^{ * }x = x{a}^{ * } \) . If \( a \in {R}^{ \odot } \), then \( {a}^{ \odot }x = x{a}^{ \odot } \) . | Proof Supoose that \( \operatorname{ind}\left( a\right) = m \) . From the condition \( {ax} = {xa},{a}^{ * }x = x{a}^{ * } \), we have \( {a}^{m}x = x{a}^{m},{\left( {a}^{m}\right) }^{ * }x = x{\left( {a}^{m}\right) }^{ * } \) . Then\n\n\[ \n{a}^{\circledR }x = {a}^{\circledR }a{a}^{\circledR }x = {a}^{\circledR }{a}^{... | Yes |
Theorem 10.2. \( {}^{\left\lbrack {11},\text{ Theorem }{4.3}\right\rbrack }\; \) Let \( a, b \in {R}^{\text{® }} \) with \( {ab} = {ba} \) and \( a{b}^{ * } = {b}^{ * }a \) . Then \( {\left( ab\right) }^{ \odot } = {a}^{ \odot }{b}^{ \odot } = {b}^{ \odot }{a}^{ \odot }. | Proof From Proposition 10.1, it follows that\n\n\[ \n{b}^{\circledR }a = a{b}^{\circledR }\text{ and }{a}^{\circledR }b = b{a}^{\circledR }. \]\n\nThe condition \( {b}^{ * }a = a{b}^{ * },{a}^{ * }{b}^{ * } = {b}^{ * }{a}^{ * } \) ensures that \( {b}^{ * }{a}^{\circledR } = {a}^{\circledR }{b}^{ * } \), which together ... | Yes |
Theorem 11.1. \( {}^{\left\lbrack {30},\text{ Theorem }{3.3}\right\rbrack } \) If \( a \in {R}^{D} \), then the following conditions are equivalent:\n\n(1) \( a \in {R}^{ \odot } \) .\n\n(2) \( a{a}^{D} \in {R}^{\{ 1,3\} } \) .\n\n(3) \( {a}^{\pi } \in {R}^{\{ 1,4\} } \) .\n\nIn this case, \( a{a}^{\circledR } \in \lef... | Proof (1) \( \Rightarrow \) (2). Suppose that \( I\left( a\right) = m \) . Then we have\n\n\[ \n{a}^{D}a{a}^{\circledR } = {a}^{D}{a}^{m + 1}{\left( {a}^{\circledR }\right) }^{m + 1} = {a}^{m}{\left( {a}^{\circledR }\right) }^{m + 1} = {a}^{\circledR },\n\]\n\n(11.71)\n\nand\n\n\[ \n{a}^{\circledR }a{a}^{D} = {a}^{\cir... | Yes |
Theorem 11.2. \( {}^{\left\lbrack {30},\text{ Theorem }{3.4}\right\rbrack } \) If \( a \in {R}^{D} \), then the following conditions are equivalent:\n\n(1) \( a \in {R}^{ \oplus } \).\n\n(2) \( a{a}^{D} \in {R}^{\{ 1,4\} } \).\n\n(3) \( {a}^{\pi } \in {R}^{\{ 1,3\} } \).\n\nIn this case, \( {a}^{\circledR }a \in \left(... | \[ {a}_{\circledR } = {\left( a{a}^{D}\right) }^{\left( 1,4\right) }{a}^{D} = \left( {1 - {a}^{\pi }{\left( {a}^{\pi }\right) }^{\left( 1,3\right) }}\right) {a}^{D}, \] for any \( {\left( a{a}^{D}\right) }^{\left( 1,4\right) } \in \left( {a{a}^{D}}\right) \{ 1,4\} \) and \( {\left( {a}^{\pi }\right) }^{\left( 1,3\right... | Yes |
Lemma 11.4. \( {}^{\left\lbrack 4,\text{ Theorem 11.4 }\right\rbrack } \) Let \( {a}_{1},{a}_{2}, d \in R \) . If \( {a}_{1},{a}_{2} \in {R}^{D} \) and \( d{a}_{1} = {a}_{2}d \) , then \( {a}_{2}^{D}d = d{a}_{1}^{D} \) . | In fact, noting that \( {\alpha a} = {a\beta } \), we have \( {\alpha }^{D}a = a{\beta }^{D} \) by Lemma 11.4. Thus,\n\n\[ {\beta }^{D} = \left( {\beta {\beta }^{D} + {\beta }^{D} - \beta {\beta }^{D}}\right) \]\n\n\[ = \left( {\beta {\beta }^{D} + \left( {1 - \beta }\right) {\beta }^{D}}\right) \]\n\n\[ = \left( {\bet... | No |
Theorem 11.6. \( {}^{\left\lbrack {30},\text{ Theorem }{3.10}\right\rbrack } \) Let \( a, b \in R \) . If \( \alpha = 1 - {ab} \in {R}^{ \odot } \), then the following conditions are equivalent:\n\n(1) \( \beta = 1 - {ba} \in {R}^{ \odot } \) .\n\n(2) \( b{\alpha }^{\pi }{ra} \in {R}^{\{ 1,4\} } \) .\n\n(3) \( v = \lef... | Proof Since \( \alpha = 1 - {ab} \in {R}^{ \odot } \), suppose that \( I\left( \alpha \right) = k \), it follows that \( \alpha \in {R}^{D} \) with \( i\left( \alpha \right) = k \) by Theorem 9.5 and \( {\alpha }^{\pi } \in {R}^{\{ 1,4\} } \) with \( 1 - \alpha {\alpha }^{\circledR } \in \left( {\alpha }^{\pi }\right) ... | Yes |
Lemma 13.4 \( {}^{\left\lbrack 5,\text{ Lemma 3.1 }\right\rbrack } \) If \( a \in R \) with \( {a}^{\circledR } \) exists and \( j \in J\left( R\right) \), then \( 1 - \sigma \) is invertible. | Proof Since\n\n\[ \sigma = \alpha {a}^{\circledR }a{\alpha }^{-1}\left( {1 - a{a}^{\circledR }}\right) \beta \]\n\n\[ = \alpha {a}^{\circledR }a\left( {1 + {a}^{\circledR }j}\right) \left( {1 - a{a}^{\circledR }}\right) \beta \]\n\n\[ = \alpha {a}^{\circledR }a\left( {1 - a{a}^{\circledR }}\right) \beta + \alpha {a}^{\... | Yes |
Theorem 13.5 \( {}^{\left\lbrack 5,\text{ Theorem 3.2}\right\rbrack } \) Suppose that \( a \in R \) with \( {a}^{\circledR } \) exists and \( j \in J\left( R\right) \) . Let \( f = a + j - \varepsilon \) . If \( j\left( {{a}^{\circledR }a - 1}\right) a = 0 \) and \( a\left( {{a}^{\circledR }a - 1}\right) j = 0 \), then... | Proof By Lemma 13.4, we know that \( 1 - \sigma \) is invertible. Since \( j \in J\left( R\right) \), we obtain that \( \gamma = \alpha \left( {1 - {a}^{\circledR }a}\right) {ja}@\beta \in J\left( R\right) \) . So, we have \( 1 - \gamma \) is invertible. Since \( {\beta }^{ * }\left( {1 - a{a}^{\circledR }}\right) j{a}... | Yes |
Proposition 14. \( {1}^{\left\lbrack {12},\text{ Proposition }1\right\rbrack } \) Let \( A \in {R}^{m \times n} \) with \( A\{ 1,4\} \neq \varnothing \), and \( T \) a square matrix with \( {T}^{k} = {PAQ} \) for some positive integer \( k \) and matrices \( P \) and \( Q \) . Suppose \( {P}^{\prime }{PA} = A = {AQ}{Q}... | Proof Applying [4, Theorem 11.6], \( {T}^{D} \) exists if and only if \( {A}^{\left( 1,4\right) }{AQTPA} + I - \) \( {A}^{\left( 1,4\right) }A \) is invertible, in which case, \( {T}^{D} = {PA}{\left\lbrack {A}^{\left( 1,4\right) }AQTPA + I - {A}^{\left( 1,4\right) }A\right\rbrack }^{-1}Q \) .\n\nIf \( A\{ 1,4\} \neq \... | Yes |
Theorem 14.3 \( {}^{\left\lbrack {12},\text{ Proposition }3\right\rbrack }\; \) Let \( A \) be regular with \( {A}^{ - } \in A\{ 1\} \), and \( T \) a square matrix with \( {T}^{k} = {PAQ} \) for some positive integer \( k \) and matrices \( P, Q \) . Suppose\n\n\[ \nU = {A}^{ - }{AQTPA} + I - {A}^{ - }A, V = {AQ}{\lef... | Proof By [4, Theorem 16.1], \( {T}^{D} \) exists with ind \( \left( T\right) \leq k, A = {P}^{\prime }{PA} = {AQ}{Q}^{\prime } \) for some matrices \( {P}^{\prime } \) and \( {Q}^{\prime } \) if and only if \( U \) is invertible. Next, we prove that \( U \) is invertible if and only if \( {XAQTPA} = A = {AQTPAY} \) for... | Yes |
Theorem 14.5 Let \( A \) be a matrix with \( A\{ 1,3\} \neq \varnothing \) and \( T \) a square matrix with \( {T}^{k} = {PAQ} \) for some positive integer \( k \), matrices \( P \) and \( Q. \) Suppose \( {P}^{\prime }{PA} = A = {AQ}{Q}^{\prime } \) for some matrix \( {P}^{\prime } \) and \( {Q}^{\prime } \), and\n\n\... | Proof Applying Theorem 9.5, \( {T}^{\circledR } \) exists if and only if \( {T}^{D} \) exists and \( {T}^{k}\{ 1,3\} \neq \varnothing \) .\n\nSince \( {P}^{\prime }{PA} = A = {AQ}{Q}^{\prime } \), we get \( {T}^{D} \) exists with \( \operatorname{ind}\left( T\right) \leq k \) if and only if\n\n\( {A}^{\left( 1,3\right)... | Yes |
Corollary 14.6 Let \( A \) be a matrix with \( A\{ 1,3\} \neq \varnothing \) and \( T \) a square matrix with \( T = {PA} \) for some matrix \( P \) . Suppose \( {P}^{\prime }{PA} = A \) for some matrix \( {P}^{\prime } \) . Then the following are equivalent:\n\n(1) \( {T}^{ \circledast } \) exists.\n\n(2) \( U \) is i... | The following result characterizes the dual pseudo core invertibility of \( T \) . | No |
Theorem 15. \( {\mathbf{1}}^{\left\lbrack {12},\text{Theorem }1\right\rbrack }\; \) Let \( T = \left\lbrack \begin{array}{ll} 0 & 0 \\ b & d \end{array}\right\rbrack \in {R}^{2 \times 2} \) . Then \( {T}^{\text{® }} \) exists if and only if \( {d}^{ \oplus } \) exists. Moreover, \( {T}^{ \oplus } = \left\lbrack \begin{... | Proof Suppose \( {d}^{ \odot } \) exists with \( I\left( d\right) = k \), then it is easy to check that \( {T}^{ \odot } \) exists\n\nwith \( {T}^{\left( \Theta \right) } = \left\lbrack \begin{matrix} 0 & 0 \\ 0 & {d}^{\left( \Theta \right) } \end{matrix}\right\rbrack \), as well as\n\n\[ I\left( T\right) = \left\{ \be... | Yes |
Theorem 15. \( {2}^{\left\lbrack {12},\text{ Theorem }2\right\rbrack } \) Let \( T = \left\lbrack \begin{array}{ll} a & 0 \\ b & d \end{array}\right\rbrack \in {R}^{2 \times 2} \) such that \( d \in {R}^{\circledR } \cap {R}_{\circledR } \) with \( \;I\left( d\right) = k,\;{a}^{k}\{ 1,4\} \neq \varnothing ,\;c = \matho... | Proof Applying Theorem 9.14, \( d \in {R}^{\circledR } \cap {R}_{\circledR } \) with \( I\left( d\right) = k \) if and only if \( {d}^{D} \) and \( {\left( {d}^{k}\right) }^{ \dagger } \) exist. In this case, \( {d}^{k}{\left( {d}^{k}\right) }^{ \dagger } = {d}^{k}{\left( {d}^{k}\right) }^{ \odot },\left( {d}^{k + 1}\r... | No |
Theorem 15. \( {3}^{\left\lbrack {12},\text{Theorem 3}\right\rbrack } \) Let \( T = \left\lbrack \begin{array}{ll} a & 0 \\ b & d \end{array}\right\rbrack \in {R}^{2 \times 2} \) such that \( a \in {R}^{ \odot } \) with \( I\left( a\right) = k,{d}^{k}\{ 1,3\} \neq \varnothing, c = \mathop{\sum }\limits_{{i + j = k - 1}... | Proof Since \( T = \left\lbrack \begin{array}{ll} a & 0 \\ b & d \end{array}\right\rbrack \), then\n\n\[ {T}^{k} = {\left\lbrack \begin{array}{ll} a & 0 \\ b & d \end{array}\right\rbrack }^{k} = \left\lbrack \begin{matrix} {a}^{k} & 0 \\ c & {d}^{k} \end{matrix}\right\rbrack \]\n\n | Yes |
Lemma 1 Let \( Z = \mathop{\sum }\limits_{{\alpha \in {G}_{4, n}}}p\left( \alpha \right) {e}_{\alpha }^{ \bullet } \) be decomposable, and \( p\left( {1,1,1,1}\right) = 1 \), its spanning matrix is\n\n\[ A = \left( \begin{matrix} 1 & {a}_{12} & \cdots & {a}_{1i} & \cdots & {a}_{1j} & \cdots & {a}_{1n} \\ 1 & {a}_{22} &... | Proof In order to prove \( B = \left( \begin{array}{l} {\alpha }_{1} \\ {\alpha }_{2} \\ {\alpha }_{3} \end{array}\right) \) is spanning matrix of decomposable element\n\n\( W = \mathop{\sum }\limits_{{\beta = \left( {i, j, k}\right) \in {G}_{3, n}}}q\left( \beta \right) {e}_{\beta }^{ \bullet } \) in \( {V}^{\left( 3\... | Yes |
Theorem 1 Let \( Z = \mathop{\sum }\limits_{{\alpha \in {G}_{4, n}}}p\left( \alpha \right) {e}_{\alpha }^{ \bullet } \) and \( A \) are similar with lemma 1. If \( {f}_{2}\left( x\right) \) (without less of generality) has one simple root and one triple root, and assume \( {a}_{12} = \) \( {a}_{22} = {a}_{32} = a,{a}_{... | Proof By assumption, \( A \) is spanning matrix of decomposable element \( Z = \mathop{\sum }\limits_{{\alpha \in {G}_{4, n}}}p\left( \alpha \right) {e}_{\alpha }^{ \bullet } \), where \n\n\[ \nA = \left( \begin{matrix} 1 & a & \cdots & {a}_{1i} & \cdots & {a}_{1j} & \cdots & {a}_{1n} \\ 1 & a & \cdots & {a}_{2i} & \cd... | Yes |
Theorem 3 Let \( Z = \mathop{\sum }\limits_{{\alpha \in {G}_{4, n}}}p\left( \alpha \right) {e}_{\alpha }^{ \bullet } \) and \( A \) are similar with lemma 1 . If \( {f}_{2}\left( x\right) \) has two double roots, assume \( {a}_{12} = {a}_{22} = a,{a}_{32} = {a}_{42} = b\left( {b \neq a}\right) \), then sub-matrix \( B ... | Proof In order to prove \( {W}_{1} = \mathop{\sum }\limits_{{\beta \in {G}_{2, n}}}S\left( \beta \right) {e}_{\beta }^{ \bullet } \) is decomposable, and \( B = \left( \begin{array}{l} {\alpha }_{1} \\ {\alpha }_{2} \end{array}\right) \) is spanning matrix of \( {W}_{1} \), we need only to prove that we always have\n\n... | No |
Lemma 7 Suppose that the condition \( \left( \mathbf{K}\right) \) holds, \( T \in \left( {0,\infty }\right) ,\bar{g},\bar{h} \in {C}^{1}\left( \left\lbrack {-\tau, T}\right\rbrack \right) ,\bar{u} \in \) \( C\left( {\bar{\mho }}_{T}\right) \cap {C}^{1,2}\left( {\mho }_{T}\right) \) satisfies \( \bar{u} \leq {u}^{ * } \... | By the condition \( \left( \mathbf{K}\right) \), a direct computation shows that for \( \left( {t, x}\right) \in {\mathcal{O}}_{{t}^{ * }} \), there exist \( {\xi }_{1} \) between \( \bar{u}\left( {t, x}\right) ,{u}_{\varepsilon }\left( {t, x}\right) \) and \( {\xi }_{2} \) between \( \bar{u}\left( {t - \tau, x}\right)... | Yes |
Lemma 8 Suppose that the condition \( \left( \mathbf{K}\right) \) holds, \( T \in \left( {0,\infty }\right) ,\bar{g},\bar{h} \in {C}^{1}\left( \left\lbrack {-\tau, T}\right\rbrack \right) ,\bar{u} \in \) \( C\left( {\bar{\mho }}_{T}\right) \cap {C}^{1,2}\left( {\mho }_{T}\right) \) with \( {\mho }_{T} = \left\{ {\left(... | A minor modification of the proof of Lemma 7 yields Lemma 8. | No |
Lemma 10 Assume that the condition \( \left( \mathbf{K}\right) \) holds. Let \( \left( {u, g, h}\right) \) be a solution of the problem (1.6). Then the following three assertions are equivalent:\n\n(i) \( {h}_{\infty } \) or \( {g}_{\infty } \) is finite;\n\n(ii) \( {h}_{\infty } - {g}_{\infty } \leq {R}^{ * } \) ;\n\n... | Proof The proof of this lemma follows that of Lemma 2.2 in [3]. | No |
Lemma 11 Assume that \( \left( \mathbf{K}\right) \) holds. Let \( \left( {u, g, h}\right) \) be a solution of (1.6). Then vanishing happens provided that \( h\left( 0\right) - g\left( 0\right) < {R}^{ * } \) and \( \parallel \phi {\parallel }_{{L}^{\infty }\left( {\left\lbrack {-\tau ,0}\right\rbrack \times \left\lbrac... | Proof Set\n\n\[ \n{h}_{0} = \frac{h\left( 0\right) - g\left( 0\right) }{2}. \]\n\nFor any \( \varepsilon > 0 \), we can find a small positive constant \( \delta \) such that\n\n\[ \n{\pi \mu \delta } \leq {\varepsilon }^{2}{h}_{0}^{2} \]\n\n\[ \nf\left( {u\left( {t, x}\right), u\left( {t - \tau, x}\right) }\right) \leq... | Yes |
Example 3.1 Let \( f\left( {x, u}\right) = {\lambda h}\left( x\right) {u}^{q - 1} \) in (1.1), where \( \lambda, q \in \mathbb{R},0 < h\left( x\right) \in {C}^{1}\left( \bar{\Omega }\right) \) and condition \( \left( V\right) \) holds, then (1.1) has no any positive solutions under assumptions in Theorem 2.1 and the co... | In fact, since \( F\left( {x,0}\right) = 0 \), we can conclude that\n\n\[ F\left( {x, u}\right) = {\int }_{0}^{u}{\lambda h}\left( x\right) {t}^{q - 1}\mathrm{\;d}t = \frac{\lambda }{q}h\left( x\right) {u}^{q} \]\n\nand\n\n\[ {F}_{1}\left( {x, u}\right) = \frac{\lambda }{q}{u}^{q}\mathop{\sum }\limits_{{i = 1}}^{n}{V}_... | Yes |
For all integers \( n \geq 1 \), let the sequence \( {P}_{n} \) be defined by (1.7). Then\n\n\[\frac{1}{{180}{\left( n + \alpha \right) }^{4}} \leq \gamma - {P}_{n} < \frac{1}{{180}{\left( n + \beta \right) }^{4}}\]\n\n(3.1)\n\nwith the best possible constants\n\n\[\alpha = \frac{1}{\sqrt[4]{{90}\ln \left( {7/3}\right)... | Proof The inequality (3.1) can be written as\n\n\[\alpha \geq \frac{1}{\sqrt[4]{{180}\left( {\frac{1}{2}\ln \left( {{n}^{2} + n + \frac{1}{3}}\right) - \psi \left( {n + 1}\right) }\right) }} - n > \beta .\]\n\nIn order to prove (3.1), we define the function \( M\left( x\right) \) by\n\n\[M\left( x\right) = {\left( u\le... | Yes |
Theorem 2 The harmonic number has the following asymptotic expansion:\n\n\[ \n{H}_{n} \sim \frac{1}{2}\ln \left( {{n}^{2} + n + \frac{1}{3}}\right) + \gamma + \mathop{\sum }\limits_{{j = 2}}^{\infty }\frac{{a}_{j}}{{\left( n + \frac{1}{2}\right) }^{2j}} \n\] | Proof Let \( m = \frac{1}{2}n\left( {n + 1}\right) \) . The asymptotic expansion (3.4) can be written as\n\n\[ \n{H}_{n} \sim \frac{1}{2}\ln \left( {2m}\right) + \frac{1}{2}\ln \left( {1 + \frac{1}{6m}}\right) + \gamma + \mathop{\sum }\limits_{{j = 2}}^{\infty }\frac{{a}_{j}}{{\left( 2m\right) }^{j}}{\left( 1 + \frac{1... | Yes |
Theorem 3 Let \( {P}_{n} \) be defined by (1.7). Then\n\n\[ \frac{\frac{1}{180}}{{\left( n + \frac{1}{2}\right) }^{4}} - \frac{\frac{17}{4536}}{{\left( n + \frac{1}{2}\right) }^{6}} + \frac{\frac{107}{25920}}{{\left( n + \frac{1}{2}\right) }^{8}} - \frac{\frac{6467}{855360}}{{\left( n + \frac{1}{2}\right) }^{10}} \]\n\... | Proof Here we only prove the first inequality in (3.18). The proof of the second inequality in (3.18) is similar, we omit them. The lower bound of (3.18) is obtained by considering the function \( G\left( x\right) \) for \( x \geq 1 \) defined by\n\n\[ G\left( x\right) = \psi \left( {x + 1}\right) - \frac{1}{2}\ln \lef... | Yes |
Theorem 2 The period of \( {D}_{p} \) is\n\n\[ \n{T}_{p} = \left( {2 - 1}\right) \left( {3 - 1}\right) \left( {5 - 1}\right) \cdots \left( {p - 1}\right) .\n\] | Proof We prove this by induction. The case for \( {D}_{3} \) has been proved. For simplicity, we try to explain the procedure by proving the case for \( {D}_{5} \) . First, we observe that\n\n\[ \n{M}_{{p}_{n + 1}} = {M}_{{p}_{n}} - \left\{ {{p}_{n + 1}h \mid h \in {M}_{{p}_{n}}}\right\} .\n\]\n\nWe show this procedure... | No |
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