Scientific Research

An Academic Publisher

Mixed-Type Reverse Order Laws for Generalized Inverses over Hilbert Space ()

**Author(s)**Leave a comment

KEYWORDS

Cite this paper

*Applied Mathematics*,

**8**, 637-644. doi: 10.4236/am.2017.85050.

1. Introduction

The reverse order law of generalized inverses plays an important role in theoretic research and numerical computations in many areas, including the singular matrix problem and optimization problem. They have attracted considerable attention since the middle 1960s, and many interesting results have been studied, see [1] - [10] .

For convenience, we firstly introduce some notations. Let H and K be infinite dimensional Hilbert spaces and $B\left(H,K\right)$ be the set of all bounded linear operators from H to K and abbreviate $B\left(K,H\right)$ to $B\left(H\right)$ if $K=H$ . For an operator $A\in B\left(H,K\right)$ , $N\left(A\right)$ and $R\left(A\right)$ are the null space and the range of A, respectively. Denote by A* the adjoint of A. Recall that $A\in B\left(H,K\right)$ has a Moore-Penrose inverse if there exists an operator $G\in B\left(K,H\right)$ satisfies the following four equations, which is said to be the Moore-Penrose conditions:

$\left(1\right)\text{\hspace{0.17em}}AGA=A;\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(2\right)\text{\hspace{0.17em}}GAG=G;\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(3\right)\text{\hspace{0.17em}}{\left(AG\right)}^{*}=AG;\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(4\right)\text{\hspace{0.17em}}{\left(GA\right)}^{*}=GA.$

If one exists, the Moore-Penrose inverse of A is unique and it is denoted by A^{+}. And let
$A\left\{i,j,\cdots ,k\right\}$ denote the set of all operator
$G\in B\left(K,H\right)$ which satisfy equations
$\left(i\right),\left(j\right),\cdots ,\left(k\right)$ from among the above Moore-Penrose equations. Such G will be called a
$\left\{i,j,k\right\}$ -inverse of A and will be denoted by
${A}^{\left(i,j,k\right)}$ . evidently,
$A\left\{1,2,3,4\right\}={A}^{+}$ when A has closed range.

For the Moore-Penrose inverse, Greville [2] gave the necessary and sufficient conditions for ${\left(AB\right)}^{+}={B}^{+}{A}^{+}$ on matrix algebra, this result was extended to bounded operators on Hilbert space by Izumino [4] . Subsequently, some researcher discussed the reverse order laws of other type generalized inverses, such

as $\left(AB\right)\theta =B\theta A\theta ,\text{\hspace{0.17em}}\theta \subset \left\{1,2,3,4\right\}$ [5] [6] [8] [10] . The mixed-type reverse-order

laws for AB like ${\left(AB\right)}^{+}={B}^{+}{\left({A}^{+}AB\right)}^{+}$ and ${\left(AB\right)}^{+}={\left({A}^{+}AB\right)}^{+}{A}^{+}$ were considered in [3] [4] when A and B are matrices. Motivated by this, Wang et al. [7] studied the mixed-type reverse-order laws for $A{B}^{\left(1,3\right)}$ . Yang and Liu [9] gave the

equivalent condition of $\left\{{\left(AB\right)}^{\left(1,2,i\right)}\right\}=\left\{{B}^{\left(1,2,i\right)}{\left(AB{B}^{\left(1,2,i\right)}\right)}^{\left(1,2,i\right)}\right\},\text{\hspace{0.17em}}\left(i=3,4\right)$ , by using

the extremal ranks of generalized Schur complements, when A and B are matrices. The mixed-type reverse order laws of ${\left(AB\right)}^{\left(1,3,4\right)}$ were discussed on operator space over Hilbert space [5] .

In this article, we study the mixed-type reverse order laws of ${\left(AB\right)}^{\left(1,2,i\right)}$ , ${\left(AB\right)}^{\left(1,3\right)}$ and ${\left(AB\right)}^{\left(1,3,4\right)}$ over infinite Hilbert space by using a block-operator matrix technique. For given A, B, it is shown that

$\left\{{\left(AB\right)}^{\left(1,3\right)}\right\}=\left\{{B}^{\left(1,3\right)}{\left(AB{B}^{\left(1,3\right)}\right)}^{\left(1,3\right)}\right\}$

and

$\left\{{\left(AB\right)}^{\left(1,2,i\right)}\right\}=\left\{{B}^{\left(1,2,i\right)}{\left(AB{B}^{\left(1,2,i\right)}\right)}^{\left(1,2,i\right)}\right\},\text{\hspace{0.17em}}\left(i=3,4\right)$

when the ranges of A, B, AB are closed. We generalized the results from [7] and [9] to the case of bounded linear operators on Hilbert spaces. Moreover, a new

equivalent condition of $\left\{{\left(AB\right)}^{\left(1,3,4\right)}\right\}=\left\{{B}^{\left(1,3,4\right)}{\left(AB{B}^{\left(1,3,4\right)}\right)}^{\left(1,3,4\right)}\right\}$ is given.

2. Main Results

To obtain our main results, we begin with some notations and lemmas. Let $A\in B\left(H,K\right)$ with closed range. It is well known that A has the following matrix decomposition

$A=\left[\begin{array}{cc}{A}_{1}& 0\\ 0& 0\end{array}\right]:\left[\begin{array}{c}R\left({A}^{*}\right)\\ N\left(A\right)\end{array}\right]\to \left[\begin{array}{c}R\left(A\right)\\ N\left({A}^{*}\right)\end{array}\right],$ (2.1)

where ${A}_{1}$ is invertible. Also, ${A}^{+}$ has the form

${A}^{+}=\left[\begin{array}{cc}{A}_{1}^{-1}& 0\\ 0& 0\end{array}\right]:\left[\begin{array}{c}R\left(A\right)\\ N\left({A}^{*}\right)\end{array}\right]\to \left[\begin{array}{c}R\left({A}^{*}\right)\\ N\left(A\right)\end{array}\right].$ (2.2)

The {1,2,3}- and {1,3,4}-inverse has also similarly matrix form.

Lemma 1 ( [5] ). Let $A\in B\left(H,K\right)$ have closed range. Then

$A\left\{1,3\right\}=\left\{\left[\begin{array}{cc}{A}_{1}^{-1}& 0\\ {X}_{3}& {X}_{\text{4}}\end{array}\right]:\left[\begin{array}{c}R\left(A\right)\\ N\left({A}^{*}\right)\end{array}\right]\to \left[\begin{array}{c}R\left({A}^{*}\right)\\ N\left(A\right)\end{array}\right]\right\}$ ,

$A\left\{1,2,3\right\}=\left\{\left[\begin{array}{cc}{A}_{1}^{-1}& 0\\ {X}_{3}& 0\end{array}\right]:\left[\begin{array}{c}R\left(A\right)\\ N\left({A}^{*}\right)\end{array}\right]\to \left[\begin{array}{c}R\left({A}^{*}\right)\\ N\left(A\right)\end{array}\right]\right\}$

and

$A\left\{1,3,4\right\}=\left\{\left[\begin{array}{cc}{A}_{1}^{-1}& 0\\ 0& {X}_{4}\end{array}\right]:\left[\begin{array}{c}R\left(A\right)\\ N\left({A}^{*}\right)\end{array}\right]\to \left[\begin{array}{c}R\left({A}^{*}\right)\\ N\left(A\right)\end{array}\right]\right\}.$

Let $A\in B\left(H,K\right)$ , $B\in B\left(L,H\right)$ and $AB\in B\left(L,K\right)$ with closed ranges. For convenience, denote by

$\{\begin{array}{l}{H}_{1}=R\left(B\right)\cap N\left(A\right),\\ {H}_{2}=R\left(B\right){\Theta}^{\perp}{H}_{1},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\\ {H}_{3}=N\left({B}^{*}\right)\cap N\left(A\right),\\ {H}_{4}=N\left({B}^{*}\right){\Theta}^{\perp}{H}_{3},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\end{array}$ $\{\begin{array}{l}{K}_{1}=R\left(AB\right)\\ {K}_{2}=R\left(A\right){\Theta}^{\perp}R\left(AB\right)\end{array}$ ,

and

${L}_{1}={B}^{+}\left({H}_{1}\right),\text{\hspace{0.17em}}{L}_{2}=R\left({B}^{*}\right){\Theta}^{\perp}{L}_{1}.$

then

$H={H}_{1}\oplus {H}_{2}\oplus {H}_{3}\oplus {H}_{4}$ , $K={K}_{1}\oplus {K}_{2}\oplus N\left({A}^{*}\right)$

and

$L={L}_{1}\oplus {L}_{2}\oplus N\left(B\right)$ .

Under these space decomposition, we get two useful representations of operators $A\in B\left(H,K\right)$ and $B\left(1,3,4\right)\}=\left\{{B}^{\left(1,3,4\right)}{\left(AB{B}^{\left(1,3,4\right)}\right)}^{\left(1,3,4\right)}\right\}$ ;

(2) $R\left(B{B}^{*}{A}^{*}\right)\subset R\left({A}^{*}\right).$

Proof We divide the proof into two cases.

Case 1. ${H}_{1}\ne \left\{0\right\}$ . Using Lemma 2, A, B have matrix forms (2.3) and (2.4), respectively. The operator AB has the matrix decomposition (2.7). Then

${A}^{*}=\left[\begin{array}{ccc}0& 0& 0\\ {A}_{12}^{\ast}& 0& 0\\ 0& 0& 0\\ {A}_{14}^{\ast}& {A}_{24}^{\ast}& 0\end{array}\right]:\left[\begin{array}{c}{K}_{1}\\ {K}_{2}\\ N\left({A}^{*}\right)\end{array}\right]\to \left[\begin{array}{c}{H}_{1}\\ {H}_{2}\\ {H}_{3}\\ {H}_{4}\end{array}\right]$ (2.16)

and

$B{B}^{*}{A}^{*}=\left[\begin{array}{ccc}{B}_{12}{B}_{12}^{\ast}{A}_{12}^{\ast}& 0& 0\\ {B}_{22}{B}_{12}^{\ast}{A}_{12}^{\ast}& 0& 0\\ 0& 0& 0\\ 0& 0& 0\end{array}\right]:\left[\begin{array}{c}{K}_{1}\\ {K}_{2}\\ N\left({A}^{*}\right)\end{array}\right]\to \left[\begin{array}{c}{H}_{1}\\ {H}_{2}\\ {H}_{3}\\ {H}_{4}\end{array}\right]$ (2.17)

by direct computation from (2.3) and (2.4). Therefore, comparing (2.16) with (2.17), it is natural that $R\left(B{B}^{*}{A}^{*}\right)\subset R\left({A}^{*}\right)$ if and only if ${B}_{12}{B}_{12}^{\ast}{A}_{12}^{\ast}=0$ . So $R\left(B{B}^{*}{A}^{*}\right)\subset R\left({A}^{*}\right)$ if and only if ${B}_{12}=0$ since ${A}_{12}$ is invertilbe.

On the other hand, it follows from Lemma 1 that

${B}^{\left(\text{134}\right)}=\left[\begin{array}{cccc}{B}_{11}^{-1}& -{B}_{11}^{-1}{B}_{12}{B}_{22}^{-1}& 0& 0\\ 0& {B}_{22}^{-1}& 0& 0\\ 0& 0& {F}_{33}& {F}_{34}\end{array}\right]:\left[\begin{array}{c}{H}_{1}\\ {H}_{2}\\ {H}_{3}\\ {H}_{4}\end{array}\right]\to \left[\begin{array}{c}{L}_{1}\\ {L}_{2}\\ N\left(B\right)\end{array}\right],$ (2.18)

and

${\left(AB\right)}^{\left(134\right)}=\left[\begin{array}{ccc}0& {M}_{12}& {M}_{13}\\ {B}_{22}^{-1}{A}_{12}^{-1}& 0& 0\\ 0& {M}_{32}& {M}_{33}\end{array}\right]:\left[\begin{array}{c}{K}_{1}\\ {K}_{2}\\ N\left({A}^{*}\right)\end{array}\right]\to \left[\begin{array}{c}{L}_{1}\\ {L}_{2}\\ N\left(B\right)\end{array}\right].$ (2.19)

where ${F}_{33},\text{\hspace{0.17em}}{F}_{34},{M}_{ij},\left(i=1,3,\text{\hspace{0.17em}}j=2,3\right)$ are arbitrary.

Combining formulae (2.7) with (2.18), it is easy to get

$AB{B}^{\left(\text{134}\right)}=\left[\begin{array}{cccc}0& {A}_{12}& 0& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 0\end{array}\right]:\left[\begin{array}{c}{H}_{1}\\ {H}_{2}\\ {H}_{3}\\ {H}_{4}\end{array}\right]\to \left[\begin{array}{c}{K}_{1}\\ {K}_{2}\\ N\left({A}^{*}\right)\end{array}\right]$ .

Using Lemma 1 again, we have

${\left(AB{B}^{\left(\text{134}\right)}\right)}^{\left(\text{134}\right)}=\left[\begin{array}{ccc}0& {G}_{12}& {G}_{13}\\ {A}_{12}^{-1}& 0& 0\\ 0& {G}_{32}& {G}_{33}\\ 0& {G}_{42}& {G}_{43}\end{array}\right]:\left[\begin{array}{c}{K}_{1}\\ {K}_{2}\\ N\left({A}^{*}\right)\end{array}\right]\to \left[\begin{array}{c}{H}_{1}\\ {H}_{2}\\ {H}_{3}\\ {H}_{4}\end{array}\right].$ (2.20)

where ${G}_{ij}\left(i=1,3,4,\text{\hspace{0.17em}}j=2,3\right)$ are arbitrary. By direct computation, it is clearly from (2.18) and (2.20) that

${B}^{\left(134\right)}{\left(AB{B}^{\left(\text{134}\right)}\right)}^{\left(\text{134}\right)}=\left[\begin{array}{ccc}-{B}_{11}^{-1}{B}_{12}{B}_{22}^{-1}{A}_{12}^{-1}& {B}_{11}^{-1}{G}_{12}& {B}_{11}^{-1}{G}_{13}\\ {B}_{22}^{-1}{A}_{12}^{-1}& 0& 0\\ 0& {F}_{33}{G}_{32}+{F}_{34}{G}_{42}& {F}_{33}{G}_{33}+{F}_{34}{G}_{43}\end{array}\right].$ (2.21)

Comparing (2.21) with (2.19), we have

$\left\{{\left(AB\right)}^{\left(1,3,4\right)}\right\}=\left\{{B}^{\left(1,3,4\right)}{\left(AB{B}^{\left(1,3,4\right)}\right)}^{\left(1,3,4\right)}\right\}$

if and only if $-{B}_{11}^{-1}{B}_{12}{B}_{22}^{-1}{A}_{12}^{-1}=0$ , that is, ${B}_{12}=0$ . Therefore, $\left\{{\left(AB\right)}^{\left(1,3,4\right)}\right\}=$ $\left\{{B}^{\left(1,3,4\right)}{\left(AB{B}^{\left(1,3,4\right)}\right)}^{\left(1,3,4\right)}\right\}$ if and only if $R\left(B{B}^{*}{A}^{*}\right)\subset R\left({A}^{*}\right)$ .

Case 2 ${H}_{1}=\left\{0\right\}$ . Then A, B have matrix forms (2.5) and (2.6), respectively. By similarly discussing to case 1 and case 2 in the proof of Theorem 3.2, it is easy

to get that $\left\{{\left(AB\right)}^{\left(1,3,4\right)}\right\}=\left\{{B}^{\left(1,3,4\right)}{\left(AB{B}^{\left(1,3,4\right)}\right)}^{\left(1,3,4\right)}\right\}$ and $R\left(B{B}^{*}{A}^{*}\right)\subset R\left({A}^{*}\right)$ al-

ways hold in this case.

So the proof is completed.

Acknowledgements

This subject is supported by NSF of China (No. 11501345) and the Natural Science Basic Research Plan of Henan Province (No. 152300410221).

Conflicts of Interest

The authors declare no conflicts of interest.

[1] |
Cvetkovic-Ilic, D.S. and Djikic, M. (2016) Various Solutions to Reverse Order Law Problems. Linear and Multilinear Algebra, 64, 1207-1219.
https://doi.org/10.1080/03081087.2015.1082956 |

[2] |
Greville, T.N.E. (1966) Note on the Generalized Inverse of a Matrix Product. Society for Industrial and Applied Mathematics Review, 8, 518-521.
https://doi.org/10.1137/1008107 |

[3] | Galperin, A.M. and Waksman, Z. (1980) On Pseudo Inverse of Operator Products. Linear Algebra and Applications, 33, 123-131. |

[4] |
Izumino, S. (1982) The Product of Operators with Closed Range and an Extension of the Reverse Order Law. Tohoku Mathematical Journal, 34, 43-52.
https://doi.org/10.2748/tmj/1178229307 |

[5] | Liu, X.J., Huang, S. and Cvetkovic-Ilic, D.S. (2012) Mixed-Type Reverse-Order Laws for {1,3,4}-Generalized Inverses over Hilbert Spaces. Applied Mathematics and Computation, 218, 8570-8577. |

[6] |
Liu, X.J., Wu, S.X. and Cvetkovic-Ilic, D.S. (2013) New Results on Reverse Order Law for {1,2,3}- and {1,2,4}-Inverses of Bounded Operators. Mathematics of Computation, 82, 1597-1607. https://doi.org/10.1090/S0025-5718-2013-02660-9 |

[7] | Wang, M., Wei, M. and Jia, Z. (2009) Mixed-Type Reverse-Order Law of . Linear Algebra and Applications, 430, 1691-1699. |

[8] | Wang, J., Zhang, H.Y. and Ji, G.X. (2010) A Generalized Reverse Order Law for the Products of Two Operators. Journal of Shaanxi Normal University (Natural Science), 38, 13-17. |

[9] |
Yang, H. and Lu, X.F. (2011) Mixed-Type Reverse-Order Laws of and . Applied Mathematics and Computation, 217, 10361-10367.
https://doi.org/10.1016/j.amc.2011.05.046 |

[10] | Zhang, H.Y. and Si, H.Y. (2016) Reverse Order Laws for {1,2,3}-Inverse of Two-Operator Product. Journal of Sichuan Normal University (Natural Science), 39, 671-677. |

Copyright © 2020 by authors and Scientific Research Publishing Inc.

This work and the related PDF file are licensed under a Creative Commons Attribution 4.0 International License.