Advances in Applied Mathematics
Vol. 07  No. 09 ( 2018 ), Article ID: 26709 , 6 pages
10.12677/AAM.2018.79133

The Representation of the Generalized Inverse of a Hamilton Matrix

Yu Guo*, Zeyuan Li, Xiaotong Liu, Wenhui He

School of Mathematical Science, Inner Mongolia University, Hohhot Inner Mongolia

Received: Aug. 15th, 2018; accepted: Aug. 30th, 2018; published: Sep. 6th, 2018

ABSTRACT

The aim of this paper is to establish an explicit representation of the Drazin inverse of a Hamilton matrix H = ( A B C A ) under certain conditions. Then we give a formula for the Drazin inverse of a Hamilton matrix when the generalized Schur complement S = A C A D B = 0 .

Keywords:Hamilton Matrix, Generalized Inverse, Drazin Inverse, Schur Complement

Hamilton矩阵广义逆的表示

郭宇*,李泽塬,刘晓彤,贺文慧

内蒙古大学,数学科学学院,内蒙古 呼和浩特

收稿日期:2018年8月15日;录用日期:2018年8月30日;发布日期:2018年9月6日

摘 要

本文首先给出了Hamilton矩阵 H = ( A B C A ) 在一定条件下Drazin逆的表达式,其次给出了当广义Schur补 S = A C A D B = 0 时Drazin逆 H D 的表达式。

关键词 :Hamilton矩阵,广义逆,Drazin逆,Schur补

Copyright © 2018 by authors and Hans Publishers Inc.

This work is licensed under the Creative Commons Attribution International License (CC BY).

http://creativecommons.org/licenses/by/4.0/

1. 引言

Hamilton体系是由19世纪爱尔兰数学家兼物理学家W. R. Hamilton从几何光学着手创建起来的理论模式,而后他将此模式创造性的应用于经典力学,得到了在经典力学范畴内的又一种力学描述形式—Hamilton力学 [1]。作为Hamilton系统中最简单且最基本的形式,有限维线性Hamilton系统所对应的系数矩阵如下:

H = ( A B C A * ) ,

其中 B , C 是Hermite矩阵, A * 是A的共轭转置,此时称H是Hamilton矩阵 [2]。Hamilton矩阵作为一种特殊的分块矩阵,在数学以及力学的很多方面都有重要的作用,如谱的计算、Riccati方程的解以及相关的不变子空间刻画等 [3]。

设矩阵 A C n × n ,A的Drazin逆是满足以下条件的唯一复矩阵 A D

A D A A D = A D , A A D = A D A , A k + 1 A D = A k ,

其中k是使得 r a n k ( A k ) = r a n k ( A k + 1 ) 成立的最小非负整数,记 k = i n d ( A ) 为A的指标 [4]。当 i n d ( A ) = 1 时,则称 A D 为A的群逆。如果A是非奇异的,则 i n d ( A ) = 0 A D = A 1 ,本文中令 A π = I A A D

Drazin逆在Markov链、奇异微分方程和差分方程、迭代方法等各领域起着重要作用 [5]。本文主要研究了Hamilton矩阵在一定条件下的Drazin逆,并给出了一些 H D 的表达式。

2. 引理

为了得到本文的主要结论,首先给出以下引理。

引理1 ( [6]). 令 P , Q C n × n ,如果 P 2 Q = 0 , P Q + Q P = 0 ,那么 ( P + Q ) D = P D + ( P + Q ) ( Q D ) 2

引理2 ( [6]). 令 P , Q C n × n ,如果 P 2 Q = 0 , Q P π = 0 ,那么 ( P + Q ) D = P D + Q ( P D ) 2 + P Q ( P D ) 3

引理3 ( [7]). 令 T C n × n ,如果 T = ( 0 B C 0 ) , B C p × ( n p ) , C C ( n p ) × p ,则 T D = ( 0 B ( C B ) D ( C B ) D C 0 )

引理4 ( [8]). 令 P , Q C n × n ,如果 P Q = 0 ,P是l-幂零的,则 ( P + Q ) D = i = 0 l 1 ( Q D ) i + 1 P i

如果 P Q = 0 ,Q是s-幂零的,则

( P + Q ) D = i = 0 s 1 Q i ( P D ) i + 1 .

引理5 ( [9]). 令 M = ( A B C D ) , A C m × m , D C p × p 。若 S = D C A D B = 0 , A π B = 0 , C A π = 0 ,则

M D = ( I C A D ) [ ( A W ) D ] 2 A ( I A D B ) , W = A A D + A D B C A D .

3. 主要结论及证明

定理1. 令 H = ( A B C A * ) ,其中A为方阵,如果 A 2 B = 0 , ( A ) 2 C = 0 , B ( A ) π = 0 , C A π = 0 ,那么

H D = ( A D B ( ( A ) D ) 2 A B ( ( A ) D ) 3 C ( A D ) 2 A C ( A D ) 3 ( A ) D ) .

证明:将矩阵H分解,得

H = ( A 0 0 A * ) + ( 0 B C 0 ) ,

P = ( A 0 0 A * ) , Q = ( 0 B C 0 ) ,

通过条件我们得到

,

,

因此,矩阵P和Q满足引理2,得到

H D = ( P + Q ) D = P D + Q ( P D ) 2 + P Q ( P D ) 3 = ( A D 0 0 ( A ) D ) + ( 0 B ( ( A ) D ) 2 C ( A D ) 2 0 ) + ( 0 A B ( ( A ) D ) 3 A C ( A D ) 3 0 ) = ( A D B ( ( A ) D ) 2 A B ( ( A ) D ) 3 C ( A D ) 2 A C ( A D ) 3 ( A ) D ) .

定理2. 令 H = ( A B C A * ) ,其中A为方阵,如果 B C A = 0 , C B A = 0 , A ( B C ) π = 0 , A * ( C B ) π = 0 ,那么

H D = ( A ( B C ) D B A * ( ( C B ) D ) 2 C B ( C B ) D ( C B ) D C A * ( C B ) D + C A B ( ( C B ) D ) 2 ) .

证明:将矩阵H分解,得

H = ( 0 B C 0 ) + ( A 0 0 A * ) ,

P = ( 0 B C 0 ) , Q = ( A 0 0 A * ) ,

通过引理3,我们可以得到

P D = ( 0 B ( C B ) D ( C B ) D C 0 ) , P π = ( ( B C ) π 0 0 ( C B ) π ) .

通过条件得

P 2 Q = ( B C A 0 0 C B A * ) = 0 ,

Q P π = ( A ( B C ) π 0 0 A * ( C B ) π ) = 0 ,

因此,矩阵P和Q满足引理2,得到

H D = ( P + Q ) D = P D + Q ( P D ) 2 + P Q ( P D ) 3 = ( 0 B ( C B ) D ( C B ) D C 0 ) + ( A ( B C ) D 0 0 A * ( C B ) D ) + ( B A * ( ( C B ) D ) 2 C 0 0 C A B ( ( C B ) D ) 2 ) = ( A ( B C ) D B A * ( ( C B ) D ) 2 C B ( C B ) D ( C B ) D C A * ( C B ) D + C A B ( ( C B ) D ) 2 ) .

定理3. 令 H = ( A B C A * ) ,其中A为方阵,如果 A 2 B = 0 , , A B = B A * , C A = A * C ,那么

H D = ( A D + A B ( ( C B ) D ) 2 C B ( C B ) D ( C B ) D C ( A * ) D A * ( C B ) D ) .

证明:将矩阵H分解,得

H = ( A 0 0 A * ) + ( 0 B C 0 ) ,

P = ( A 0 0 A * ) , Q = ( 0 B C 0 ) ,

通过条件我们得到

P 2 Q = ( 0 A 2 B ( A * ) 2 C 0 ) = 0 ,

P Q + Q P = ( 0 A B B A * A * C + C A 0 ) = 0 ,

因此,矩阵P和Q满足引理1,得到

H D = ( P + Q ) D = P D + ( P + Q ) ( Q D ) 2 = ( A D 0 0 ( A * ) D ) + ( A B C A * ) ( 0 B ( C B ) D ( C B ) D C 0 ) 2 = ( A D + A B ( ( C B ) D ) 2 C B ( C B ) D ( C B ) D C ( A * ) D A * ( C B ) D ) .

下面我们给出A在矩阵H的广义Schur补 S = A * C A D B = 0 H D 的表达式。

定理4. 令 H = ( A B C A * ) ,其中A为方阵,如果 S = A * C A D B = 0 , A π B C = 0 , C A π B = 0 , A B B A * = 0 ,那么

H D = ( A B C A * ) E 2 ,

其中

E l = F l + i = 1 k F i + l ( A i A π 0 C A i 1 A π 0 ) , l 1 ,

F l = ( Q 2 D ) l = ( I C A D ) [ ( A W ) D ] l + 1 A 2 A D ( I A D B ) ,

W = A A D + A D B C A D .

证明:由 S = A * C A D B = 0 可得

H = ( A B C C A D B ) = ( 0 A π B 0 0 ) + ( A A A D B C C A D B ) ,

P = ( 0 A π B 0 0 ) , Q = ( A A A D B C C A D B ) ,

通过条件可得 P Q + Q P = 0 , P 2 = 0 , P D = 0 。由引理,我们得到

H D = ( P + Q ) ( Q D ) 2 , (1)

接下来我们求 Q D

我们将Q分解为 Q = Q 1 + Q 2 ,其中

Q 1 = ( A A π 0 C A π 0 ) , Q 2 = ( A 2 A D A A D B C A A D C A D B ) ,

可以得到 Q 1 k + 1 阶幂零的, k = i n d ( A ) ,并且 Q 1 Q 2 = 0 ,由引理4可得

Q D = i = 0 k ( Q 2 D ) i + 1 Q 1 i ,

其中 Q 1 i = ( A i A π 0 C A i 1 A π 0 ) , i 1

下面求 Q 2 D ,由引理5,

, W = A A D + A D B C A D ,

再由 A 2 A D ( I A D B ) ( I C A D ) = A W ,可以得到

( Q 2 D ) i = ( I C A D ) [ ( A W ) D ] i + 1 A 2 A D ( I A D B ) , W = A A D + A D B C A D ,

E = Q D , F = Q 2 D ,则

E l = F l + i = 1 k F i + l ( A i A π 0 C A i 1 A π 0 ) , l 1 ,

E = Q D 代入(1),证毕。

基金项目

内蒙古大学校级大学生创新创业训练计划项目(项目编号:201711198)。

文章引用

郭宇,李泽塬,刘晓彤,贺文慧. Hamilton矩阵广义逆的表示
The Representation of the Generalized Inverse of a Hamilton Matrix[J]. 应用数学进展, 2018, 07(09): 1147-1152. https://doi.org/10.12677/AAM.2018.79133

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  10. NOTES

    *通讯作者。

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