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Journal of Applied Mathematics
VolumeΒ 2012Β (2012), Article IDΒ 504650, 17 pages
http://dx.doi.org/10.1155/2012/504650
Research Article

The Group Involutory Matrix of the Combinations of Two Idempotent Matrices

1College of Mathematics and Computer Science, Guangxi University for Nationalities, Nanning 530006, China
2College of Mathematics and Computer Science, Bijie University, Guizhou 551700, China
3School of Mathematical Sciences, Monash University, Clayton Cambus, VIC 3800, Australia

Received 19 December 2011; Accepted 16 March 2012

Academic Editor: MehmetΒ Sezer

Copyright Β© 2012 Lingling Wu et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract

We discuss the following problem: when π‘Žπ‘ƒ+𝑏𝑄+𝑐𝑃𝑄+𝑑𝑄𝑃+𝑒𝑃𝑄𝑃+𝑓𝑄𝑃𝑄+𝑔𝑃𝑄𝑃𝑄 of idempotent matrices 𝑃 and 𝑄, where π‘Ž,𝑏,𝑐,𝑑,𝑒,𝑓,π‘”βˆˆβ„‚ and π‘Žβ‰ 0,𝑏≠0, is group involutory.

1. Introduction

Throughout this paper ℂ𝑛×𝑛 stands for the set of 𝑛×𝑛 complex matrices. Let π΄βˆˆβ„‚π‘›Γ—π‘›. 𝐴 is said to be idempotent if 𝐴2=𝐴. 𝐴 is said to be group invertible if there exists an π‘‹βˆˆβ„‚π‘›Γ—π‘› such that𝐴𝑋𝐴=𝐴,𝑋𝐴𝑋=𝑋,𝐴𝑋=𝑋𝐴(1.1) hold. If such an 𝑋 exists, then it is unique, denoted by 𝐴𝑔, and called the group inverse of 𝐴. It is well known that the group inverse of a square matrix 𝐴 exists if and only if rank(𝐴2)=rank(𝐴) (see, e.g., [1] for details). Clearly, not every matrix is group invertible. But the group inverse of every idempotent matrix exists and is this matrix itself.

Recall that a matrix 𝐴 with the group inverse is said to be group involutory if 𝐴𝑔=𝐴. 𝐴 is the group involutory matrix if and only if it is tripotent, that is, satisfies 𝐴3=𝐴 (see [2]). Thus, for a nonzero idempotent matrix 𝑃 and a nonzero scalar π‘Ž, π‘Žπ‘ƒ is a group involutory matrix if and only if either π‘Ž=1 or π‘Ž=βˆ’1.

Recently, some properties of linear combinations of idempotents or projections are widely discussed (see, e.g., [3–12] and the literature mentioned below). In [13], authors established a complete solution to the problem of when a linear combination of two different projectors is also a projector. In [14], authors considered the following problem: when a linear combination of nonzero different idempotent matrices is the group involutory matrix. In [15], authors provided the complete list of situations in which a linear combination of two idempotent matrices is the group involutory matrix. In [16], authors discussed the group inverse of π‘Žπ‘ƒ+𝑏𝑄+𝑐𝑃𝑄+𝑑𝑄𝑃+𝑒𝑃𝑄𝑃+𝑓𝑄𝑃𝑄+𝑔𝑃𝑄𝑃𝑄 of idempotent matrices 𝑃 and 𝑄, where π‘Ž,𝑏,𝑐,𝑑,𝑒,𝑓,π‘”βˆˆβ„‚ with π‘Ž,𝑏≠0, deduced its explicit expressions, and some necessary and sufficient conditions for the existence of the group inverse of π‘Žπ‘ƒ+𝑏𝑄+𝑐𝑃𝑄.

In this paper, we will investigate the following problem: when π‘Žπ‘ƒ+𝑏𝑄+𝑐𝑃𝑄+𝑑𝑄𝑃+𝑒𝑃𝑄𝑃+𝑓𝑄𝑃𝑄+𝑔𝑃𝑄𝑃𝑄 is group involutory. To this end, we need the results below.

Lemma 1.1 (see [16, Theorems 2.1 and  2.4]). Let 𝑃,π‘„βˆˆβ„‚π‘›Γ—π‘› be two different nonzero idempotent matrices. Suppose (𝑃𝑄)2=(𝑄𝑃)2. Then for any scalars π‘Ž,𝑏,𝑐,𝑑,𝑒,𝑓,𝑔, where π‘Ž,𝑏≠0 and πœƒ=π‘Ž+𝑏+𝑐+𝑑+𝑒+𝑓+𝑔, π‘Žπ‘ƒ+𝑏𝑄+𝑐𝑃𝑄+𝑑𝑄𝑃+𝑒𝑃𝑄𝑃+𝑓𝑄𝑃𝑄+𝑔(𝑃𝑄)2 is group invertible, and
(i) if πœƒβ‰ 0, then ξ€·π‘Žπ‘ƒ+𝑏𝑄+𝑐𝑃𝑄+𝑑𝑄𝑃+𝑒𝑃𝑄𝑃+𝑓𝑄𝑃𝑄+𝑔(𝑃𝑄)2𝑔=1π‘Ž1𝑃+𝑏1π‘„βˆ’π‘Ž+1𝑏+𝑐1π‘Žπ‘π‘ƒπ‘„βˆ’π‘Ž+1𝑏+𝑑+ξ‚€2π‘Žπ‘π‘„π‘ƒπ‘Ž+1𝑏+𝑐+𝑑+π‘Žπ‘π‘π‘‘βˆ’π‘π‘’π‘Ž2𝑏1𝑃𝑄𝑃+π‘Ž+2𝑏+𝑐+𝑑+π‘Žπ‘π‘π‘‘βˆ’π‘Žπ‘“π‘Žπ‘2ξ‚Άβˆ’ξ‚΅2π‘„π‘ƒπ‘„π‘Ž+2𝑏+𝑐+𝑑+π‘Žπ‘π‘π‘‘βˆ’π‘π‘’π‘Ž2𝑏+π‘π‘‘βˆ’π‘Žπ‘“π‘Žπ‘2βˆ’1πœƒξ‚Άπ‘ƒπ‘„π‘ƒπ‘„;(1.2)
(ii) if πœƒ=0, then ξ€·π‘Žπ‘ƒ+𝑏𝑄+𝑐𝑃𝑄+𝑑𝑄𝑃+𝑒𝑃𝑄𝑃+𝑓𝑄𝑃𝑄+𝑔(𝑃𝑄)2𝑔=1π‘Ž1𝑃+𝑏1π‘„βˆ’π‘Ž+1𝑏+𝑐1π‘Žπ‘π‘ƒπ‘„βˆ’π‘Ž+1𝑏+𝑑+ξ‚€2π‘Žπ‘π‘„π‘ƒπ‘Ž+1𝑏+𝑐+𝑑+π‘Žπ‘π‘π‘‘βˆ’π‘π‘’π‘Ž2𝑏1𝑃𝑄𝑃+π‘Ž+2𝑏+𝑐+𝑑+π‘Žπ‘π‘π‘‘βˆ’π‘Žπ‘“π‘Žπ‘2ξ‚Άβˆ’ξ‚΅2π‘„π‘ƒπ‘„π‘Ž+2𝑏+𝑐+𝑑+π‘Žπ‘π‘π‘‘βˆ’π‘π‘’π‘Ž2𝑏+π‘π‘‘βˆ’π‘Žπ‘“π‘Žπ‘2ξ‚Ά(𝑃𝑄)2.(1.3)

Lemma 1.2 (see [16, Theorem  3.1]). Let 𝑃,π‘„βˆˆβ„‚π‘›Γ—π‘› be two different nonzero idempotent matrices. Suppose (𝑄𝑃)2=0. Then for any scalars π‘Ž,𝑏,𝑐,𝑑,𝑒,𝑓, and 𝑔, where π‘Ž,𝑏≠0, π‘Žπ‘ƒ+𝑏𝑄+𝑐𝑃𝑄+𝑑𝑄𝑃+𝑒𝑃𝑄𝑃+𝑓𝑄𝑃𝑄+𝑔(𝑃𝑄)2 is group invertible, and ξ€·π‘Žπ‘ƒ+𝑏𝑄+𝑐𝑃𝑄+𝑑𝑄𝑃+𝑒𝑃𝑄𝑃+𝑓𝑄𝑃𝑄+𝑔(𝑃𝑄)2𝑔=1π‘Ž1𝑃+𝑏1π‘„βˆ’π‘Ž+1𝑏+𝑐1π‘Žπ‘π‘ƒπ‘„βˆ’π‘Ž+1𝑏+𝑑+ξ‚€2π‘Žπ‘π‘„π‘ƒπ‘Ž+1𝑏+𝑐+𝑑+π‘Žπ‘π‘π‘‘βˆ’π‘π‘’π‘Ž2𝑏1𝑃Q𝑃+π‘Ž+2𝑏+𝑐+𝑑+π‘Žπ‘π‘π‘‘βˆ’π‘Žπ‘“π‘Žπ‘2ξ‚Άβˆ’ξ‚΅2π‘„π‘ƒπ‘„π‘Ž+2𝑏+2𝑐+𝑑+𝑔+π‘Žπ‘π‘π‘‘βˆ’π‘π‘’βˆ’π‘π‘’π‘Ž2𝑏+π‘π‘‘βˆ’π‘Žπ‘“βˆ’π‘π‘“π‘Žπ‘2+𝑐2π‘‘π‘Ž2𝑏2ξ‚Ά(𝑃𝑄)2.(1.4)

2. Main Results

In this section, we will research when some combination of two nonzero idempotent matrices is a group involutory matrix.

First, we will discuss some situations lying in the category of (𝑃𝑄)2=(𝑄𝑃)2.

Theorem 2.1. Let 𝑃,π‘„βˆˆβ„‚π‘›Γ—π‘› be two different nonzero idempotent matrices with (𝑃𝑄)2=(𝑄𝑃)2, and let 𝐴 be a combination of the form 𝐴=π‘Žπ‘ƒ+𝑏𝑄+𝑐𝑃𝑄+𝑑𝑄𝑃+𝑒𝑃𝑄𝑃+𝑓𝑄𝑃𝑄+𝑔𝑃𝑄𝑃𝑄,(2.1) where π‘Ž,𝑏,𝑐,𝑑,𝑒,𝑓,π‘”βˆˆβ„‚ with π‘Ž,𝑏≠0. Denote πœƒ=π‘Ž+𝑏+𝑐+𝑑+𝑒+𝑓+𝑔. Then the following list comprises characteristics of all cases where 𝐴 is the group involutory matrix:
(a) the cases denoted by (π‘Ž1)∼(π‘Ž3), in which 𝑃𝑄=𝑄𝑃,(2.2) and any of the following sets of additional conditions hold:(a1)either π‘Ž=1 or π‘Ž=βˆ’1, either πœƒ=1 or πœƒ=βˆ’1 or πœƒ=0, and 𝑄=𝑃𝑄;(a2)either 𝑏=1 or 𝑏=βˆ’1, either πœƒ=1 or πœƒ=βˆ’1 or πœƒ=0, and 𝑃=𝑃𝑄;(a3)either π‘Ž=1 or π‘Ž=βˆ’1, either 𝑏=1 or 𝑏=βˆ’1, either πœƒ=1 or πœƒ=βˆ’1 or πœƒ=0 or 𝑃𝑄=0.
(b) the cases denoted by (𝑏1)∼(𝑏6), in which 𝑃𝑄≠𝑄𝑃,𝑃𝑄𝑃=𝑄𝑃𝑄,(2.3) and any of the following sets of additional conditions hold:(b1)π‘Ž=Β±1,𝑏=βˆ“1, either πœƒ=1 or πœƒ=βˆ’1 or πœƒ=0 or 𝑃𝑄𝑃=0; (b2)π‘Ž=𝑏=Β±1,𝑐=𝑑=βˆ“1, either πœƒ=1 or πœƒ=βˆ’1 or πœƒ=0 or 𝑃𝑄𝑃=0;(b3)π‘Ž=𝑏=Β±1,𝑐=βˆ“1, either πœƒ=1 or πœƒ=βˆ’1 or πœƒ=0, and 𝑄𝑃=𝑃𝑄𝑃;(b4)π‘Ž=𝑏=Β±1,𝑑=βˆ“1, either πœƒ=1 or πœƒ=βˆ’1 or πœƒ=0, and 𝑃𝑄=𝑃𝑄𝑃;(b5)π‘Ž=𝑏=Β±1, 𝑐=βˆ“1, and 𝑄𝑃=0;(b6)π‘Ž=𝑏=Β±1, 𝑑=βˆ“1, and 𝑃𝑄=0,
(c) the cases denoted by (𝑐1)∼(𝑐18), in which 𝑃𝑄𝑃≠𝑄𝑃𝑄,𝑃𝑄𝑃𝑄=𝑄𝑃𝑄𝑃,(2.4) and any of the following sets of additional conditions hold:(c1)π‘Ž=Β±1,𝑏=βˆ“1,𝑐+𝑑+2𝑒±𝑐𝑑=Β±1, either πœƒ=1 or πœƒ=βˆ’1, and 𝑄𝑃𝑄=𝑃𝑄𝑃𝑄; (c2)π‘Ž=𝑏=𝑒=Β±1,𝑐=𝑑=βˆ“1, either πœƒ=1 or πœƒ=βˆ’1, and 𝑄𝑃𝑄=𝑃𝑄𝑃𝑄;(c3)π‘Ž=Β±1,𝑏=βˆ“1,𝑐+𝑑+2π‘“βˆ“π‘π‘‘=βˆ“1, either πœƒ=1 or πœƒ=βˆ’1, and 𝑃𝑄𝑃=𝑃𝑄𝑃𝑄;(c4)π‘Ž=𝑏=𝑓=Β±1,𝑐=𝑑=βˆ“1, either πœƒ=1 or πœƒ=βˆ’1, and 𝑃𝑄𝑃=𝑃𝑄𝑃𝑄;(c5)π‘Ž=Β±1,𝑏=βˆ“1,𝑐+𝑑+2𝑒±𝑐𝑑=Β±1,𝑐+𝑑+2π‘“βˆ“π‘π‘‘=βˆ“1, either 𝑔=1 or 𝑔=βˆ’1;(c6)π‘Ž=𝑏=𝑒=𝑓=Β±1,𝑐=𝑑=βˆ“1, either 𝑔=βˆ“1 or 𝑔=βˆ“3;(c7)π‘Ž=Β±1,𝑏=βˆ“1,𝑐+𝑑+2𝑒±𝑐𝑑=Β±1, and 𝑄𝑃𝑄=0;(c8)π‘Ž=𝑏=𝑒=Β±1,𝑐=𝑑=βˆ“1, and 𝑄𝑃𝑄=0;(c9)π‘Ž=Β±1,𝑏=βˆ“1,𝑐+𝑑+2π‘“βˆ“π‘π‘‘=βˆ“1, and 𝑃𝑄𝑃=0;(c10)π‘Ž=𝑏=𝑓=Β±1,𝑐=𝑑=βˆ“1, and 𝑃𝑄𝑃=0;(c11)π‘Ž=Β±1,𝑏=βˆ“1,𝑐+𝑑+2𝑒±𝑐𝑑=Β±1,𝑐+𝑑+2π‘“βˆ“π‘π‘‘=βˆ“1, and 𝑃𝑄𝑃𝑄=0;(c12)π‘Ž=𝑏=𝑒=𝑓=Β±1,𝑐=𝑑=βˆ“1, and 𝑃𝑄𝑃𝑄=0;(c13)π‘Ž=Β±1,𝑏=βˆ“1,2𝑒+𝑐+𝑑±𝑐𝑑=Β±1,πœƒ=0, and 𝑄𝑃𝑄=𝑃𝑄𝑃𝑄;(c14)π‘Ž=𝑏=𝑒=Β±1,𝑐=𝑑=βˆ“1,πœƒ=0, and 𝑄𝑃𝑄=𝑃𝑄𝑃𝑄;(c15)π‘Ž=Β±1,𝑏=βˆ“1,2𝑓+𝑐+π‘‘βˆ“π‘π‘‘=βˆ“1,πœƒ=0, and 𝑃𝑄𝑃=𝑃𝑄𝑃𝑄;(c16)π‘Ž=𝑏=𝑓=Β±1,𝑐=𝑑=βˆ“1,πœƒ=0, and 𝑃𝑄𝑃=𝑃𝑄𝑃𝑄;(c17)π‘Ž=Β±1,𝑏=βˆ“1,2𝑒+𝑐+𝑑±𝑐𝑑=Β±1,2𝑓+𝑐+π‘‘βˆ“π‘π‘‘=βˆ“1,𝑔=0;(c18)π‘Ž=𝑏=𝑒=𝑓=Β±1,𝑐=𝑑=βˆ“1,𝑔=βˆ“2.

Proof. Obviously, the condition (2.2) implies that the group inverse of 𝐴 exists and is of the form (1.2) when πœƒβ‰ 0 or the form (1.3) when πœƒ=0 by Lemma 1.1. So do the conditions (2.2), (2.3), and (2.4). We will straightforwardly show that a matrix 𝐴 of the form (2.1) is the group involutory matrix if and only if π΄βˆ’π΄π‘”=0.
(a)  Under the condition (2.2), 𝐴=π‘Žπ‘ƒ+𝑏𝑄+πœ‡π‘ƒπ‘„, where πœ‡=𝑐+𝑑+𝑒+𝑓+𝑔.
(1) If πœƒβ‰ 0, then 𝐴𝑔=1π‘Ž1𝑃+𝑏1𝑄+πœƒβˆ’1π‘Žβˆ’1𝑏𝑃𝑄,(2.5) and so π΄βˆ’π΄π‘”=ξ‚€1π‘Žβˆ’π‘Žξ‚ξ‚€1𝑃+π‘βˆ’π‘ξ‚ξ‚€1𝑄+πœ‡βˆ’πœƒ+1π‘Ž+1𝑏𝑃𝑄=0.(2.6) Multiplying (2.6) by 𝑃 and 𝑄, respectively, leads to ξ‚€1π‘Žβˆ’π‘Žξ‚ξ‚€1𝑃+π‘βˆ’π‘ξ‚ξ‚€1𝑃𝑄+πœ‡βˆ’πœƒ+1π‘Ž+1𝑏1𝑃𝑄=0,π‘Žβˆ’π‘Žξ‚ξ‚€1𝑃𝑄+π‘βˆ’π‘ξ‚ξ‚€1𝑄+πœ‡βˆ’πœƒ+1π‘Ž+1𝑏𝑃𝑄=0,(2.7) and then ξ‚€1π‘Žβˆ’π‘Žξ‚ξ‚€1𝑃+π‘βˆ’π‘ξ‚ξ‚€1𝑃𝑄=π‘Žβˆ’π‘Žξ‚ξ‚€1𝑃𝑄+π‘βˆ’π‘ξ‚π‘„.(2.8) Multiplying the above equation, respectively, by 𝑃 and by 𝑄, we get ξ‚€1π‘Žβˆ’π‘Žξ‚ξ‚€1(π‘ƒβˆ’π‘ƒπ‘„)=0,π‘βˆ’π‘ξ‚(π‘„βˆ’π‘ƒπ‘„)=0.(2.9) Thus, since 𝑃≠𝑄, we have three situations: 𝑃=𝑃𝑄 and 𝑏=π‘βˆ’1; π‘Ž=π‘Žβˆ’1 and 𝑄=𝑃𝑄; π‘Ž=π‘Žβˆ’1 and 𝑏=π‘βˆ’1.
When 𝑄=𝑃𝑄 and π‘Ž=π‘Žβˆ’1, (2.6) becomes (πœƒβˆ’πœƒβˆ’1)𝑄=0 and then πœƒ=Β±1. Therefore, we obtain (π‘Ž1) except the situation πœƒ=0. Similarly, when 𝑏=π‘βˆ’1 and 𝑃=𝑃𝑄, we have (π‘Ž2) except the situation πœƒ=0. When π‘Ž=π‘Žβˆ’1 and 𝑏=π‘βˆ’1, (2.6) becomes (πœƒβˆ’πœƒβˆ’1)𝑃𝑄=0 and then πœƒ=Β±1 or 𝑃𝑄=0. Therefore, we obtain (π‘Ž3) except the situation πœƒ=0.
(2) If πœƒ=0, then 𝐴𝑔=1π‘Ž1𝑃+𝑏1π‘„βˆ’π‘Ž+1𝑏𝑃𝑄,(2.10) and then π΄βˆ’π΄π‘”=ξ‚€1π‘Žβˆ’π‘Žξ‚ξ‚€1𝑃+π‘βˆ’π‘ξ‚ξ‚€1𝑄+πœ‡+π‘Ž+1𝑏𝑃𝑄=0.(2.11) Analogous to the process of reaching (2.9) in (a)(1), we have ξ‚€1π‘βˆ’π‘ξ‚ξ‚€1(π‘„βˆ’π‘ƒπ‘„)=0,π‘Žβˆ’π‘Žξ‚(π‘ƒβˆ’π‘ƒπ‘„)=0.(2.12) Thus, we have three situations: 𝑃=𝑃𝑄 and 𝑏=π‘βˆ’1; π‘Ž=π‘Žβˆ’1 and 𝑄=𝑃𝑄; π‘Ž=π‘Žβˆ’1 and 𝑏=π‘βˆ’1, since 𝑃≠𝑄. Similar to the argument in (π‘Ž)(1), substituting them, respectively, into (2.11), we can obtain the situation πœƒ=0, respectively, in (π‘Ž1),(π‘Ž2), and (π‘Ž3).
(b) Under the condition (2.3), 𝐴=π‘Žπ‘ƒ+𝑏𝑄+𝑐𝑃𝑄+𝑑𝑄𝑃+πœˆπ‘ƒπ‘„π‘ƒ, where 𝜈=𝑒+𝑓+𝑔.
(1) If πœƒβ‰ 0, then 𝐴𝑔=1π‘Ž1𝑃+𝑏1π‘„βˆ’π‘Ž+1𝑏+𝑐1π‘Žπ‘π‘ƒπ‘„βˆ’π‘Ž+1𝑏+𝑑+ξ‚€1π‘Žπ‘π‘„π‘ƒπ‘Ž+1𝑏+𝑐+𝑑+1π‘Žπ‘πœƒξ‚π‘ƒπ‘„π‘ƒ,(2.13) and so π΄βˆ’π΄π‘”=ξ‚€1π‘Žβˆ’π‘Žξ‚ξ‚€1𝑃+π‘βˆ’π‘ξ‚ξ‚€1𝑄+𝑐+π‘Ž+1𝑏+𝑐+ξ‚€1π‘Žπ‘π‘ƒπ‘„π‘‘+π‘Ž+1𝑏+𝑑1π‘Žπ‘π‘„π‘ƒ+πœˆβˆ’π‘Žβˆ’1π‘βˆ’π‘+π‘‘βˆ’1π‘Žπ‘πœƒξ‚π‘ƒπ‘„π‘ƒ=0.(2.14) Multiplying the above equation, respectively, on the two sides by 𝑃 yields ξ‚€10=π‘Žβˆ’π‘Žξ‚ξ‚€1𝑃+𝑐+𝑏+π‘Ž+π‘ξ‚ξ‚€π‘π‘Žπ‘π‘ƒπ‘„+𝜈+π‘‘βˆ’βˆ’1π‘Žπ‘πœƒξ‚ξ‚€1𝑃𝑄𝑃,(2.15)0=π‘Žβˆ’π‘Žξ‚ξ‚€1𝑃+𝑏+𝑑+π‘Ž+π‘‘ξ‚ξ‚€π‘‘π‘Žπ‘π‘„π‘ƒ+𝜈+π‘βˆ’βˆ’1π‘Žπ‘πœƒξ‚π‘ƒπ‘„π‘ƒ.(2.16) Multiplying (2.15) on the left sides by 𝑄 and (2.16) on the right sides by 𝑄, by (2.3), we have ξ‚€1π‘Žβˆ’π‘Žξ‚ξ‚€1𝑄𝑃+𝑏+𝑐+𝑑+𝜈+π‘Žβˆ’1πœƒξ‚ξ‚€1𝑄𝑃𝑄=0,π‘Žβˆ’π‘Žξ‚ξ‚€1𝑃𝑄+𝑏+𝑐+𝑑+𝜈+π‘Žβˆ’1πœƒξ‚π‘„π‘ƒπ‘„=0,(2.17) and then (π‘Žβˆ’π‘Žβˆ’1)(π‘„π‘ƒβˆ’π‘ƒπ‘„)=0. Since 𝑄𝑃≠𝑃𝑄, π‘Ž=π‘Žβˆ’1. Similarly, 𝑏=π‘βˆ’1.
Substituting π‘Ž=π‘Žβˆ’1 inside (2.17) yields (πœƒβˆ’πœƒβˆ’1)𝑄𝑃𝑄=0 and then πœƒ=πœƒβˆ’1 or 𝑄𝑃𝑄=0. We will discuss the remainder for detail as follows:
When π‘Ž=π‘Žβˆ’1, 𝑏=π‘βˆ’1, (2.14) becomesξ‚€10=𝑐+π‘Ž+1𝑏+𝑐1π‘Žπ‘π‘ƒπ‘„+𝑑+π‘Ž+1𝑏+𝑑+ξ‚€1π‘Žπ‘π‘„π‘ƒπœˆβˆ’π‘Žβˆ’1π‘βˆ’π‘+π‘‘βˆ’1π‘Žπ‘πœƒξ‚π‘ƒπ‘„π‘ƒ,(2.18)
(i) if π‘Ž+𝑏=0, then 1𝑐+π‘Ž+1𝑏+𝑐1π‘Žπ‘=0,𝑑+π‘Ž+1𝑏+π‘‘π‘Žπ‘=0,(2.19) and so it follows from (2.18) that ξ‚€1πœƒβˆ’πœƒξ‚ξ‚€1𝑃𝑄𝑃=𝜈+𝑐+π‘‘βˆ’πœƒξ‚π‘ƒπ‘„π‘ƒ=0.(2.20) Therefore, either πœƒ=πœƒβˆ’1 or 𝑃𝑄𝑃=0 implies that (2.18) holds, namely, (2.14) holds. Thus, we have (𝑏1) except the situation πœƒ=0.
(ii) if π‘Ž=𝑏, then (2.18) becomes ξ‚€10=(2𝑐+2π‘Ž)𝑃𝑄+(2𝑑+2π‘Ž)𝑄𝑃+2πœˆβˆ’πœƒβˆ’πœƒξ‚π‘ƒπ‘„π‘ƒ.(2.21) Multiplying the above equation, respectively, on the right side by 𝑃 and on the left side by 𝑄, we have ξ‚€10=(2𝑐+2π‘Ž)𝑃𝑄+𝜈+π‘‘βˆ’π‘βˆ’πœƒξ‚ξ‚€1𝑃𝑄𝑃,(2.22)0=(2𝑑+2π‘Ž)𝑄𝑃+𝜈+π‘βˆ’π‘‘βˆ’πœƒξ‚π‘ƒπ‘„π‘ƒ.(2.23) So if πœƒ=πœƒβˆ’1, then the two equations above (2.22) and (2.23) become, respectively, (𝑐+π‘Ž)(π‘ƒπ‘„βˆ’π‘ƒπ‘„π‘ƒ)=0,(𝑑+π‘Ž)(π‘„π‘ƒβˆ’π‘ƒπ‘„π‘ƒ)=0.(2.24) Or if 𝑃𝑄𝑃=0, then (2.22) and (2.23) become, respectively, (𝑐+π‘Ž)𝑃𝑄=0,(𝑑+π‘Ž)𝑄𝑃=0.(2.25) Since 𝑃𝑄≠𝑄𝑃, it follows from (2.24) and (2.25) that we have the six situations: πœƒ=πœƒβˆ’1 and 𝑐=𝑑=βˆ’π‘Ž; πœƒ=πœƒβˆ’1, 𝑐=βˆ’π‘Ž and 𝑄𝑃=𝑃𝑄𝑃; πœƒ=πœƒβˆ’1, 𝑑=βˆ’π‘Ž, and 𝑃𝑄=𝑃𝑄𝑃; 𝑐=βˆ’π‘Ž and 𝑄𝑃=0; 𝑑=βˆ’π‘Ž and 𝑃𝑄=0; 𝑐=𝑑=βˆ’π‘Ž and 𝑃𝑄𝑃=0. Thus, we have (𝑏2)∼(𝑏4) except the situation πœƒ=0, and (𝑏5) and (𝑏6).
(2) If πœƒ=0, then 𝐴𝑔=1π‘Ž1𝑃+𝑏1π‘„βˆ’π‘Ž+1𝑏+𝑐1π‘Žπ‘π‘ƒπ‘„βˆ’π‘Ž+1𝑏+𝑑1π‘Žπ‘π‘„π‘ƒ+π‘Ž+1𝑏+𝑐+π‘‘ξ‚π‘Žπ‘π‘ƒπ‘„π‘ƒ,(2.26) and then π΄βˆ’π΄π‘”=ξ‚€1π‘Žβˆ’π‘Žξ‚ξ‚€1𝑃+π‘βˆ’π‘ξ‚ξ‚€1𝑄+𝑐+π‘Ž+1𝑏+𝑐+ξ‚€1π‘Žπ‘π‘ƒπ‘„π‘‘+π‘Ž+1𝑏+𝑑1π‘Žπ‘π‘„π‘ƒ+πœˆβˆ’π‘Žβˆ’1π‘βˆ’π‘+π‘‘ξ‚π‘Žπ‘π‘ƒπ‘„π‘ƒ=0.(2.27) Analogous to the process in (b)(1), using (2.27) we can obtain ξ‚€1π‘Žβˆ’π‘Žξ‚ξ‚€1π‘„π‘ƒβˆ’π‘Žβˆ’π‘Žξ‚ξ‚€1𝑃𝑄𝑃=0,π‘Žβˆ’π‘Žξ‚ξ‚€1π‘ƒπ‘„βˆ’π‘Žβˆ’π‘Žξ‚π‘ƒπ‘„π‘ƒ=0.(2.28) Thus, since 𝑃𝑄≠𝑄𝑃, 𝑃𝑄≠𝑃𝑄𝑃 and/or 𝑄𝑃≠𝑃𝑄𝑃 and then π‘Ž=π‘Žβˆ’1. Similarly, 𝑏=π‘βˆ’1. Hence, π‘Ž=±𝑏.
(i)  If π‘Ž=βˆ’π‘, then 1𝑐+π‘Ž+1𝑏+𝑐1π‘Žπ‘=0,𝑑+π‘Ž+1𝑏+𝑑1π‘Žπ‘=0,πœˆβˆ’π‘Žβˆ’1π‘βˆ’π‘+π‘‘π‘Žπ‘=βˆ’2(π‘Ž+𝑏)=0.(2.29) Thus, (2.27) holds. Hence we have the situation πœƒ=0 in (𝑏1).
(ii) If π‘Ž=𝑏, then (2.27) becomes (𝑐+π‘Ž)𝑃𝑄+(𝑑+π‘Ž)𝑄𝑃+πœˆπ‘ƒπ‘„π‘ƒ=0.(2.30) Multiplying the above equation on the left side, respectively, by 𝑃 and by 𝑄, we have (𝑐+π‘Ž)(π‘ƒπ‘„βˆ’π‘ƒπ‘„π‘ƒ)=0,(𝑑+π‘Ž)(π‘„π‘ƒβˆ’π‘ƒπ‘„π‘ƒ)=0.(2.31) Thus, 𝑐=𝑑=βˆ’π‘Ž; 𝑐=βˆ’π‘Ž and 𝑄𝑃=𝑃𝑄𝑃; 𝑑=βˆ’π‘Ž and 𝑃𝑄=𝑃𝑄𝑃. Hence, we have the situation πœƒ=0, respectively, in (𝑏2),(𝑏3), and (𝑏4).
(c) Under the condition (2.4), 𝐴=π‘Žπ‘ƒ+𝑏𝑄+𝑐𝑃𝑄+𝑑𝑄𝑃+𝑒𝑃𝑄𝑃+𝑓𝑄𝑃𝑄𝑃+𝑔𝑃𝑄𝑃𝑄.(2.32)
(1)  If πœƒβ‰ 0, then 𝐴𝑔=1π‘Ž1𝑃+𝑏1π‘„βˆ’π‘Ž+1𝑏+𝑐1π‘Žπ‘π‘ƒπ‘„βˆ’π‘Ž+1𝑏+𝑑+ξ‚€2π‘Žπ‘π‘„π‘ƒπ‘Ž+1𝑏+𝑐+𝑑+π‘Žπ‘π‘π‘‘βˆ’π‘π‘’π‘Ž2𝑏1𝑃𝑄𝑃+π‘Ž+2𝑏+𝑐+𝑑+π‘Žπ‘π‘dβˆ’π‘Žπ‘“π‘Žπ‘2ξ‚Άβˆ’ξ‚΅2π‘„π‘ƒπ‘„π‘Ž+2𝑏+𝑐+𝑑+π‘Žπ‘π‘π‘‘βˆ’π‘π‘’π‘Ž2𝑏+π‘π‘‘βˆ’π‘Žπ‘“π‘Žπ‘2βˆ’1πœƒξ‚Άπ‘ƒπ‘„π‘ƒπ‘„,(2.33) and so π΄βˆ’π΄π‘”=ξ‚€1π‘Žβˆ’π‘Žξ‚ξ‚€1𝑃+π‘βˆ’π‘ξ‚ξ‚€1𝑄+𝑐+π‘Ž+1𝑏+𝑐1π‘Žπ‘π‘ƒπ‘„+𝑑+π‘Ž+1𝑏+𝑑+ξ‚€2π‘Žπ‘π‘„π‘ƒπ‘’βˆ’π‘Žβˆ’1π‘βˆ’π‘+π‘‘βˆ’π‘Žπ‘π‘π‘‘βˆ’π‘π‘’π‘Ž2𝑏+ξ‚΅1π‘ƒπ‘„π‘ƒπ‘“βˆ’π‘Žβˆ’2π‘βˆ’π‘+π‘‘βˆ’π‘Žπ‘π‘π‘‘βˆ’π‘Žπ‘“π‘Žπ‘2ξ‚Ά+ξ‚΅2𝑄𝑃𝑄𝑔+π‘Ž+2𝑏+𝑐+𝑑+π‘Žπ‘π‘π‘‘βˆ’π‘π‘’π‘Ž2𝑏+π‘π‘‘βˆ’π‘Žπ‘“π‘Žπ‘2βˆ’1πœƒξ‚Άπ‘ƒπ‘„π‘ƒπ‘„=0.(2.34) If 𝑃𝑄=0, then 𝑄𝑃𝑄=0=𝑃𝑄𝑃 and so it contradicts (2.4). Thus 𝑃𝑄≠0. Similarly, 𝑄𝑃≠0.
Multiplying (2.34) on the left side by 𝑄𝑃 yields ξ‚€1π‘Žβˆ’π‘Žξ‚ξ‚€1𝑄𝑃+𝑏+𝑐+π‘Ž+π‘ξ‚ξ‚€π‘π‘Žπ‘π‘„π‘ƒπ‘„+𝑑+𝑒+𝑓+π‘”βˆ’βˆ’1π‘Žπ‘πœƒξ‚π‘ƒπ‘„π‘ƒπ‘„=0.(2.35) Multiplying the above equation, respectively, on the left side by 𝑃 and on the right side by 𝑃𝑄 yields, by (2.4), ξ‚€10=π‘Žβˆ’π‘Žξ‚ξ‚€1𝑃𝑄𝑃+π‘Ž1βˆ’π‘Ž+πœƒβˆ’πœƒξ‚ξ‚€1𝑃𝑄𝑃𝑄,(2.36)0=π‘Žβˆ’π‘Žξ‚ξ‚€1𝑄𝑃𝑄+π‘Ž1βˆ’π‘Ž+πœƒβˆ’πœƒξ‚π‘ƒπ‘„π‘ƒπ‘„.(2.37) Since 𝑃𝑄𝑃≠𝑄𝑃𝑄, π‘Ž=π‘Žβˆ’1 by (2.36) and (2.37). Similarly, we can gain 𝑏=π‘βˆ’1. Substituting π‘Ž=π‘Žβˆ’1 inside (2.36) yields πœƒ=πœƒβˆ’1 or 𝑃𝑄𝑃𝑄=0.
(i)  Consider the case of π‘Ž=π‘Žβˆ’1, 𝑏=π‘βˆ’1 and πœƒ=πœƒβˆ’1.
Substituting π‘Ž=π‘Žβˆ’1, 𝑏=π‘βˆ’1, and πœƒ=πœƒβˆ’1 inside (2.35) yields ξ‚€π‘π‘Ž+𝑏+𝑐+ξ‚π‘Žπ‘(π‘„π‘ƒπ‘„βˆ’π‘ƒπ‘„π‘ƒπ‘„)=0.(2.38) Similarly, we have ξ‚€π‘‘π‘Ž+𝑏+𝑑+ξ‚π‘Žπ‘(π‘ƒπ‘„π‘ƒβˆ’π‘ƒπ‘„π‘ƒπ‘„)=0.(2.39)
If 𝑃𝑄𝑃=𝑃𝑄𝑃𝑄, then 𝑄𝑃𝑄≠𝑃𝑄𝑃𝑄 by the hypothesis 𝑃𝑄𝑃≠𝑄𝑃𝑄 and so π‘Ž+𝑏+𝑐+𝑐/π‘Žπ‘=0 by (2.38). Multiplying (2.34) on the right side by 𝑄 yields ξ‚€π‘Ž+𝑐+𝑑+2π‘“βˆ’π‘π‘‘π‘Žξ‚(π‘„π‘ƒπ‘„βˆ’π‘ƒπ‘„π‘ƒπ‘„)=0.(2.40) Thus, π‘Ž+𝑐+𝑑+2π‘“βˆ’π‘π‘‘/π‘Ž=0 and then (2.14) becomes ξ‚€π‘‘π‘Ž+𝑏+𝑑+ξ‚ξ‚΅π‘Žπ‘π‘„π‘ƒ+π‘“βˆ’π‘Žβˆ’2π‘βˆ’π‘+π‘‘βˆ’π‘Žπ‘π‘π‘‘βˆ’π‘Žπ‘“π‘Žξ‚Ά+𝑄𝑃𝑄𝑏+𝑒+𝑔+π‘π‘‘βˆ’π‘Žπ‘“π‘Žξ‚Άβˆ’πœƒπ‘ƒπ‘„π‘ƒ=0.(2.41) Multiplying the above equation on the right side by 𝑃 yields ξ‚€π‘‘π‘Ž+𝑏+𝑑+ξ‚π‘Žπ‘(π‘„π‘ƒβˆ’π‘ƒπ‘„π‘ƒ)=0.(2.42) Assume 𝑃𝑄=𝑃𝑄𝑃. Then 𝑄𝑃𝑄=𝑄𝑃𝑄𝑃=𝑃𝑄𝑃𝑄=𝑃𝑄=𝑃𝑄𝑃 and it contradicts the hypothesis 𝑃𝑄𝑃≠𝑄𝑃𝑄. Thus, π‘Ž+𝑏+𝑑+𝑑/π‘Žπ‘=0.
Similarly, if 𝑄𝑃𝑄=𝑃𝑄𝑃𝑄, then we can obtain π‘‘π‘Ž+𝑏+𝑑+π‘Žπ‘=0, 𝑏+𝑐+𝑑+2π‘’βˆ’π‘π‘‘/𝑏=0, and π‘Ž+𝑏+𝑐+𝑐/π‘Žπ‘=0.
Obviously, if 𝑄𝑃𝑄≠𝑄𝑃𝑄𝑃 and 𝑄𝑃𝑄≠𝑃𝑄𝑃𝑄, we have π‘Ž+𝑏+𝑑+𝑑/π‘Žπ‘=0, π‘Ž+𝑏+𝑐+𝑐/π‘Žπ‘=0, 𝑏+𝑐+𝑑+2π‘’βˆ’π‘π‘‘/𝑏=0, and π‘Ž+𝑐+𝑑+2π‘“βˆ’π‘π‘‘/π‘Ž=0.
Next, we calculate these scalars. If π‘Ž+𝑏=0, then π‘Ž+𝑏+𝑐+𝑐/π‘Žπ‘=0 for any 𝑐 and π‘Ž+𝑏+𝑑+𝑑/π‘Žπ‘=0 for any 𝑑, and so 𝑐,𝑑,𝑒 are chosen to satisfy 𝑏+𝑐+𝑑+2π‘’βˆ’π‘π‘‘/𝑏=0. Similarly 𝑐,𝑑,𝑓 are chosen to satisfy π‘Ž+𝑐+𝑑+2π‘“βˆ’π‘π‘‘/π‘Ž=0.
If π‘Ž=𝑏, then 𝑐=𝑑=βˆ’π‘Ž, and 𝑒=π‘Ž by solving 𝑏+𝑐+𝑑+2π‘’βˆ’π‘π‘‘/𝑏=0, and 𝑓=π‘Ž by solving π‘Ž+𝑐+𝑑+2π‘“βˆ’π‘π‘‘/π‘Ž=0.
Note that 𝑏+𝑐+𝑑+2π‘’βˆ’π‘π‘‘/𝑏=0 and π‘Ž+𝑐+𝑑+2π‘“βˆ’π‘π‘‘/π‘Ž=0 imply 𝑔=πœƒβˆ’(π‘Ž+𝑏). Hence, we have (𝑐1)∼(𝑐6).
(ii) Consider the case of π‘Ž=π‘Žβˆ’1, 𝑏=π‘βˆ’1, and 𝑃𝑄𝑃𝑄=0.
Multiplying (2.34), respectively, on the right side by 𝑄𝑃 and on the left side by 𝑃𝑄 yieldsξ‚€1𝑐+π‘Ž+1𝑏+𝑐1π‘Žπ‘π‘„π‘ƒπ‘„=0,𝑑+π‘Ž+1𝑏+π‘‘ξ‚π‘Žπ‘π‘ƒπ‘„π‘ƒ=0.(2.43)
If 𝑄𝑃𝑄=0, then 𝑃𝑄𝑃≠0 and so a+𝑏+𝑑+𝑑/π‘Žπ‘=0 and (2.34) becomes ξ‚€10=𝑐+π‘Ž+1𝑏+𝑐2π‘Žπ‘π‘ƒπ‘„+π‘’βˆ’π‘Žβˆ’1π‘βˆ’π‘+π‘‘βˆ’π‘Žπ‘π‘π‘‘βˆ’π‘π‘’π‘Ž2𝑏𝑃𝑄𝑃.(2.44) Multiplying (2.44) on right side by 𝑄 yields ξ‚€1𝑐+π‘Ž+1𝑏+π‘ξ‚π‘Žπ‘π‘ƒπ‘„=0.(2.45) Since 𝑃𝑄≠0, π‘Ž+𝑏+𝑐+𝑐/π‘Žπ‘=0 and then (2.44) becomes ξ‚€2𝑒+𝑏+𝑐+π‘‘βˆ’π‘π‘‘π‘ξ‚π‘ƒπ‘„π‘ƒ.(2.46) Thus, 2𝑒+𝑏+𝑐+π‘‘βˆ’π‘π‘‘/𝑏=0.
If 𝑃𝑄𝑃=0, then we, similarly, have π‘Ž+𝑏+𝑐+𝑐/π‘Žπ‘=0, π‘Ž+𝑏+𝑑+𝑑/π‘Žπ‘=0, and 2𝑓+π‘Ž+𝑐+π‘‘βˆ’π‘π‘‘/π‘Ž=0.
If 𝑃𝑄𝑃≠0 and 𝑄𝑃𝑄≠0, then, multiplying (2.34), on the right side by 𝑄 and on the left side by 𝑃 yields π‘Ž+𝑏+𝑐+𝑐/π‘Žπ‘=0, and multiplying (2.34) on the right side by 𝑃 and on the left side by 𝑄 yields π‘Ž+𝑏+𝑑+𝑑/π‘Žπ‘=0. Thus, (2.34) becomes ξ‚€2π‘’βˆ’π‘Žβˆ’1π‘βˆ’π‘+π‘‘βˆ’π‘Žπ‘π‘π‘‘βˆ’π‘π‘’π‘Ž2𝑏1𝑃𝑄𝑃+π‘“βˆ’π‘Žβˆ’2π‘βˆ’π‘+π‘‘βˆ’π‘Žπ‘π‘π‘‘βˆ’π‘Žπ‘“π‘Žπ‘2𝑄𝑃𝑄=0.(2.47) Multiplying the equation above on the right side, respectively, by 𝑃 and by 𝑄 yields 2𝑒+𝑏+𝑐+π‘‘βˆ’π‘π‘‘π‘=0,2𝑓+π‘Ž+𝑐+π‘‘βˆ’π‘π‘‘π‘Ž=0.(2.48) As the argument above in (i), we have (𝑐7)∼(𝑐12).
(2) If πœƒ=0, then 𝐴𝑔=1π‘Ž1𝑃+𝑏1π‘„βˆ’π‘Ž+1𝑏+𝑐1π‘Žπ‘π‘ƒπ‘„βˆ’π‘Ž+1𝑏+𝑑+ξ‚€2π‘Žπ‘π‘„π‘ƒπ‘Ž+1𝑏+𝑐+𝑑+π‘Žπ‘π‘π‘‘βˆ’π‘π‘’π‘Ž2𝑏1𝑃𝑄𝑃+π‘Ž+2𝑏+𝑐+𝑑+π‘Žπ‘π‘π‘‘βˆ’π‘Žπ‘“π‘Žπ‘2ξ‚Άβˆ’ξ‚΅2π‘„π‘ƒπ‘„π‘Ž+2𝑏+𝑐+𝑑+π‘Žπ‘π‘π‘‘βˆ’π‘π‘’π‘Ž2𝑏+π‘π‘‘βˆ’π‘Žπ‘“π‘Žπ‘2𝑃𝑄𝑃𝑄,(2.49) and so π΄βˆ’π΄π‘”=ξ‚€1π‘Žβˆ’π‘Žξ‚ξ‚€1𝑃+π‘βˆ’π‘ξ‚ξ‚€1𝑄+𝑐+π‘Ž+1𝑏+𝑐+ξ‚€1π‘Žπ‘π‘ƒπ‘„π‘‘+π‘Ž+1𝑏+𝑑2π‘Žπ‘π‘„π‘ƒ+π‘’βˆ’π‘Žβˆ’1π‘βˆ’π‘+π‘‘βˆ’π‘Žπ‘π‘π‘‘βˆ’π‘π‘’π‘Ž2𝑏+ξ‚΅1π‘ƒπ‘„π‘ƒπ‘“βˆ’π‘Žβˆ’2π‘βˆ’π‘+π‘‘βˆ’π‘Žπ‘π‘π‘‘βˆ’π‘Žπ‘“π‘Žπ‘2ξ‚Ά+ξ‚΅2𝑄𝑃𝑄𝑔+π‘Ž+2𝑏+𝑐+𝑑+π‘Žπ‘π‘π‘‘βˆ’π‘π‘’π‘Ž2𝑏+π‘π‘‘βˆ’π‘Žπ‘“π‘Žπ‘2𝑃𝑄𝑃𝑄=0.(2.50) Analogous to the process in (c)(1), using (2.50), we can get ξ‚€1π‘Žβˆ’π‘Žξ‚ξ‚€1(π‘ƒπ‘„π‘ƒβˆ’π‘ƒπ‘„π‘ƒπ‘„)=0,π‘Žβˆ’π‘Žξ‚(π‘„π‘ƒπ‘„βˆ’π‘ƒπ‘„π‘ƒπ‘„)=0.(2.51)
Thus, since 𝑃𝑄𝑃≠𝑄𝑃𝑄, 𝑃𝑄𝑃≠𝑃𝑄𝑃𝑄 and/or 𝑄𝑃𝑄≠𝑃𝑄𝑃𝑄 and then π‘Ž=π‘Žβˆ’1. Similarly, 𝑏=π‘βˆ’1. Therefore, multiplying (2.50) on the right side by 𝑄 and on the left side by 𝑃 yields ξ‚€π‘π‘Ž+𝑏+𝑐+ξ‚π‘Žπ‘(π‘ƒπ‘„βˆ’π‘ƒπ‘„π‘ƒπ‘„)=0.(2.52) Multiplying (2.50) on the right side by 𝑃 and on the left side by 𝑄 yields ξ‚€π‘‘π‘Ž+𝑏+𝑑+ξ‚π‘Žπ‘(π‘„π‘ƒβˆ’π‘ƒπ‘„π‘ƒπ‘„)=0.(2.53) Since 𝑃𝑄≠𝑃𝑄𝑃𝑄 and 𝑄𝑃≠𝑃𝑄𝑃𝑄, π‘Ž+𝑏+𝑐+𝑐/π‘Žπ‘=0 and π‘Ž+𝑏+𝑑+𝑑/π‘Žπ‘=0. Multiplying (2.50) on the left side, respectively, by 𝑃 and by 𝑄 yields ξ‚€2𝑒+𝑏+𝑐+π‘‘βˆ’π‘π‘‘π‘ξ‚ξ‚€(π‘ƒπ‘„π‘ƒβˆ’π‘ƒπ‘„π‘ƒπ‘„)=0,2𝑓+π‘Ž+𝑐+π‘‘βˆ’π‘π‘‘π‘Žξ‚(π‘„π‘ƒπ‘„βˆ’π‘ƒπ‘„π‘ƒπ‘„)=0.(2.54) Thus, we have 2𝑒+𝑏+𝑐+π‘‘βˆ’π‘π‘‘/𝑏=0 and 𝑄𝑃𝑄=𝑃𝑄𝑃𝑄; 2𝑓+π‘Ž+𝑐+π‘‘βˆ’π‘π‘‘/π‘Ž=0 and 𝑃𝑄𝑃=𝑃𝑄𝑃𝑄; 2𝑒+𝑏+𝑐+π‘‘βˆ’π‘π‘‘/𝑏=0 and 2𝑓+π‘Ž+𝑐+π‘‘βˆ’π‘π‘‘/π‘Ž=0.
Note that 2𝑒+𝑏+𝑐+π‘‘βˆ’π‘π‘‘/𝑏=0 and 2𝑓+π‘Ž+𝑐+π‘‘βˆ’π‘π‘‘/π‘Ž=0 imply 𝑔=βˆ’(π‘Ž+𝑏) by πœƒ=0. As the argument above in (c)(1), we have (𝑐13)∼(𝑐18).

Remark 2.2. Clearly, [15, (a) and (b) in Theorem] are the special cases in Theorem 2.1.

Example 2.3. Let βŽ›βŽœβŽœβŽœβŽœβŽœβŽœβŽœβŽœβŽœβŽœβŽœβŽœβŽœβŽœβŽœβŽœβŽœβŽœβŽœβŽβŽžβŽŸβŽŸβŽŸβŽŸβŽŸβŽŸβŽŸβŽŸβŽŸβŽŸβŽŸβŽŸβŽŸβŽŸβŽŸβŽŸβŽŸβŽŸβŽŸβŽ βŽ›βŽœβŽœβŽœβŽœβŽœβŽœβŽœβŽœβŽœβŽœβŽœβŽœβŽœβŽœβŽœβŽœβŽœβŽœβŽœβŽβŽžβŽŸβŽŸβŽŸβŽŸβŽŸβŽŸβŽŸβŽŸβŽŸβŽŸβŽŸβŽŸβŽŸβŽŸβŽŸβŽŸβŽŸβŽŸβŽŸβŽ π‘ƒ=1000000000010000000000100000000001000000000010000000000000000000000000000000000000000000000000000000,𝑄=010001000000000000000000000100000000000000001000000100010000000000000000000001000010000βˆ’1100000000000.(2.55) Then they, obviously, are idempotent, and (𝑃𝑄)2=(𝑄𝑃)2 but 𝑃𝑄𝑃≠𝑄𝑃𝑄. By Theorem 2.1(𝑐5), 7𝐴=π‘ƒβˆ’π‘„+2𝑃𝑄+2π‘„π‘ƒβˆ’21π‘ƒπ‘„π‘ƒβˆ’2𝑄𝑃𝑄+𝑃𝑄𝑃𝑄(2.56) is the group involutory matrix, namely, 𝐴=𝐴𝑔, since 2+2+2βˆ—(βˆ’7/2)+2βˆ—2=1 and 2+2+2βˆ—(βˆ’1/2)βˆ’2βˆ—2=βˆ’1. By Theorem 2.1(𝑐17), π‘ƒβˆ’π‘„+π‘ƒπ‘„βˆ’2𝑄𝑃+2π‘ƒπ‘„π‘ƒβˆ’π‘„π‘ƒπ‘„(2.57) is group involutory since 1βˆ’2+2βˆ—2+1βˆ—(βˆ’2)=1 and 1βˆ’2+2βˆ—(βˆ’1)βˆ’1βˆ—(βˆ’2)=βˆ’1.

Next, we will study the situation (𝑃𝑄)2=0 or (𝑄𝑃)2=0.

Theorem 2.4. Let 𝑃,π‘„βˆˆβ„‚π‘›Γ—π‘› be two different nonzero idempotent matrices, and let 𝐴 be a combination of the form 𝐴=π‘Žπ‘ƒ+𝑏𝑄+𝑐𝑃𝑄+𝑑𝑄𝑃+𝑒𝑃𝑄𝑃+𝑓𝑄𝑃𝑄+𝑔𝑃𝑄𝑃𝑄,(2.58) where π‘Ž,𝑏,𝑐,𝑑,𝑒,𝑓,π‘”βˆˆβ„‚ with π‘Ž,𝑏≠0. Suppose that 𝑃𝑄𝑃𝑄≠0,𝑄𝑃𝑄𝑃=0,(2.59) and any of the following sets of additional conditions hold:

(𝑑1)π‘Ž=𝑏=Β±1, c=𝑑=βˆ“1, 𝑒=𝑓=Β±1,𝑔=βˆ“1;

(𝑑2)π‘Ž=Β±1, 𝑏=βˆ“1, 2𝑒+𝑐+𝑑±𝑐𝑑=Β±1, 2𝑓+𝑐+π‘‘βˆ“π‘π‘‘=βˆ“1.

Then 𝐴 is the group involutory matrix.

Proof. By Lemma 1.2, 0=π΄βˆ’π΄π‘”=ξ‚€1π‘Žβˆ’π‘Žξ‚ξ‚€1𝑃+π‘βˆ’π‘ξ‚ξ‚€1𝑄+𝑐+π‘Ž+1𝑏+𝑐1π‘Žπ‘π‘ƒπ‘„+𝑑+π‘Ž+1𝑏+𝑑+ξ‚€2π‘Žπ‘π‘„π‘ƒπ‘’βˆ’π‘Žβˆ’1π‘βˆ’π‘+π‘‘βˆ’π‘Žπ‘π‘π‘‘βˆ’π‘π‘’π‘Ž2𝑏+ξ‚΅1π‘ƒπ‘„π‘ƒπ‘“βˆ’π‘Žβˆ’2π‘βˆ’π‘+π‘‘βˆ’π‘Žπ‘π‘π‘‘βˆ’π‘Žπ‘“π‘Žπ‘2ξ‚Ά+ξ‚΅2𝑄𝑃𝑄𝑔+π‘Ž+2𝑏+2𝑐+𝑑+𝑔+π‘Žπ‘π‘π‘‘βˆ’π‘π‘’βˆ’π‘π‘’π‘Ž2𝑏+π‘π‘‘βˆ’π‘Žπ‘“βˆ’π‘π‘“π‘Žπ‘2+𝑐2π‘‘π‘Ž2𝑏2ξ‚Ά(𝑃𝑄)2.(2.60) Since 𝑃𝑄𝑃𝑄≠0, multiplying (2.60), respectively, on the right side and on the right side by 𝑃𝑄𝑃𝑄 yields ξ‚€1π‘Žβˆ’π‘Žξ‚ξ‚€1𝑃𝑄𝑃𝑄=0,π‘βˆ’π‘ξ‚π‘ƒπ‘„π‘ƒπ‘„=0,(2.61) and so π‘Ž=π‘Žβˆ’1 and 𝑏=π‘βˆ’1. Substituting them inside (2.60), we get ξ‚€10=𝑐+π‘Ž+1𝑏+𝑐1π‘Žπ‘π‘ƒπ‘„+𝑑+π‘Ž+1𝑏+𝑑+ξ‚€2π‘Žπ‘π‘„π‘ƒπ‘’βˆ’π‘Žβˆ’1π‘βˆ’π‘+π‘‘βˆ’π‘Žπ‘π‘π‘‘βˆ’π‘π‘’π‘Ž2𝑏+ξ‚΅1π‘ƒπ‘„π‘ƒπ‘“βˆ’π‘Žβˆ’2π‘βˆ’π‘+π‘‘βˆ’π‘Žπ‘π‘π‘‘βˆ’π‘Žπ‘“π‘Žπ‘2ξ‚Ά+ξ‚΅2𝑄𝑃𝑄𝑔+π‘Ž+2𝑏+2𝑐+𝑑+𝑔+π‘Žπ‘π‘π‘‘βˆ’π‘π‘’βˆ’π‘π‘’π‘Ž2𝑏+π‘π‘‘βˆ’π‘Žπ‘“βˆ’π‘π‘“π‘Žπ‘2+𝑐2π‘‘π‘Ž2𝑏2𝑃𝑄𝑃𝑄.(2.62) Multiplying (2.62) on the left side by 𝑃𝑄𝑃 yields ξ‚€1𝑐+π‘Ž+1𝑏+π‘ξ‚π‘Žπ‘π‘ƒπ‘„π‘ƒπ‘„=0,(2.63) and then 1𝑐+π‘Ž+1𝑏+π‘π‘Žπ‘=0.(2.64) So (2.62) becomes ξ‚€10=𝑑+π‘Ž+1𝑏+𝑑2π‘Žπ‘π‘„π‘ƒ+π‘’βˆ’π‘Žβˆ’1π‘βˆ’π‘+π‘‘βˆ’π‘Žπ‘π‘π‘‘βˆ’π‘π‘’π‘Ž2𝑏+ξ‚΅1π‘ƒπ‘„π‘ƒπ‘“βˆ’π‘Žβˆ’2π‘βˆ’π‘+π‘‘βˆ’π‘Žπ‘π‘π‘‘βˆ’π‘Žπ‘“π‘Žπ‘2ξ‚Ά+ξ‚΅2𝑄𝑃𝑄𝑔+π‘Ž+2𝑏+2𝑐+𝑑+𝑔+π‘Žπ‘π‘π‘‘βˆ’π‘π‘’βˆ’π‘π‘’π‘Ž2𝑏+π‘π‘‘βˆ’π‘Žπ‘“βˆ’π‘π‘“π‘Žπ‘2+𝑐2π‘‘π‘Ž2𝑏2𝑃𝑄𝑃𝑄.(2.65) Multiplying (2.65) on the left side by 𝑃𝑄 and on the right side by 𝑃 yields ξ‚€1𝑑+π‘Ž+1𝑏+π‘‘ξ‚π‘Žπ‘π‘ƒπ‘„P𝑄=0.(2.66) Therefore, 1𝑑+π‘Ž+1𝑏+π‘‘π‘Žπ‘=0.(2.67) Similarly, we can obtain 20=π‘’βˆ’π‘Žβˆ’1π‘βˆ’π‘+π‘‘βˆ’π‘Žπ‘π‘π‘‘βˆ’π‘π‘’π‘Ž2𝑏,10=π‘“βˆ’π‘Žβˆ’2π‘βˆ’π‘+π‘‘βˆ’π‘Žπ‘π‘π‘‘βˆ’π‘Žπ‘“π‘Žπ‘2,20=𝑔+π‘Ž+2𝑏+2𝑐+𝑑+𝑔+π‘Žπ‘π‘π‘‘βˆ’π‘π‘’βˆ’π‘π‘’π‘Ž2𝑏+π‘π‘‘βˆ’π‘Žπ‘“βˆ’π‘π‘“π‘Žπ‘2+𝑐2π‘‘π‘Ž2𝑏2.(2.68) By (2.64) and (2.67), we can obtain 1𝑏+𝑐+𝑑+2π‘’βˆ’π‘π‘‘π‘1=0,π‘Ž+𝑐+𝑑+2π‘“βˆ’π‘π‘‘π‘Ž=0.(2.69) Since π‘Ž=π‘Žβˆ’1 and 𝑏=π‘βˆ’1, π‘Ž=±𝑏. If π‘Ž=βˆ’π‘, then (2.64) holds for any 𝑐, (2.67) holds for any 𝑑, and, for any 𝑐,𝑑,𝑒,𝑓 satisfying (2.69) and any 𝑔, 2𝑔+π‘Ž+2𝑏+2𝑐+𝑑+𝑔+π‘Žπ‘π‘π‘‘βˆ’π‘π‘’βˆ’π‘π‘’π‘Ž2𝑏+π‘π‘‘βˆ’π‘Žπ‘“βˆ’π‘π‘“π‘Žπ‘2+𝑐2π‘‘π‘Ž2𝑏2=𝑐2π‘π‘‘βˆ’2π‘βˆ’π‘‘βˆ’(𝑒+𝑓)+π‘Ž(π‘’βˆ’π‘“)=𝑐2π‘π‘‘βˆ’2π‘βˆ’π‘‘+(𝑐+𝑑)+π‘Žξ‚€1π‘Žβˆ’π‘π‘‘π‘Žξ‚=0.(2.70) If π‘Ž=𝑏, then, by (2.64) ~ (2.69), 𝑐=𝑑=βˆ’π‘Ž and 𝑒=𝑓=π‘Ž and so 𝑔=βˆ’π‘Ž from (2.68).
Hence, we have (𝑑1) and (𝑑2).

Example 2.5. Let βŽ›βŽœβŽœβŽœβŽœβŽœβŽœβŽβŽžβŽŸβŽŸβŽŸβŽŸβŽŸβŽŸβŽ βŽ›βŽœβŽœβŽœβŽœβŽœβŽœβŽβŽžβŽŸβŽŸβŽŸβŽŸβŽŸβŽŸβŽ π‘ƒ=1000010000000000,𝑄=00101001001010βˆ’11.(2.71) Obviously they are idempotent, and (𝑄𝑃)2=0 but (𝑃𝑄)2β‰ 0. By Theorem 2.4(𝑑2), 5π‘ƒβˆ’π‘„+2π‘ƒπ‘„βˆ’2𝑄𝑃+25π‘ƒπ‘„π‘ƒβˆ’2π‘„π‘ƒπ‘„βˆ’2𝑃𝑄𝑃𝑄(2.72) is group involutory since 2βˆ’2+2βˆ—(5/2)+2βˆ—(βˆ’2)=1 and 2βˆ’2+2βˆ—(βˆ’5/2)βˆ’2βˆ—(βˆ’2)=βˆ’1.

Similarly, we have the following result.

Theorem 2.6. Let 𝑃,π‘„βˆˆβ„‚π‘›Γ—π‘› be two different nonzero idempotent matrices, and let 𝐴 be a combination of the form 𝐴=π‘Žπ‘ƒ+𝑏𝑄+𝑐𝑃𝑄+𝑑𝑄𝑃+𝑒𝑃𝑄𝑃+𝑓𝑄𝑃𝑄+β„Žπ‘„π‘ƒπ‘„π‘ƒ,(2.73) where π‘Ž,𝑏,𝑐,𝑑,𝑒,𝑓,β„Žβˆˆβ„‚ with π‘Ž,𝑏≠0. Suppose that 𝑄𝑃𝑄𝑃≠0,𝑃𝑄𝑃𝑄=0,(2.74) and any of the following sets of additional conditions hold:(e1)π‘Ž=𝑏=Β±1, 𝑐=𝑑=βˆ“1, 𝑒=𝑓=Β±1,β„Ž=βˆ“1; (e2)π‘Ž=Β±1, 𝑏=βˆ“1, 2𝑒+𝑐+𝑑±𝑐𝑑=Β±1, 2𝑓+𝑐+π‘‘βˆ“π‘π‘‘=βˆ“1.Then 𝐴 is the group involutory matrix.

Acknowledgment

This work was supported by the National Natural Science Foundation of China (11061005) and the Ministry of Education Science anf Technology Key Project (210164) and Grants (HCIC201103) of Guangxi Key Laborarory of Hybrid Computational and IC Design Analysis Open Fund.

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