The Group Involutory Matrix of the Combinations of Two Idempotent Matrices

Abstract

We discuss the following problem: when 𝑎𝑃+𝑏𝑄+𝑐𝑃𝑄+𝑑𝑄𝑃+𝑒𝑃𝑄𝑃+𝑓𝑄𝑃𝑄+𝑔𝑃𝑄𝑃𝑄 of idempotent matrices 𝑃 and 𝑄, where 𝑎,𝑏,𝑐,𝑑,𝑒,𝑓,𝑔∈ℂ and 𝑎≠0,𝑏≠0, is group involutory.

1. Introduction

Throughout this paper ℂ^𝑛×𝑛 stands for the set of 𝑛×𝑛 complex matrices. Let 𝐴∈ℂ^𝑛×𝑛. 𝐴 is said to be idempotent if 𝐴²=𝐴. 𝐴 is said to be group invertible if there exists an 𝑋∈ℂ^𝑛×𝑛 such that𝐴𝑋𝐴=𝐴,𝑋𝐴𝑋=𝑋,𝐴𝑋=𝑋𝐴(1.1) hold. If such an 𝑋 exists, then it is unique, denoted by 𝐴_𝑔, 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 𝐴_𝑔=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 𝑎, 𝑎𝑃 is a group involutory matrix if and only if either 𝑎=1 or 𝑎=−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 𝑎𝑃+𝑏𝑄+𝑐𝑃𝑄+𝑑𝑄𝑃+𝑒𝑃𝑄𝑃+𝑓𝑄𝑃𝑄+𝑔𝑃𝑄𝑃𝑄 of idempotent matrices 𝑃 and 𝑄, where 𝑎,𝑏,𝑐,𝑑,𝑒,𝑓,𝑔∈ℂ with 𝑎,𝑏≠0, deduced its explicit expressions, and some necessary and sufficient conditions for the existence of the group inverse of 𝑎𝑃+𝑏𝑄+𝑐𝑃𝑄.

In this paper, we will investigate the following problem: when 𝑎𝑃+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,𝑄∈ℂ^𝑛×𝑛 be two different nonzero idempotent matrices. Suppose (𝑃𝑄)²=(𝑄𝑃)². Then for any scalars 𝑎,𝑏,𝑐,𝑑,𝑒,𝑓,𝑔, where 𝑎,𝑏≠0 and 𝜃=𝑎+𝑏+𝑐+𝑑+𝑒+𝑓+𝑔, 𝑎𝑃+𝑏𝑄+𝑐𝑃𝑄+𝑑𝑄𝑃+𝑒𝑃𝑄𝑃+𝑓𝑄𝑃𝑄+𝑔(𝑃𝑄)² is group invertible, and
(i) if 𝜃≠0, then 𝑎𝑃+𝑏𝑄+𝑐𝑃𝑄+𝑑𝑄𝑃+𝑒𝑃𝑄𝑃+𝑓𝑄𝑃𝑄+𝑔(𝑃𝑄)²_𝑔=1+𝑎𝑃2+1𝑏𝑄−𝑎1+1𝑏+𝑎𝑐𝑏𝑃𝑄−𝑎1+1𝑏+𝑎𝑑𝑏𝑄𝑃−𝑎2+1𝑏+𝑐𝑎+𝑏𝑑+𝑐𝑑𝑎−²𝑏𝑏𝑒𝑃𝑄𝑃+𝑎1+2𝑏+𝑐𝑎+𝑏𝑑+𝑐𝑑𝑎−𝑏²𝑎𝑓𝑄𝑃𝑄𝑎+2𝑏+𝑐𝑎+𝑏𝑑+𝑐𝑑𝑎−²𝑏𝑏𝑒+𝑐𝑑𝑎−𝑏²𝑎𝑓−1𝜃𝑃𝑄𝑃𝑄;(1.2)
(ii) if 𝜃=0, then 𝑎𝑃+𝑏𝑄+𝑐𝑃𝑄+𝑑𝑄𝑃+𝑒𝑃𝑄𝑃+𝑓𝑄𝑃𝑄+𝑔(𝑃𝑄)²_𝑔=1+𝑎𝑃2+1𝑏𝑄−𝑎1+1𝑏+𝑎𝑐𝑏𝑃𝑄−𝑎1+1𝑏+𝑎𝑑𝑏𝑄𝑃−𝑎2+1𝑏+𝑐𝑎+𝑏𝑑+𝑐𝑑𝑎−²𝑏𝑏𝑒𝑃𝑄𝑃+𝑎1+2𝑏+𝑐𝑎+𝑏𝑑+𝑐𝑑𝑎−𝑏²𝑎𝑓𝑄𝑃𝑄𝑎+2𝑏+𝑐𝑎+𝑏𝑑+𝑐𝑑𝑎−²𝑏𝑏𝑒+𝑐𝑑𝑎−𝑏²𝑎𝑓(𝑃𝑄)².(1.3)

Lemma 1.2 (see [16, Theorem  3.1]). Let 𝑃,𝑄∈ℂ^𝑛×𝑛 be two different nonzero idempotent matrices. Suppose (𝑄𝑃)²=0. Then for any scalars 𝑎,𝑏,𝑐,𝑑,𝑒,𝑓, and 𝑔, where 𝑎,𝑏≠0, 𝑎𝑃+𝑏𝑄+𝑐𝑃𝑄+𝑑𝑄𝑃+𝑒𝑃𝑄𝑃+𝑓𝑄𝑃𝑄+𝑔(𝑃𝑄)² is group invertible, and 𝑎𝑃+𝑏𝑄+𝑐𝑃𝑄+𝑑𝑄𝑃+𝑒𝑃𝑄𝑃+𝑓𝑄𝑃𝑄+𝑔(𝑃𝑄)²_𝑔=1+𝑎𝑃2+1𝑏𝑄−𝑎1+1𝑏+𝑎𝑐𝑏𝑃𝑄−𝑎1+1𝑏+𝑎𝑑𝑏𝑄𝑃−𝑎2+1𝑏+𝑐𝑎+𝑏𝑑+𝑐𝑑𝑎−²𝑏𝑏𝑒𝑃Q𝑃+𝑎1+2𝑏+𝑐𝑎+𝑏𝑑+𝑐𝑑𝑎−𝑏²𝑎𝑓𝑄𝑃𝑄𝑎+2𝑏+2𝑐+𝑎𝑑𝑏+𝑔+𝑐𝑑−𝑎𝑏²𝑒𝑏−𝑐𝑒+𝑐𝑑−𝑎𝑎𝑏𝑓²−𝑐𝑓+𝑐²𝑑𝑎²𝑏²(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,𝑄∈ℂ^𝑛×𝑛 be two different nonzero idempotent matrices with (𝑃𝑄)²=(𝑄𝑃)², and let 𝐴 be a combination of the form 𝐴=𝑎𝑃+𝑏𝑄+𝑐𝑃𝑄+𝑑𝑄𝑃+𝑒𝑃𝑄𝑃+𝑓𝑄𝑃𝑄+𝑔𝑃𝑄𝑃𝑄,(2.1) where 𝑎,𝑏,𝑐,𝑑,𝑒,𝑓,𝑔∈ℂ with 𝑎,𝑏≠0. Denote 𝜃=𝑎+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 (𝑎₁)∼(𝑎₃), in which 𝑃𝑄=𝑄𝑃,(2.2) and any of the following sets of additional conditions hold:(a₁)either 𝑎=1 or 𝑎=−1, either 𝜃=1 or 𝜃=−1 or 𝜃=0, and 𝑄=𝑃𝑄;(a₂)either 𝑏=1 or 𝑏=−1, either 𝜃=1 or 𝜃=−1 or 𝜃=0, and 𝑃=𝑃𝑄;(a₃)either 𝑎=1 or 𝑎=−1, either 𝑏=1 or 𝑏=−1, either 𝜃=1 or 𝜃=−1 or 𝜃=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)𝑎=±1,𝑏=∓1, either 𝜃=1 or 𝜃=−1 or 𝜃=0 or 𝑃𝑄𝑃=0; (b₂)𝑎=𝑏=±1,𝑐=𝑑=∓1, either 𝜃=1 or 𝜃=−1 or 𝜃=0 or 𝑃𝑄𝑃=0;(b₃)𝑎=𝑏=±1,𝑐=∓1, either 𝜃=1 or 𝜃=−1 or 𝜃=0, and 𝑄𝑃=𝑃𝑄𝑃;(b₄)𝑎=𝑏=±1,𝑑=∓1, either 𝜃=1 or 𝜃=−1 or 𝜃=0, and 𝑃𝑄=𝑃𝑄𝑃;(b₅)𝑎=𝑏=±1, 𝑐=∓1, and 𝑄𝑃=0;(b₆)𝑎=𝑏=±1, 𝑑=∓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)𝑎=±1,𝑏=∓1,𝑐+𝑑+2𝑒±𝑐𝑑=±1, either 𝜃=1 or 𝜃=−1, and 𝑄𝑃𝑄=𝑃𝑄𝑃𝑄; (c₂)𝑎=𝑏=𝑒=±1,𝑐=𝑑=∓1, either 𝜃=1 or 𝜃=−1, and 𝑄𝑃𝑄=𝑃𝑄𝑃𝑄;(c₃)𝑎=±1,𝑏=∓1,𝑐+𝑑+2𝑓∓𝑐𝑑=∓1, either 𝜃=1 or 𝜃=−1, and 𝑃𝑄𝑃=𝑃𝑄𝑃𝑄;(c₄)𝑎=𝑏=𝑓=±1,𝑐=𝑑=∓1, either 𝜃=1 or 𝜃=−1, and 𝑃𝑄𝑃=𝑃𝑄𝑃𝑄;(c₅)𝑎=±1,𝑏=∓1,𝑐+𝑑+2𝑒±𝑐𝑑=±1,𝑐+𝑑+2𝑓∓𝑐𝑑=∓1, either 𝑔=1 or 𝑔=−1;(c₆)𝑎=𝑏=𝑒=𝑓=±1,𝑐=𝑑=∓1, either 𝑔=∓1 or 𝑔=∓3;(c₇)𝑎=±1,𝑏=∓1,𝑐+𝑑+2𝑒±𝑐𝑑=±1, and 𝑄𝑃𝑄=0;(c₈)𝑎=𝑏=𝑒=±1,𝑐=𝑑=∓1, and 𝑄𝑃𝑄=0;(c₉)𝑎=±1,𝑏=∓1,𝑐+𝑑+2𝑓∓𝑐𝑑=∓1, and 𝑃𝑄𝑃=0;(c₁₀)𝑎=𝑏=𝑓=±1,𝑐=𝑑=∓1, and 𝑃𝑄𝑃=0;(c₁₁)𝑎=±1,𝑏=∓1,𝑐+𝑑+2𝑒±𝑐𝑑=±1,𝑐+𝑑+2𝑓∓𝑐𝑑=∓1, and 𝑃𝑄𝑃𝑄=0;(c₁₂)𝑎=𝑏=𝑒=𝑓=±1,𝑐=𝑑=∓1, and 𝑃𝑄𝑃𝑄=0;(c₁₃)𝑎=±1,𝑏=∓1,2𝑒+𝑐+𝑑±𝑐𝑑=±1,𝜃=0, and 𝑄𝑃𝑄=𝑃𝑄𝑃𝑄;(c₁₄)𝑎=𝑏=𝑒=±1,𝑐=𝑑=∓1,𝜃=0, and 𝑄𝑃𝑄=𝑃𝑄𝑃𝑄;(c₁₅)𝑎=±1,𝑏=∓1,2𝑓+𝑐+𝑑∓𝑐𝑑=∓1,𝜃=0, and 𝑃𝑄𝑃=𝑃𝑄𝑃𝑄;(c₁₆)𝑎=𝑏=𝑓=±1,𝑐=𝑑=∓1,𝜃=0, and 𝑃𝑄𝑃=𝑃𝑄𝑃𝑄;(c₁₇)𝑎=±1,𝑏=∓1,2𝑒+𝑐+𝑑±𝑐𝑑=±1,2𝑓+𝑐+𝑑∓𝑐𝑑=∓1,𝑔=0;(c₁₈)𝑎=𝑏=𝑒=𝑓=±1,𝑐=𝑑=∓1,𝑔=∓2.

Proof. Obviously, the condition (2.2) implies that the group inverse of 𝐴 exists and is of the form (1.2) when 𝜃≠0 or the form (1.3) when 𝜃=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 𝐴−𝐴_𝑔=0.
(a)  Under the condition (2.2), 𝐴=𝑎𝑃+𝑏𝑄+𝜇𝑃𝑄, where 𝜇=𝑐+𝑑+𝑒+𝑓+𝑔.
(1) If 𝜃≠0, then 𝐴_𝑔=1𝑎𝑃+1𝑏𝑄+𝜃1−𝑎1−1𝑏𝑃𝑄,(2.5) and so 𝐴−𝐴_𝑔=𝑎−1𝑎𝑃+𝑏−1𝑏𝑄+𝜇−𝜃1+𝑎1+1𝑏𝑃𝑄=0.(2.6) Multiplying (2.6) by 𝑃 and 𝑄, respectively, leads to 𝑎𝑎−−1𝑎1𝑃+𝑏−1𝑏𝑃𝑄+𝜇−𝜃1+𝑎1+1𝑏𝑃𝑄=0,𝑎𝑃𝑄+𝑏−1𝑏𝑄+𝜇−𝜃1+𝑎1+1𝑏𝑃𝑄=0,(2.7) and then 𝑎−𝑎1𝑃+𝑏−1𝑏𝑃𝑄=𝑎−𝑎1𝑃𝑄+𝑏−1𝑏𝑄.(2.8) Multiplying the above equation, respectively, by 𝑃 and by 𝑄, we get 𝑎−𝑎1(𝑃−𝑃𝑄)=0,𝑏−1𝑏(𝑄−𝑃𝑄)=0.(2.9) Thus, since 𝑃≠𝑄, we have three situations: 𝑃=𝑃𝑄 and 𝑏=𝑏⁻¹; 𝑎=𝑎⁻¹ and 𝑄=𝑃𝑄; 𝑎=𝑎⁻¹ and 𝑏=𝑏⁻¹.
When 𝑄=𝑃𝑄 and 𝑎=𝑎⁻¹, (2.6) becomes (𝜃−𝜃⁻¹)𝑄=0 and then 𝜃=±1. Therefore, we obtain (𝑎₁) except the situation 𝜃=0. Similarly, when 𝑏=𝑏⁻¹ and 𝑃=𝑃𝑄, we have (𝑎₂) except the situation 𝜃=0. When 𝑎=𝑎⁻¹ and 𝑏=𝑏⁻¹, (2.6) becomes (𝜃−𝜃⁻¹)𝑃𝑄=0 and then 𝜃=±1 or 𝑃𝑄=0. Therefore, we obtain (𝑎₃) except the situation 𝜃=0.
(2) If 𝜃=0, then 𝐴_𝑔=1𝑎𝑃+1𝑏𝑄−𝑎1+1𝑏𝑃𝑄,(2.10) and then 𝐴−𝐴_𝑔=𝑎−1𝑎𝑃+𝑏−1𝑏𝑄+𝜇+𝑎1+1𝑏𝑃𝑄=0.(2.11) Analogous to the process of reaching (2.9) in (a)(1), we have 𝑏−1𝑏(𝑄−𝑃𝑄)=0,𝑎−𝑎1(𝑃−𝑃𝑄)=0.(2.12) Thus, we have three situations: 𝑃=𝑃𝑄 and 𝑏=𝑏⁻¹; 𝑎=𝑎⁻¹ and 𝑄=𝑃𝑄; 𝑎=𝑎⁻¹ and 𝑏=𝑏⁻¹, since 𝑃≠𝑄. Similar to the argument in (𝑎)(1), substituting them, respectively, into (2.11), we can obtain the situation 𝜃=0, respectively, in (𝑎₁),(𝑎₂), and (𝑎₃).
(b) Under the condition (2.3), 𝐴=𝑎𝑃+𝑏𝑄+𝑐𝑃𝑄+𝑑𝑄𝑃+𝜈𝑃𝑄𝑃, where 𝜈=𝑒+𝑓+𝑔.
(1) If 𝜃≠0, then 𝐴_𝑔=1𝑎+𝑃+11𝑏𝑄−𝑎1+1𝑏+𝑎𝑐𝑏𝑃𝑄−𝑎1+1𝑏+𝑎𝑑𝑏𝑄𝑃𝑎+1𝑏+𝑐𝑎+𝑏𝑑+𝜃1𝑃𝑄𝑃,(2.13) and so 𝐴−𝐴_𝑔=𝑎−1+𝑑+𝑎1𝑃+𝑏−1𝑏𝑄+𝑐+𝑎1+1𝑏+𝑎𝑐𝑏𝑃𝑄𝑎+1𝑏+𝑎𝑑𝑏𝑄𝑃+𝜈−𝑎1−1𝑏−𝑐𝑎+𝑏𝑑−𝜃1𝑃𝑄𝑃=0.(2.14) Multiplying the above equation, respectively, on the two sides by 𝑃 yields 0=𝑎−10=𝑎−𝑎1𝑃+𝑐+𝑏+𝑎1+𝑎𝑐𝑏𝑃𝑄+𝜈+𝑑−𝑎𝑐𝑏−𝜃1𝑃𝑄𝑃,(2.15)𝑎𝑃+𝑏+𝑑+𝑎1+𝑎𝑑𝑏𝑄𝑃+𝜈+𝑐−𝑎𝑑𝑏−𝜃1𝑃𝑄𝑃.(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 𝑎𝑎−−𝑎11𝑄𝑃+𝑏+𝑐+𝑑+𝜈+𝑎1−𝜃1𝑄𝑃𝑄=0,𝑎𝑃𝑄+𝑏+𝑐+𝑑+𝜈+𝑎1−𝜃1𝑄𝑃𝑄=0,(2.17) and then (𝑎−𝑎⁻¹)(𝑄𝑃−𝑃𝑄)=0. Since 𝑄𝑃≠𝑃𝑄, 𝑎=𝑎⁻¹. Similarly, 𝑏=𝑏⁻¹.
Substituting 𝑎=𝑎⁻¹ inside (2.17) yields (𝜃−𝜃⁻¹)𝑄𝑃𝑄=0 and then 𝜃=𝜃⁻¹ or 𝑄𝑃𝑄=0. We will discuss the remainder for detail as follows:
When 𝑎=𝑎⁻¹, 𝑏=𝑏⁻¹, (2.14) becomes0=𝑐+1+𝜈𝑎−+11𝑏+𝑎𝑐𝑏𝑃𝑄+𝑑+𝑎1+1𝑏+𝑎𝑑𝑏𝑄𝑃𝑎−1𝑏−𝑐𝑎+𝑏𝑑−𝜃1𝑃𝑄𝑃,(2.18)
(i) if 𝑎+𝑏=0, then 𝑐+1𝑎+1𝑏+𝑎𝑐𝑏=0,𝑑+𝑎1+1𝑏+𝑎𝑑𝑏=0,(2.19) and so it follows from (2.18) that 𝜃−𝜃1𝑃𝑄𝑃=𝜈+𝑐+𝑑−𝜃1𝑃𝑄𝑃=0.(2.20) Therefore, either 𝜃=𝜃⁻¹ or 𝑃𝑄𝑃=0 implies that (2.18) holds, namely, (2.14) holds. Thus, we have (𝑏₁) except the situation 𝜃=0.
(ii) if 𝑎=𝑏, then (2.18) becomes 0=(2𝑐+2𝑎)𝑃𝑄+(2𝑑+2𝑎)𝑄𝑃+2𝜈−𝜃−1𝜃𝑃𝑄𝑃.(2.21) Multiplying the above equation, respectively, on the right side by 𝑃 and on the left side by 𝑄, we have 0=(2𝑐+2𝑎)𝑃𝑄+𝜈+𝑑−𝑐−10=(2𝑑+2𝑎)𝑄𝑃+𝜈+𝑐−𝑑−𝜃1𝑃𝑄𝑃,(2.22)𝜃𝑃𝑄𝑃.(2.23) So if 𝜃=𝜃⁻¹, then the two equations above (2.22) and (2.23) become, respectively, (𝑐+𝑎)(𝑃𝑄−𝑃𝑄𝑃)=0,(𝑑+𝑎)(𝑄𝑃−𝑃𝑄𝑃)=0.(2.24) Or if 𝑃𝑄𝑃=0, then (2.22) and (2.23) become, respectively, (𝑐+𝑎)𝑃𝑄=0,(𝑑+𝑎)𝑄𝑃=0.(2.25) Since 𝑃𝑄≠𝑄𝑃, it follows from (2.24) and (2.25) that we have the six situations: 𝜃=𝜃⁻¹ and 𝑐=𝑑=−𝑎; 𝜃=𝜃⁻¹, 𝑐=−𝑎 and 𝑄𝑃=𝑃𝑄𝑃; 𝜃=𝜃⁻¹, 𝑑=−𝑎, and 𝑃𝑄=𝑃𝑄𝑃; 𝑐=−𝑎 and 𝑄𝑃=0; 𝑑=−𝑎 and 𝑃𝑄=0; 𝑐=𝑑=−𝑎 and 𝑃𝑄𝑃=0. Thus, we have (𝑏₂)∼(𝑏₄) except the situation 𝜃=0, and (𝑏₅) and (𝑏₆).
(2) If 𝜃=0, then 𝐴_𝑔=1𝑎𝑃+1𝑏𝑄−𝑎1+1𝑏+𝑎𝑐𝑏𝑃𝑄−𝑎1+1𝑏+𝑎𝑑𝑏𝑄𝑃+𝑎1+1𝑏+𝑐𝑎+𝑏𝑑𝑃𝑄𝑃,(2.26) and then 𝐴−𝐴_𝑔=𝑎−1+𝑑𝑎+1𝑃+𝑏−1𝑏𝑄+𝑐+𝑎1+1𝑏+𝑎𝑐𝑏𝑃𝑄𝑎+1𝑏+𝑎𝑑𝑏𝑄𝑃+𝜈−𝑎1−1𝑏−𝑐𝑎+𝑏𝑑𝑃𝑄𝑃=0.(2.27) Analogous to the process in (b)(1), using (2.27) we can obtain 𝑎𝑎−−𝑎11𝑄𝑃−𝑎−𝑎1𝑃𝑄𝑃=0,𝑎𝑃𝑄−𝑎−𝑎1𝑃𝑄𝑃=0.(2.28) Thus, since 𝑃𝑄≠𝑄𝑃, 𝑃𝑄≠𝑃𝑄𝑃 and/or 𝑄𝑃≠𝑃𝑄𝑃 and then 𝑎=𝑎⁻¹. Similarly, 𝑏=𝑏⁻¹. Hence, 𝑎=±𝑏.
(i)  If 𝑎=−𝑏, then 𝑐+1𝑑+𝑎1+1𝑏+𝑎𝑐𝑏=0,𝜈−1𝑎+1𝑏+𝑎𝑑𝑏=0,𝑎−1𝑏−𝑐𝑎+𝑏𝑑=−2(𝑎+𝑏)=0.(2.29) Thus, (2.27) holds. Hence we have the situation 𝜃=0 in (𝑏₁).
(ii) If 𝑎=𝑏, then (2.27) becomes (𝑐+𝑎)𝑃𝑄+(𝑑+𝑎)𝑄𝑃+𝜈𝑃𝑄𝑃=0.(2.30) Multiplying the above equation on the left side, respectively, by 𝑃 and by 𝑄, we have (𝑐+𝑎)(𝑃𝑄−𝑃𝑄𝑃)=0,(𝑑+𝑎)(𝑄𝑃−𝑃𝑄𝑃)=0.(2.31) Thus, 𝑐=𝑑=−𝑎; 𝑐=−𝑎 and 𝑄𝑃=𝑃𝑄𝑃; 𝑑=−𝑎 and 𝑃𝑄=𝑃𝑄𝑃. Hence, we have the situation 𝜃=0, respectively, in (𝑏₂),(𝑏₃), and (𝑏₄).
(c) Under the condition (2.4), 𝐴=𝑎𝑃+𝑏𝑄+𝑐𝑃𝑄+𝑑𝑄𝑃+𝑒𝑃𝑄𝑃+𝑓𝑄𝑃𝑄𝑃+𝑔𝑃𝑄𝑃𝑄.(2.32)
(1)  If 𝜃≠0, then 𝐴_𝑔=1+𝑎𝑃2+1𝑏𝑄−𝑎1+1𝑏+𝑎𝑐𝑏𝑃𝑄−𝑎1+1𝑏+𝑎𝑑𝑏𝑄𝑃−𝑎2+1𝑏+𝑐𝑎+𝑏𝑑+𝑐𝑑𝑎−²𝑏𝑏𝑒𝑃𝑄𝑃+𝑎1+2𝑏+𝑐𝑎+𝑏𝑑+𝑐d𝑎−𝑏²𝑎𝑓𝑄𝑃𝑄𝑎+2𝑏+𝑐𝑎+𝑏𝑑+𝑐𝑑𝑎−²𝑏𝑏𝑒+𝑐𝑑𝑎−𝑏²𝑎𝑓−1𝜃𝑃𝑄𝑃𝑄,(2.33) and so 𝐴−𝐴_𝑔=𝑎−1+𝑒−𝑎2𝑃+𝑏−1𝑏𝑄+𝑐+𝑎1+1𝑏+𝑎𝑐𝑏𝑃𝑄+𝑑+𝑎1+1𝑏+𝑎𝑑𝑏𝑄𝑃+𝑓−𝑎1−1𝑏−𝑐𝑎+𝑏𝑑−𝑐𝑑𝑎−²𝑏𝑏𝑒𝑃𝑄𝑃+𝑔+2𝑎−2𝑏−𝑐𝑎+𝑏𝑑−𝑐𝑑𝑎−𝑏²𝑎𝑓𝑄𝑃𝑄𝑎+2𝑏+𝑐𝑎+𝑏𝑑+𝑐𝑑𝑎−²𝑏𝑏𝑒+𝑐𝑑𝑎−𝑏²𝑎𝑓−1𝜃𝑃𝑄𝑃𝑄=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 𝑎−𝑎1𝑄𝑃+𝑏+𝑐+𝑎1+𝑎𝑐𝑏𝑄𝑃𝑄+𝑑+𝑒+𝑓+𝑔−𝑎𝑐𝑏−𝜃1𝑃𝑄𝑃𝑄=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), 0=𝑎−10=𝑎−𝑎1𝑃𝑄𝑃+𝑎1−𝑎+𝜃−𝜃1𝑃𝑄𝑃𝑄,(2.36)𝑎𝑄𝑃𝑄+𝑎1−𝑎+𝜃−𝜃1𝑃𝑄𝑃𝑄.(2.37) Since 𝑃𝑄𝑃≠𝑄𝑃𝑄, 𝑎=𝑎⁻¹ by (2.36) and (2.37). Similarly, we can gain 𝑏=𝑏⁻¹. Substituting 𝑎=𝑎⁻¹ inside (2.36) yields 𝜃=𝜃⁻¹ or 𝑃𝑄𝑃𝑄=0.
(i)  Consider the case of 𝑎=𝑎⁻¹, 𝑏=𝑏⁻¹ and 𝜃=𝜃⁻¹.
Substituting 𝑎=𝑎⁻¹, 𝑏=𝑏⁻¹, and 𝜃=𝜃⁻¹ inside (2.35) yields 𝑎+𝑏+𝑐+𝑎𝑐𝑏(𝑄𝑃𝑄−𝑃𝑄𝑃𝑄)=0.(2.38) Similarly, we have 𝑎+𝑏+𝑑+𝑎𝑑𝑏(𝑃𝑄𝑃−𝑃𝑄𝑃𝑄)=0.(2.39)
If 𝑃𝑄𝑃=𝑃𝑄𝑃𝑄, then 𝑄𝑃𝑄≠𝑃𝑄𝑃𝑄 by the hypothesis 𝑃𝑄𝑃≠𝑄𝑃𝑄 and so 𝑎+𝑏+𝑐+𝑐/𝑎𝑏=0 by (2.38). Multiplying (2.34) on the right side by 𝑄 yields 𝑎+𝑐+𝑑+2𝑓−𝑐𝑎𝑑(𝑄𝑃𝑄−𝑃𝑄𝑃𝑄)=0.(2.40) Thus, 𝑎+𝑐+𝑑+2𝑓−𝑐𝑑/𝑎=0 and then (2.14) becomes 𝑎++𝑏𝑏++𝑑𝑒++𝑎𝑔𝑑𝑏+𝑐𝑄𝑑𝑃−+𝑎𝑓𝑓−𝑎−2𝑏−𝑐𝑎+𝑏𝑑−𝑐𝑑−𝑎𝑎𝑓𝑄𝑃𝑄𝑎−𝜃𝑃𝑄𝑃=0.(2.41) Multiplying the above equation on the right side by 𝑃 yields 𝑎+𝑏+𝑑+𝑎𝑑𝑏(𝑄𝑃−𝑃𝑄𝑃)=0.(2.42) Assume 𝑃𝑄=𝑃𝑄𝑃. Then 𝑄𝑃𝑄=𝑄𝑃𝑄𝑃=𝑃𝑄𝑃𝑄=𝑃𝑄=𝑃𝑄𝑃 and it contradicts the hypothesis 𝑃𝑄𝑃≠𝑄𝑃𝑄. Thus, 𝑎+𝑏+𝑑+𝑑/𝑎𝑏=0.
Similarly, if 𝑄𝑃𝑄=𝑃𝑄𝑃𝑄, then we can obtain 𝑎+𝑏+𝑑+𝑑𝑎𝑏=0, 𝑏+𝑐+𝑑+2𝑒−𝑐𝑑/𝑏=0, and 𝑎+𝑏+𝑐+𝑐/𝑎𝑏=0.
Obviously, if 𝑄𝑃𝑄≠𝑄𝑃𝑄𝑃 and 𝑄𝑃𝑄≠𝑃𝑄𝑃𝑄, we have 𝑎+𝑏+𝑑+𝑑/𝑎𝑏=0, 𝑎+𝑏+𝑐+𝑐/𝑎𝑏=0, 𝑏+𝑐+𝑑+2𝑒−𝑐𝑑/𝑏=0, and 𝑎+𝑐+𝑑+2𝑓−𝑐𝑑/𝑎=0.
Next, we calculate these scalars. If 𝑎+𝑏=0, then 𝑎+𝑏+𝑐+𝑐/𝑎𝑏=0 for any 𝑐 and 𝑎+𝑏+𝑑+𝑑/𝑎𝑏=0 for any 𝑑, and so 𝑐,𝑑,𝑒 are chosen to satisfy 𝑏+𝑐+𝑑+2𝑒−𝑐𝑑/𝑏=0. Similarly 𝑐,𝑑,𝑓 are chosen to satisfy 𝑎+𝑐+𝑑+2𝑓−𝑐𝑑/𝑎=0.
If 𝑎=𝑏, then 𝑐=𝑑=−𝑎, and 𝑒=𝑎 by solving 𝑏+𝑐+𝑑+2𝑒−𝑐𝑑/𝑏=0, and 𝑓=𝑎 by solving 𝑎+𝑐+𝑑+2𝑓−𝑐𝑑/𝑎=0.
Note that 𝑏+𝑐+𝑑+2𝑒−𝑐𝑑/𝑏=0 and 𝑎+𝑐+𝑑+2𝑓−𝑐𝑑/𝑎=0 imply 𝑔=𝜃−(𝑎+𝑏). Hence, we have (𝑐₁)∼(𝑐₆).
(ii) Consider the case of 𝑎=𝑎⁻¹, 𝑏=𝑏⁻¹, and 𝑃𝑄𝑃𝑄=0.
Multiplying (2.34), respectively, on the right side by 𝑄𝑃 and on the left side by 𝑃𝑄 yields𝑑𝑐++𝑎11+1𝑏+𝑎𝑐𝑏𝑄𝑃𝑄=0,𝑎+1𝑏+𝑎𝑑𝑏𝑃𝑄𝑃=0.(2.43)
If 𝑄𝑃𝑄=0, then 𝑃𝑄𝑃≠0 and so a+𝑏+𝑑+𝑑/𝑎𝑏=0 and (2.34) becomes 0=𝑐+1𝑎+1𝑏+𝑎𝑐𝑏𝑃𝑄+𝑒−𝑎2−1𝑏−𝑐𝑎+𝑏𝑑−𝑐𝑑𝑎−²𝑏𝑏𝑒𝑃𝑄𝑃.(2.44) Multiplying (2.44) on right side by 𝑄 yields 𝑐+𝑎1+1𝑏+𝑎𝑐𝑏𝑃𝑄=0.(2.45) Since 𝑃𝑄≠0, 𝑎+𝑏+𝑐+𝑐/𝑎𝑏=0 and then (2.44) becomes 2𝑒+𝑏+𝑐+𝑑−𝑐𝑏𝑑𝑃𝑄𝑃.(2.46) Thus, 2𝑒+𝑏+𝑐+𝑑−𝑐𝑑/𝑏=0.
If 𝑃𝑄𝑃=0, then we, similarly, have 𝑎+𝑏+𝑐+𝑐/𝑎𝑏=0, 𝑎+𝑏+𝑑+𝑑/𝑎𝑏=0, and 2𝑓+𝑎+𝑐+𝑑−𝑐𝑑/𝑎=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 𝑎+𝑏+𝑐+𝑐/𝑎𝑏=0, and multiplying (2.34) on the right side by 𝑃 and on the left side by 𝑄 yields 𝑎+𝑏+𝑑+𝑑/𝑎𝑏=0. Thus, (2.34) becomes 𝑒−𝑎2−1𝑏−𝑐𝑎+𝑏𝑑−𝑐𝑑𝑎−²𝑏𝑏𝑒𝑃𝑄𝑃+𝑓−𝑎1−2𝑏−𝑐𝑎+𝑏𝑑−𝑐𝑑𝑎−𝑏²𝑎𝑓𝑄𝑃𝑄=0.(2.47) Multiplying the equation above on the right side, respectively, by 𝑃 and by 𝑄 yields 2𝑒+𝑏+𝑐+𝑑−𝑐𝑑𝑏=0,2𝑓+𝑎+𝑐+𝑑−𝑐𝑎𝑑=0.(2.48) As the argument above in (i), we have (𝑐₇)∼(𝑐₁₂).
(2) If 𝜃=0, then 𝐴_𝑔=1+𝑎𝑃2+1𝑏𝑄−𝑎1+1𝑏+𝑎𝑐𝑏𝑃𝑄−𝑎1+1𝑏+𝑎𝑑𝑏𝑄𝑃−𝑎2+1𝑏+𝑐𝑎+𝑏𝑑+𝑐𝑑𝑎−²𝑏𝑏𝑒𝑃𝑄𝑃+𝑎1+2𝑏+𝑐𝑎+𝑏𝑑+𝑐𝑑𝑎−𝑏²𝑎𝑓𝑄𝑃𝑄𝑎+2𝑏+𝑐𝑎+𝑏𝑑+𝑐𝑑𝑎−²𝑏𝑏𝑒+𝑐𝑑𝑎−𝑏²𝑎𝑓𝑃𝑄𝑃𝑄,(2.49) and so 𝐴−𝐴_𝑔=𝑎−1+𝑑+𝑎1𝑃+𝑏−1𝑏𝑄+𝑐+𝑎1+1𝑏+𝑎𝑐𝑏𝑃𝑄+𝑓−𝑎1+1𝑏+𝑎𝑑𝑏𝑄𝑃+𝑒−𝑎2−1𝑏−𝑐𝑎+𝑏𝑑−𝑐𝑑𝑎−²𝑏𝑏𝑒𝑃𝑄𝑃+𝑔+2𝑎−2𝑏−𝑐𝑎+𝑏𝑑−𝑐𝑑𝑎−𝑏²𝑎𝑓𝑄𝑃𝑄𝑎+2𝑏+𝑐𝑎+𝑏𝑑+𝑐𝑑𝑎−²𝑏𝑏𝑒+𝑐𝑑𝑎−𝑏²𝑎𝑓𝑃𝑄𝑃𝑄=0.(2.50) Analogous to the process in (c)(1), using (2.50), we can get 𝑎𝑎−−1𝑎1(𝑃𝑄𝑃−𝑃𝑄𝑃𝑄)=0,𝑎(𝑄𝑃𝑄−𝑃𝑄𝑃𝑄)=0.(2.51)
Thus, since 𝑃𝑄𝑃≠𝑄𝑃𝑄, 𝑃𝑄𝑃≠𝑃𝑄𝑃𝑄 and/or 𝑄𝑃𝑄≠𝑃𝑄𝑃𝑄 and then 𝑎=𝑎⁻¹. Similarly, 𝑏=𝑏⁻¹. Therefore, multiplying (2.50) on the right side by 𝑄 and on the left side by 𝑃 yields 𝑎+𝑏+𝑐+𝑎𝑐𝑏(𝑃𝑄−𝑃𝑄𝑃𝑄)=0.(2.52) Multiplying (2.50) on the right side by 𝑃 and on the left side by 𝑄 yields 𝑎+𝑏+𝑑+𝑎𝑑𝑏(𝑄𝑃−𝑃𝑄𝑃𝑄)=0.(2.53) Since 𝑃𝑄≠𝑃𝑄𝑃𝑄 and 𝑄𝑃≠𝑃𝑄𝑃𝑄, 𝑎+𝑏+𝑐+𝑐/𝑎𝑏=0 and 𝑎+𝑏+𝑑+𝑑/𝑎𝑏=0. Multiplying (2.50) on the left side, respectively, by 𝑃 and by 𝑄 yields 22𝑓𝑒++𝑎𝑏++𝑐𝑐++𝑑𝑑−−𝑐𝑐𝑑𝑏𝑑(𝑃𝑄𝑃−𝑃𝑄𝑃𝑄)=0,𝑎(𝑄𝑃𝑄−𝑃𝑄𝑃𝑄)=0.(2.54) Thus, we have 2𝑒+𝑏+𝑐+𝑑−𝑐𝑑/𝑏=0 and 𝑄𝑃𝑄=𝑃𝑄𝑃𝑄; 2𝑓+𝑎+𝑐+𝑑−𝑐𝑑/𝑎=0 and 𝑃𝑄𝑃=𝑃𝑄𝑃𝑄; 2𝑒+𝑏+𝑐+𝑑−𝑐𝑑/𝑏=0 and 2𝑓+𝑎+𝑐+𝑑−𝑐𝑑/𝑎=0.
Note that 2𝑒+𝑏+𝑐+𝑑−𝑐𝑑/𝑏=0 and 2𝑓+𝑎+𝑐+𝑑−𝑐𝑑/𝑎=0 imply 𝑔=−(𝑎+𝑏) by 𝜃=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 𝑃=⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝1000000000010000000000100000000001000000000010000000000000000000000000000000000000000000000000000000⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠,𝑄=⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝0000000000100001000000000000100000000000000010000010000100000000000000−001000010100000000100000000000⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠.(2.55) Then they, obviously, are idempotent, and (𝑃𝑄)²=(𝑄𝑃)² but 𝑃𝑄𝑃≠𝑄𝑃𝑄. By Theorem 2.1(𝑐₅), 𝐴=𝑃−𝑄+2𝑃𝑄+2𝑄𝑃−72𝑃𝑄𝑃−21𝑄𝑃𝑄+𝑃𝑄𝑃𝑄(2.56) is the group involutory matrix, namely, 𝐴=𝐴_𝑔, since 2+2+2∗(−7/2)+2∗2=1 and 2+2+2∗(−1/2)−2∗2=−1. By Theorem 2.1(𝑐₁₇), 𝑃−𝑄+𝑃𝑄−2𝑄𝑃+2𝑃𝑄𝑃−𝑄𝑃𝑄(2.57) is group involutory since 1−2+2∗2+1∗(−2)=1 and 1−2+2∗(−1)−1∗(−2)=−1.

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

Theorem 2.4. Let 𝑃,𝑄∈ℂ^𝑛×𝑛 be two different nonzero idempotent matrices, and let 𝐴 be a combination of the form 𝐴=𝑎𝑃+𝑏𝑄+𝑐𝑃𝑄+𝑑𝑄𝑃+𝑒𝑃𝑄𝑃+𝑓𝑄𝑃𝑄+𝑔𝑃𝑄𝑃𝑄,(2.58) where 𝑎,𝑏,𝑐,𝑑,𝑒,𝑓,𝑔∈ℂ with 𝑎,𝑏≠0. Suppose that 𝑃𝑄𝑃𝑄≠0,𝑄𝑃𝑄𝑃=0,(2.59) and any of the following sets of additional conditions hold:

(𝑑₁)𝑎=𝑏=±1, c=𝑑=∓1, 𝑒=𝑓=±1,𝑔=∓1;

(𝑑₂)𝑎=±1, 𝑏=∓1, 2𝑒+𝑐+𝑑±𝑐𝑑=±1, 2𝑓+𝑐+𝑑∓𝑐𝑑=∓1.

Then 𝐴 is the group involutory matrix.

Proof. By Lemma 1.2, 0=𝐴−𝐴_𝑔=𝑎−1+𝑒−𝑎2𝑃+𝑏−1𝑏𝑄+𝑐+𝑎1+1𝑏+𝑎𝑐𝑏𝑃𝑄+𝑑+𝑎1+1𝑏+𝑎𝑑𝑏𝑄𝑃+𝑓−𝑎1−1𝑏−𝑐𝑎+𝑏𝑑−𝑐𝑑𝑎−²𝑏𝑏𝑒𝑃𝑄𝑃+𝑔+2𝑎−2𝑏−𝑐𝑎+𝑏𝑑−𝑐𝑑𝑎−𝑏²𝑎𝑓𝑄𝑃𝑄𝑎+2𝑏+2𝑐+𝑎𝑑𝑏+𝑔+𝑐𝑑−𝑎𝑏²𝑒𝑏−𝑐𝑒+𝑐𝑑−𝑎𝑎𝑏𝑓²−𝑐𝑓+𝑐²𝑑𝑎²𝑏²(𝑃𝑄)².(2.60) Since 𝑃𝑄𝑃𝑄≠0, multiplying (2.60), respectively, on the right side and on the right side by 𝑃𝑄𝑃𝑄 yields 𝑎−𝑎1𝑃𝑄𝑃𝑄=0,𝑏−1𝑏𝑃𝑄𝑃𝑄=0,(2.61) and so 𝑎=𝑎⁻¹ and 𝑏=𝑏⁻¹. Substituting them inside (2.60), we get 0=𝑐+1+𝑒−𝑎2+1𝑏+𝑎𝑐𝑏𝑃𝑄+𝑑+𝑎1+1𝑏+𝑎𝑑𝑏𝑄𝑃+𝑓−𝑎1−1𝑏−𝑐𝑎+𝑏𝑑−𝑐𝑑𝑎−²𝑏𝑏𝑒𝑃𝑄𝑃+𝑔+2𝑎−2𝑏−𝑐𝑎+𝑏𝑑−𝑐𝑑𝑎−𝑏²𝑎𝑓𝑄𝑃𝑄𝑎+2𝑏+2𝑐+𝑎𝑑𝑏+𝑔+𝑐𝑑−𝑎𝑏²𝑒𝑏−𝑐𝑒+𝑐𝑑−𝑎𝑎𝑏𝑓²−𝑐𝑓+𝑐²𝑑𝑎²𝑏²𝑃𝑄𝑃𝑄.(2.62) Multiplying (2.62) on the left side by 𝑃𝑄𝑃 yields 𝑐+𝑎1+1𝑏+𝑎𝑐𝑏𝑃𝑄𝑃𝑄=0,(2.63) and then 𝑐+1𝑎+1𝑏+𝑎𝑐𝑏=0.(2.64) So (2.62) becomes 0=𝑑+1+𝑓−𝑎+11𝑏+𝑎𝑑𝑏𝑄𝑃+𝑒−𝑎2−1𝑏−𝑐𝑎+𝑏𝑑−𝑐𝑑𝑎−²𝑏𝑏𝑒𝑃𝑄𝑃+𝑔+2𝑎−2𝑏−𝑐𝑎+𝑏𝑑−𝑐𝑑𝑎−𝑏²𝑎𝑓𝑄𝑃𝑄𝑎+2𝑏+2𝑐+𝑎𝑑𝑏+𝑔+𝑐𝑑−𝑎𝑏²𝑒𝑏−𝑐𝑒+𝑐𝑑−𝑎𝑎𝑏𝑓²−𝑐𝑓+𝑐²𝑑𝑎²𝑏²𝑃𝑄𝑃𝑄.(2.65) Multiplying (2.65) on the left side by 𝑃𝑄 and on the right side by 𝑃 yields 𝑑+𝑎1+1𝑏+𝑎𝑑𝑏𝑃𝑄P𝑄=0.(2.66) Therefore, 𝑑+1𝑎+1𝑏+𝑎𝑑𝑏=0.(2.67) Similarly, we can obtain 0=𝑒−20=𝑓−1𝑎−1𝑏−𝑐𝑎+𝑏𝑑−𝑐𝑑𝑎−²𝑏𝑏𝑒,𝑎−2𝑏−𝑐𝑎+𝑏𝑑−𝑐𝑑𝑎−𝑏²𝑎𝑓0=𝑔+2,𝑎+2𝑏+2𝑐+𝑎𝑑𝑏+𝑔+𝑐𝑑−𝑎𝑏²𝑒𝑏−𝑐𝑒+𝑐𝑑−𝑎𝑎𝑏𝑓²−𝑐𝑓+𝑐²𝑑𝑎²𝑏².(2.68) By (2.64) and (2.67), we can obtain 1𝑏+𝑐+𝑑+2𝑒−𝑐𝑏𝑑=0,𝑎1+𝑐+𝑑+2𝑓−𝑐𝑎𝑑=0.(2.69) Since 𝑎=𝑎⁻¹ and 𝑏=𝑏⁻¹, 𝑎=±𝑏. If 𝑎=−𝑏, 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, 𝑔+2𝑎+2𝑏+2𝑐+𝑎𝑑𝑏+𝑔+𝑐𝑑−𝑎𝑏²𝑒𝑏−𝑐𝑒+𝑐𝑑−𝑎𝑎𝑏𝑓²−𝑐𝑓+𝑐²𝑑𝑎²𝑏²=𝑐²𝑑−2𝑐−𝑑−(𝑒+𝑓)+𝑐=𝑐₂𝑎(𝑒−𝑓)𝑑−2𝑐−𝑑+(𝑐+𝑑)+𝑐𝑎𝑎1−𝑐𝑎𝑑=0.(2.70) If 𝑎=𝑏, then, by (2.64) ~ (2.69), 𝑐=𝑑=−𝑎 and 𝑒=𝑓=𝑎 and so 𝑔=−𝑎 from (2.68).
Hence, we have (𝑑₁) and (𝑑₂).

Example 2.5. Let 𝑃=⎛⎜⎜⎜⎜⎜⎜⎝1000010000000000⎞⎟⎟⎟⎟⎟⎟⎠,𝑄=⎛⎜⎜⎜⎜⎜⎜⎝01010000−10110101⎞⎟⎟⎟⎟⎟⎟⎠.(2.71) Obviously they are idempotent, and (𝑄𝑃)²=0 but (𝑃𝑄)²≠0. By Theorem 2.4(𝑑₂), 𝑃−𝑄+2𝑃𝑄−2𝑄𝑃+52𝑃𝑄𝑃−52𝑄𝑃𝑄−2𝑃𝑄𝑃𝑄(2.72) is group involutory since 2−2+2∗(5/2)+2∗(−2)=1 and 2−2+2∗(−5/2)−2∗(−2)=−1.

Similarly, we have the following result.

Theorem 2.6. Let 𝑃,𝑄∈ℂ^𝑛×𝑛 be two different nonzero idempotent matrices, and let 𝐴 be a combination of the form 𝐴=𝑎𝑃+𝑏𝑄+𝑐𝑃𝑄+𝑑𝑄𝑃+𝑒𝑃𝑄𝑃+𝑓𝑄𝑃𝑄+ℎ𝑄𝑃𝑄𝑃,(2.73) where 𝑎,𝑏,𝑐,𝑑,𝑒,𝑓,ℎ∈ℂ with 𝑎,𝑏≠0. Suppose that 𝑄𝑃𝑄𝑃≠0,𝑃𝑄𝑃𝑄=0,(2.74) and any of the following sets of additional conditions hold:(e₁)𝑎=𝑏=±1, 𝑐=𝑑=∓1, 𝑒=𝑓=±1,ℎ=∓1; (e₂)𝑎=±1, 𝑏=∓1, 2𝑒+𝑐+𝑑±𝑐𝑑=±1, 2𝑓+𝑐+𝑑∓𝑐𝑑=∓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.