# The Group Involutory Matrix of the Combinations of Two Idempotent Matrices

Authors: Lingling Wu; Xiaoji Liu; Yaoming Yu
Journal: Journal of Applied Mathematics
 (2012)

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## Abstract

We discuss the following problem: whenaP+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.


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## Body

## 1. Introduction

Throughout this paperℂn×n stands for the set of n×n complex matrices. Let A∈ℂn×n. A is said to be idempotent if A2=A. A is said to be group invertible if there exists an X∈ℂn×n such that(1.1)AXA=A,XAX=X,AX=XA
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 matrixA 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 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: whenaP+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]).
LetP,Q∈ℂn×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) ifθ≠0, then
(1.2)(aP+bQ+cPQ+dQP+ePQP+fQPQ+g(PQ)2)g=1aP+1bQ-(1a+1b+cab)PQ-(1a+1b+dab)QP+(2a+1b+c+dab+cd-bea2b)PQP+(1a+2b+c+dab+cd-afab2)QPQ-(2a+2b+c+dab+cd-bea2b+cd-afab2-1θ)PQPQ;
(ii)  ifθ=0, then
(1.3)(aP+bQ+cPQ+dQP+ePQP+fQPQ+g(PQ)2)g=1aP+1bQ-(1a+1b+cab)PQ-(1a+1b+dab)QP+(2a+1b+c+dab+cd-bea2b)PQP+(1a+2b+c+dab+cd-afab2)QPQ-(2a+2b+c+dab+cd-bea2b+cd-afab2)(PQ)2.Lemma 1.2 (see [16, Theorem  3.1]).
LetP,Q∈ℂn×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
(1.4)(aP+bQ+cPQ+dQP+ePQP+fQPQ+g(PQ)2)g=1aP+1bQ-(1a+1b+cab)PQ-(1a+1b+dab)QP+(2a+1b+c+dab+cd-bea2b)PQP+(1a+2b+c+dab+cd-afab2)QPQ-(2a+2b+2c+d+gab+cd-be-cea2b+cd-af-cfab2+c2da2b2)(PQ)2.


## 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.
LetP,Q∈ℂn×n be two different nonzero idempotent matrices with (PQ)2=(QP)2, and let A be a combination of the form
(2.1)A=aP+bQ+cPQ+dQP+ePQP+fQPQ+gPQPQ,
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
(2.2)PQ=QP,
and any of the following sets of additional conditions hold:(a1)
eithera=1 or a=-1, either θ=1 or θ=-1 or θ=0, and Q=PQ;(a2)
eitherb=1 or b=-1, either θ=1 or θ=-1 or θ=0, and P=PQ;(a3)
eithera=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
(2.3)PQ≠QP,PQP=QPQ,
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
(2.4)PQP≠QPQ,PQPQ=QPQP,
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-Ag=0.
(a)  Under the condition (2.2), A=aP+bQ+μPQ, where μ=c+d+e+f+g.
(1) Ifθ≠0, then
(2.5)Ag=1aP+1bQ+(1θ-1a-1b)PQ,
and so
(2.6)A-Ag=(a-1a)P+(b-1b)Q+(μ-1θ+1a+1b)PQ=0.
Multiplying (2.6) by P and Q, respectively, leads to
(2.7)(a-1a)P+(b-1b)PQ+(μ-1θ+1a+1b)PQ=0,(a-1a)PQ+(b-1b)Q+(μ-1θ+1a+1b)PQ=0,
and then
(2.8)(a-1a)P+(b-1b)PQ=(a-1a)PQ+(b-1b)Q.
Multiplying the above equation, respectively, by P and by Q, we get
(2.9)(a-1a)(P-PQ)=0,(b-1b)(Q-PQ)=0.
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.
WhenQ=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) Ifθ=0, then
(2.10)Ag=1aP+1bQ-(1a+1b)PQ,
and then
(2.11)A-Ag=(a-1a)P+(b-1b)Q+(μ+1a+1b)PQ=0.
Analogous to the process of reaching (2.9) in (a)(1), we have
(2.12)(b-1b)(Q-PQ)=0,(a-1a)(P-PQ)=0.
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) Ifθ≠0, then
(2.13)Ag=1aP+1bQ-(1a+1b+cab)PQ-(1a+1b+dab)QP+(1a+1b+c+dab+1θ)PQP,
and so
(2.14)A-Ag=(a-1a)P+(b-1b)Q+(c+1a+1b+cab)PQ+(d+1a+1b+dab)QP+(ν-1a-1b-c+dab-1θ)PQP=0.
Multiplying the above equation, respectively, on the two sides by P yields
(2.15)0=(a-1a)P+(c+b+1a+cab)PQ+(ν+d-cab-1θ)PQP,(2.16)0=(a-1a)P+(b+d+1a+dab)QP+(ν+c-dab-1θ)PQP.
Multiplying (2.15) on the left sides by Q and (2.16) on the right sides by Q, by (2.3), we have
(2.17)(a-1a)QP+(b+c+d+ν+1a-1θ)QPQ=0,(a-1a)PQ+(b+c+d+ν+1a-1θ)QPQ=0,
and then (a-a-1)(QP-PQ)=0. Since QP≠PQ, a=a-1. Similarly, b=b-1.
Substitutinga=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:
Whena=a-1, b=b-1, (2.14) becomes(2.18)0=(c+1a+1b+cab)PQ+(d+1a+1b+dab)QP+(ν-1a-1b-c+dab-1θ)PQP,
(i) ifa+b=0, then
(2.19)c+1a+1b+cab=0,d+1a+1b+dab=0,
and so it follows from (2.18) that
(2.20)(θ-1θ)PQP=(ν+c+d-1θ)PQP=0.
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) ifa=b, then (2.18) becomes
(2.21)0=(2c+2a)PQ+(2d+2a)QP+(2ν-θ-1θ)PQP.
Multiplying the above equation, respectively, on the right side by P and on the left side by Q, we have
(2.22)0=(2c+2a)PQ+(ν+d-c-1θ)PQP,(2.23)0=(2d+2a)QP+(ν+c-d-1θ)PQP.
So if θ=θ-1, then the two equations above (2.22) and (2.23) become, respectively,
(2.24)(c+a)(PQ-PQP)=0,(d+a)(QP-PQP)=0.
Or if PQP=0, then (2.22) and (2.23) become, respectively,
(2.25)(c+a)PQ=0,(d+a)QP=0.
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) Ifθ=0, then
(2.26)Ag=1aP+1bQ-(1a+1b+cab)PQ-(1a+1b+dab)QP+(1a+1b+c+dab)PQP,
and then
(2.27)A-Ag=(a-1a)P+(b-1b)Q+(c+1a+1b+cab)PQ+(d+1a+1b+dab)QP+(ν-1a-1b-c+dab)PQP=0.
Analogous to the process in (b)(1), using (2.27) we can obtain
(2.28)(a-1a)QP-(a-1a)PQP=0,(a-1a)PQ-(a-1a)PQP=0.
Thus, since PQ≠QP, PQ≠PQP and/or QP≠PQP and then a=a-1. Similarly, b=b-1. Hence, a=±b.
(i)  Ifa=-b, then
(2.29)c+1a+1b+cab=0,d+1a+1b+dab=0,ν-1a-1b-c+dab=-2(a+b)=0.
Thus, (2.27) holds. Hence we have the situation θ=0 in (b1).
(ii) Ifa=b, then (2.27) becomes
(2.30)(c+a)PQ+(d+a)QP+νPQP=0.
Multiplying the above equation on the left side, respectively, by P and by Q, we have
(2.31)(c+a)(PQ-PQP)=0,(d+a)(QP-PQP)=0.
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),
(2.32)A=aP+bQ+cPQ+dQP+ePQP+fQPQP+gPQPQ.
(1)  Ifθ≠0, then
(2.33)Ag=1aP+1bQ-(1a+1b+cab)PQ-(1a+1b+dab)QP+(2a+1b+c+dab+cd-bea2b)PQP+(1a+2b+c+dab+cd-afab2)QPQ-(2a+2b+c+dab+cd-bea2b+cd-afab2-1θ)PQPQ,
and so
(2.34)A-Ag=(a-1a)P+(b-1b)Q+(c+1a+1b+cab)PQ+(d+1a+1b+dab)QP+(e-2a-1b-c+dab-cd-bea2b)PQP+(f-1a-2b-c+dab-cd-afab2)QPQ+(g+2a+2b+c+dab+cd-bea2b+cd-afab2-1θ)PQPQ=0.
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
(2.35)(a-1a)QP+(b+c+1a+cab)QPQ+(d+e+f+g-cab-1θ)PQPQ=0.
Multiplying the above equation, respectively, on the left side by P and on the right side by PQ yields, by (2.4),
(2.36)0=(a-1a)PQP+(1a-a+θ-1θ)PQPQ,(2.37)0=(a-1a)QPQ+(1a-a+θ-1θ)PQPQ.
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 ofa=a-1, b=b-1 and θ=θ-1.
Substitutinga=a-1, b=b-1, and θ=θ-1 inside (2.35) yields
(2.38)(a+b+c+cab)(QPQ-PQPQ)=0.
Similarly, we have
(2.39)(a+b+d+dab)(PQP-PQPQ)=0.
IfPQP=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
(2.40)(a+c+d+2f-cda)(QPQ-PQPQ)=0.
Thus, a+c+d+2f-cd/a=0 and then (2.14) becomes
(2.41)(a+b+d+dab)QP+(f-a-2b-c+dab-cd-afa)QPQ+(b+e+g+cd-afa-θ)PQP=0.
Multiplying the above equation on the right side by P yields
(2.42)(a+b+d+dab)(QP-PQP)=0.
Assume PQ=PQP. Then QPQ=QPQP=PQPQ=PQ=PQP and it contradicts the hypothesis PQP≠QPQ. Thus, a+b+d+d/ab=0.
Similarly, ifQPQ=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, ifQPQ≠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. Ifa+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.
Ifa=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 thatb+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 ofa=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(2.43)(c+1a+1b+cab)QPQ=0,(d+1a+1b+dab)PQP=0.
IfQPQ=0, then PQP≠0 and so a+b+d+d/ab=0 and (2.34) becomes
(2.44)0=(c+1a+1b+cab)PQ+(e-2a-1b-c+dab-cd-bea2b)PQP.
Multiplying (2.44) on right side by Q yields
(2.45)(c+1a+1b+cab)PQ=0.
Since PQ≠0, a+b+c+c/ab=0 and then (2.44) becomes
(2.46)(2e+b+c+d-cdb)PQP.
Thus, 2e+b+c+d-cd/b=0.
IfPQP=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.
IfPQP≠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
(2.47)(e-2a-1b-c+dab-cd-bea2b)PQP+(f-1a-2b-c+dab-cd-afab2)QPQ=0.
Multiplying the equation above on the right side, respectively, by P and by Q yields
(2.48)2e+b+c+d-cdb=0,2f+a+c+d-cda=0.
As the argument above in (i), we have (c7)~(c12).
(2) Ifθ=0, then
(2.49)Ag=1aP+1bQ-(1a+1b+cab)PQ-(1a+1b+dab)QP+(2a+1b+c+dab+cd-bea2b)PQP+(1a+2b+c+dab+cd-afab2)QPQ-(2a+2b+c+dab+cd-bea2b+cd-afab2)PQPQ,
and so
(2.50)A-Ag=(a-1a)P+(b-1b)Q+(c+1a+1b+cab)PQ+(d+1a+1b+dab)QP+(e-2a-1b-c+dab-cd-bea2b)PQP+(f-1a-2b-c+dab-cd-afab2)QPQ+(g+2a+2b+c+dab+cd-bea2b+cd-afab2)PQPQ=0.
Analogous to the process in (c)(1), using (2.50), we can get
(2.51)(a-1a)(PQP-PQPQ)=0,(a-1a)(QPQ-PQPQ)=0.
Thus, sincePQP≠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
(2.52)(a+b+c+cab)(PQ-PQPQ)=0.
Multiplying (2.50) on the right side by P and on the left side by Q yields
(2.53)(a+b+d+dab)(QP-PQPQ)=0.
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
(2.54)(2e+b+c+d-cdb)(PQP-PQPQ)=0,(2f+a+c+d-cda)(QPQ-PQPQ)=0.
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 that2e+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(2.55)P=(1000000000010000000000100000000001000000000010000000000000000000000000000000000000000000000000000000),Q=(010001000000000000000000000100000000000000001000000100010000000000000000000001000010000-1100000000000).
Then they, obviously, are idempotent, and (PQ)2=(QP)2 but PQP≠QPQ. By Theorem 2.1(c5),
(2.56)A=P-Q+2PQ+2QP-72PQP-12QPQ+PQPQ
is the group involutory matrix, namely, A=Ag, since 2+2+2*(-7/2)+2*2=1 and 2+2+2*(-1/2)-2*2=-1. By Theorem 2.1(c17),
(2.57)P-Q+PQ-2QP+2PQP-QPQ
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.
LetP,Q∈ℂn×n be two different nonzero idempotent matrices, and let A be a combination of the form
(2.58)A=aP+bQ+cPQ+dQP+ePQP+fQPQ+gPQPQ,
where a,b,c,d,e,f,g∈ℂ with a,b≠0. Suppose that
(2.59)PQPQ≠0,QPQP=0,
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.ThenA is the group involutory matrix.Proof.
By Lemma1.2,
(2.60)0=A-Ag=(a-1a)P+(b-1b)Q+(c+1a+1b+cab)PQ+(d+1a+1b+dab)QP+(e-2a-1b-c+dab-cd-bea2b)PQP+(f-1a-2b-c+dab-cd-afab2)QPQ+(g+2a+2b+2c+d+gab+cd-be-cea2b+cd-af-cfab2+c2da2b2)(PQ)2.
Since PQPQ≠0, multiplying (2.60), respectively, on the right side and on the right side by PQPQ yields
(2.61)(a-1a)PQPQ=0,(b-1b)PQPQ=0,
and so a=a-1 and b=b-1. Substituting them inside (2.60), we get
(2.62)0=(c+1a+1b+cab)PQ+(d+1a+1b+dab)QP+(e-2a-1b-c+dab-cd-bea2b)PQP+(f-1a-2b-c+dab-cd-afab2)QPQ+(g+2a+2b+2c+d+gab+cd-be-cea2b+cd-af-cfab2+c2da2b2)PQPQ.
Multiplying (2.62) on the left side by PQP yields
(2.63)(c+1a+1b+cab)PQPQ=0,
and then
(2.64)c+1a+1b+cab=0.
So (2.62) becomes
(2.65)0=(d+1a+1b+dab)QP+(e-2a-1b-c+dab-cd-bea2b)PQP+(f-1a-2b-c+dab-cd-afab2)QPQ+(g+2a+2b+2c+d+gab+cd-be-cea2b+cd-af-cfab2+c2da2b2)PQPQ.
Multiplying (2.65) on the left side by PQ and on the right side by P yields
(2.66)(d+1a+1b+dab)PQPQ=0.
Therefore,
(2.67)d+1a+1b+dab=0.
Similarly, we can obtain
(2.68)0=e-2a-1b-c+dab-cd-bea2b,0=f-1a-2b-c+dab-cd-afab2,0=g+2a+2b+2c+d+gab+cd-be-cea2b+cd-af-cfab2+c2da2b2.
By (2.64) and (2.67), we can obtain
(2.69)1b+c+d+2e-cdb=0,1a+c+d+2f-cda=0.
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,
(2.70)g+2a+2b+2c+d+gab+cd-be-cea2b+cd-af-cfab2+c2da2b2=c2d-2c-d-(e+f)+ca(e-f)=c2d-2c-d+(c+d)+ca(1a-cda)=0.
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(2.71)P=(1000010000000000),Q=(00101001001010-11).
Obviously they are idempotent, and (QP)2=0 but (PQ)2≠0. By Theorem 2.4(d2),
(2.72)P-Q+2PQ-2QP+52PQP-52QPQ-2PQPQ
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.
LetP,Q∈ℂn×n be two different nonzero idempotent matrices, and let A be a combination of the form
(2.73)A=aP+bQ+cPQ+dQP+ePQP+fQPQ+hQPQP,
where a,b,c,d,e,f,h∈ℂ with a,b≠0. Suppose that
(2.74)QPQP≠0,PQPQ=0,
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.



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*Source: 504650-2012-05-17.xml*