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Linear Algebra and its Applications 437 (2012) 376–387
Contents lists available at SciVerse ScienceDirect
Linear Algebra and its Applications
journal homepage: www.elsevier .com/locate/ laa
On invertibility of combinations of k-potent operators
Chunyuan Denga,1, Dragana S. Cvetkovic-Ilic b,∗,2, Yimin Wei c,d,3
aSchool of Mathematics Science, South China Normal University, Guangzhou 510631, PR China
bDepartment of Mathematics, Faculty of Sciences and Mathematics, University of Niš, Višegradska 33, 18000 Niš, Serbia
cSchool of Mathematical Sciences, Fudan University, Shanghai 200433, PR China
dKey Laboratory of Mathematics for Nonlinear Sciences, Fudan University, Ministry of Education, PR China
A R T I C L E I N F O A B S T R A C T
Article history:
Received 18 January 2012
Accepted 13 February 2012
Available online 10 March 2012
Submitted by R. Brualdi
AMS classification:
15A09
47A05
Keywords:
Inverse
Group inverse
Linear combination of k-potents
Some properties of combinations c1P1 + c2P2 − c3Ps1P
k−1−s2 , where
P1 and P2 are two different nonzero k-potents, c1, c2 and c3 are three
nonzerocomplexnumbers andpositive integersk ≥ 2and s ≤ k−1,
are obtained. Furthermore, the invertibility, the group invertibility
and the k-potency of the linear combinations of k-potents are inves-
tigated, under certain commutativity properties imposed on them.
© 2012 Elsevier Inc. All rights reserved.
1. Introduction
Let H and K be separable, infinite dimensional, complex Hilbert spaces. We denote the set of all
bounded linear operators from H into K by B(H,K) and by B(H) when H = K. For A ∈ B(H,K), letσ(A), R(A) and N (A) be the spectrum, the range and the null space of A, respectively. The identity
operator on the closed subspace U is denoted by IU or I if there does not exist confusion. Recall that
an operator P ∈ B(H) is idempotent if P2 = P, tripotent if P3 = P, k-potent if Pk = P, where k ≥ 2
∗ Corresponding author.
E-mail addresses: [email protected], [email protected] (C. Deng), [email protected], [email protected] (D.S. Cvetkovic-Ilic),
[email protected], [email protected] (Y. Wei).1 Supported by the National Natural Science Foundation of China under Grant 11171222 and the Doctoral Program of the Ministry
of Education under Grant 20094407120001.2 Supported by Grant No. 174007 of the Ministry of Science and Technological Development, Republic of Serbia.3 Supported by theNational Natural Science Foundation of China underGrant 10871051, ShanghaiMunicipal Education Committee
and Doctoral Program of the Ministry of Education under Grant 20090071110003.
0024-3795/$ - see front matter © 2012 Elsevier Inc. All rights reserved.
http://dx.doi.org/10.1016/j.laa.2012.02.011
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C. Deng et al. / Linear Algebra and its Applications 437 (2012) 376–387 377
is a positive integer. For T ∈ B(H), the group inverse [10,11,15,21,26] of T is the unique (if it exists)
element T# ∈ B(H) such that
TT# = T#T, T#TT# = T#, T = TT#T .
If T is group invertible, thenR(T) is closed and the spectral idempotent Tπ is given by Tπ = I − TT#.The operator matrix form of a group invertible operator T with respect to the space decomposition
H = R(T) ⊕ N (T) is given by T = T1 ⊕ 0, where T1 is invertible in B(R(T)) and in that case
T# = T−11 ⊕ 0. We will use the usual notation P = I − P and P0 = I for any operator P ∈ B(H). For a
positive integer k ≥ 2, the set of kth complex roots of 1 shall be denoted by
�k = {ω0k , ω
1k , . . . , ω
k−1k }, where ωk = exp(2π i/k), i = √−1.
We also use
μ# =⎧⎨⎩ μ−1 if μ �= 0
0 if μ = 0, for arbitrary μ ∈ C.
Linear combinations of k-potents have been considered in recent years [1–9,14,16–23,27,25–31]. For
the cases: (1) P1 and P2 are idempotents; (2) P1 is idempotent and P2 is tripotent; (3) P1 and P2 are
commuting tripotent matrices; (4) P1 is idempotent and P2 is k-potent, the problem of characterizing
when a linear combination c1P1 + c2P2 is an idempotent matrix (or a tripotent matrix, or a group
involutory matrix) was studied in [1–7,23,27], respectively. Benítez and Thome [8] extended gener-
alized projectors to k-generalized projectors and listed all situations when a linear combination of
commuting k-generalized projectors is a k-generalized projector.
The purpose of this paper is to characterize the invertibility, the group invertibility and the k-
potency of the combinations of k-potents and their products under certain commutativity properties
imposed on some k-potents. Let P1 and P2 be nonzero k-potents, c1, c2 ∈ C\{0}, c3 ∈ C and let s be
a positive integer such that 0 < s < k − 1. We will investigate properties of the combinations of the
form c1P1 + c2P2 − c3Ps1P
k−1−s2 . Firstly, we introduce some canonical decompositions.
Lemma 1.1. If P ∈ B(H) is a k-potent (k � 2), thenR(P) is closed and Pk−1 is idempotent with
R(P) = R(P2) = · · · = R(Pk), N (P) = N (P2) = · · · = N (Pk).
Using the space decompositionH = R(P) ⊕ N (P), P can be written as 2 × 2 operator matrix form
P =⎛⎝ A 0
0 0
⎞⎠ , where A ∈ B(R(P)) is invertible and Ak−1 = I. (1)
Also, the following result holds:
Lemma 1.2 [12, Theorem 2.1]. P ∈ B(H) is a k-potent if and only if
(i) σ(P) ⊆ {0} ∪ �k−1,
(ii) There exists an invertible operator S such that
SPS−1 = ∑λ∈σ(P)
⊕λE(λ),
where ⊕ denotes the orthogonal direct sum and E(λ) are orthogonal projections such that∑λ∈σ(P) E(λ) = I, E(λi)E(λj) = E(λj)E(λi) = 0, λi, λj ∈ σ(P), λi �= λj .
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For an arbitrary finite commuting nonzero k-potent family {Pi}mi=1, we deduce that σ
(m∏i=1
Pi
)⊆
{0}∪�k−1 and σ
(c1In + c2
m∏i=1
Pi
)⊆ {c1 + c2λ : λ ∈ {0}∪�k−1}. Note that, for arbitrary λ ∈ �k−1
and c1, c2 ∈ C\{0}, c1 + c2λ �= 0 ⇐⇒ λ �= − c1c2
. Hence,
c1In + c2
m∏i=1
Pi is invertible ⇐⇒(− c1
c2
)k−1
�= 1.
Any k-potent P is group invertible. If P2 = P, then P# = P. If Pk = P with k > 2, then P# = Pk−2.
Moreover, we have the following observation.
Lemma 1.3. Let T1, T2, T3 ∈ B(H) be group invertible and commuting operators. Then
e1 = (I − Tπ1 )(I − Tπ
2 )(I − Tπ3 ), e2 = (I − Tπ
1 )(I − Tπ2 )Tπ
3 ,
e3 = (I − Tπ1 )Tπ
2 (I − Tπ3 ), e4 = (I − Tπ
1 )Tπ2 Tπ
3 , e5 = Tπ1 (I − Tπ
2 )(I − Tπ3 ), (2)
e6 = Tπ1 (I − Tπ
2 )Tπ3 , e7 = Tπ
1 Tπ2 (I − Tπ
3 ), e8 = Tπ1 Tπ
2 Tπ3
are idempotents with
H =8⊕
i=1
R(ei),8∑
i=1
ei = I, eiej = ejei = 0, i �= j, i, j = 1, 8. (3)
2. Characterizations of commuting k-potents
In this section, wewill characterize the invertibility, the group invertibility and idempotency of the
linear combinationsof group invertible andk-potentoperatorsunder certain commutativityproperties
imposed on them. We always assume F denotes a commuting family of nonzero k-potents and P1, P2and P3 are nonzero k-potents which are not scalar multiples of each other.
Theorem 2.1. Let T1, T2, T3 ∈ B(H) be group invertible and commuting operators and ei, i = 1, 8 be
given by (2). Let � be a linear combination of the form
� = aT1 + bT2 + cT3, a, b ∈ C\{0}, c ∈ C.
Then � is invertible if and only if
I + T#1
(b
aT2 + c
aT3
)e1, I + b
aT#1 T2e2, I + c
aT#1 T3e3, I + c
bT#2 T3e5 (4)
are invertible and e8 = 0.
Proof. Let ei, i = 1, 8 be defined as in (2). Using the space decompositionH = 8⊕i=1
R(ei), we know ei
is a block diagonal operator having the identity I in the ith block and zero otherwise, and
T1 = X1 ⊕ X2 ⊕ X3 ⊕ X4 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0,
T2 = Y1 ⊕ Y2 ⊕ 0 ⊕ 0 ⊕ Y5 ⊕ Y6 ⊕ 0 ⊕ 0,
T3 = Z1 ⊕ 0 ⊕ Z3 ⊕ 0 ⊕ Z5 ⊕ 0 ⊕ Z7 ⊕ 0,
(5)
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C. Deng et al. / Linear Algebra and its Applications 437 (2012) 376–387 379
where Xi, Yj , Zl , i ∈ {1, 2, 3, 4}, j ∈ {1, 2, 5, 6} and l ∈ {1, 3, 5, 7} are invertible. Hence,
T#1 = X
−11 ⊕ X
−12 ⊕ X
−13 ⊕ X
−14 ⊕ 0 ⊕ 0 ⊕ 0 ⊕ 0,
T#2 = Y
−11 ⊕ Y
−12 ⊕ 0 ⊕ 0 ⊕ Y
−15 ⊕ Y
−16 ⊕ 0 ⊕ 0,
T#3 = Z
−11 ⊕ 0 ⊕ Z
−13 ⊕ 0 ⊕ Z
−15 ⊕ 0 ⊕ Z
−17 ⊕ 0.
(6)
Now, we obtain
� = aT1 + bT2 + cT3 = (aX1 + bY1 + cZ1)e1 + (aX2 + bY2)e2 + (aX3 + cZ3)e3
+aX4e4 + (bY5 + cZ5)e5 + bY6e6 + cZ7e7 + 0e8. (7)
Note that
I + T#1 ( b
aT2 + c
aT3)e1 = (I + b
aX
−11 Y1 + c
aX
−11 Z1) ⊕ I ⊕ I ⊕ I ⊕ I ⊕ I ⊕ I ⊕ I,
I + baT#1 T2e2 = I ⊕ (I + b
aX
−12 Y2) ⊕ I ⊕ I ⊕ I ⊕ I ⊕ I ⊕ I,
I + caT#1 T3e3 = I ⊕ I ⊕ (I + c
aX
−13 Z3) ⊕ I ⊕ I ⊕ I ⊕ I ⊕ I,
I + cbT#2 T3e5 = I ⊕ I ⊕ I ⊕ I ⊕ (I + c
bY
−15 Z5) ⊕ I ⊕ I ⊕ I.
(8)
By (7) and (8), we get that � is an invertible if and only if e8 = 0 and (4) holds. �
Corollary 2.1. Let k-potents P1, P2, P3 ∈ F and � be a linear combination of the form
� = aP1 + bP2 + cP3, a, b ∈ C\{0}, c ∈ C
with (− ba)k−1 �= 1, (− c
a)k−1 �= 1 and (− c
b)k−1 �= 1. Then � is invertible if and only if I + P#1 ( b
aP2 +
caP3)e1 is invertible and e8 = 0.
Proof. We use the notions from the proof of Theorem 2.1. Since (− ba)k−1 �= 1 and P#1 P2e2 is k-potent,
we have − ab
/∈ �k−1 ∪ {0} and σ(P#1 P2e2) ⊂ �k−1 ∪ {0}. Hence, − abI − P#1 P2e2 is invertible, i.e.,
I + baP#1 P2e2 is invertible. Similarly, we have that I + c
aP#1 P3e3, I + c
bP#2 P3e5 are invertible. The result
follows immediately by Theorem 2.1. �
Let T3 = 0 in Theorem 2.1. Then ei = 0, i ∈ {1, 3, 5, 7} and ej, j ∈ {2, 4, 6, 8} can be rewritten as
e′1 = (I − Tπ1 )(I − Tπ
2 ), e′2 = (I − Tπ1 )Tπ
2 , e′3 = Tπ1 (I − Tπ
2 ), e′4 = Tπ1 Tπ
2 , (9)
respectively. The following results are derived by Theorem 2.1.
Corollary 2.2. Let P1, P2 ∈ B(H) be group invertible and commuting operators and let e′i , i = 1, 4 be
defined as in (9) and �′ be a linear combination of the form
�′ = aP1 + bP2, a, b ∈ C\{0}.�′ is invertible if and only if I + b
aP#1 P2e
′1 is invertible and e′4 = 0. In addition, if P1, P2 are k-potents and
(− ba)k−1 �= 1, then
�′ is invertible ⇐⇒ e′4 = 0 ⇐⇒ N (P1) ∩ N (P2) = {0}. (10)
Proof. Let P3 = 0 in Theorem 2.1 and c = 0 in Corollary 2.1. �
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Remark 1. In Corollary 2.2, by the proof of Theorem 2.1, P1, P2 and �′ reduce as
P1 = X2 ⊕ X4 ⊕ 0 ⊕ 0, P2 = Y2 ⊕ 0 ⊕ Y6 ⊕ 0, �′ = (aX2 + bY2) ⊕ aX4 ⊕ bY6 ⊕ 0
with respect to the space decomposition H = ⊕4i=1R(e′i), respectively. In the case when P1, P2 are
k-potents,
e′4 = 0 ⇐⇒ Pk−11 + (I − P
k−11 )P2 is invertible ⇐⇒ P1 + (I − P
k−11 )P2 is invertible.
One special case for tripotent operators P1, P2, a = 1 and b = −1 in Corollary 2.2 has been proved in
the paper of Liu et al. [18, Theorem 2.1].
Corollary 2.3 [18, Theorem 2.1]. Let P1, P2 ∈ Cn×n be two commuting tripotents. Then P1 − P2 is
nonsingular if and only if In − P1P2 and P1 + (In − P21)P2 are nonsingular.
The following results generalize [23, Theorem 3.2] to k-potents, and we should remark that our
proof is much simpler.
Theorem 2.2. Let k-potents P1, P2, P3 ∈ F and ei, i = 1, 8 be defined as in (2). Let � be a linear
combination of the form
� = aP1 + bP2 + cP3, a, b, c ∈ C.
Then � is always group invertible. In particular, if k = 2, then � is m-potent if and only if⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩
(a + b + c)m = a + b + c, when e1 �= 0,
(a + b)m = a + b, when e2 �= 0,
(a + c)m = a + c, when e3 �= 0,
(b + c)m = b + c, when e5 �= 0,
am = a, when e4 �= 0,
bm = b, when e6 �= 0,
cm = c, when e7 �= 0
(11)
and
�# = (a + b + c)#e1 + (a + b)#e2 + (a + c)#e3 + a#e4 + (b + c)#e5 + b#e6 + c#e7. (12)
Proof. Weuse thenotions fromtheproof of Theorem2.1. Remark that in the caseof k-potents, I−Pπi =
Pk−1i , i = 1, 3. By (7), we know that
� = (aX1+bY1+cZ1)e1+(aX2+bY2)e2+(aX3+cZ3)e3+aX4e4+(bY5+cZ5)e5+bY6e6+cZ7e7,
where k-potents Xi, Yj and Zl satisfy Xk−1i = I, Y
k−1j = I, Z
k−1l = I, i ∈ {1, 2, 3, 4}, j ∈ {1, 2, 5, 6} and
l ∈ {1, 3, 5, 7}. By Lemma 1.2, a k-potent operator is diagonalizable. Since X1, Y1, Z1 (resp. X2 and Y2;
or X3 and Z3; or Y5 and Z5) are k-potents andmutually commutative, they are simultaneously diagonal-
izable. Hence there is an invertible operator S1 such that S1X1S−11 , S1Y1S
−11 and S1Z1S
−11 are all diagonal
operators and their diagonal entries belong to �k−1 with proper multiplicities. Thus aX1 + bY1 + cZ1is group invertible. Similarly, aX2 + bY2, aX3 + cZ3 and gY5 + cZ5 are group invertible, so � is group
invertible. If k = 2, then Xi, Yj and Zl in (7) are identity operators, so (11) and (12) hold. �
It is interesting to note that all sorts of situations in [23, Theorem 3.2] are some special cases of
(11) in Theorem 2.2. Furthermore, Theorem 2.2 also improves some recent results [1, Theorem,13,
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C. Deng et al. / Linear Algebra and its Applications 437 (2012) 376–387 381
Theorem 2.41], with brief proof. The following theorem in the matrix case was for the first time given
in [1, Theorem]:
Corollary 2.4. Let two different nonzero idempotents P1, P2 ∈ F , �′ be their linear combination of the
form
�′ = aP1 + bP2, a, b ∈ C\{0}.�′ is idempotent if and only if any one of the following items holds:
(i) a = 1, b = 1, P1P2 = 0;(ii) a = 1, b = −1, P1P2 = 0;(iii) a = −1, b = 1, P1P2 = 0.
Proof. Take m = 2 and c = 0 in Theorem 2.2. �
3. Properties of combinations of the form c1P1 + c2P2 − c3Ps1P
k−1−s2
Let P1, P2 be k-potents.Wefirst show that the kernel of c1P1+c2P2−(c1+c2)Pk−11 P2 is independent
of the choice of c1, c2 ∈ C\{0}.Theorem 3.1. Let P1, P2 be k-potents and c1, c2 ∈ C\{0}. Then the following statements hold:
(i) N (P1 − P2) = N (c1P1 + c2P2 − (c1 + c2)Pk−11 P2).
(ii) If P1, P2 are commutative, then for every positive integer 0 < s < k,
N (P1 − P2) = N (c1P1 + c2P2 − (c1 + c2)Ps1P
k−s2 ) ∩ N (Ps1P
k−s2 − P2).
Proof. (i) By Lemma 1.1, P1 and P2 are given by
P1 =⎛⎝ A 0
0 0
⎞⎠ , P2 =⎛⎝ X Y
Z W
⎞⎠ , (13)
where A ∈ B(R(P1)) is invertible and Ak−1 = I. Now, we have that
P1 − P2 =⎛⎝ A − X −Y
−Z −W
⎞⎠and
c1P1 + c2P2 − (c1 + c2)Pk−11 P2 =
⎛⎝ c1(A − X) −c1Y
c2Z c2W
⎞⎠ ,
which immediately shows that N (P1 − P2) = N (c1P1 + c2P2 − (c1 + c2)Pk−11 P2).
(ii) Since P1 and P2 are commutative, they are simultaneous diagonalizable, so it is sufficient to
prove that for p, q, r : (�k−1 ∪ {0}) × (�k−1 ∪ {0}) −→ C, given by
p(λ, μ) = λ − μ, q(λ, μ) = c1λ + c2μ − (c1 + c2)λsμk−s, r(λ, μ) = λsμk−s − μ,
the following holds:
p(λ, μ) = 0 ⇔ q(λ, μ) = 0 and r(λ, μ) = 0. (14)
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By distinguishing four cases: (a) λ = μ = 0, (b) λ = 0, μ �= 0, (c) λ �= 0, μ = 0, (d) λ �= 0, μ �= 0,
it is evident that (14) holds, so we get that (ii) holds. �
Let P1 and P2 be nonzero k-potents. By (13), we get that P1P2 = 0 implies P2 =⎛⎝ 0 0
Z W
⎞⎠, so we
get c1P1 + c2P2 =⎛⎝ c1A 0
c2Z c2W
⎞⎠ and P1 + P2 − P2Pk−11 =
⎛⎝ A 0
0 W
⎞⎠ . Thus, for every c1, c2 ∈ C\{0},
c1P1 + c2P2 is invertible ⇐⇒ P1 + P2 − P2Pk−11 is invertible.
As for the case P1P2 �= 0, we have the following theorem, which is a generalization of Theorem 2.5 in
[18].
Theorem 3.2. Let P1, P2 be nonzero k-potents with P1P2 �= 0 and k ≥ 3. For every c1, c2 ∈ C\{0},c3 ∈ C and positive integer 0 < s < k − 1:
(i) If Pk−12 P1 = P
k−11 P2 and (c1 + c2)
k−1 �= ck−13 , then
c1P1 + c2P2 − c3Ps1P
k−1−s2 is invertible ⇐⇒ P1 + P2 − P2P
k−11 is invertible.
(ii) If Pk−12 P1 = P
k−11 P2 and (c1 + c2)
k−1 = ck−13 , then at least one of the following
c1P1 + c2P2 − c3Ps1P
k−1−s2 and
k−2∑i=1
ck−2−i3
(c1P1 + c2P2
)i + ck−23 Ps1P
k−1−s2
is not invertible.
Proof. (i) By (13), if Pk−12 P1 = P
k−11 P2, we deduce that
Y = 0, Xk−1A = X �= 0,k−2∑i=0
WiZXk−2−i = 0.
Since P2 is k-potent and
P2 =⎛⎝ X 0
Z W
⎞⎠ =⎛⎜⎜⎝ Xk 0
k−1∑i=0
WiZXk−1−i Wk
⎞⎟⎟⎠ = Pk2,
X andW are k-potents and Z = k−1∑i=0
WiZXk−1−i. So X2 = XA,
Z =k−1∑i=0
WiZXk−1−i =⎛⎝k−2∑
i=0
WiZXk−2−i
⎞⎠ X + Wk−1Z = Wk−1Z (15)
and
Z =k−1∑i=0
WiZXk−1−i = ZXk−1 + W
⎛⎝k−2∑i=0
WiZXk−2−i
⎞⎠ = ZXk−1. (16)
Note that
P1 + P2 − P2Pk−11 =
⎛⎝ A 0
0 W
⎞⎠ . (17)
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Since R(P1) = R(X) ⊕ N (X) and N (P1) = R(W) ⊕ N (W), P1 and P2 can be written as the 4 × 4
operator matrices:
P1 =
⎛⎜⎜⎜⎜⎜⎜⎝A3 A4 0 0
A2 A1 0 0
0 0 0 0
0 0 0 0
⎞⎟⎟⎟⎟⎟⎟⎠ , P2 =
⎛⎜⎜⎜⎜⎜⎜⎝X1 0 0 0
0 0 0 0
Z1 Z2 W1 0
Z3 Z4 0 0
⎞⎟⎟⎟⎟⎟⎟⎠ , (18)
where Xk−11 = I and W
k−11 = I. From X2 = XA, we have that A3 = X1 and A4 = 0. Note that
A =⎛⎝ X1 0
A2 A1
⎞⎠ is invertible and Ak−1 = I. So A1 is invertible and Ak−11 = I. By (15) and (16), we get
Wk−1Z =⎛⎝ I 0
0 0
⎞⎠ ⎛⎝ Z1 Z2
Z3 Z4
⎞⎠ =⎛⎝ Z1 Z2
Z3 Z4
⎞⎠ = ZXk−1 =⎛⎝ Z1 Z2
Z3 Z4
⎞⎠ ⎛⎝ I 0
0 0
⎞⎠ .
It follows that Z2 = 0, Z3 = 0, and Z4 = 0. Hence, by (18),
c1P1 + c2P2 − c3Ps1P
k−1−s2 =
⎛⎜⎜⎜⎜⎜⎜⎜⎜⎝
(c1 + c2)X1 − c3I 0 0 0
c1A2 − c3s−1∑i=0
Ai1A2X
k−2−i1 c1A1 0 0
c2Z1 0 c2W1 0
0 0 0 0
⎞⎟⎟⎟⎟⎟⎟⎟⎟⎠, (19)
where A1, X1 andW1 are invertible k-potents. From((c1 + c2)X1 − c3I
)·([
(c1 + c2)X1
]k−2 + [(c1 + c2)X1
]k−3c3
+[(c1 + c2)X1
]k−4c23 + · · · + c
k−23 I
)=
((c1 + c2)
k−1 − ck−13
)I
(20)
and (c1 + c2)k−1 �= c
k−13 we get that (c1 + c2)X1 − c3I is always invertible. Now (19) implies that
c1P1 + c2P2 − c3Ps1P
k−1−s2 is invertible if and only if N (W) = {0}. By (17),
N (W) = {0} ⇐⇒ W is invertible ⇐⇒ P1 + P2 − P2Pk−11 is invertible.
Hence c1P1 + c2P2 − c3Ps1P
k−1−s2 is invertible if and only if P1 + P2 − P2P
k−11 is invertible.
(ii) If (c1 + c2)k−1 = c
k−13 , by (20) it follows that at least one of the following operators (c1 +
c2)X1 − c3I and
[(c1 + c2)X1]k−2 + [(c1 + c2)X1]
k−3 c3 + [(c1 + c2)X1]k−4 c23 + · · · + c
k−23 I
is not invertible , which implies that one of the following operators
c1P1 + c2P2 − c3Ps1P
k−1−s2 and
k−2∑i=1
ck−2−i3
(c1P1 + c2P2
)i + ck−23 Ps1P
k−1−s2
is not invertible. �
Remark 2. (1) In Theorem 3.2, item (i), the condition that c1P1 + c2P2 − c3Ps1P
k−1−s2 is invertible can
be replaced by the condition that c1P1+c2P2−c3I is invertible. Also, in the same theorem, item (ii), the
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384 C. Deng et al. / Linear Algebra and its Applications 437 (2012) 376–387
condition that at least oneof theoperators c1P1+c2P2−c3Ps1P
k−1−s2 and
∑k−2i=1 c
k−2−i3
(c1P1 + c2P2
)i+ck−23 Ps1P
k−1−s2 is not invertible can be replaced by the condition that at least one of the operators
c1P1 + c2P2 − c3I and∑k−2
i=0 ck−2−i3
(c1P1 + c2P2
)iis not invertible.
(2) In Theorem3.2, item (i), if P2 is invertible, then the second and the forth rows and columns in (18)
and (19)will disappear, i.e., P1 =⎛⎝ X1 0
0 0
⎞⎠ and P2 =⎛⎝ X1 0
Z1 W1
⎞⎠ . In this case, it is clear that P1 +P2 −
P2Pk−11 =
⎛⎝ X1 0
0 W1
⎞⎠ is invertible. Then c1P1 + c2P2 − c3Ps1P
k−1−s2 =
⎛⎝ (c1 + c2)X1 − c3I 0
c2Z1 c2W1
⎞⎠is always invertible. Note that(
(c1 + c2)k−1 − c
k−13
)((c1 + c2)X1 − c3I
)−1
= [(c1 + c2)X1]k−2 + [(c1 + c2)X1]k−3c3 + [(c1 + c2)X1]k−4c23 + · · · + ck−23 I
= ck−2−i3
k−2∑i=0
((c1 + c2)X1
)iand (
c1P1 + c2P2 − c3Ps1P
k−1−s2
)−1 =⎛⎝ (
(c1 + c2)X1 − c3I)−1
0
−W−11 Z1
((c1 + c2)X1 − c3I
)−1 1c2W
−11
⎞⎠
=⎛⎝ I 0
−W−11 Z1 I
⎞⎠ ⎛⎝ ((c1 + c2)X1 − c3I
)−10
0 1c2W
−11
⎞⎠ .
Now, by computation we get that(c1P1 + c2P2 − c3P
s1P
k−1−s2
)−1
=[I + P
k−22 (P1 − P2)P
k−11
]·[
ck−2−i3
(c1 + c2)k−1 − ck−13
Pk−11
k−2∑i=0
((c1 + c2)P1
)i+ 1
c2Pk−22 (I − P
k−11 )
]. (21)
(3) If P1 is invertible in Theorem 3.2, item (i), P1, P2 can be represented by P1 =⎛⎝ X Y
Z W
⎞⎠ and
P2 =⎛⎝ A 0
0 0
⎞⎠, where Ak−1 = I and A ∈ B(R(P2)). The condition Pk−11 P2 = P
k−12 P1 implies that
Y = 0 and X = A. The invertibility of P1 implies that A andW are invertible k-potents. Now, it is clear
that P1 + P2 − P2Pk−11 =
⎛⎝ A 0
Z W
⎞⎠ = P1 is invertible. Then
c1P1 + c2P2 − c3Ps1P
k−1−s2 =
⎛⎜⎝ (c1 + c2)A − c3I 0
c1Z − c3DAk−1−s c1W
⎞⎟⎠
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C. Deng et al. / Linear Algebra and its Applications 437 (2012) 376–387 385
is invertible, where Ps1 =⎛⎝ As 0
D Ws
⎞⎠, for some operator D. Since (c1 + c2)k−1 �= c
k−13 , we get that
(c1 + c2)A − c3I is invertible and, by a direct computation we get(c1P1 + c2P2 − c3P
s1P
k−1−s2
)−1
= 1c1Pk−21 (I − P
k−12 ) +
[I − 1
c1Pk−21 (I − P
k−12 )(c1P1 − c3P
s1P
k−1−s2 )
]×
[ck−2−i3
(c1+c2)k−1−ck−13
Pk−12
k−2∑i=0
((c1 + c2)P2
)i].
(22)
In particular, in the case when one of the operators P1 and P2 is invertible and k = 3, if we apply
Remark 2, we get [18, Theorem 2.5].
Corollary 3.1. Let P1, P2 be nonzero tripotents such that one of them is invertible, P21P2 = P22P1 and
c1, c2 ∈ C\{0}, c3 ∈ C. If (c1 + c2)2 = c23 , then c1P1 + c2P2 − c3P1P2 or c1P1 + c2P2 + c3P1P2 is not
invertible. If (c1 + c2)2 �= c23 , then c1P1 + c2P2 − c3P1P2 is invertible. Furthermore, if P1 is invertible, then(
c1P1 + c2P2 − c3P1P2
)−1
= 1c1
(P1 − P1P
22
)+ 1
(c1+c2)2−c23
(I − 1
c1
(P1 − P1P
22
)(c1P1−c3P1P2
))(c3P
22 + (c1+c2)P2
).
If P2 is invertible, then(c1P1 + c2P2 − c3P1P2
)−1 =[I + P2(P1 − P2)P
21
]×
[1
(c1 + c2)2 − c23
(c3P
21 + (c1 + c2)P1
)+ 1
c2P2(I − P21)
].
Let P1, P2 satisfy the conditions in Corollary 3.1. In [18, Theorem 2.5], the authors got two repre-
sentations of(c1P1 + c2P2 − c3P1P2
)−1as follows:
(1) if P1 is nonsingular, then[(c1 + c2)
2 − c23
](c1P1 + c2P2 − c3P1P2
)−1
= (c1 + c2)P1 + c3P22 + c
−11 c2c3
(P22 − P1P2
)+c
−11 c23
(P2 − P1P
22
)+ c
−11
(c22 + c1c2 − c23
)(P1 − P1P
22
).
(23)
(2) If P2 is nonsingular, then[(c1 + c2)
2 − c23
](c1P1 + c2P2 − c3P1P2
)−1
= (c1 + c2)P1 − c3
(2P21 − P2P1
)+ c
−12
(c21 + c1c2 − c23
)(P2 − P2P
21
).
(24)
We remark that the formulas (23) and (24) are equivalent to the formulas in Corollary 3.1. It is clear
that the formulas in Corollary 3.1 are the particular case of the formulas (21) and (22), which are valid
for any k-potents k ≥ 3 and any positive integer 0 < s < k − 1. Moreover, in Corollary 3.1, if c3 = 0,
then Corollary 3.1 reduces as the following one.
Corollary 3.2 [9, Theorem 3.1]. Let P1, P2 be nonzero tripotents such that one of them is invertible,
P21P2 = P22P1 and c1, c2 ∈ C\{0}. Then c1P1 + c2P2 is invertible if and only if c1 + c2 �= 0. In this case, if
Author's personal copy
386 C. Deng et al. / Linear Algebra and its Applications 437 (2012) 376–387
P1 is nonsingular, then(c1P1 + c2P2
)−1 = 1
c1 + c2
(P1 + c2c
−11 P1(I − P22)
).
If P2 is nonsingular, then(c1P1 + c2P2
)−1 = 1
c1 + c2
(P2 + c1c
−12 P2(I − P21)
).
Proof. We only consider the case when P2 is invertible. The remaining case can be proved in the same
way. The invertibility of P2 implies that P22 = I, P1 = P21P2 and P21 = P1P2. By Corollary 3.1,(c1P1 + c2P2
)−1 =[I + P2(P1 − P2)P
21
]×
[1
(c1+c2)2(c1 + c2)P1 + 1
c2P2(I − P21)
]= 1
c1+c2
[I − P21 + P2P1
]×
[P1 + c1+c2
c2P2(I − P21)
]= 1
c1+c2
[P2P
21 + c1+c2
c2P2(I − P21)
]= 1
c1+c2
[P2 + c1c
−12 P2(I − P21)
].
Let P1 and P2 be two k-potents (k ≥ 3) with Pk−11 P2 = P
k−12 P1, c1, c2 ∈ C\{0}, c1 + c2 �= 0. Note that
P#1 = Pk−21 and P#2 = P
k−22 . By (19) we get
c1P1 + c2P2 =
⎛⎜⎜⎜⎜⎜⎜⎝(c1 + c2)X1 0 0 0
c1A2 c1A1 0 0
c2Z1 0 c2W1 0
0 0 0 0
⎞⎟⎟⎟⎟⎟⎟⎠ . �
Theorem 3.3. Let P1, P2 be nonzero k-potents, k ≥ 3 such that Pk−11 P2 = P
k−12 P1. Let c1, c2 ∈ C\{0},
c3 ∈ C be such that c1 + c2 �= c3 and let s be a positive integer such that 0 < s < k − 1. Then
c1P1 + c2P2 − c3Ps1P
k−s2 is invertible ⇐⇒ P1 + P2 − P2P
k−11 is invertible.
Proof. If Pk−11 P2 = P
k−12 P1, by the proof of item (i) Theorem 3.2, we get
c1P1 + c2P2 − c3Ps1P
k−s2 =
⎛⎜⎜⎜⎜⎜⎜⎜⎝
(c1 + c2 − c3)X1 0 0 0
c1A2 − c3s−1∑i=0
Ai1A2X
k−1−i1 c1A1 0 0
c2Z1 0 c2W1 0
0 0 0 0
⎞⎟⎟⎟⎟⎟⎟⎟⎠ , (25)
whereA1, X1 andW1 are invertiblek-potentoperators. Since c1+c2 �= c3,weget c1P1+c2P2−c3Ps1P
k−s2
is invertible if and only if N (W) = {0} if and only ifW is invertible if and only if P1 + P2 − P2Pk−11 is
invertible. �
Acknowledgements
The authors would like to thank the anonymous reviewer for his/her very useful comments that
helped to improve the presentation of this paper.
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C. Deng et al. / Linear Algebra and its Applications 437 (2012) 376–387 387
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