# On the Open Problem Related to Rank Equalities for the Sum of Finitely Many Idempotent Matrices and Its Applications

Authors: Mei-xiang Chen; Qing-hua Chen; Qiao-xin Li; Zhong-peng Yang
Journal: The Scientific World Journal
 (2014)

---

## Abstract

Tian and Styan have shown many rank equalities for the sum of two and three idempotent matrices and pointed out that rank equalities for the sumP
1
+
⋯
+
P
k with P
1
,
…
,
P
k be idempotent (k
>
3) are still open. In this paper, by using block Gaussian elimination, we obtained rank equalities for the sum of finitely many idempotent matrices and then solved the open problem mentioned above. Extensions to scalar-potent matrices and some related matrices are also included.


---

## Body

## 1. Introduction

LetC
m
×
n and G
L
n
(
C
) be the sets of m
×
n complex matrices and n
×
n nonsingular matrices, respectively. The n
×
n identity matrix is denoted by I
n or simply by I if the size is immaterial. Let Z
+ be the set of all the positive integer numbers. The symbols r
(
A
) and A
T stand for the rank and transpose of A
∈
C
m
×
n, respectively, while tr
⁡
A denotes the trace of a square matrix A. A matrix A
∈
C
n
×
n is said to beidempotent, if A
2
=
A, andscalar-potent (determined by λ), if A
2
=
λ
A, for some (
0
≠
)
λ
∈
C (see, e.g., [1]). When λ
=
1, it coincides with the definition of an idempotent matrix.As one of the fundamental building blocks in matrix theory, idempotent matrices are very useful in many contexts and have been extensively studied in the literature (see, e.g., [1–6]). Here we focus on the research on the rank of the sum of idempotent matrices.Gröss and Trenkler have studied rank of the sum of two idempotent matrices (see [3, Theorem 3]). Also, Tian and Styan have shown a rank equality for two idempotent matrices as follows.Proposition 1 (see [1, Theorem 2.4] and [2, Theorem 2.1]).
LetP
,
Q
∈
C
n
×
n be idempotent. Then
(1)
r
(
P
+
Q
)
=
r
(
P
Q
Q
0
)
-
r
(
Q
)
.Tian and Styan have extended the rank equality for the sum of idempotent matrices to the scalar-potent matrices (see, e.g., [1]).Proposition 2 (see [1, P110]).
LetP
,
Q
∈
C
n
×
n be scalar-potent matrices determined by nonzero complexes λ
,
μ. Then
(2)
r
(
μ
P
+
λ
Q
)
=
r
(
P
Q
Q
0
)
-
r
(
Q
)
,
h
P
2
=
λ
P
,
Q
2
=
μ
Q
,
λ
μ
≠
0
.Later, Tian and Styan considered the rank equality for the sum of three idempotent matrices in [2] as follows.Proposition 3 (see [2, P95]).
LetP
1
,
P
2
,
P
3
∈
C
n
×
n be idempotent. Then
(3)
r
(
P
1
+
P
2
+
P
3
)
=
r
(
1
2
P
1
P
2
P
3
P
2
0
P
2
P
3
P
3
P
3
P
2
0
)
-
r
(
P
2
)
-
r
(
P
3
)
.By (3), Tian and Styan have induced many useful results, for example, if P
1, P
2, P
3 are idempotent and P
1
+
P
2
+
P
3
=
0, then P
1
=
P
2
=
P
3
=
0. The literatures [2, 4–6] show that establishing various kinds of rank equalities for k idempotent matrices is interesting. Tian and Styan pointed out that rank equalities for the sum P
1
+
⋯
+
P
k with P
1
,
…
,
P
k be idempotent (k
>
3) are still open (see [2, P95]).In this paper, by applying block Gaussian elimination, rank equalities for the sum of finitely many idempotent matrices are obtained. These results generalize (3) and solve the open problem proposed by Tian and Styan (see, e.g., [2]). Also, new rank equalities for finitely many idempotent matrices are given. The rank equality (3) is generalized to k scalar-potent matrices as well.


## 2. Main Results

Before showing main results, we need some preparations.Lemma 4.
LetA
1
,
…
,
A
k
∈
C
n
×
m. Then
(4)
r
(
A
1
0
⋯
0
A
1
0
A
2
⋯
0
A
2
⋯
⋯
⋱
⋯
⋯
0
0
⋯
A
k
A
k
A
1
A
2
⋯
A
k
0
)
=
∑
i
=
1
k
‍
r
(
A
i
)
+
r
(
∑
i
=
1
k
‍
A
i
)
,
for any k
∈
Z
+.Proof.
Let(5)
G
=
(
I
n
0
⋯
0
0
0
I
n
⋯
0
0
⋯
⋯
⋱
⋯
⋯
0
0
⋯
I
n
0
-
I
n
-
I
n
⋯
-
I
n
I
n
)
∈
C
n
(
k
+
1
)
×
n
(
k
+
1
)
,
S
=
(
I
m
0
⋯
0
-
I
m
0
I
m
⋯
0
-
I
m
⋯
⋯
⋱
⋯
⋯
0
0
⋯
I
m
-
I
m
0
0
⋯
0
I
m
)
∈
C
m
(
k
+
1
)
×
m
(
k
+
1
)
.
It is evident that G and S are nonsingular.
By calculation,(6)
G
(
A
1
0
⋯
0
A
1
0
A
2
⋯
0
A
2
⋯
⋯
⋱
⋯
⋯
0
0
⋯
A
k
A
k
A
1
A
2
⋯
A
k
0
)
S
=
diag
⁡
(
A
1
,
…
,
A
k
,
-
∑
i
=
1
k
‍
A
i
)
;
since G and S are nonsingular, hence
(7)
r
(
A
1
0
⋯
0
A
1
0
A
2
⋯
0
A
2
⋯
⋯
⋱
⋯
⋯
0
0
⋯
A
k
A
k
A
1
A
2
⋯
A
k
0
)
=
r
(
diag
⁡
(
A
1
,
…
,
A
k
,
-
∑
i
=
1
k
‍
A
i
)
)
=
∑
i
=
1
k
‍
r
(
A
i
)
+
r
(
∑
i
=
1
k
‍
A
i
)
.
This completes the proof.The proof method of Lemma4 is inspired by Marsaglia and Styan [5, Theorem 9]. By (4), we get the rank equality for the sum of finitely many idempotent matrices; it is different from the one of three idempotent matrices (3) given by Tian and Styan. Consequently, to find the generalization of Proposition 3 and solve the open problem given by Tian and Styan (see, e.g., [2]), it is necessary to seek a new method different from Lemma 4.Lemma 5 (see [7, Problem 4.9]).
LetP
∈
C
n
×
n be idempotent. Then r
(
P
)
=
tr
⁡
P.In this section, from now on, forA
1
,
A
2
,
…
,
A
k
∈
C
n
×
n, one denotes
(8)
W
(
A
1
,
A
2
,
…
,
A
k
)
=
(
1
2
A
1
A
2
A
3
⋯
A
k
-
1
A
k
A
2
0
A
2
A
3
⋯
A
2
A
k
-
1
A
2
A
k
A
3
A
3
A
2
0
⋯
A
3
A
k
-
1
A
3
A
k
⋮
⋮
⋮
⋱
⋮
⋮
A
k
-
1
A
k
-
1
A
2
A
k
-
1
A
3
⋯
0
A
k
-
1
A
k
A
k
A
k
A
2
A
k
A
3
⋯
A
k
A
k
-
1
0
)
∈
C
n
k
×
n
k
.Theorem 6.
For anyk
∈
Z
+, let P
1
,
…
,
P
k
∈
C
n
×
n be idempotent. Then
(9)
r
(
∑
i
=
1
k
‍
P
i
)
=
r
(
W
(
P
1
,
…
,
P
k
)
)
-
∑
i
=
2
k
‍
r
(
P
i
)
=
r
(
W
(
P
1
,
…
,
P
k
)
)
-
∑
i
=
2
k
‍
tr
⁡
P
i
=
r
(
W
(
P
1
,
…
,
P
k
)
)
-
tr
⁡
(
∑
i
=
2
k
‍
P
i
)
.Proof.
From Lemma4, it follows that
(10)
r
(
P
1
0
⋯
0
P
1
0
P
2
⋯
0
P
2
⋯
⋯
⋱
⋯
⋯
0
0
⋯
P
k
P
k
P
1
P
2
⋯
P
k
0
)
=
∑
i
=
1
k
‍
r
(
P
i
)
+
r
(
∑
i
=
1
k
‍
P
i
)
.
On the other hand, by block Gaussian elimination, we will see that(11)
r
(
P
1
0
⋯
0
P
1
0
P
2
⋯
0
P
2
⋯
⋯
⋱
⋯
⋯
0
0
⋯
P
k
P
k
P
1
P
2
⋯
P
k
0
)
=
r
(
W
(
P
1
,
…
,
P
k
)
)
+
r
(
P
1
)
.
In fact, let us write the matrix(12)
(
P
1
0
⋯
0
P
1
0
P
2
⋯
0
P
2
⋯
⋯
⋱
⋯
⋯
0
0
⋯
P
k
P
k
P
1
P
2
⋯
P
k
0
)
as the quadripartitioned matrix
(13)
(
M
11
M
12
M
21
0
)
∈
C
n
(
k
+
1
)
×
n
(
k
+
1
)
,
where
(14)
M
11
=
diag
⁡
(
P
1
,
…
,
P
k
)
∈
C
n
k
×
n
k
,
M
12
=
(
P
1
T
,
…
,
P
k
T
)
T
∈
C
n
k
×
n
,
M
21
=
(
P
1
,
…
,
P
k
)
∈
C
n
×
n
k
.
By (11) and (13), it suffices to show
(15)
r
(
M
11
M
12
M
21
0
)
=
r
(
W
(
P
1
,
…
,
P
k
)
)
+
r
(
P
1
)
.
Direct calculations to (13) show that
(16)
(
I
n
k
W
12
0
I
n
)
(
M
11
M
12
M
21
0
)
=
(
M
11
+
W
12
M
21
M
12
M
21
0
)
,
where
(17)
W
12
=
(
0
,
-
P
2
T
,
…
,
-
P
k
T
)
T
∈
C
n
k
×
n
,
(
I
n
k
W
12
0
I
n
)
∈
G
L
n
(
k
+
1
)
(
C
)
.
If we define X
11
=
M
11
+
W
12
M
21, by (14) and (17), we get
(18)
X
11
=
(
P
1
0
⋯
0
0
-
P
2
P
1
0
⋯
-
P
2
P
k
-
1
-
P
2
P
k
⋯
⋯
⋱
⋯
⋯
-
P
k
-
1
P
1
-
P
k
-
1
P
2
⋯
0
-
P
k
-
1
P
k
-
P
k
P
1
-
P
k
P
2
⋯
-
P
k
P
k
-
1
0
)
.
Moreover, let
(19)
S
21
=
(
-
1
2
E
n
,
0
,
…
,
0
)
,
G
21
=
(
P
1
,
0
,
…
,
0
)
,
R
12
=
(
-
1
2
P
1
T
,
0
,
…
,
0
)
T
;
then
(20)
(
I
n
k
0
S
21
I
n
)
,
(
I
n
k
0
G
21
I
n
)
,
(
I
n
k
R
12
0
I
n
)
are
nonsingular
.
By (14) and (19), we see that
(21)
X
11
+
M
12
G
21
=
(
2
P
1
0
⋯
0
0
0
0
⋯
-
P
2
P
k
-
1
-
P
2
P
k
⋯
⋯
⋯
⋯
⋯
0
-
P
k
-
1
P
2
⋯
0
-
P
k
-
1
P
k
0
-
P
k
P
2
⋯
-
P
k
P
k
-
1
0
)
∈
C
n
k
×
n
k
.
Then by applying (14) and (19) yields
(22)
(
X
11
+
M
12
G
21
)
R
12
+
M
12
=
(
0
,
P
2
T
,
…
,
P
k
T
)
T
∈
C
n
k
×
n
.
Thus,
(23)
S
21
X
11
+
M
21
+
S
21
M
12
G
21
=
S
21
(
X
11
+
M
12
G
21
)
+
M
21
=
(
-
1
2
I
,
0
,
…
,
0
)
×
(
2
P
1
0
⋯
0
0
0
0
⋯
-
P
2
P
k
-
1
-
P
2
P
k
⋯
⋯
⋯
⋯
⋯
0
-
P
k
-
1
P
2
⋯
0
-
P
k
-
1
P
k
0
-
P
k
P
2
⋯
-
P
k
P
k
-
1
0
)
+
(
P
1
,
P
2
,
…
,
P
k
)
=
(
0
,
P
2
,
…
,
P
k
)
.
Hence it follows from (14) and (19) that
(24)
(
S
21
X
11
+
M
21
+
S
21
M
12
G
21
)
R
12
+
S
21
M
12
=
-
1
2
P
1
.
Consequently, from (16)–(24), it follows that(25)
(
I
n
k
0
S
21
I
n
)
(
I
n
k
W
12
0
I
n
)
(
M
11
M
12
M
21
0
)
(
I
n
k
0
G
21
I
n
)
(
I
n
k
R
12
0
I
n
)
=
(
I
n
k
0
S
21
I
n
)
(
X
11
M
12
M
21
0
)
(
I
n
k
0
G
21
I
n
)
(
I
n
k
R
12
0
I
n
)
=
(
X
11
+
M
12
G
21
(
X
11
+
M
12
G
21
)
R
12
+
M
12
S
21
X
11
+
M
21
+
S
21
M
12
G
21
(
S
21
X
11
+
M
21
+
S
21
M
12
G
21
)
R
12
+
S
21
M
12
)
=
(
2
P
1
0
⋯
0
0
0
0
0
⋯
-
P
2
P
k
-
1
-
P
2
P
k
P
2
⋯
⋯
⋱
⋯
⋯
0
-
P
k
-
1
P
2
⋯
0
-
P
k
-
1
P
k
P
k
-
1
0
-
P
k
P
2
⋯
-
P
k
P
k
-
1
0
P
k
0
P
2
⋯
P
k
-
1
P
k
-
1
2
P
)
=
(
2
P
1
0
0
(
0
⋯
-
P
2
P
k
-
1
-
P
2
P
k
P
2
⋯
⋱
⋯
⋯
-
P
k
-
1
P
2
⋯
0
-
P
k
-
1
P
k
P
k
-
1
-
P
k
P
2
⋯
-
P
k
P
k
-
1
0
P
k
P
2
⋯
P
k
-
1
P
k
-
1
2
P
)
)
=
diag
⁡
(
2
P
1
,
Z
)
,where
(26)
Z
=
(
(
0
⋯
-
P
2
P
k
-
1
-
P
2
P
k
⋯
⋯
⋯
⋯
-
P
k
-
1
P
2
⋯
0
-
P
k
-
1
P
k
-
P
k
P
2
⋯
-
P
k
P
k
-
1
0
)
(
P
2
⋮
P
k
-
1
P
k
)
(
P
2
,
…
,
P
k
-
1
,
P
k
)
-
1
2
P
1
)
=
(
Z
11
Z
12
Z
21
-
1
2
P
1
)
∈
C
n
k
×
n
k
.
Since(
I
n
k
0
s
21
I
n
), (
I
n
k
W
12
0
I
n
), (
I
n
k
0
G
21
I
n
) and (
I
n
k
R
12
0
I
n
) are nonsingular, we get(27)
r
(
M
11
M
12
M
21
0
)
=
r
(
P
1
)
+
r
(
Z
11
Z
12
Z
21
-
1
2
P
1
)
.
Also, it is easy to verify that
(28)
(
0
I
n
-
I
n
k
0
)
(
Z
11
Z
12
Z
21
-
1
2
P
1
)
(
0
I
n
k
-
I
n
0
)
=
(
1
2
P
1
Z
21
Z
12
-
Z
11
)
=
W
(
P
1
,
…
,
P
k
)
.
Since(
0
I
n
-
I
n
k
0
) and (
0
I
n
k
-
I
n
0
) are nonsingular, by (13) and (26)–(28), we obtain
(29)
r
(
M
11
M
12
M
21
0
)
=
r
(
P
1
)
+
r
(
Z
)
=
r
(
P
1
)
+
r
(
W
(
P
1
,
…
,
P
k
)
)
.
Combining (13) with (29) together with Lemma 5 yields the desired results.Whenk
=
3, Theorem 6 leads to Proposition 3 at once, and when k
=
2, it leads to Proposition 1; for the idempotent matrices P and Q, it follows that
(30)
(
2
I
n
0
0
I
n
)
W
(
P
,
Q
)
(
I
n
0
0
1
2
I
n
)
=
(
2
I
n
0
0
I
n
)
(
1
2
P
Q
Q
0
)
(
I
n
0
0
1
2
I
n
)
=
(
P
Q
Q
0
)
.For the sum of two idempotent matrices, Tian and Styan have given out many rank equalities (see [1, Theorem 2.4] and [2, Theorems 2.1, 2.3, 2.4, and 2.7]). Let P
,
P
1
,
…
,
P
k
∈
C
n
×
n be idempotent; using Theorem 6 together with [6, Theorem 6] and [6, (26)] yields the equalities as follows:
(31)
r
(
∑
i
=
1
k
‍
P
i
)
=
r
(
W
(
P
1
,
…
,
P
k
)
)
+
r
(
P
1
)
+
r
(
P
1
-
P
2
P
1
-
P
3
⋮
P
1
-
P
k
)
-
r
(
P
1
P
2
⋮
P
k
)
-
r
(
P
1
P
2
P
1
P
3
⋮
⋱
P
1
P
k
)
,
(32)
r
(
P
+
∑
i
=
1
k
‍
P
i
)
=
r
(
W
(
P
,
P
1
,
…
,
P
k
)
)
+
r
(
P
)
.
+
r
(
P
-
k
P
1
P
⋯
P
P
P
-
k
P
2
⋯
P
⋮
⋮
⋱
⋮
P
P
⋯
P
-
k
P
k
)
-
r
(
P
P
⋯
P
P
1
P
2
…
P
k
)
-
r
(
P
P
1
P
P
2
⋮
⋱
P
P
k
)
.Theorem6 together with (31) and (32) indicates that the sum of k
(
≥
3
) idempotent matrices has various kinds of rank equalities, as shown in the discussions in the literature [1, 2].In view of (11), by applying Lemma 5, we see the following.Corollary 7.
For anyk
∈
Z
+, let P
1
,
…
,
P
k
∈
C
n
×
n be idempotent. Then
(33)
r
(
P
1
0
⋯
0
P
1
0
P
2
⋯
0
P
2
⋯
⋯
⋯
⋯
⋯
0
0
⋯
P
k
P
k
P
1
P
2
⋯
P
k
0
)
-
r
(
W
(
P
1
,
…
,
P
k
)
)
=
r
(
P
1
)
=
tr
⁡
P
1
.This immediately implies that the difference of the ranks of two block matrices in the left side of (33) is always equal to r
(
P
1
) or tr
⁡
P
1, independently on the choice of k, when k
≥
3.


## 3. The Rank Formulas for the Sum ofk Scalar-Potent Matrices and Applications

Theorem6 can easily be extended to scalar-potent matrices; in fact,
(34)
(
1
λ
i
P
i
)
2
=
1
λ
i
P
i
,
r
(
1
λ
i
P
i
)
=
r
(
P
i
)
,
h
h
h
h
h
h
h
h
h
h
h
h
P
i
2
=
λ
i
P
i
,
i
=
1
,
…
,
k
.
So (
1
/
λ
i
)
P
i is idempotent.Theorem 8.
For any givenk
∈
Z
+, let P
i
∈
C
n
×
n be scalar-potent (determined by λ
i
(
≠
0
)), i
=
1
,
…
,
k. Then
(35)
r
[
∑
i
=
1
k
‍
(
∏
j
≠
i
‍
λ
j
)
P
i
]
=
r
(
W
(
1
λ
1
P
1
,
P
2
,
…
,
P
k
)
)
-
∑
i
=
2
k
‍
r
(
P
i
)
=
r
(
W
(
1
λ
1
P
1
,
P
2
,
…
,
P
k
)
)
-
∑
i
=
2
k
‍
1
λ
i
tr
⁡
P
i
.Proof.
By (8) and (34), we get(36)
W
(
1
λ
1
P
1
,
…
,
1
λ
k
P
k
)
=
(
1
2
λ
1
P
1
1
λ
2
P
2
1
λ
3
P
3
⋯
1
λ
k
-
1
P
k
-
1
1
λ
k
P
k
1
λ
2
P
2
0
1
λ
2
λ
3
P
2
P
3
⋯
1
λ
2
λ
k
-
1
P
2
P
k
-
1
1
λ
2
λ
k
P
2
P
k
1
λ
3
P
3
1
λ
3
λ
2
P
3
P
2
0
⋯
1
λ
3
λ
k
-
1
P
3
P
k
-
1
1
λ
3
λ
k
P
3
P
k
⋮
⋮
⋮
⋱
⋮
⋮
1
λ
k
-
1
P
k
-
1
1
λ
k
-
1
λ
2
P
k
-
1
P
2
1
λ
k
-
1
λ
3
P
k
-
1
P
3
⋯
0
1
λ
k
-
1
λ
k
P
k
-
1
P
k
1
λ
k
P
k
1
λ
k
λ
2
P
k
P
2
1
λ
k
λ
3
P
k
P
3
⋯
1
λ
k
λ
k
-
1
P
k
P
k
-
1
0
)
.
On the other hand, using (36), we obtain
(37)
W
(
1
λ
1
P
1
,
P
2
,
…
,
P
k
)
=
G
W
(
1
λ
1
P
1
,
…
,
1
λ
k
P
k
)
G
,
with G
=
diag
⁡
(
I
n
,
λ
2
I
2
,
…
,
λ
k
I
n
). Since G is nonsingular, by (37), we can write
(38)
r
(
W
(
1
λ
1
P
1
,
P
2
,
…
,
P
k
)
)
=
r
(
W
(
1
λ
1
P
1
,
1
λ
2
P
2
,
…
,
1
λ
k
P
k
)
)
.
From (34), (
1
/
λ
i
)
P
i is idempotent. Using Theorem 6 together with (38) yields the equality
(39)
r
(
∑
i
=
1
k
‍
1
λ
i
P
i
)
=
r
(
W
(
1
λ
1
P
1
,
1
λ
2
P
2
,
…
,
1
λ
k
P
k
)
)
-
∑
i
=
2
k
‍
r
(
1
λ
i
P
i
)
=
r
(
W
(
1
λ
1
P
1
,
P
2
,
…
,
P
k
)
)
-
∑
i
=
2
k
‍
r
(
1
λ
i
P
i
)
.
We note that
(40)
r
(
∑
i
=
1
k
‍
1
λ
i
P
i
)
=
r
[
(
∏
j
=
1
k
‍
λ
j
)
(
∑
i
=
1
k
‍
1
λ
i
P
i
)
]
=
r
[
∑
i
=
1
k
‍
(
∏
j
≠
i
‍
λ
j
)
P
i
]
.
From (39) and (40), we get the desired result since r
(
P
i
)
=
r
(
(
1
/
λ
i
)
P
i
)
=
(
1
/
λ
i
)
tr
⁡
P
i.Whenk
=
2, this leads immediately to Proposition 2, since it can be written as
(41)
(
2
λ
I
n
0
0
I
n
)
(
1
2
λ
P
Q
Q
0
)
(
I
n
0
0
1
2
λ
I
n
)
=
(
P
Q
Q
0
)
,
with P
2
=
λ
P,  Q
2
=
μ
Q, and  λ
μ
≠
0.For any given idempotent matrixP, Farebrother and Trenkler [8] denoted the set of generalized quadratic matrices as
(42)
Ω
n
(
P
)
=
{
A
∈
C
n
×
n
:
A
2
=
α
A
+
β
P
,
A
P
=
P
A
=
A
,
∀
α
,
β
∈
C
}
.
If P
=
I, it coincides with the definition of a quadratic matrix (see, e.g., [9]). In view of [10, Lemma 1] and [11, Lemma 2.2], we conclude that (42) can be expressed equivalently as
(43)
Ω
n
(
P
)
=
{
A
∈
C
n
×
n
:
(
A
-
a
P
)
(
A
-
b
P
)
=
0
,
A
P
=
P
A
=
A
,
∀
a
,
b
∈
C
A
∈
C
n
×
n
:
(
A
-
a
P
)
(
A
-
b
P
)
=
0
,
}
.
If A
∈
Ω
n
(
P
), then from (42) and (43), we see that
(44)
a
=
1
2
(
α
+
α
2
+
4
β
)
,
b
=
1
2
(
α
-
α
2
+
4
β
)
.Lemma 9.
For any given idempotent matrixP, if A
∈
Ω
n
(
P
) satisfies (
A
-
a
P
)
(
A
-
b
P
)
=
0 with a
≠
b, then A
-
a
P is a scalar-potent matrix determined by b
-
a.Proof.
For the matrixP, there exists a nonsingular matrix S such that P
=
S
diag
⁡
(
I
k
,
0
)
S
-
1. From A
P
=
P
A, we can write A
=
S
diag
⁡
(
A
1
,
A
2
)
S
-
1 being A
1
∈
C
k
×
k. From A
P
=
A, we get A
2
=
0; namely, A
=
S
diag
⁡
(
A
1
,
0
)
S
-
1. We have (
A
1
-
a
I
k
)
(
A
1
-
b
I
k
)
=
0. It is seen from the fact that a matrix is diagonalizable if and only if its minimal polynomial has simple roots (see [12, Corollary 3.3.10]). Thus, there exists a nonsingular matrix S
1 such that A
1
=
S
1
diag
⁡
(
a
I
r
,
b
I
k
-
r
)
S
1
-
1. Let R
=
diag
⁡
(
S
1
,
I
); then R is nonsingular and
(45)
P
=
R
diag
⁡
(
I
r
,
I
k
-
r
,
0
)
R
-
1
,
A
=
R
diag
⁡
(
a
I
r
,
b
I
k
-
r
,
0
)
R
-
1
.
Hence
(46)
A
-
a
P
=
R
diag
⁡
(
0
,
(
b
-
a
)
I
k
-
r
,
0
)
R
-
1
=
(
b
-
a
)
R
diag
⁡
(
0
,
I
n
-
r
,
0
)
R
-
1
.
Now, it is evident that (
A
-
a
P
)
2
=
(
b
-
a
)
(
A
-
a
P
).Theorem 10.
For any given idempotent matricesP
1
,
…
,
P
k
∈
C
n
×
n and any k
∈
Z
+, if A
i
∈
Ω
n
(
P
i
) satisfies (
A
i
-
a
i
P
i
)
(
A
i
-
b
i
P
i
)
=
0 with a
i
≠
b
i, i
=
1
,
…
,
k, then
(47)
r
[
∑
i
=
1
k
‍
(
∏
j
≠
i
‍
(
b
j
-
a
j
)
)
(
A
i
-
a
i
P
i
)
]
=
r
(
W
(
1
b
1
-
a
1
(
A
1
-
a
1
P
1
)
,
A
2
-
a
2
P
2
,
…
,
A
k
-
a
k
P
k
)
)
-
∑
i
=
2
k
‍
r
(
A
i
-
a
i
P
i
)
=
r
(
W
(
1
b
1
-
a
1
(
A
1
-
a
1
P
1
)
,
A
2
-
a
2
P
2
,
…
,
A
k
-
a
k
P
k
)
)
-
∑
i
=
2
k
‍
1
b
i
-
a
i
tr
⁡
(
A
i
-
a
i
P
i
)
.Proof.
For the idempotent matricesP
i, by applying Lemma 9, we see that A
i
-
a
i
P
i is a scalar-potent matrix determined by b
i
-
a
i; then results follow from Theorem 8.



---

*Source: 702413-2014-07-10.xml*