Partial orders based on the CS decomposition
UDC 512.5 A new decomposition for square matrices is given by using two known matrix decompositions, a new characterization of the core-EP order is obtained by using this new matrix decomposition. Also, we will use a matrix decomposition to investigate the minus, star, shar...
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Ukrains’kyi Matematychnyi Zhurnal| _version_ | 1860512227196403712 |
|---|---|
| author | Xu, S. Z. Chen, J. L. Benítez, J. Xu, S. Z. Chen, J. L. Benítez, J. |
| author_facet | Xu, S. Z. Chen, J. L. Benítez, J. Xu, S. Z. Chen, J. L. Benítez, J. |
| author_sort | Xu, S. Z. |
| baseUrl_str | https://umj.imath.kiev.ua/index.php/umj/oai |
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| datestamp_date | 2022-03-26T11:02:02Z |
| description | UDC 512.5
A new decomposition for square matrices is given by using two known matrix decompositions, a new characterization of the core-EP order is obtained by using this new matrix decomposition. Also, we will use a matrix decomposition to investigate the minus, star, sharp and core partial orders in the setting of complex matrices.
|
| doi_str_mv | 10.37863/umzh.v72i8.6025 |
| first_indexed | 2026-03-24T03:25:26Z |
| format | Article |
| fulltext |
DOI: 10.37863/umzh.v72i8.6025
UDC 512.5
S. Z. Xu (Huaiyin Inst. Technology, China),
J. L. Chen (School Math., Southeast Univ., Nanjing, China),
J. Benı́tez (Univ. Politècnica de València, Inst. Mat. Multidisciplinar, València, Spain)
PARTIAL ORDERS BASED ON THE CS DECOMPOSITION*
ЧАСТКОВI ВПОРЯДКУВАННЯ НА ОСНОВI CS-РОЗКЛАДУ
A new decomposition for square matrices is given by using two known matrix decompositions, a new characterization
of the core-EP order is obtained by using this new matrix decomposition. Also, we will use a matrix decomposition to
investigate the minus, star, sharp and core partial orders in the setting of complex matrices.
За допомогою двох вiдомих розкладiв матриць отримано новий розклад для квадратних матриць, а за допомо-
гою цього нового розкладу отримано нову характеристику core-EP-впорядкування. Також використано матричний
розклад для вивчення часткового порядку типу “minus”, “star”, “sharp” та “core” для комплексних матриць.
1. Introduction. Let \BbbC m\times n denotes the set of all m \times n complex matrices. Let A\ast , \scrR (A),
\scrN (A) and \mathrm{r}\mathrm{k}(A) denote the conjugate transpose, column space, null space and rank of A \in \BbbC m\times n,
respectively. For A \in \BbbC m\times n, if X \in \BbbC n\times m satisfies
AXA = A, XAX = X, (AX)\ast = AX and (XA)\ast = XA,
then X is called a Moore – Penrose inverse of A. If such a matrix X exists, then it is unique and
denoted by A\dagger . Many existence criteria and properties of the Moore – Penrose inverse can be found
in [1, 6, 8, 9]. If AXA = A holds, then X (and denoted by A - ) is called an inner inverse of A and
the set of all inner inverses of A is denoted by A\{ 1\} .
Let A \in \BbbC n\times n. It can be easily proved that the set of elements X \in \BbbC n\times n such that
AXA = A, XAX = X and AX = XA
is empty or a singleton. If this set is a singleton, its unique element is called the group inverse of A
and denoted by A\#.
The core inverse for a complex matrix was introduced by Baksalary and Trenkler [4]. Let
A \in \BbbC n\times n. A matrix X \in \BbbC n\times n is called a core inverse of A, if it satisfies AX = PA and
\scrR (X) \subseteq \scrR (A), here PA denotes the orthogonal projector onto \scrR (A). If such a matrix exists, then
it is unique and denoted by A #\bigcirc . For a square complex matrix A, one has that A is core invertible,
A is group invertible, and \mathrm{r}\mathrm{k}(A) = \mathrm{r}\mathrm{k}(A2) are three equivalent conditions (see [1]). We denote
\BbbC CM
n = \{ A \in \BbbC n\times n | \mathrm{r}\mathrm{k}(A) = \mathrm{r}\mathrm{k}(A2)\} . The core partial order for a complex matrix was also
introduced in [4]. For A \in \BbbC CM
n and B \in \BbbC n\times n, the binary relation A
#\bigcirc
\leq B is defined as follows:
A
#\bigcirc
\leq B \leftrightarrow A #\bigcirc A = A #\bigcirc B \mathrm{a}\mathrm{n}\mathrm{d} AA #\bigcirc = BA #\bigcirc .
* This research is supported by the National Natural Science Foundation of China (No. 11771076). The first author is
supported by the National Science Foundation of Jiangsu, Province of China (No. BK20191047) and the National Science
Foundation of Jiangsu Education Committee (No. 19KJB110005).
c\bigcirc S. Z. XU, J. L. CHEN, J. BENITEZ, 2020
ISSN 1027-3190. Укр. мат. журн., 2020, т. 72, № 8 1119
1120 S. Z. XU, J. L. CHEN, J. BENITEZ
In [4] (Theorem 6), it is proved that the core partial order is a matrix partial order. Baksalary and
Trenkler [4] gave several characterizations and various relationships between the matrix core partial
order and other matrix partial orders by using the decomposition of Hartwig and Spindelböck [9].
Let us recall some other well-known partial orders in \BbbC n\times n. For A,B \in \BbbC n\times n,
the left star partial order A \ast \leq B : A\ast A = A\ast B and \scrR (A) \subseteq \scrR (B) [3];
the star partial order A
\ast
\leq B : A\ast A = A\ast B and AA\ast = BA\ast [7];
the minus partial order A
-
\leq B : A - A = A - B and AA - = BA - , where A - denotes some
inner inverse of A [8];
the sharp partial order A
\#
\leq B : A\#A = A\#B and AA\# = BA\# [12].
In addition, \bfone n and \bfzero n will denote the n\times 1 column vectors all of whose components are 1 and
0, respectively. 0m\times n (abbr., 0) denotes the zero matrix of size m \times n. If \scrS is a subspace of \BbbC n,
then P\scrS stands for the orthogonal projector onto the subspace \scrS . A matrix A \in \BbbC n\times n is called an
EP matrix if \scrR (A) = \scrR (A\ast ), A is called projection if A\ast = A = A2 and A is unitary if AA\ast = In,
where In denotes the identity matrix of size n. Let A \in \BbbC n\times n, the smallest integer k such that
\mathrm{r}\mathrm{k}(Ak) = \mathrm{r}\mathrm{k}(Ak+1) is called the index of A and denoted by \mathrm{i}\mathrm{n}\mathrm{d}(A) = k.
2. Preliminaries. A related decomposition of the matrix decomposition of Hartwig and Spin-
delböck [9] was given in [1] (Theorem 2.1) by Benı́tez. In [2] it can be found a simpler proof of this
decomposition. Let us start this section with the concept of principal angles.
Definition 2.1 [18]. Let \scrS 1 and \scrS 2 be two nontrivial subspaces of \BbbC n. We define the principal
angles \theta 1, . . . , \theta r \in [0, \pi /2] between \scrS 1 and \scrS 2 by
\mathrm{c}\mathrm{o}\mathrm{s} \theta i = \sigma i(P\scrS 1P\scrS 2),
for i = 1, . . . , r, where r = \mathrm{m}\mathrm{i}\mathrm{n}\{ \mathrm{d}\mathrm{i}\mathrm{m}\scrS 1, \mathrm{d}\mathrm{i}\mathrm{m}\scrS 2\} . The real numbers \sigma i(P\scrS 1P\scrS 2) \geq 0 are the
singular values of P\scrS 1P\scrS 2 .
The following theorem can be found in [1] (Theorem 2.1).
Theorem 2.1. Let A \in \BbbC n\times n, r = \mathrm{r}\mathrm{k}(A), and let \theta 1, . . . , \theta p be the principal angles between
\scrR (A) and \scrR (A\ast ) belonging to ]0, \pi /2[. Denote by x and y the multiplicities of the angles 0 and
\pi /2 as a canonical angle between \scrR (A) and \scrR (A\ast ), respectively. There exists a unitary matrix
Y \in \BbbC n\times n such that
A = Y
\biggl[
MC MS
0 0
\biggr]
Y \ast , (2.1)
where M \in \BbbC r\times r is nonsingular,
C = \mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}(\bfzero y, \mathrm{c}\mathrm{o}\mathrm{s} \theta 1, . . . , \mathrm{c}\mathrm{o}\mathrm{s} \theta p,\bfone x),
S =
\biggl[
\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}(\bfone y, \mathrm{s}\mathrm{i}\mathrm{n} \theta 1, . . . , \mathrm{s}\mathrm{i}\mathrm{n} \theta p) 0p+y,n - (r+p+y)
0x,p+y 0x,n - (r+p+y)
\biggr]
,
and r = y+ p+ x. Furthermore, x and y+ n - r are the multiplicities of the singular values 1 and
0 in P\scrR (A)P\scrR (A\ast ), respectively. We call (2.1) as the CS decomposition of A.
ISSN 1027-3190. Укр. мат. журн., 2020, т. 72, № 8
PARTIAL ORDERS BASED ON THE CS DECOMPOSITION 1121
In this decomposition, one has C2 + SS\ast = Ir. Recall that A\dagger always exists. We have that A\#
exists if and only if C is nonsingular [1] (Theorem 3.7). The following equalities hold:
A\dagger = Y
\biggl[
CM - 1 0
S\ast M - 1 0
\biggr]
Y \ast , A\# = Y
\biggl[
C - 1M - 1 C - 1M - 1C - 1S
0 0
\biggr]
Y \ast .
From A #\bigcirc = A\#AA\dagger , we obtain A #\bigcirc = Y
\biggl[
C - 1M - 1 0
0 0
\biggr]
Y \ast , AA #\bigcirc = Y
\biggl[
Ir 0
0 0
\biggr]
Y \ast
and A #\bigcirc A = Y
\biggl[
Ir C - 1S
0 0
\biggr]
Y \ast . We have
AA #\bigcirc - A #\bigcirc A = Y
\biggl[
Ir 0
0 0
\biggr]
Y \ast - Y
\biggl[
Ir C - 1S
0 0
\biggr]
Y \ast = Y
\biggl[
0 - C - 1S
0 0
\biggr]
Y \ast . (2.2)
Thus,
| AA #\bigcirc - A #\bigcirc A| =
\bigm| \bigm| \bigm| \bigm| Y \biggl[
0 - C - 1S
0 0
\biggr]
Y \ast
\bigm| \bigm| \bigm| \bigm| = 0.
Therefore, AA #\bigcirc - A #\bigcirc A is always singular and \mathrm{r}\mathrm{k}(AA #\bigcirc - A #\bigcirc A) = \mathrm{r}\mathrm{k}(C - 1S) = \mathrm{r}\mathrm{k}(S) < n.
From (2.2), we have that A is an EP matrix if and only if S = 0, that is all the canonical angles
between \scrR (A) and \scrR (A\ast ) are 0. This result also can be found in [1] (Theorem 3.7).
Proposition 2.1. If A \in \BbbC n\times n is core invertible and A has the form (2.1), then AA #\bigcirc - A #\bigcirc A
is always singular with \mathrm{r}\mathrm{k}(AA #\bigcirc - A #\bigcirc A) = \mathrm{r}\mathrm{k}(S) < n.
In [19] (Theorem 3.1), the authors proved the following lemma for an element in a ring with
involution.
Lemma 2.1. Let A \in \BbbC n\times n. Then A is core invertible with A #\bigcirc = X if and only if (AX)\ast =
= AX, XA2 = A and AX2 = X.
Proposition 2.2. Let A, B, U \in \BbbC n\times n with A = UBU\ast , where B is core invertible and U is
unitary. Then A is core invertible. In this case, one has A #\bigcirc = UB #\bigcirc U\ast .
Proof. Let X = UB #\bigcirc U\ast , we have
AX = AUB #\bigcirc U\ast = UBU\ast UB #\bigcirc U\ast = UBB #\bigcirc U\ast is Hermitian,
XA2 = UB #\bigcirc U\ast (UBU\ast )2 = UB #\bigcirc (B)2U\ast = UBU\ast = A,
AX2 = UAU\ast (UB #\bigcirc U\ast )2 = UB(B #\bigcirc )2U\ast = UB #\bigcirc U\ast = X.
Thus, A #\bigcirc = UB #\bigcirc U\ast in view of Lemma 2.1.
Recently, Wang introduced a new decomposition for square matrices, named the Core-EP decom-
position in [17] (Theorem 2.1).
Lemma 2.2. Let A \in \BbbC n\times n with \mathrm{i}\mathrm{n}\mathrm{d}(A) = k. Then A can be written as
A = A1 +A2, (2.3)
in which
(1) A1 \in \BbbC CM
n ;
(2) Ak
2 = 0;
ISSN 1027-3190. Укр. мат. журн., 2020, т. 72, № 8
1122 S. Z. XU, J. L. CHEN, J. BENITEZ
(3) A\ast
1A2 = A2A1 = 0.
We call the equality (2.3) as the Core-EP decomposition of A.
Definition 2.2 ([14], Definition 3.1). Let A \in \BbbC n\times n with \mathrm{i}\mathrm{n}\mathrm{d}(A) = k. A matrix X \in \BbbC n\times n is
called a core-EP inverse of A if X is an outer inverse of A and satisfying
\scrR (X) = \scrR (X\ast ) = \scrR (Ak).
If such X exists, then it is unique and denoted by A \dagger \bigcirc .
3. A matrix decomposition related the CS decomposition and the core-EP decomposition.
Theorem 3.1. Let A \in \BbbC n\times n with \mathrm{i}\mathrm{n}\mathrm{d}(A) = k and r = \mathrm{r}\mathrm{k}(A). Then there exists a unitary
matrix U \in \BbbC n\times n such that
A = U
\biggl[
MC MS
0 D4
\biggr]
U\ast , (3.1)
where M and C are both nonsingular, D4 is nilpotent, C2 +SS\ast = Ir and matrices C and S have
the form after equality (2.1).
Proof. From Lemma 2.2, we have
A = A1 +A2,
in which A1 \in \BbbC CM
n , Ak
2 = 0 and A\ast
1A2 = A2A1 = 0. Now, utilizing the decomposition in
Theorem 2.1 to A1, there exists a unitary matrix U such that
A1 = U
\biggl[
MC MS
0 0
\biggr]
U\ast ,
in which M is nonsingular. We also have that C is nonsingular in view of A1 \in \BbbC CM
n and
[1] (Theorem 3.7). Have in mind C is Hermitian. If we let A2 = U
\biggl[
D1 D2
D3 D4
\biggr]
U\ast , where
D1 \in \BbbC r\times r, then
A\ast
1A2 = U
\biggl[
CM\ast D1 CM\ast D2
S\ast M\ast D1 S\ast M\ast D2
\biggr]
U\ast and A2A1 = U
\biggl[
D1MC D1MS
D3MC D3MS
\biggr]
U\ast . (3.2)
From (3.2) and A\ast
1A2 = A2A1 = 0, we get CM\ast D1 = 0, CM\ast D2 = 0 and D3MC = 0. The
nonsingularity of C and M implies that D1, D2 and D3 are zero matrices. Thus,
A = A1 +A2 = U
\biggl[
MC MS
0 D4
\biggr]
U\ast .
The equality Ak
2 = 0 implies that D4 is nilpotent.
Theorem 3.1 is proved.
Note that the decomposition in Theorem 3.1 has the same form as Schur form, but the decompo-
sition in Theorem 3.1 seems easier to handle. Have in mind that M and C are both nonsingular, C
is diagonal and real, D4 is nilpotent and C2 + SS\ast = Ir by Theorem 3.1.
Since C is nonsingular, we obtain
ISSN 1027-3190. Укр. мат. журн., 2020, т. 72, № 8
PARTIAL ORDERS BASED ON THE CS DECOMPOSITION 1123
A\#
1 = U
\biggl[
C - 1M - 1 C - 1M - 1C - 1S
0 0
\biggr]
U\ast
by [1] (Theorem 3.7). It is evident that A #\bigcirc
1 = U
\biggl[
C - 1M - 1 0
0 0
\biggr]
U\ast . From [17] (Theorem 3.2),
we get A \dagger \bigcirc = A #\bigcirc
1 = U
\biggl[
C - 1M - 1 0
0 0
\biggr]
U\ast .
In the following theorem, we will use the matrix decomposition in Theorem 3.1 to investigate the
core-EP order, which was introduced by Wang in [17], defined as follows: for matrices A,B \in \BbbC n\times n
A
\dagger \bigcirc
\leq B \leftrightarrow A \dagger \bigcirc A = A \dagger \bigcirc B and AA \dagger \bigcirc = BA \dagger \bigcirc .
Theorem 3.2. Let A,B \in \BbbC n\times n. Assume that A has the form (3.1). Then A
\dagger \bigcirc
\leq B if and only if
B - A can be written as
B - A = U
\biggl[
0 0
0 B4
\biggr]
U\ast for some B4 \in \BbbC (n - r)\times (n - r). (3.3)
Proof. Let B - A = U
\biggl[
B1 B2
B3 B4
\biggr]
U\ast , where B1 \in \BbbC r\times r. Note that A
\dagger \bigcirc
\leq B if and only if
A \dagger \bigcirc (B - A) = (B - A)A \dagger \bigcirc = 0, where
A \dagger \bigcirc (B - A) = U
\biggl[
C - 1M - 1B1 C - 1M - 1B2
0 0
\biggr]
U\ast , (3.4)
(B - A)A \dagger \bigcirc = U
\biggl[
B1C
- 1M - 1 0
B3C
- 1M - 1 0
\biggr]
U\ast . (3.5)
Assume A
\dagger \bigcirc
\leq B. From (3.4) and (3.5), we get C - 1M - 1B1 = 0, C - 1M - 1B2 = 0 and B3C
- 1M - 1 =
= 0, thus from the nonsingularity of C and M, we obtain that B1, B2 and B3 are zero matrices.
Therefore, we get (3.3). To prove the opposite implication, it is easy to check that (B - A)A \dagger \bigcirc =
= A \dagger \bigcirc (B - A) = 0, that is A
\dagger \bigcirc
\leq B.
Theorem 3.2 is proved.
4. Core, star, group and minus partial order. In this section, we will consider the core, star,
group and minus partial orders by using Theorem 2.1.
Lemma 4.1 ([12], Lemma 2.2). Let A \in \BbbC n\times n be group invertible. Then A
\#
\leq B if and only if
A2 = AB = BA.
Lemma 4.2 ([20], [Theorem 3.2). Let A,B \in \BbbC n\times n be two core invertible matrices. Then
A
#\bigcirc
\leq B if and only if A \ast \leq B and B #\bigcirc AA #\bigcirc = A #\bigcirc .
An equivalent form of the minus partial order is the following statement: for complex matrix
case can be found in [5, 15] and for element in rings case can be found in [11].
Lemma 4.3. Let A,B \in \BbbC n\times n. Then the following statements are equivalent:
(1) B
-
\leq A;
(2) B = AA - B = BA - A = BA - B for some A - \in A\{ 1\} ;
(3) B = AA - B = BA - A = BA - B for all A - \in A\{ 1\} .
ISSN 1027-3190. Укр. мат. журн., 2020, т. 72, № 8
1124 S. Z. XU, J. L. CHEN, J. BENITEZ
The following lemma was proved in the more general setting of rings with involution in [16]
(Theorem 4.10).
Lemma 4.4. Let A,B \in \BbbC n\times n. If A,B are both core invertible and B
-
\leq A, then B
#\bigcirc
\leq A if
and only if A #\bigcirc BA #\bigcirc = B #\bigcirc .
Theorem 4.1. Let A,B \in \BbbC n\times n. Assume that A has the form (2.1). If A is core invertible, then
the following statements are equivalent:
(1) A
#\bigcirc
\leq B;
(2) B - A can be written as
B - A = Y
\biggl[
0 0
0 B4
\biggr]
Y \ast , B4 \in \BbbC (n - r)\times (n - r); (4.1)
(3) AA #\bigcirc (B - A) = 0 and (B - A)AA #\bigcirc = 0;
(4) B = A+ (In - AA #\bigcirc )X(In - AA #\bigcirc ) for some matrix X \in \BbbC n\times n.
Proof. (1) \leftrightarrow (2). Since A is core invertible and core invertibility of a matrix gives the group
invertibility of such matrix, hence C is nonsingular. Note that
A #\bigcirc = A\#AA\dagger = Y
\biggl[
C - 1M - 1 0
0 0
\biggr]
Y \ast .
If we let B - A = Y
\biggl[
B1 B2
B3 B4
\biggr]
Y \ast , where B1 \in \BbbC r\times r, and suppose that A
#\bigcirc
\leq B, then A
#\bigcirc
\leq B
if and only of A #\bigcirc (B - A) = (B - A)A #\bigcirc = 0 and
A #\bigcirc (B - A) = Y
\biggl[
C - 1M - 1B1 C - 1M - 1B2
0 0
\biggr]
Y \ast , (4.2)
(B - A)A #\bigcirc = Y
\biggl[
B1C
- 1M - 1 0
B3C
- 1M - 1 0
\biggr]
Y \ast . (4.3)
From (4.2) and (4.3), we get C - 1M - 1B1 = 0, C - 1M - 1B2 = 0 and B3C
- 1M - 1 = 0. Thus, from
the nonsingularity of C and M, we obtain that B1, B2 and B3 are zero matrices. Conversely, if we
have (4.1), it is easy to check that AA #\bigcirc B = A, which is equivalent to A #\bigcirc A = A #\bigcirc B. And we have
AA #\bigcirc = BA #\bigcirc in a similar way.
(2) \Rightarrow (3). It is easy to check that AA #\bigcirc (B - A) = 0 and (B - A)AA #\bigcirc = 0 by AA #\bigcirc =
= Y
\biggl[
Ir 0
0 0
\biggr]
Y \ast .
(3) \Rightarrow (2). If we let B - A = Y
\biggl[
X1 X2
X3 X4
\biggr]
Y \ast , then AA #\bigcirc (B - A) = 0 implies that X1
and X2 are zero matrices and (B - A)AA #\bigcirc = 0 implies that X1 and X3 are zero matrices. Thus
we have the form in (4.1).
(2) \Rightarrow (4). Note that (4.1) can be written as
B - A = Y
\biggl[
0 0
0 In - r
\biggr] \biggl[
0 0
0 B4
\biggr] \biggl[
0 0
0 In - r
\biggr]
Y \ast =
ISSN 1027-3190. Укр. мат. журн., 2020, т. 72, № 8
PARTIAL ORDERS BASED ON THE CS DECOMPOSITION 1125
= Y
\biggl\{ \biggl( \biggl[
Ir 0
0 In - r
\biggr]
-
\biggl[
Ir 0
0 0
\biggr] \biggr) \biggl[
0 0
0 B4
\biggr] \biggl( \biggl[
Ir 0
0 In - r
\biggr]
-
\biggl[
Ir 0
0 0
\biggr] \biggr) \biggr\}
Y \ast .
Therefore, B - A = (In - AA #\bigcirc )X(In - AA #\bigcirc ) for some matrix X \in \BbbC n\times n.
(4) \Rightarrow (3) is trivial.
Theorem 4.1 is proved.
Remark 4.1. If A \in \BbbC n\times n is core invertible, then A #\bigcirc = A \dagger \bigcirc by [14] (Theorem 3.8). Thus, the
equivalence between (1) and (2) in Theorem 4.1 also can be got by Theorem 3.2.
Theorem 4.2. Let A,B \in \BbbC n\times n. Assume that A has the form (2.1). If A is group invertible,
then we have the following two parts.
Part (I). The following statements are equivalent:
(1) A
\ast
\leq B;
(2) B - A can be written as
B - A = Y
\biggl[
0 0
- B4S
\ast C - 1 B4
\biggr]
Y \ast , B4 \in \BbbC (n - r)\times (n - r); (4.4)
(3) AA\dagger (B - A) = 0 and (B - A)AA\dagger = (B - A)[AA\dagger - (AA\#)\ast ];
(4) B = A+ (In - AA\dagger )X(In - AA\dagger )(In - A\#A)\ast for some matrix X \in \BbbC n\times n.
Part (II). The following statements are equivalent:
(1) A
\#
\leq B;
(2) B - A can be written as
B - A = Y
\biggl[
0 - C - 1SB4
0 B4
\biggr]
Y \ast , B4 \in \BbbC (n - r)\times (n - r); (4.5)
(3) AA\#(B - A) = 0 and (B - A)AA\# = 0;
(4) there exists a projection Q such that QA = 0 and B = A + (In - A\#A)QXQ for some
matrix X \in \BbbC n\times n.
Proof. Part (I).
(1) \leftrightarrow (2). Since A is group invertible, we get that C is nonsingular. Let
B - A = Y
\biggl[
B1 B2
B3 B4
\biggr]
Y \ast
and suppose A
\ast
\leq B. We marked with \bigstar , the entries that we are not interest in. Since A
\ast
\leq B if and
only if A\ast (B - A) = (B - A)A\ast = 0 and
A\ast (B - A) = Y
\biggl[
CM\ast B1 CM\ast B2
\bigstar \bigstar
\biggr]
Y \ast , (4.6)
(B - A)A\ast = Y
\biggl[
\bigstar 0
B3CM\ast +B4S
\ast M\ast 0
\biggr]
Y \ast , (4.7)
thus, from (4.6) and (4.7), we get CM\ast B1 = 0, CM\ast B2 = 0 and B3CM\ast + B4S
\ast M\ast = 0.
The nonsingularity of C and M leads to B1 and B2 are zero matrices and B3 = - B4S
\ast C - 1.
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1126 S. Z. XU, J. L. CHEN, J. BENITEZ
Conversely, by B - A = Y
\biggl[
0 0
- B4S
\ast C - 1 B4
\biggr]
Y \ast we have
(B - A)A\ast = Y
\biggl[
0 0
- B4S
\ast C - 1 B4
\biggr] \biggl[
CM\ast 0
S\ast M\ast 0
\biggr]
Y \ast = 0, (4.8)
A\ast (B - A) = Y
\biggl[
CM\ast 0
S\ast M\ast 0
\biggr] \biggl[
0 0
- B4S
\ast C - 1 B4
\biggr]
Y \ast = 0. (4.9)
From (4.8) and (4.9), we get BA\ast = AA\ast and A\ast B = A\ast A. That is A
\ast
\leq B.
(2) \Rightarrow (3). Since we have A\dagger = Y
\biggl[
CM - 1 0
S\ast M - 1 0
\biggr]
Y \ast , so, AA\dagger = Y
\biggl[
Ir 0
0 0
\biggr]
Y \ast .
Observe that
(AA\#)\ast =
\biggl(
Y
\biggl[
Ir C - 1S
0 0
\biggr]
Y \ast
\biggr) \ast
= Y
\biggl[
Ir 0
S\ast C - 1 0
\biggr]
Y \ast ,
thus,
AA\dagger - (AA\#)\ast = Y
\biggl[
Ir 0
0 0
\biggr]
Y \ast - Y
\biggl[
Ir 0
S\ast C - 1 0
\biggr]
Y \ast = Y
\biggl[
0 0
- S\ast C - 1 0
\biggr]
Y \ast .
Therefore, it is easy to check that AA\dagger (B - A) = 0 and (B - A)AA\dagger = (B - A)[AA\dagger - (AA\#)\ast ].
(3) \Rightarrow (2). Let B - A = Y
\biggl[
X1 X2
X3 X4
\biggr]
Y \ast , then AA\dagger (B - A) = 0 implies that X1 and X2
are zeros and (B - A)AA\dagger = (B - A)Y
\biggl[
0 0
- S\ast C - 1 0
\biggr]
Y \ast implies X3 = - X4S
\ast C - 1. Thus,
we have the form in (4.4).
(2) \Rightarrow (4). Note that (4.4) can be written as
B - A = Y
\biggl[
0 0
0 B4
\biggr] \biggl[
0 0
- S\ast C - 1 In - r
\biggr]
Y \ast =
= Y
\biggl\{ \biggl[
0 0
0 B4
\biggr] \biggl( \biggl[
Ir 0
0 In - r
\biggr]
-
\biggl[
Ir C - 1S
0 0
\biggr] \biggr) \ast \biggr\}
Y \ast =
= Y
\biggl[
0 0
0 B4
\biggr]
Y \ast
\Bigl(
In - A\#A
\Bigr) \ast
.
Therefore, B - A = (In - AA\dagger )X(In - AA\dagger )
\bigl(
In - A\#A
\bigr) \ast
for some matrix X \in \BbbC n\times n.
(4) \Rightarrow (1). Since B - A = (In - AA\dagger )X(In - AA\dagger )(In - A\#A)\ast for some matrix X \in \BbbC n\times n,
thus, it is not difficult to verify that A\ast (B - A) = 0 = (B - A)A\ast by A\ast (In - AA\dagger ) = 0 and
(In - AA\#)\ast A\ast = 0.
Part (II).
(1) \Rightarrow (2). Since A is group invertible, we get that C is nonsingular. If we let B - A =
= Y
\biggl[
B1 B2
B3 B4
\biggr]
Y \ast , where B1 \in \BbbC r\times r, and suppose A
\#
\leq B, then AB = A2 = BA by
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PARTIAL ORDERS BASED ON THE CS DECOMPOSITION 1127
Lemma 4.1. It is obvious that AB = A2 = BA if and only if (B - A)A = 0 = A(B - A). We
marked with \bigstar the entries that we are not interest in
(B - A)A = Y
\biggl[
B1 B2
B3 B4
\biggr] \biggl[
MC MS
0 0
\biggr]
Y \ast = Y
\biggl[
B1MC \bigstar
B3MC \bigstar
\biggr]
Y \ast , (4.10)
A(B - A) = Y
\biggl[
\bigstar MCB2 +MSB4
0 0
\biggr]
Y \ast . (4.11)
From (4.10) and (4.11), we get B3MC = 0, B1MC = 0 and 0 = MCB2 + MSB4. Thus, from
the nonsingularity of C and M, we obtain that B1 and B3 are zero matrices and B2 = - C - 1SB4.
(2) \Rightarrow (3). Since we have
A\# = Y
\biggl[
C - 1M - 1 C - 1M - 1C - 1S
0 0
\biggr]
Y \ast , so AA\# = Y
\biggl[
Ir C - 1S
0 0
\biggr]
Y \ast .
It is easy to check that AA\#(B - A) = 0 and (B - A)AA\# = 0.
(3) \Rightarrow (1). Since AA\#(B - A) = 0 and (B - A)AA\# = 0 are equivalent to AA\#B = A
and BAA\# = A, respectively, we get A\#B = A\#A and AA\# = BA\# by multiplying A\# on the
left-hand side of AA\#B = A and multiplying A\# on the right-hand side of BAA\# = A. That is
A
\#
\leq B by the definition of the sharp star partial order.
(2) \Rightarrow (4). Note that (4.5) can be written as
B - A = Y
\biggl[
0 - C - 1S
0 In - r
\biggr] \biggl[
0 0
0 B4
\biggr]
Y \ast =
= Y
\biggl\{ \biggl( \biggl[
Ir 0
0 In - r
\biggr]
-
\biggl[
Ir C - 1S
0 0
\biggr] \biggr) \biggl[
0 0
0 B4
\biggr] \biggr\}
Y \ast .
Therefore, B - A = (In - A\#A)(In - AA\dagger )X(In - AA\dagger ) for some matrix X \in \BbbC n\times n.
(4) \Rightarrow (1). Multiplying by A on the left-hand side of B - A = (In - A\#A)QXQ, we have
A2 = AB, and multiplying by A on the right-hand side of B - A = (In - A\#A)QXQ, we obtain
A2 = BA. Thus, A
\#
\leq B by Lemma 4.1.
Theorem 4.2 is proved.
Let A,B \in \BbbC n\times n and let A be a group invertible matrix. If A has the form (2.1) and let
B - A = Y
\biggl[
B1 B2
B3 B4
\biggr]
Y \ast . From part (I) in Theorem 4.2, we get A
\ast
\leq B if and only if B - A
can be written as
B - A = Y
\biggl[
0 0
- B4S
\ast C - 1 B4
\biggr]
Y \ast , B4 \in \BbbC (n - r)\times (n - r).
Similarly, from part (II) in Theorem 4.2, we have A
\#
\leq B if and only if B - A can be written as
B - A = Y
\biggl[
0 - C - 1SB4
0 B4
\biggr]
Y \ast , B4 \in \BbbC (n - r)\times (n - r).
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1128 S. Z. XU, J. L. CHEN, J. BENITEZ
Thus B4S
\ast = 0 and SB4 = 0 implies A
\#
\leq B if and only if A
\ast
\leq B. It is obvious that A\# - A #\bigcirc =
= Y
\biggl[
0 C - 1M - 1C - 1S
0 0
\biggr]
Y \ast . Therefore, A is an EP matrix if and only if S = 0, that is all
the canonical angles between \scrR (A) and \scrR (A\ast ) are 0. This result can be found in [1] (Theorem 3.7).
Thus if all the canonical angles between \scrR (A) and \scrR (A\ast ) are 0, then A
\#
\leq B if and only if A
\ast
\leq B.
But in general, the condition, B4S
\ast = 0 and SB4 = 0 is weaker than A is an EP matrix.
Theorem 4.3. Let A,B \in \BbbC n\times n. Assume that A is group invertible.
(1) If (AA\dagger - AA\#)B(In - AA\dagger ) = 0 and (AA\dagger - AA\#)B\ast (In - AA\dagger ) = 0, then A
\#
\leq B if
and only if A
\ast
\leq B.
(2) If A
\#
\leq B and A
\ast
\leq B, then (AA\dagger - AA\#)B(In - AA\dagger ) = 0 and (AA\dagger - AA\#)B\ast (In -
- AA\dagger ) = 0.
Proof. Let us write A as in Theorem 2.1 and B - A = Y
\biggl[
B1 B2
B3 B4
\biggr]
Y \ast . Since A is group
invertible, matrix C is nonsingular.
(1). The equality (AA\dagger - AA\#)B(In - AA\dagger ) = 0 can be rewritten as (AA\dagger - AA\#)(B -
- A)(In - AA\dagger ) = 0. Now,
0=(AA\dagger - AA\#)(B - A)(In - AA\dagger )=Y
\biggl[
0 - C - 1S
0 0
\biggr] \biggl[
B1 B2
B3 B4
\biggr] \biggl[
0 0
0 In - r
\biggr]
Y \ast =
= Y
\biggl[
0 - C - 1SB4
0 0
\biggr]
Y \ast
implies SB4 = 0.
Now, (In - AA\dagger )B(AA\dagger - AA\#)\ast = 0 is equivalent to (In - AA\dagger )(B - A)(AA\dagger - AA\#)\ast = 0.
Hence,
0=(In - AA\dagger )(B - A)(AA\dagger - AA\#)\ast =Y
\biggl[
0 0
0 In - r
\biggr] \biggl[
B1 B2
B3 B4
\biggr] \biggl[
0 0
- S\ast C - 1 0
\biggr]
Y \ast =
= Y
\biggl[
0 0
0 In - r
\biggr] \biggl[
- B2S
\ast C - 1 0
- B4S
\ast C - 1 0
\biggr]
Y \ast = Y
\biggl[
0 0
- B4S
\ast C - 1 0
\biggr]
Y \ast
implies B4S
\ast = 0. By Theorem 4.2, we obtain
A
\ast
\leq B \Leftarrow \Rightarrow B1 = 0, B2 = 0, B3 = B4S
\ast C - 1 \Leftarrow \Rightarrow
\Leftarrow \Rightarrow B1 = 0, B3 = 0, B2 = - C - 1SB4 \Leftarrow \Rightarrow A
\#
\leq B.
(2). By Theorem 4.2 we have B1 = 0, B2 = 0, B3 = 0, 0 = B4S
\ast , and SB4 = 0, and the
computations made in the previous item show that (AA\dagger - AA\#)B(In - AA\dagger ) = 0 and (AA\dagger -
- AA\#)B\ast (In - AA\dagger ) = 0.
Theorem 4.3 is proved.
Corollary 4.1. If A is an EP matrix, then A
\#
\leq B if and only if A
\ast
\leq B.
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PARTIAL ORDERS BASED ON THE CS DECOMPOSITION 1129
Note that Corollary 4.1 is a consequence of Theorem 4.3. There exists a another method to prove
this result as follows. In [4] (Theorem 7), Baksalary and Trenker proved that for complex matrices
A and B, if A is an EP matrix, then A
#\bigcirc
\leq B if and only if A
\ast
\leq B. In [13] (Theorem 3.3), Mailk
proved that for complex matrices A and B, if A is an EP matrix, then A
#\bigcirc
\leq B if and only if A
\#
\leq B.
Thus, if A is an EP matrix, then A
\#
\leq B if and only if A
\ast
\leq B.
Theorem 4.4. Let A,B \in \BbbC n\times n are core invertible. Then A
#\bigcirc
\leq B if and only if A \ast \leq B and
\scrR (A) \subseteq \scrN (B #\bigcirc - A #\bigcirc ).
Proof. Assume that A has the form (2.1). Since A is core invertible and a core invertible matrix
is group invertible, we get C is nonsingular. Let B #\bigcirc = Y
\biggl[
F1 F2
F3 F4
\biggr]
Y \ast . It is enough to prove
that B #\bigcirc AA #\bigcirc = A #\bigcirc if and only if \scrR (A) \subseteq \scrN (B #\bigcirc - A #\bigcirc ) by Lemma 4.2. If B #\bigcirc AA #\bigcirc = A #\bigcirc ,
then (B #\bigcirc - A #\bigcirc )(In - AA #\bigcirc ) = B #\bigcirc - A #\bigcirc , and, thus, exists X \in \BbbC n\times n such that B #\bigcirc - A #\bigcirc =
= X(In - AA #\bigcirc ), and B #\bigcirc - A #\bigcirc = X(In - AA #\bigcirc ) implies (B #\bigcirc - A #\bigcirc )\ast = (In - AA #\bigcirc )X\ast .
Hence, \scrR [(B #\bigcirc - A #\bigcirc )\ast ] \subseteq \scrR (In - AA #\bigcirc ). But \scrR [(B #\bigcirc - A #\bigcirc )\ast ] = [\scrN (B #\bigcirc - A #\bigcirc )]\bot and, by using
that AA #\bigcirc is the orthogonal projector onto \scrR (A), we have \scrR (In - AA #\bigcirc ) = \scrR (AA #\bigcirc )\bot = \scrR (A)\bot .
Therefore, [\scrN (B #\bigcirc - A #\bigcirc )]\bot \subseteq \scrR (A)\bot , hence \scrR (A) \subseteq \scrN (B #\bigcirc - A #\bigcirc ).
Conversely, if \scrR (A) \subseteq \scrN (B #\bigcirc - A #\bigcirc ), then \scrR [(B #\bigcirc - A #\bigcirc )\ast ] = [\scrN (B #\bigcirc - A #\bigcirc )]\bot \subseteq [\scrR (A)]\bot =
= \scrR (AA #\bigcirc )\bot = \scrR (In - AA #\bigcirc ), hence, B #\bigcirc - A #\bigcirc = X \prime (In - AA #\bigcirc ) for some matrix X \prime . Therefore,
B #\bigcirc AA #\bigcirc = [A #\bigcirc +X \prime (In - AA #\bigcirc )]AA #\bigcirc = A #\bigcirc AA #\bigcirc = A #\bigcirc .
Theorem 4.4 is proved.
Let A,B \in \BbbC n\times n. To study a partial order between A and B, we have two directions. One is to
use the matrix decomposition of A; another is to use the matrix decomposition of B.
Theorem 4.5. Let A,B \in \BbbC n\times n and A be group invertible. Assume that A has the form (2.1).
Then B
-
\leq A if and only if B can be written as
B = Y
\biggl[
B1 B1C
- 1S
0 0
\biggr]
Y \ast , C - 1M - 1 \in B1\{ 1\} . (4.12)
Moreover, if B
-
\leq A, then B1 is core invertible and B #\bigcirc = Y
\biggl[
B #\bigcirc
1 0
0 0
\biggr]
Y \ast .
Proof. Since A is group invertible, we have that C is nonsingular. Let B = Y
\biggl[
B1 B2
B3 B4
\biggr]
Y \ast
with B1 \in \BbbC r\times r. If B
-
\leq A, then B = AA #\bigcirc B = BA #\bigcirc A = BA #\bigcirc B by Lemma 4.3. From
AA #\bigcirc B = Y
\biggl[
B1 B2
0 0
\biggr]
Y \ast and B = AA #\bigcirc B, we get that B3 and B4 are zero matrices. From
BA #\bigcirc A = Y
\biggl[
B1 B1C
- 1S
0 0
\biggr]
Y \ast and B = BA #\bigcirc A, we get B2 = B1C
- 1S. From
BA #\bigcirc B = Y
\biggl[
B1 B2
B3 B4
\biggr] \biggl[
C - 1M - 1 0
0 0
\biggr] \biggl[
B1 B2
B3 B4
\biggr]
Y \ast =
= Y
\biggl[
B1C
- 1M - 1B1 B1C
- 1M - 1B2
0 0
\biggr]
Y \ast
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1130 S. Z. XU, J. L. CHEN, J. BENITEZ
and BA #\bigcirc B = B, we get B1C
- 1M - 1B1 = B1. Thus B has the form in (4.12).
For the opposite implication, it is easy to check that B = AA #\bigcirc B = BA #\bigcirc A = BA #\bigcirc B, which
gives B
-
\leq A by Lemma 4.3.
Since B can be written as
B = Y
\biggl[
B1 B1C
- 1S
0 0
\biggr]
Y \ast .
The group invertibility of B leads to the core invertibility of B and B1 by [10] (Theorem 1). It is
easy to verify that B #\bigcirc = Y
\biggl[
B #\bigcirc
1 0
0 0
\biggr]
Y \ast by Proposition 2.1 and Lemma 2.1.
Theorem 4.5 is proved.
Theorem 4.5 will be useful in the next theorem. In the following, we will answer the question,
when the minus partial order is the core partial order.
Theorem 4.6. Let A,B \in \BbbC n\times n be core invertible. Assume that A has the form (2.1). Then
B
#\bigcirc
\leq A if and only if B can be written as
B = Y
\biggl[
B1 B1C
- 1S
0 0
\biggr]
Y \ast , C - 1M - 1 \in B1\{ 1\} (4.13)
and B1 = MCB #\bigcirc
1 MC.
Proof. Since A is core invertible and the core invertibility is equivalent to the group in-
vertibility, we get that C is nonsingular. Let B = Y
\biggl[
B1 B2
B3 B4
\biggr]
Y \ast , where B1 \in \BbbC r\times r.
Suppose B
#\bigcirc
\leq A. Since B
#\bigcirc
\leq A implies B
-
\leq A, Theorem 4.5 and Lemma 4.4 imply B =
= Y
\biggl[
B1 B1C
- 1S
0 0
\biggr]
Y \ast , C - 1M - 1 \in B1\{ 1\} and B #\bigcirc = Y
\biggl[
B #\bigcirc
1 0
0 0
\biggr]
Y \ast . Since B
#\bigcirc
\leq A
implies A #\bigcirc BA #\bigcirc = B #\bigcirc by Lemma 4.4,
A #\bigcirc BA #\bigcirc = Y
\biggl[
C - 1M - 1 0
0 0
\biggr] \biggl[
B1 B2
B3 B4
\biggr] \biggl[
C - 1M - 1 0
0 0
\biggr]
Y \ast =
= Y
\biggl[
C - 1M - 1B1C
- 1M - 1 0
0 0
\biggr]
Y \ast . (4.14)
From (4.14) and A #\bigcirc BA #\bigcirc = B #\bigcirc (by Lemma 4.4), we get
C - 1M - 1B1C
- 1M - 1 = B #\bigcirc
1 .
That is B1 = MCB #\bigcirc
1 MC.
The opposite implication is trivial by Lemma 4.4 and Theorem 4.5.
Theorem 4.6 is proved.
5. Core invertibility under the core partial order. In [12] (Theorem 2.2), Mitra has shown
that for matrices A,B \in \BbbC n\times n, if A
\ast
\leq B, then B\dagger - A\dagger = (B - A)\dagger . A natural question is that if A
and B - A are core invertible and A
#\bigcirc
\leq B, is B core invertible? Moreover, if B is core invertible,
do we have B #\bigcirc - A #\bigcirc = (B - A) #\bigcirc ? In the following theorem, we will answer this question.
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PARTIAL ORDERS BASED ON THE CS DECOMPOSITION 1131
Theorem 5.1. Let A,B \in \BbbC n\times n. If A and B - A are both core invertible and A
#\bigcirc
\leq B, then B
is core invertible. In this case
B #\bigcirc = A #\bigcirc + (B - A) #\bigcirc - A #\bigcirc A(B - A) #\bigcirc .
Moreover, if (AA\dagger - AA\#)B(In - AA\dagger ) = 0, then
(1) (B - A) #\bigcirc = B #\bigcirc - A #\bigcirc ;
(2) (B - A)
#\bigcirc
\leq B.
Proof. Let us write A as in (2.1). From Theorem 4.1, we have
B - A = Y
\biggl[
0 0
0 B4
\biggr]
Y \ast . (5.1)
Since A and B - A are core invertible, we get that C is nonsingular and B4 is core invertible in
view of the Proposition 2.1. The equality (5.1) gives that B = Y
\biggl[
MC MS
0 B4
\biggr]
Y \ast .
Let
X = Y
\biggl[
C - 1M - 1 - C - 1SB #\bigcirc
4
0 B #\bigcirc
4
\biggr]
Y \ast ,
then we have
BX = Y
\biggl[
Ir 0
0 B4B
#\bigcirc
4
\biggr]
Y \ast is Hermitian,
XB2 = Y
\biggl[
C - 1M - 1 - C - 1SB #\bigcirc
4
0 B #\bigcirc
4
\biggr] \biggl[
(MC)2 MCMS +MSB4
0 B2
4
\biggr]
Y \ast = B,
BX2 = Y
\biggl[
Ir 0
0 B4B
#\bigcirc
4
\biggr] \biggl[
C - 1M - 1 - C - 1SB #\bigcirc
4
0 B #\bigcirc
4
\biggr]
Y \ast = X.
Thus, B #\bigcirc = X in view of Lemma 2.1.
That is, we have B #\bigcirc = Y
\biggl[
C - 1M - 1 - C - 1SB #\bigcirc
4
0 B #\bigcirc
4
\biggr]
Y \ast . The equality (5.1) gives that
(B - A) #\bigcirc = Y
\biggl[
0 0
0 B #\bigcirc
4
\biggr]
Y \ast in view of the Proposition 2.1. Thus, B #\bigcirc = A #\bigcirc + (B - A) #\bigcirc +
+ Y
\biggl[
0 - C - 1SB #\bigcirc
4
0 0
\biggr]
Y \ast .
Having in mind A #\bigcirc = Y
\biggl[
C - 1M - 1 0
0 0
\biggr]
Y \ast . Finally, since we have A #\bigcirc A = AA\# and
Y
\biggl[
0 C - 1SB #\bigcirc
4
0 0
\biggr]
Y \ast = Y
\biggl[
Ir C - 1S
0 0
\biggr] \biggl[
0 0
0 B #\bigcirc
4
\biggr]
Y \ast = AA\#(B - A) #\bigcirc .
Thus, B #\bigcirc = A #\bigcirc + (B - A) #\bigcirc - A #\bigcirc A(B - A) #\bigcirc .
Theorem 5.1 is proved.
Corollary 5.1. Let A,B \in \BbbC n\times n. If A and B - A are both core invertible, A
#\bigcirc
\leq B, then
A2 = AB if and only if B #\bigcirc - A #\bigcirc = (B - A) #\bigcirc .
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1132 S. Z. XU, J. L. CHEN, J. BENITEZ
Proof. Assume that A has the form (2.1). Since B #\bigcirc - A #\bigcirc = (B - A) #\bigcirc if and only if
A #\bigcirc A(B - A) #\bigcirc = 0 by Theorem 5.1 and A #\bigcirc A(B - A) #\bigcirc = Y
\biggl[
0 C - 1SB #\bigcirc
4
0 0
\biggr]
Y \ast , thus it is
sufficient to prove that A2 = AB if and only if Y
\biggl[
0 C - 1SB #\bigcirc
4
0 0
\biggr]
Y \ast = 0. It is equivalent to
show that AA\#(B - A) #\bigcirc = 0 if and only if A2 = AB. Since we get B - A = (B - A) #\bigcirc (B - A)2
and (B - A) #\bigcirc = (B - A)((B - A) #\bigcirc )2, thus,
AA\#(B - A) #\bigcirc = 0 \leftrightarrow A(B - A) #\bigcirc = 0 \leftrightarrow A(B - A) = 0.
Let us write A as in Theorem 2.1 and B - A = Y
\biggl[
B1 B2
B3 B4
\biggr]
Y \ast . We have that B1,
B2 and B3 are zero matrices by Theorem 4.1. From the proof of Theorem 4.3, the condition
(AA\dagger - AA\#)B(In - AA\dagger ) = 0 implies SB4 = 0.
The part (1) is obvious by
A(B - A) = Y
\biggl[
MC MS
0 0
\biggr] \biggl[
0 0
0 B4
\biggr]
Y \ast = 0.
To prove the part (2). It is sufficient to show that (B - A)\ast (B - A) = (B - A)\ast B and
(B - A)2 = B(B - A) by [16] (Theorem 2.4):
(B - A)\ast (B - A) = Y
\biggl[
0 0
0 B\ast
4
\biggr] \biggl[
0 0
0 B4
\biggr]
Y \ast = Y
\biggl[
0 0
0 B\ast
4B4
\biggr]
Y \ast , (5.2)
(B - A)\ast B = Y
\biggl[
0 0
0 B\ast
4
\biggr] \biggl[
MC MS
0 B4
\biggr]
Y \ast = Y
\biggl[
0 0
0 B\ast
4B4
\biggr]
Y \ast , (5.3)
(B - A)2 = Y
\biggl[
0 0
0 B4
\biggr] \biggl[
0 0
0 B4
\biggr]
Y \ast = Y
\biggl[
0 0
0 B2
4
\biggr]
Y \ast , (5.4)
B(B - A) = Y
\biggl[
MC MS
0 B4
\biggr] \biggl[
0 0
0 B4
\biggr]
Y \ast = Y
\biggl[
0 0
0 B2
4
\biggr]
Y \ast . (5.5)
From (5.2), (5.3), (5.4) and (5.5) we get (B - A)
#\bigcirc
\leq B.
Corollary 5.1 is proved.
In [1] (Theorem 3.7), the author proved that if A is an EP matrix, then S = 0.
Corollary 5.2. Let A,B \in \BbbC n\times n. If A and B - A are core invertible, A
#\bigcirc
\leq B and A is an EP
matrix, then B #\bigcirc - A #\bigcirc = (B - A) #\bigcirc .
References
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Received 12.06.17,
after revision — 07.06.18
ISSN 1027-3190. Укр. мат. журн., 2020, т. 72, № 8
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| id | umjimathkievua-article-6025 |
| institution | Ukrains’kyi Matematychnyi Zhurnal |
| keywords_txt_mv | keywords |
| language | English |
| last_indexed | 2026-03-24T03:25:26Z |
| publishDate | 2020 |
| publisher | Institute of Mathematics, NAS of Ukraine |
| record_format | ojs |
| resource_txt_mv | umjimathkievua/21/da1e86e106c2adddeaab0dd31a4d3f21.pdf |
| spelling | umjimathkievua-article-60252022-03-26T11:02:02Z Partial orders based on the CS decomposition Partial orders based on the CS decomposition Xu, S. Z. Chen, J. L. Benítez, J. Xu, S. Z. Chen, J. L. Benítez, J. UDC 512.5 A new decomposition for square matrices is given by using two known matrix decompositions,&nbsp;a new characterization of the core-EP order is obtained by using this new matrix decomposition.&nbsp;Also, we will use a matrix decomposition to investigate the minus, star, sharp and core partial orders in the setting of complex matrices. УДК 512.5 Частковi впорядкування на основi CS-розкладу За допомогою двох відомих розкладів&nbsp; матриць отримано новий розклад для квадратних матриць, а за допомогою цього нового розкладу отримано нову характеристику core-EP-впорядкування.&nbsp;Також &nbsp; використано матричний розклад для вивчення часткового порядку типу “minus”, “star”, “sharp” та “core” для комплексних матриць.&nbsp; Institute of Mathematics, NAS of Ukraine 2020-08-18 Article Article application/pdf https://umj.imath.kiev.ua/index.php/umj/article/view/6025 10.37863/umzh.v72i8.6025 Ukrains’kyi Matematychnyi Zhurnal; Vol. 72 No. 8 (2020); 1119-1133 Український математичний журнал; Том 72 № 8 (2020); 1119-1133 1027-3190 en https://umj.imath.kiev.ua/index.php/umj/article/view/6025/8743 |
| spellingShingle | Xu, S. Z. Chen, J. L. Benítez, J. Xu, S. Z. Chen, J. L. Benítez, J. Partial orders based on the CS decomposition |
| title | Partial orders based on the CS decomposition |
| title_alt | Partial orders based on the CS decomposition |
| title_full | Partial orders based on the CS decomposition |
| title_fullStr | Partial orders based on the CS decomposition |
| title_full_unstemmed | Partial orders based on the CS decomposition |
| title_short | Partial orders based on the CS decomposition |
| title_sort | partial orders based on the cs decomposition |
| url | https://umj.imath.kiev.ua/index.php/umj/article/view/6025 |
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