Solutions of Sylvester equation in $C^*$-modular operators
UDC 517.9 We study the solvability of the Sylvester equation $AX + Y B = C$ and the operator equation $AXD + FY B = C$ in the general setting of the adjointable operators between Hilbert $C^*$ -modules. Based on the Moore – Penrose inverses of the associated operators, we propose necessary and suffi...
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Institute of Mathematics, NAS of Ukraine
2021
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Ukrains’kyi Matematychnyi Zhurnal| _version_ | 1860506981356273664 |
|---|---|
| author | Moghani, Z. Niazi Mohammadzadeh Karizaki, M. Khanehgir, M. Moghani, Z. Niazi Mohammadzadeh Karizaki, M. Khanehgir, M. Moghani, Z. Niazi Mohammadzadeh Karizaki, M. Khanehgir, M. |
| author_facet | Moghani, Z. Niazi Mohammadzadeh Karizaki, M. Khanehgir, M. Moghani, Z. Niazi Mohammadzadeh Karizaki, M. Khanehgir, M. Moghani, Z. Niazi Mohammadzadeh Karizaki, M. Khanehgir, M. |
| author_sort | Moghani, Z. Niazi |
| baseUrl_str | https://umj.imath.kiev.ua/index.php/umj/oai |
| collection | OJS |
| datestamp_date | 2025-03-31T08:48:21Z |
| description | UDC 517.9
We study the solvability of the Sylvester equation $AX + Y B = C$ and the operator equation $AXD + FY B = C$ in the general setting of the adjointable operators between Hilbert $C^*$ -modules. Based on the Moore – Penrose inverses of the associated operators, we propose necessary and sufficient conditions for the existence of solutions to these equations, and obtain the general expressions of the solutions in the solvable cases. We also provide an approach to the study of the positive solutions for a special case of Lyapunov equation. |
| doi_str_mv | 10.37863/umzh.v73i3.152 |
| first_indexed | 2026-03-24T02:02:03Z |
| format | Article |
| fulltext |
DOI: 10.37863/umzh.v73i3.152
UDC 517.9
Z. Niazi Moghani (Dep. Math., Mashhad Branch, Islamic Azad. Univ., Iran),
M. Mohammadzadeh Karizaki (Dep. Comput. Eng., Univ. Torbat Heydarieh, Iran),
M. Khanehgir (Dep. Math., Mashhad Branch, Islamic Azad. Univ., Iran)
SOLUTIONS OF SYLVESTER EQUATION IN \bfitC \ast -MODULAR OPERATORS
РОЗВ’ЯЗКИ РIВНЯННЯ СIЛЬВЕСТРА ДЛЯ \bfitC \ast -МОДУЛЬНИХ ОПЕРАТОРIВ
We study the solvability of the Sylvester equation AX + Y B = C and the operator equation AXD + FY B = C in
the general setting of the adjointable operators between Hilbert C\ast -modules. Based on the Moore – Penrose inverses of
the associated operators, we propose necessary and sufficient conditions for the existence of solutions to these equations,
and obtain the general expressions of the solutions in the solvable cases. We also provide an approach to the study of the
positive solutions for a special case of Lyapunov equation.
Розглянуто розв’язнiсть рiвняння Сiльвестра AX + Y B = C та операторного рiвняння AXD + FY B = C при
загальних умовах сумiжностi операторiв мiж гiльбертовими C\ast -модулями. На основi обернених Мура – Пенроуза
для пов’язаних операторiв отримано необхiднi та достатнi умови iснування розв’язкiв цих рiвнянь, а також загальнi
вирази для розв’язкiв у випадку, коли вони iснують. Крiм того, запропоновано пiдхiд до вивчення додатних розв’язкiв
у спецiальному випадку рiвняння Ляпунова.
1. Introduction. In the year 2001, the Sylvester equation CX - XAT = B was studied for
matrices by [12]. Thereafter, more general equation AX - XF = BY was considered in [24]. The
generalized Sylvester equation AV +BW = EV J+R with unknown matrices V and W, has many
applications in linear systems theory [7, 19]. One special and important case is the Lyapunov matrix
equation AX +XTC = B, which has important applications in the control theory and robust fault
detection [8]. In 2007, Piao et al. [18] studied this equation when A and C are square matrices
with different dimensions. Mor et al. [16] obtained the explicit solution and eigenvalue bounds
for the Lyapunov matrix equation ATP + PA = - BBT and determined the number of positive
eigenvalues of the positive semidefinite solution through the controllability matrix. The other special
case of interest is the equation A\ast X + X\ast A = B. It was studied for matrices by Braden [3], and
for the Hilbert space operators by Djordjević [6]. Fang and Yu [9] investigated the solvability of
the operator equations A\ast X +X\ast A = C and A\ast XB + B\ast X\ast A = C for adjointable operators on
Hilbert C\ast -modules whose ranges may not be closed. On the other hand, Mousavi et al. [17] studied
the operator equations AX + Y B = C and AXA\ast +BY B\ast = C in Hilbert C\ast -modules.
In this paper, by using the matrix forms of adjointable operators between Hilbert C\ast -modules,
we first propose necessary and sufficient conditions for the existence of solutions to the Sylvester
equation
AX + Y B = C, (1.1)
and then with the help of that we describe necessary and sufficient conditions for the existence of
solution to the operator equation
AXD + FY B = C, (1.2)
c\bigcirc Z. NIAZI MOGHANI, M. MOHAMMADZADEH KARIZAKI, M. KHANEHGIR, 2021
354 ISSN 1027-3190. Укр. мат. журн., 2021, т. 73, № 3
SOLUTIONS OF SYLVESTER EQUATION IN C\ast -MODULAR OPERATORS 355
when A, B, D and F are bounded adjointable operators with closed range between Hilbert C\ast -
modules. Moreover, we obtain the explicit solutions to these operator equations in the solvable cases.
The results obtained in this paper generalize earlier results due to Djordjević [6].
It is a very active topic to study positive solutions to matrix equations or positive solutions to
operator equations. The positive solutions to the operator equations AX = C and XB = D were
studied by Dajić et al. [5] for Hilbert space operators. In 2008, Xu [20] considered the Hermitian
and positive solutions to these equations in Hilbert C\ast -modules setting. Also, Hermitian positive
semidefinite solution to the matrix equation AXB = C was studied by Khatri and Mitra [11] and
then by Zhang [23] in 2004. Cvetković – Ilić and Koliha [4] described the positive solution to the
special case AXA\ast = B for elements of C\ast -algebras. The other purpose of this work, is to provide
an approach to the study of the positive solutions to the operator equation
AX +X\ast A\ast = B, (1.3)
in the framework of Hilbert C\ast -modules. The paper is organized as follows. In Section 2, we recall
some knowledge about the Hilbert C\ast -modules. For this purpose, we use [13 – 15, 21]. In Section
3, we study the general solutions to the operator equation (1.1). Based on the Moore – Penrose
inverses of the associated operators, in Section 4 we give the necessary and sufficient conditions
for the existence of a solution to the operator equation (1.2), and provide a formula for the general
solution to this operator equation. Finally, in Section 5, we find the positive solutions of Eq. (1.3) for
adjointable operators over Hilbert C\ast -modules, where X is the unknown operator.
2. Preliminaries. Throughout this paper, \scrA denotes a C\ast -algebra. An inner-product \scrA -
module is a linear space \scrX which is a right \scrA -module, together with a map (x, y) \mapsto \rightarrow
\bigl\langle
x, y
\bigr\rangle
:
\scrX \times \scrX \rightarrow \scrA such that for any x, y, z \in \scrX , \alpha , \beta \in \BbbC and a \in \scrA , the following conditions hold:
(i) \langle x, \alpha y + \beta z\rangle = \alpha \langle x, y\rangle + \beta \langle x, z\rangle ;
(ii) \langle x, ya\rangle = \langle x, y\rangle a;
(iii) \langle y, x\rangle = \langle x, y\rangle \ast ;
(iv) \langle x, x\rangle \geq 0 and \langle x, x\rangle = 0 \Leftarrow \Rightarrow x = 0.
An inner-product \scrA -module \scrX which is complete with respect to the induced norm \| x\| =
\sqrt{}
\| \langle x, x\rangle \|
for any x \in \scrX is called a (right) Hilbert \scrA -module. A closed submodule \scrM of a Hilbert \scrA -module
\scrX is said to be orthogonally complemented if \scrX = \scrM \oplus \scrM \bot , where
\scrM \bot = \{ x \in \scrX : \langle x, y\rangle = 0 for any y \in \scrM \} .
Now, suppose that \scrX and \scrY are two Hilbert \scrA -modules, let \scrL (\scrX ,\scrY ) be the set of operators T :
\scrX \rightarrow \scrY for which there is an operator T \ast : \scrY \rightarrow \scrX such that
\langle Tx, y\rangle = \langle x, T \ast y\rangle for any x \in \scrX and y \in \scrY .
It is known that any element T \in \scrL (\scrX ,\scrY ) must be a bounded linear operator, which is also \scrA -linear
in the sense that T (xa) = (Tx)a for x \in \scrX and a \in \scrA . We call \scrL (\scrX ,\scrY ), the set of adjointable
operators from \scrX to \scrY . For any T \in \scrL (\scrX ,\scrY ), the null and the range space of T are denoted by
\mathrm{k}\mathrm{e}\mathrm{r}(T ) and \mathrm{r}\mathrm{a}\mathrm{n}(T ), respectively. In the case \scrX = \scrY ,\scrL (\scrX ,\scrX ) which is abbreviated to \scrL (\scrX ), is
a C\ast -algebra. Let \scrL (\scrX )sa be the set of self-adjoint elements and \scrL (\scrX )+ be the set of positive
elements in \scrL (\scrX ), respectively. The identity operator on \scrX is denoted by 1\scrX or 1 if there is no
ambiguity.
ISSN 1027-3190. Укр. мат. журн., 2021, т. 73, № 3
356 Z. NIAZI MOGHANI, M. MOHAMMADZADEH KARIZAKI, M. KHANEHGIR
Lemma 2.1 [13]. Let \scrX be a Hilbert \scrA -module and T \in \scrL (\scrX ). Then T \in \scrL (\scrX )+ if and only
if \langle Tx, x\rangle \geq 0 for all x in \scrX .
Theorem 2.1 [13]. Suppose that \scrX and \scrY are Hilbert \scrA -modules and T \in \scrL (\scrX ,\scrY ) has closed
range. Then:
(i) \mathrm{k}\mathrm{e}\mathrm{r} (T ) is orthogonally complemented in \scrX , with complement \mathrm{r}\mathrm{a}\mathrm{n} (T \ast );
(ii) \mathrm{r}\mathrm{a}\mathrm{n} (T ) is orthogonally complemented in \scrY , with complement \mathrm{k}\mathrm{e}\mathrm{r} (T \ast );
(iii) the map T \ast \in \scrL (\scrY ,\scrX ) has closed range.
Boichuk and Samoilenko suggested some methods for the construction of the Moore – Penrose
pseudoinverse for the original linear Fredholm operators in Banach and Hilbert spaces [2]. Boichuk
and Pokutnyi studied perturbation theory of operator equations in the Fréchet and Hilbert spaces
by using the notion of strong generalized inverse operators in [1]. Xu and Sheng [21] showed that
an adjointable operator between two Hilbert \scrA -modules admits a bounded Moore – Penrose inverse
if and only if it has closed range. The Moore – Penrose inverse T \dagger of T is the unique element in
\scrL (\scrY ,\scrX ) which satisfies the following conditions:
TT \dagger T = T, T \dagger TT \dagger = T \dagger , (TT \dagger )\ast = TT \dagger , (T \dagger T )\ast = T \dagger T.
From these conditions we obtain that (T \dagger )\ast = (T \ast )\dagger , TT \dagger and T \dagger T are orthogonal projections, in
the sense that they are self-adjoint idempotent operators. Furthermore, we have
\mathrm{r}\mathrm{a}\mathrm{n}(T ) = \mathrm{r}\mathrm{a}\mathrm{n}(TT \dagger ), \mathrm{r}\mathrm{a}\mathrm{n}(T \dagger ) = \mathrm{r}\mathrm{a}\mathrm{n}(T \dagger T ) = \mathrm{r}\mathrm{a}\mathrm{n}(T \ast ),
\mathrm{k}\mathrm{e}\mathrm{r}(T ) = \mathrm{k}\mathrm{e}\mathrm{r}(T \dagger T ), \mathrm{k}\mathrm{e}\mathrm{r}(T \dagger ) = \mathrm{k}\mathrm{e}\mathrm{r}(TT \dagger ) = \mathrm{k}\mathrm{e}\mathrm{r}(T \ast ).
It is well-known, that T \in \scrL (\scrX ,\scrY ) is regular if there exists S \in \scrL (\scrY ,\scrX ) such that TST = T. Note
that if T is regular, then T \dagger exists (see [10], Theorem 6). If in addition, \scrX = \scrY and T \geq 0, then
TT \dagger = T \dagger T and T \dagger \geq 0.
Remark 2.1. Let \scrX and \scrY be Hilbert \scrA -modules, we use the notation \scrX \oplus \scrY to denote the direct
sum of \scrX and \scrY , which is also a Hilbert \scrA -module whose \scrA -valued inner product is given by\Biggl\langle \Biggl(
x1
y1
\Biggr)
,
\Biggl(
x2
y2
\Biggr) \Biggr\rangle
= \langle x1, x2\rangle + \langle y1, y2\rangle
for xi \in \scrX and yi \in \scrY , i = 1, 2. To simplify the notation, we use x\oplus y to denote
\biggl(
x
y
\biggr)
\in \scrX \oplus \scrY .
A matrix form of a bounded adjointable operator T \in \scrL (\scrX ,\scrY ) can be induced by some natural
decompositions of Hilbert C\ast -modules. Indeed, if \scrM and \scrN are closed orthogonally complemented
submodules of \scrX and \scrY , respectively, and \scrX = \scrM \oplus \scrM \bot , \scrY = \scrN \oplus \scrN \bot , then T can be written
as the following 2\times 2 matrix
T =
\Biggl[
T1 T2
T3 T4
\Biggr]
,
where T1 = P\scrN TP\scrM \in \scrL (\scrM ,\scrN ), T2 = P\scrN T (1 - P\scrM ) \in \scrL (\scrM \bot ,\scrN ), T3 = (1 - P\scrN )TP\scrM \in
\in \scrL (\scrM ,\scrN \bot ), T4 = (1 - P\scrN )T (1 - P\scrM ) \in \scrL (\scrM \bot ,\scrN \bot ) and P\scrM and P\scrN denote the projections
corresponding to \scrM and \scrN , respectively.
The proof of the following lemma can be found in [14].
ISSN 1027-3190. Укр. мат. журн., 2021, т. 73, № 3
SOLUTIONS OF SYLVESTER EQUATION IN C\ast -MODULAR OPERATORS 357
Lemma 2.2 [14]. Suppose that \scrX and \scrY are Hilbert \scrA -modules and T \in \scrL (\scrX ,\scrY ) has closed
range. Then T has the following matrix decomposition with respect to the orthogonal decompositions
of closed submodules \scrX = \mathrm{r}\mathrm{a}\mathrm{n}(T \ast )\oplus \mathrm{k}\mathrm{e}\mathrm{r}(T ) and \scrY = \mathrm{r}\mathrm{a}\mathrm{n}(T )\oplus \mathrm{k}\mathrm{e}\mathrm{r}(T \ast ):
T =
\Biggl[
T1 0
0 0
\Biggr]
:
\Biggl[
\mathrm{r}\mathrm{a}\mathrm{n}(T \ast )
\mathrm{k}\mathrm{e}\mathrm{r}(T )
\Biggr]
\rightarrow
\Biggl[
\mathrm{r}\mathrm{a}\mathrm{n}(T )
\mathrm{k}\mathrm{e}\mathrm{r}(T \ast )
\Biggr]
,
where T1 is invertible. Moreover,
T \dagger =
\Biggl[
T - 1
1 0
0 0
\Biggr]
:
\Biggl[
\mathrm{r}\mathrm{a}\mathrm{n}(T )
\mathrm{k}\mathrm{e}\mathrm{r}(T \ast )
\Biggr]
\rightarrow
\Biggl[
\mathrm{r}\mathrm{a}\mathrm{n}(T \ast )
\mathrm{k}\mathrm{e}\mathrm{r}(T )
\Biggr]
.
Remark 2.2 [22]. Suppose that \scrX is a Hilbert \scrA -module and P \in \scrL (\scrX ) is a projection. Let
\scrX 1 = P\scrX and \scrX 2 = (1 - P )\scrX and let T : \scrX 1\oplus \scrX 2 \rightarrow \scrX , T (\xi \oplus \eta ) = \xi +\eta for \xi \in \scrX 1 and \eta \in \scrX 2.
Then T is invertible with T - 1(x) = P (x) \oplus (1 - P )(x) for x \in \scrX , and it is easy to verify that
T \ast = T - 1 so that T is a unitary. Furthermore, for any self-adjoint element T of \scrL (\scrX ), T can be
decomposed as
T = T11 + T12 + T \ast
12 + T22,
where T11 = PTP, T12 = PT (1 - P ) and T22 = (1 - P )T (1 - P ). By Lemma 2.1 we know that
T \geq 0 \Leftarrow \Rightarrow \langle (T11 + T12 + T \ast
12 + T22)(\xi + \eta ), \xi + \eta \rangle \geq 0 \forall \xi \in \scrX 1, \eta \in \scrX 2
\Leftarrow \Rightarrow \langle T11\xi , \xi \rangle + \langle T12\eta , \xi \rangle + \langle T \ast
12\xi , \eta \rangle + \langle T22\eta , \eta \rangle \geq 0 \forall \xi \in \scrX 1, \eta \in \scrX 2
\Leftarrow \Rightarrow
\Biggl\langle \Biggl(
T11 T12
T \ast
12 T22
\Biggr)
(\xi \oplus \eta ), (\xi \oplus \eta )
\Biggr\rangle
\geq 0 \forall \xi \in \scrX 1, \eta \in \scrX 2
\Leftarrow \Rightarrow
\Biggl(
T11 T12
T \ast
12 T22
\Biggr)
\geq 0.
Lemma 2.3 [21]. Let \scrX 1 and \scrX 2 be Hilbert \scrA -modules and T =
\biggl(
T11 T12
T \ast
12 T22
\biggr)
be a self-
adjoint element of \scrL (\scrX 1 \oplus \scrX 2) with Tij \in \scrL (\scrX j \oplus \scrX i), i, j = 1, 2. Suppose that \mathrm{r}\mathrm{a}\mathrm{n}(T11) is closed.
Then T \geq 0 if and only if the following three conditions are satisfied:
T11 \geq 0, T12 = T11T
\dagger
11T12, T22 - T \ast
12T
\dagger
11T12 \geq 0.
Lemma 2.4 [15]. Let \scrX ,\scrY ,\scrZ be Hilbert \scrA -modules. Also, let T \in \scrL (\scrY ,\scrZ ) and S \in \scrL (\scrX ,\scrY )
have closed ranges, and A \in \scrL (\scrX ,\scrZ ). Then the equation
TXS = A, X \in \scrL (\scrY ), (2.1)
has a solution if and only if TT \dagger AS\dagger S = A. In this case, any solution of Eq. (2.1) has the form
X = T \dagger AS\dagger .
ISSN 1027-3190. Укр. мат. журн., 2021, т. 73, № 3
358 Z. NIAZI MOGHANI, M. MOHAMMADZADEH KARIZAKI, M. KHANEHGIR
Theorem 2.2 [15]. Let \scrX ,\scrY ,\scrZ be Hilbert \scrA -modules, S \in \scrL (\scrX ,\scrY ) and T \in \scrL (\scrZ ,\scrY ) be
invertible operators and A \in \scrL (\scrY ). Then the following statements are equivalent:
(a) there exists a solution X \in \scrL (\scrX ,\scrZ ) to the operator equation TXS\ast + SX\ast T \ast = A;
(b) A = A\ast .
If (a) or (b) is satisfied, then any solution to
TXS\ast + SX\ast T \ast = A, X \in \scrL (\scrX ,\scrZ ),
has the form
X =
1
2
T - 1A(S\ast ) - 1 - T - 1Z(S\ast ) - 1,
where Z \in \scrL (\scrY ) satisfies Z\ast = - Z.
3. Operator equation \bfitA \bfitX + \bfitY \bfitB = \bfitC . In this section, we study the solvability of operator
equation (1.1) in the general context of the Hilbert C\ast -modules and present a general solution of it.
Theorem 3.1. Let \scrX ,\scrY be Hilbert \scrA -modules,A \in \scrL (\scrX ,\scrY ) and B \in \scrL (\scrY ,\scrX ) be invertible
and C \in \scrL (\scrY ). Then the Eq. (1.1) has a solution (X,Y ) \in \scrL (\scrY ,\scrX ) \times \scrL (\scrX ,\scrY ). In the case any
solution of the Eq. (1.1) is represented by
X =
1
2
A - 1C +
1
2
WB, (3.1)
Y =
1
2
CB - 1 - 1
2
AW, (3.2)
where W \in \scrL (\scrX ) is arbitrary.
Proof. It is easy to see that operators X and Y of the forms (3.1) and (3.2) are a solution of
Eq. (1.1). On the other hand, let X be any solution of Eq. (1.1) then X = A - 1C - A - 1Y B and
A - 1Y = A - 1CB - 1 - XB - 1. We have
X =
1
2
A - 1C +
\biggl(
1
2
A - 1CB - 1 - A - 1Y
\biggr)
B =
=
1
2
A - 1C +
\biggl(
1
2
[A - 1Y +XB - 1] - A - 1Y
\biggr)
B =
=
1
2
A - 1C +
\biggl(
1
2
(XB - 1 - A - 1Y )
\biggr)
B =
1
2
A - 1C +
1
2
WB,
where W = XB - 1 - A - 1Y. Also, Y = CB - 1 - AXB - 1 and XB - 1 = A - 1CB - 1 - A - 1Y and
so we can write
Y =
1
2
CB - 1 +A
\biggl(
1
2
A - 1CB - 1 - XB - 1
\biggr)
=
=
1
2
CB - 1 +A
\biggl(
1
2
[XB - 1 +A - 1Y ] - XB - 1
\biggr)
=
=
1
2
CB - 1 +A
\biggl(
1
2
(A - 1Y - XB - 1)
\biggr)
=
1
2
CB - 1 - 1
2
AW.
Theorem 3.1 is proved.
Now, we solve Eq. (1.1) in the case when A and B have closed ranges.
ISSN 1027-3190. Укр. мат. журн., 2021, т. 73, № 3
SOLUTIONS OF SYLVESTER EQUATION IN C\ast -MODULAR OPERATORS 359
Theorem 3.2. Let \scrX and \scrY be Hilbert \scrA -modules and A \in \scrL (\scrX ,\scrY ) and B \in \scrL (\scrY ,\scrX ) have
closed ranges and \mathrm{r}\mathrm{a}\mathrm{n}(A) = \mathrm{r}\mathrm{a}\mathrm{n}(B\ast ) and \mathrm{r}\mathrm{a}\mathrm{n}(A\ast ) = \mathrm{r}\mathrm{a}\mathrm{n}(B) and C \in \scrL (\scrY ). Then the following
statements are equivalent:
(a) there exists a solution (X,Y ) \in \scrL (\scrY ,\scrX )\times \scrL (\scrX ,\scrY ) of Eq. (1.1);
(b) (1 - AA\dagger )C(1 - B\dagger B) = 0.
If (a) or (b) is satisfied, then any solution (X,Y ) of equation (1.1) has the form
X =
1
2
A\dagger C +
1
2
A\dagger C(1 - B\dagger B) +
1
2
WB + (1 - A\dagger A)Z,
Y =
1
2
AA\dagger CB\dagger + (1 - AA\dagger )CB\dagger - 1
2
AWBB\dagger + V (1 - BB\dagger ),
where Z \in \scrL (\scrY ,\scrX ), V \in \scrL (\scrX ,\scrY ) are arbitrary operators and W \in \scrL (\scrX ) so that (1 -
- AA\dagger )WBB\dagger = 0.
Proof. (a) \Rightarrow (b). Suppose that (X,Y ) is a solution of Eq. (1.1). Thus we have
(1 - AA\dagger )C(1 - B\dagger B) = (1 - AA\dagger )(AX + Y B)(1 - B\dagger B) =
= (1 - AA\dagger )AX(1 - B\dagger B) + (1 - AA\dagger )Y B(1 - B\dagger B) = 0.
(b) \Rightarrow (a). Suppose (1 - AA\dagger )C(1 - B\dagger B) = 0. Since \mathrm{r}\mathrm{a}\mathrm{n}(\mathrm{A}) and \mathrm{r}\mathrm{a}\mathrm{n}(B) are closed, then ap-
plying Theorem 2.1 we get \scrX = \mathrm{r}\mathrm{a}\mathrm{n}(A\ast )\oplus \mathrm{k}\mathrm{e}\mathrm{r}(A) and \scrY = \mathrm{r}\mathrm{a}\mathrm{n}(A)\oplus \mathrm{k}\mathrm{e}\mathrm{r}(A\ast ). By our assumptions
and using Lemma 2.2, operators A and B have the matrix forms
A =
\Biggl[
A1 0
0 0
\Biggr]
:
\Biggl[
\mathrm{r}\mathrm{a}\mathrm{n}(A\ast )
\mathrm{k}\mathrm{e}\mathrm{r}(A)
\Biggr]
\rightarrow
\Biggl[
\mathrm{r}\mathrm{a}\mathrm{n}(A)
\mathrm{k}\mathrm{e}\mathrm{r}(A\ast )
\Biggr]
and
B =
\Biggl[
B1 0
0 0
\Biggr]
:
\Biggl[
\mathrm{r}\mathrm{a}\mathrm{n}(A)
\mathrm{k}\mathrm{e}\mathrm{r}(A\ast )
\Biggr]
\rightarrow
\Biggl[
\mathrm{r}\mathrm{a}\mathrm{n}(A\ast )
\mathrm{k}\mathrm{e}\mathrm{r}(A)
\Biggr]
,
where A1 and B1 are invertible. Now, condition (b) implies that C has the form
C =
\Biggl[
C1 C2
C3 0
\Biggr]
:
\Biggl[
\mathrm{r}\mathrm{a}\mathrm{n}(A)
\mathrm{k}\mathrm{e}\mathrm{r}(A\ast )
\Biggr]
\rightarrow
\Biggl[
\mathrm{r}\mathrm{a}\mathrm{n}(A)
\mathrm{k}\mathrm{e}\mathrm{r}(A\ast )
\Biggr]
.
Let us assume that the operators X and Y have the following matrix forms:
X =
\Biggl[
X1 X2
X3 X4
\Biggr]
:
\Biggl[
\mathrm{r}\mathrm{a}\mathrm{n}(A)
\mathrm{k}\mathrm{e}\mathrm{r}(A\ast )
\Biggr]
\rightarrow
\Biggl[
\mathrm{r}\mathrm{a}\mathrm{n}(A\ast )
\mathrm{k}\mathrm{e}\mathrm{r}(A)
\Biggr]
and
Y =
\Biggl[
Y1 Y2
Y3 Y4
\Biggr]
:
\Biggl[
\mathrm{r}\mathrm{a}\mathrm{n}(A\ast )
\mathrm{k}\mathrm{e}\mathrm{r}(A)
\Biggr]
\rightarrow
\Biggl[
\mathrm{r}\mathrm{a}\mathrm{n}(A)
\mathrm{k}\mathrm{e}\mathrm{r}(A\ast )
\Biggr]
.
Then from AX + Y B = C it follows that
ISSN 1027-3190. Укр. мат. журн., 2021, т. 73, № 3
360 Z. NIAZI MOGHANI, M. MOHAMMADZADEH KARIZAKI, M. KHANEHGIR\Biggl[
A1 0
0 0
\Biggr] \Biggl[
X1 X2
X3 X4
\Biggr]
+
\Biggl[
Y1 Y2
Y3 Y4
\Biggr] \Biggl[
B1 0
0 0
\Biggr]
=
=
\Biggl[
A1X1 + Y1B1 A1X2
Y3B1 0
\Biggr]
=
\Biggl[
C1 C2
C3 0
\Biggr]
.
That is,
A1X1 + Y1B1 = C1, (3.3)
A1X2 = C2, (3.4)
Y3B1 = C3. (3.5)
According Theorem 3.1, Eq. (3.3) has a solution (X1, Y1) of the form X1 =
1
2
A - 1
1 C1 +
1
2
W1B1,
Y1 =
1
2
C1B
- 1
1 - 1
2
A1W1, where W1 \in \scrL (\mathrm{r}\mathrm{a}\mathrm{n}(\mathrm{A}\ast )) is arbitrary. In view of (3.4) and (3.5) we
deduce X2 = A - 1
1 C2 and Y3 = C3B
- 1
1 . Hence,
X =
\left[ 12A - 1
1 C1 +
1
2
W1B1 A - 1
1 C2
X3 X4
\right]
and
Y =
\left[ 12C1B
- 1
1 - 1
2
A1W1 Y2
C3B
- 1
1 Y4
\right] ,
where X3, X4, Y2 and Y4 can be taken arbitrary. Let
Z =
\Biggl[
Z1 Z2
X3 X4
\Biggr]
:
\Biggl[
\mathrm{r}\mathrm{a}\mathrm{n}(A)
\mathrm{k}\mathrm{e}\mathrm{r}(A\ast )
\Biggr]
\rightarrow
\Biggl[
\mathrm{r}\mathrm{a}\mathrm{n}(A\ast )
\mathrm{k}\mathrm{e}\mathrm{r}(A)
\Biggr]
and
V =
\Biggl[
V1 Y2
V3 Y4
\Biggr]
:
\Biggl[
\mathrm{r}\mathrm{a}\mathrm{n}(A\ast )
\mathrm{k}\mathrm{e}\mathrm{r}(A)
\Biggr]
\rightarrow
\Biggl[
\mathrm{r}\mathrm{a}\mathrm{n}(A)
\mathrm{k}\mathrm{e}\mathrm{r}(A\ast )
\Biggr]
.
From the condition (1 - A\dagger A)WBB\dagger = 0 we derive that W has the following matrix form:
W =
\Biggl[
W1 W2
0 W4
\Biggr]
:
\Biggl[
\mathrm{r}\mathrm{a}\mathrm{n}(A\ast )
\mathrm{k}\mathrm{e}\mathrm{r}(A)
\Biggr]
\rightarrow
\Biggl[
\mathrm{r}\mathrm{a}\mathrm{n}(A\ast )
\mathrm{k}\mathrm{e}\mathrm{r}(A)
\Biggr]
.
Thus, we have
1
2
A\dagger C =
\left[ 12A - 1
1 C1
1
2
A - 1
1 C2
0 0
\right] , 1
2
A\dagger C(1 - B\dagger B) =
\left[ 0 1
2
A - 1
1 C2
0 0
\right] ,
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SOLUTIONS OF SYLVESTER EQUATION IN C\ast -MODULAR OPERATORS 361
1
2
WB =
\left[ 12W1B1 0
0 0
\right] , (1 - A\dagger A)Z =
\Biggl[
0 0
X3 X4
\Biggr]
.
Consequently,
X =
1
2
A\dagger C +
1
2
A\dagger C(1 - B\dagger B) +
1
2
WB + (1 - A\dagger A)Z.
On the other hand,
1
2
AA\dagger CB\dagger =
\left[ 12C1B
- 1
1 0
0 0
\right] , - 1
2
AWBB\dagger =
\left[ - 1
2
A1W1 0
0 0
\right] ,
(1 - AA\dagger )CB\dagger =
\Biggl[
0 0
C3B
- 1
1 0
\Biggr]
, V (1 - BB\dagger ) =
\Biggl[
0 Y2
0 Y4
\Biggr]
.
Then
Y =
1
2
AA\dagger CB\dagger + (1 - AA\dagger )CB\dagger - 1
2
AWBB\dagger + V (1 - BB\dagger ).
Theorem 3.2 is proved.
4. Operator equation \bfitA \bfitX \bfitD + \bfitF \bfitY \bfitB = \bfitC . In this section, we study the general solutions
to Eq. (1.2) below in the general context of the Hilbert C\ast -modules. The necessary and sufficient
conditions for the existence of a solution are given and the set of solutions are completely described.
Theorem 4.1. Let \scrX ,\scrY ,\scrZ ,\scrW ,\scrV ,\scrK be Hilbert \scrA -modules, A \in \scrL (\scrZ ,\scrK ), B \in \scrL (\scrX ,\scrW ),
D \in \scrL (\scrX ,\scrY ), F \in \scrL (\scrV ,\scrK ) be invertible and C \in \scrL (\scrX ,\scrK ). Then the operator equation (1.2) has
a solution (X,Y ) \in \scrL (\scrY ,\scrZ )\times \scrL (\scrW ,\scrV ). In the case any solution of the Eq. (1.2) is represented by
X =
1
2
A - 1CD - 1 - A - 1KD - 1 (4.1)
and
Y =
1
2
F - 1CB - 1 + F - 1KB - 1, (4.2)
where K \in \scrL (\scrX ,\scrK ) is arbitrary.
Proof. We show that if the Eq. (1.2) has a solution (X,Y ) \in \scrL (\scrY ,\scrZ ) \times \scrL (\scrW ,\scrV ), then it
must be of the forms (4.1) and (4.2). Let T =
\biggl[
A 0
0 B\ast
\biggr]
: \scrZ \oplus \scrW \rightarrow \scrK \oplus \scrX , \^Y =
\biggl[
0 X
Y \ast 0
\biggr]
:
\scrV \oplus \scrY \rightarrow \scrZ \oplus \scrW , S =
\biggl[
F 0
0 D\ast
\biggr]
: \scrV \oplus \scrY \rightarrow \scrK \oplus \scrX and N =
\biggl[
0 C
C\ast 0
\biggr]
: \scrX \oplus \scrX \rightarrow \scrK \oplus \scrK . By
T \^Y S\ast + S( \^Y )\ast T \ast = N, we have
T \^Y S\ast + S( \^Y )\ast T \ast =
\Biggl[
A 0
0 B\ast
\Biggr] \Biggl[
0 X
Y \ast 0
\Biggr] \Biggl[
F \ast 0
0 D
\Biggr]
+
+
\Biggl[
F 0
0 D\ast
\Biggr] \Biggl[
0 Y
X\ast 0
\Biggr] \Biggl[
A\ast 0
0 B
\Biggr]
=
\Biggl[
0 C
C\ast 0
\Biggr]
,
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362 Z. NIAZI MOGHANI, M. MOHAMMADZADEH KARIZAKI, M. KHANEHGIR
or, equivalently, \Biggl[
0 AXD + FY B
B\ast Y \ast F \ast +D\ast X\ast A\ast 0
\Biggr]
=
\Biggl[
0 C
C\ast 0
\Biggr]
.
Therefore AXD + FY B = C. Obviously, by our assumptions T and S are invertible and from
Theorem 2.2, it follows that \^Y has the following representation:
\^Y =
1
2
T - 1N(S\ast ) - 1 - T - 1Z(S\ast ) - 1, (4.3)
where Z \in \scrL (\scrK \oplus \scrX ) satisfies Z\ast = - Z. Let Z =
\biggl[
Z1 Z2
Z3 Z4
\biggr]
. From (4.3), we have
\Biggl[
0 X
Y \ast 0
\Biggr]
=
1
2
\Biggl[
A - 1 0
0 (B\ast ) - 1
\Biggr] \Biggl[
0 C
C\ast 0
\Biggr] \Biggl[
(F \ast ) - 1 0
0 D - 1
\Biggr]
-
-
\Biggl[
A - 1 0
0 (B\ast ) - 1
\Biggr] \Biggl[
Z1 Z2
Z3 Z4
\Biggr] \Biggl[
(F \ast ) - 1 0
0 D - 1
\Biggr]
.
It enforces that
X =
1
2
A - 1CD - 1 - A - 1Z2D
- 1,
Y \ast =
1
2
(B\ast ) - 1C\ast (F \ast ) - 1 - (B\ast ) - 1Z3(F
\ast ) - 1,
A - 1Z1(F
\ast ) - 1 = (B\ast ) - 1Z4D
- 1 = 0.
From invertibility of the operators A, B, D and F it follows that Z1 = 0 and Z4 = 0. On the other
hand, from Z\ast = - Z, we get Z\ast
3 = - Z2. For K = Z2, we have
X =
1
2
A - 1CD - 1 - A - 1KD - 1
and
Y =
1
2
F - 1CB - 1 + F - 1KB - 1.
Theorem 4.1 is proved.
Now, we are ready to state our main result of this section.
Theorem 4.2. Suppose that \scrX and \scrY are Hilbert \scrA -modules, A,F \in \scrL (\scrX ,\scrY ), B,D \in
\scrL (\scrY ,\scrX ) and A, B, D, F have closed ranges and C \in \scrL (\scrY ) such that \mathrm{r}\mathrm{a}\mathrm{n}(F ) = \mathrm{r}\mathrm{a}\mathrm{n}(D\ast ),
\mathrm{r}\mathrm{a}\mathrm{n}(F \ast ) = \mathrm{r}\mathrm{a}\mathrm{n}(D), and
FF \dagger A(1 - DD\dagger ) = 0, (1 - FF \dagger )ADD\dagger = 0,
F \dagger FB(1 - D\dagger D) = 0, (1 - F \dagger F )BD\dagger D = 0,
D\dagger DB\dagger BA = D\dagger DA, D\dagger DAA\dagger B\ast = D\dagger DB\ast ,
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SOLUTIONS OF SYLVESTER EQUATION IN C\ast -MODULAR OPERATORS 363
F \dagger FBB\dagger A\ast = F \dagger FA\ast , F \dagger FA\dagger AB = F \dagger FB,
F \dagger CB\dagger B(1 - D\dagger D) = F \dagger C(1 - D\dagger D),
(1 - FF \dagger )AA\dagger CD\dagger = (1 - FF \dagger )CD\dagger .
Then the following statements are equivalent:
(a) there exists a solution (X,Y ) \in \scrL (\scrX )\times \scrL (\scrX ) to Eq. (1.2);
(b) (1 - AA\dagger )C(1 - B\dagger B) = 0 and (1 - FF \dagger )C(1 - D\dagger D) = 0.
If (a) or (b) is satisfied, then any solution to Eq. (1.2) has the form
X =
1
2
F \dagger FA\dagger CD\dagger + (1 - F \dagger F )A\dagger CD\dagger +
1
2
F \dagger FA\dagger C(1 - B\dagger B)D\dagger +
+
1
2
F \dagger FWBD\dagger + F \dagger F (1 - A\dagger A)ZD\dagger + U(1 - DD\dagger )
and
Y =
1
2
F \dagger AA\dagger CB\dagger DD\dagger + F \dagger (1 - AA\dagger )CB\dagger DD\dagger + F \dagger CB\dagger (1 - DD\dagger ) -
- 1
2
F \dagger AWBB\dagger DD\dagger + F \dagger V (1 - BB\dagger )DD\dagger + (1 - F \dagger F )T,
where Z \in \scrL (\scrY ,\scrX ) and T,U, V,W \in \scrL (\scrX ) are arbitrary operators, so that F \dagger F (1 -
- AA\dagger )WBB\dagger DD\dagger = 0 .
Proof. (a) \Rightarrow (b). Suppose that (X,Y ) \in \scrL (\scrX )\times \scrL (\scrX ) is a solution of Eq. (1.2). Then
(1 - AA\dagger )C(1 - B\dagger B) = (1 - AA\dagger )(AXD + FY B)(1 - B\dagger B) =
= (1 - AA\dagger )AXD(1 - B\dagger B) + (1 - AA\dagger )FY B(1 - B\dagger B) = 0
and
(1 - FF \dagger )C(1 - D\dagger D) = (1 - FF \dagger )(AXD + FY B)(1 - D\dagger D) =
= (1 - FF \dagger )AXD(1 - D\dagger D) + (1 - FF \dagger )FY B(1 - D\dagger D) = 0.
(b) \Rightarrow (a). Suppose that (X,Y ) \in \scrL (\scrX ) \times \scrL (\scrX ) is a solution of Eq. (1.2). Since \mathrm{r}\mathrm{a}\mathrm{n}(\mathrm{D}) and
\mathrm{r}\mathrm{a}\mathrm{n}(F ) are closed, we have \scrX = \mathrm{r}\mathrm{a}\mathrm{n}(D\ast ) \oplus \mathrm{k}\mathrm{e}\mathrm{r}(D) and \scrY = \mathrm{r}\mathrm{a}\mathrm{n}(F ) \oplus \mathrm{k}\mathrm{e}\mathrm{r}(F \ast ). Applying our
hypotheses A,B,D and F have the following matrix forms:
A =
\Biggl[
A1 A2
A3 A4
\Biggr]
:
\Biggl[
\mathrm{r}\mathrm{a}\mathrm{n}(D)
\mathrm{k}\mathrm{e}\mathrm{r}(D\ast )
\Biggr]
\rightarrow
\Biggl[
\mathrm{r}\mathrm{a}\mathrm{n}(F )
\mathrm{k}\mathrm{e}\mathrm{r}(F \ast )
\Biggr]
,
B =
\Biggl[
B1 B2
B3 B4
\Biggr]
:
\Biggl[
\mathrm{r}\mathrm{a}\mathrm{n}(D\ast )
\mathrm{k}\mathrm{e}\mathrm{r}(D)
\Biggr]
\rightarrow
\Biggl[
\mathrm{r}\mathrm{a}\mathrm{n}(F \ast )
\mathrm{k}\mathrm{e}\mathrm{r}(F )
\Biggr]
,
D =
\Biggl[
D1 0
0 0
\Biggr]
:
\Biggl[
\mathrm{r}\mathrm{a}\mathrm{n}(D\ast )
\mathrm{k}\mathrm{e}\mathrm{r}(D)
\Biggr]
\rightarrow
\Biggl[
\mathrm{r}\mathrm{a}\mathrm{n}(D)
\mathrm{k}\mathrm{e}\mathrm{r}(D\ast )
\Biggr]
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364 Z. NIAZI MOGHANI, M. MOHAMMADZADEH KARIZAKI, M. KHANEHGIR
and
F =
\Biggl[
F1 0
0 0
\Biggr]
:
\Biggl[
\mathrm{r}\mathrm{a}\mathrm{n}(F \ast )
\mathrm{k}\mathrm{e}\mathrm{r}(F )
\Biggr]
\rightarrow
\Biggl[
\mathrm{r}\mathrm{a}\mathrm{n}(F )
\mathrm{k}\mathrm{e}\mathrm{r}(F \ast )
\Biggr]
.
Hence D1 and F1 are invertible by Lemma 2.2. Also, we can assume that unknown operators X
and Y have the matrix forms
X =
\Biggl[
X1 X2
X3 X4
\Biggr]
:
\Biggl[
\mathrm{r}\mathrm{a}\mathrm{n}(D)
\mathrm{k}\mathrm{e}\mathrm{r}(D\ast )
\Biggr]
\rightarrow
\Biggl[
\mathrm{r}\mathrm{a}\mathrm{n}(D)
\mathrm{k}\mathrm{e}\mathrm{r}(D\ast )
\Biggr]
and
Y =
\Biggl[
Y1 Y2
Y3 Y4
\Biggr]
:
\Biggl[
\mathrm{r}\mathrm{a}\mathrm{n}(F \ast )
\mathrm{k}\mathrm{e}\mathrm{r}(F )
\Biggr]
\rightarrow
\Biggl[
\mathrm{r}\mathrm{a}\mathrm{n}(F \ast )
\mathrm{k}\mathrm{e}\mathrm{r}(F )
\Biggr]
.
From condition (1 - FF \dagger )C(1 - D\dagger D) = 0 it obtains C has the form
C =
\Biggl[
C1 C2
C3 0
\Biggr]
:
\Biggl[
\mathrm{r}\mathrm{a}\mathrm{n}(D\ast )
\mathrm{k}\mathrm{e}\mathrm{r}(D)
\Biggr]
\rightarrow
\Biggl[
\mathrm{r}\mathrm{a}\mathrm{n}(F )
\mathrm{k}\mathrm{e}\mathrm{r}(F \ast )
\Biggr]
.
Conditions FF \dagger A(1 - DD\dagger ) = 0 and (1 - FF \dagger )ADD\dagger = 0 imply that A has the matrix form
A =
\Biggl[
A1 0
0 A4
\Biggr]
:
\Biggl[
\mathrm{r}\mathrm{a}\mathrm{n}(D)
\mathrm{k}\mathrm{e}\mathrm{r}(D\ast )
\Biggr]
\rightarrow
\Biggl[
\mathrm{r}\mathrm{a}\mathrm{n}(F )
\mathrm{k}\mathrm{e}\mathrm{r}(F \ast )
\Biggr]
,
and conditions F \dagger FB(1 - D\dagger D) = 0 and (1 - F \dagger F )BD\dagger D = 0 imply that B have the matrix form
B =
\Biggl[
B1 0
0 B4
\Biggr]
:
\Biggl[
\mathrm{r}\mathrm{a}\mathrm{n}(D\ast )
\mathrm{k}\mathrm{e}\mathrm{r}(D)
\Biggr]
\rightarrow
\Biggl[
\mathrm{r}\mathrm{a}\mathrm{n}(F \ast )
\mathrm{k}\mathrm{e}\mathrm{r}(F )
\Biggr]
.
By replacing these matrix forms in the operator equation AXD + FY B = C we get\Biggl[
A1 0
0 A4
\Biggr] \Biggl[
X1 X2
X3 X4
\Biggr] \Biggl[
D1 0
0 0
\Biggr]
+
\Biggl[
F1 0
0 0
\Biggr] \Biggl[
Y1 Y2
Y3 Y4
\Biggr] \Biggl[
B1 0
0 B4
\Biggr]
=
=
\Biggl[
A1X1D1 + F1Y1B1 F1Y2B4
A4X3D1 0
\Biggr]
=
\Biggl[
C1 C2
C3 0
\Biggr]
.
That is,
A1X1D1 + F1Y1B1 = C1, (4.4)
F1Y2B4 = C2, (4.5)
A4X3D1 = C3. (4.6)
Taking X0 = X1D1 and Y0 = F1Y1, then equation (4.4) becomes
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SOLUTIONS OF SYLVESTER EQUATION IN C\ast -MODULAR OPERATORS 365
A1X0 + Y0B1 = C1. (4.7)
Due to the fact that D\dagger DB\dagger BA = D\dagger DA and the above matrix forms we have\Biggl[
1 0
0 0
\Biggr] \Biggl[
B\dagger
1B1 0
0 B\dagger
4B4
\Biggr] \Biggl[
A1 0
0 A4
\Biggr]
=
\Biggl[
1 0
0 0
\Biggr] \Biggl[
A1 0
0 A4
\Biggr]
,
which gives us the identity B\dagger
1B1A1 = A1. Hence, \mathrm{r}\mathrm{a}\mathrm{n}(A1) \subseteq \mathrm{r}\mathrm{a}\mathrm{n}(B\ast
1). Similarly, applying condi-
tion D\dagger DAA\dagger B\ast = D\dagger DB\ast we find\Biggl[
1 0
0 0
\Biggr] \Biggl[
A1A
\dagger
1 0
0 A4A
\dagger
4
\Biggr] \Biggl[
B\ast
1 0
0 B\ast
4
\Biggr]
=
\Biggl[
1 0
0 0
\Biggr] \Biggl[
B\ast
1 0
0 B\ast
4
\Biggr]
.
It in turn gives us A1A
\dagger
1B
\ast
1 = B\ast
1 and so \mathrm{r}\mathrm{a}\mathrm{n}(B\ast
1) \subseteq \mathrm{r}\mathrm{a}\mathrm{n}(\mathrm{A}1). Therefore,\mathrm{r}\mathrm{a}\mathrm{n}(A1) = \mathrm{r}\mathrm{a}\mathrm{n}(B\ast
1).
Taking into account conditions F \dagger FBB\dagger A\ast = F \dagger FA\ast and F \dagger FA\dagger AB = F \dagger FB and by using a
similar method we yield \mathrm{r}\mathrm{a}\mathrm{n}(A\ast
1) = \mathrm{r}\mathrm{a}\mathrm{n}(B1), too. Also, by multiplication FF \dagger on the left and D\dagger D
on the right to the identity (1 - AA\dagger )C(1 - B\dagger B) = 0, we obtain (1 - A1A
\dagger
1)C1(1 - B\dagger
1B1) = 0.
Now, all conditions of Theorem 3.2 hold and so Eq. (4.7) is solvable and any solution of it can be
represented in the following forms:
X0 =
1
2
A\dagger
1C1 +
1
2
A\dagger
1C1(1 - B\dagger
1B1) +
1
2
W1B1 + (1 - A\dagger
1A1)Z1,
Y0 =
1
2
A1A
\dagger
1C1B
\dagger
1 + (1 - A1A
\dagger
1)C1B
\dagger
1 -
1
2
A1W1B1B1\dagger + V1(1 - B1B
\dagger
1),
(4.8)
where Z1 \in \scrL (\mathrm{r}\mathrm{a}\mathrm{n}(F ), \mathrm{r}\mathrm{a}\mathrm{n}(D)),V1 \in \scrL (\mathrm{r}\mathrm{a}\mathrm{n}(D), \mathrm{r}\mathrm{a}\mathrm{n}(F )) and W1 \in \scrL (\mathrm{r}\mathrm{a}\mathrm{n}(D)) that W1 satisfies
in (1 - A1A
\dagger
1)W1B1B
\dagger
1 = 0. This last identity implies that F \dagger F (1 - AA\dagger )WBB\dagger DD\dagger = 0. Since
D1 and F1 are invertible, then from (4.8) we deduce
X1 =
1
2
A\dagger
1C1D
- 1
1 +
1
2
A\dagger
1C1(1 - B\dagger
1B1)D
- 1
1 +
1
2
W1B1D
- 1
1 + (1 - A\dagger
1A1)Z1D
- 1
1 ,
Y1 =
1
2
F - 1
1 A1A
\dagger
1C1B
\dagger
1 + F - 1
1 (1 - A1A
\dagger
1)C1B
\dagger
1 -
1
2
F - 1
1 A1W1B1B1\dagger + F - 1
1 V1(1 - B1B
\dagger
1).
The assumptions F \dagger CB\dagger B(1 - D\dagger D) = F \dagger C(1 - D\dagger D) and (1 - FF \dagger )AA\dagger CD\dagger = (1 - FF \dagger )CD\dagger
give us F - 1
1 C2B
\dagger
4B4 = F - 1
1 C2 and A4(A4)
\dagger C3D
- 1
1 = C3D
- 1
1 , respectively. With the aid of these
facts and by using Lemma 2.4 it yields that the operator equations (4.5) and (4.6) are solvable, and
further we have Y2 = F - 1
1 C2(B4)
\dagger and X3 = (A4)
\dagger C3D
- 1
1 . Hence,
X =
\left[ 12A\dagger
1C1D
- 1
1 +
1
2
A\dagger
1C1(1 - B\dagger
1B1)D
- 1
1 +
1
2
W1B1D
- 1
1 + (1 - A\dagger
1A1)Z1D
- 1
1 X2
(A4)
\dagger C3D
- 1
1 X4
\right]
and
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366 Z. NIAZI MOGHANI, M. MOHAMMADZADEH KARIZAKI, M. KHANEHGIR
Y =
=
\left[ 12F - 1
1 A1A
\dagger
1C1B
\dagger
1+F - 1
1 (1 - A1A
\dagger
1)C1B
\dagger
1 -
1
2
F - 1
1 A1W1B1B
\dagger
1+F - 1
1 V1(1 - B1B
\dagger
1) F - 1
1 C2B
\dagger
4
Y3 Y4
\right] ,
where X2,X4,Y3 and Y4 can be taken arbitrary. On the other hand, from the assumptions \mathrm{r}\mathrm{a}\mathrm{n}(F ) =
= \mathrm{r}\mathrm{a}\mathrm{n}(D\ast ) and \mathrm{r}\mathrm{a}\mathrm{n}(F \ast ) = \mathrm{r}\mathrm{a}\mathrm{n}(D) we observe that the operators W,Z,U, V and T have the
following matrix forms:
W =
\Biggl[
W1 W2
W3 W4
\Biggr]
:
\Biggl[
\mathrm{r}\mathrm{a}\mathrm{n}(D)
\mathrm{k}\mathrm{e}\mathrm{r}(D\ast )
\Biggr]
\rightarrow
\Biggl[
\mathrm{r}\mathrm{a}\mathrm{n}(D)
\mathrm{k}\mathrm{e}\mathrm{r}(D\ast )
\Biggr]
,
Z =
\Biggl[
Z1 Z2
Z3 Z4
\Biggr]
:
\Biggl[
\mathrm{r}\mathrm{a}\mathrm{n}(F )
\mathrm{k}\mathrm{e}\mathrm{r}(F \ast )
\Biggr]
\rightarrow
\Biggl[
\mathrm{r}\mathrm{a}\mathrm{n}(D)
\mathrm{k}\mathrm{e}\mathrm{r}(D\ast )
\Biggr]
,
U =
\Biggl[
U1 X2
U3 X4
\Biggr]
:
\Biggl[
\mathrm{r}\mathrm{a}\mathrm{n}(D)
\mathrm{k}\mathrm{e}\mathrm{r}(D\ast )
\Biggr]
\rightarrow
\Biggl[
\mathrm{r}\mathrm{a}\mathrm{n}(D)
\mathrm{k}\mathrm{e}\mathrm{r}(D\ast )
\Biggr]
,
V =
\Biggl[
V1 V2
V3 V4
\Biggr]
:
\Biggl[
\mathrm{r}\mathrm{a}\mathrm{n}(D)
\mathrm{k}\mathrm{e}\mathrm{r}(D\ast )
\Biggr]
\rightarrow
\Biggl[
\mathrm{r}\mathrm{a}\mathrm{n}(F )
\mathrm{k}\mathrm{e}\mathrm{r}(F \ast )
\Biggr]
and
T =
\Biggl[
T1 T2
Y3 Y4
\Biggr]
:
\Biggl[
\mathrm{r}\mathrm{a}\mathrm{n}(D)
\mathrm{k}\mathrm{e}\mathrm{r}(D\ast )
\Biggr]
\rightarrow
\Biggl[
\mathrm{r}\mathrm{a}\mathrm{n}(D)
\mathrm{k}\mathrm{e}\mathrm{r}(D\ast )
\Biggr]
.
Hence, we obtain
1
2
F \dagger FA\dagger CD\dagger =
\left[ 12A\dagger
1C1D
- 1
1 0
0 0
\right] ,
(1 - F \dagger F )A\dagger CD\dagger =
\Biggl[
0 0
A\dagger
4C3D
- 1
1 0
\Biggr]
,
1
2
F \dagger FA\dagger C(1 - B\dagger B)D\dagger =
\left[ 12A\dagger
1C1(1 - B\dagger
1B1)D
- 1
1 0
0 0
\right] ,
1
2
F \dagger FWBD\dagger =
\left[ 12W1B1D
- 1
1 0
0 0
\right] ,
F \dagger F (1 - A\dagger A)ZD\dagger =
\Biggl[
(1 - A\dagger
1A1)Z1D
- 1
1 0
0 0
\Biggr]
,
ISSN 1027-3190. Укр. мат. журн., 2021, т. 73, № 3
SOLUTIONS OF SYLVESTER EQUATION IN C\ast -MODULAR OPERATORS 367
U(1 - DD\dagger ) =
\Biggl[
0 X2
0 X4
\Biggr]
.
From these we get
X =
1
2
F \dagger FA\dagger CD\dagger + (1 - F \dagger F )A\dagger CD\dagger +
1
2
F \dagger FA\dagger C(1 - B\dagger B)D\dagger +
1
2
F \dagger FWBD\dagger +
+F \dagger F (1 - A\dagger A)ZD\dagger + U(1 - DD\dagger ).
Also, we have
1
2
F \dagger AA\dagger CB\dagger DD\dagger =
\left[ 12F - 1
1 A1A
\dagger
1C1B
\dagger
1 0
0 0
\right] ,
F \dagger (1 - AA\dagger )CB\dagger DD\dagger =
\Biggl[
F - 1
1 (1 - A1A
\dagger
1)C1B
\dagger
1 0
0 0
\Biggr]
,
F \dagger CB\dagger (1 - DD\dagger ) =
\Biggl[
0 F - 1
1 C2B
\dagger
4
0 0
\Biggr]
,
- 1
2
F \dagger AWBB\dagger DD\dagger =
\left[ - 1
2
F - 1
1 A1W1B1B
\dagger
1 0
0 0
\right] ,
F \dagger V (1 - BB\dagger )DD\dagger =
\Biggl[
F - 1
1 V1(1 - B1B
\dagger
1) 0
0 0
\Biggr]
,
(1 - F \dagger F )T =
\Biggl[
0 0
Y3 Y4
\Biggr]
.
Accordingly, we derive
Y =
1
2
F \dagger AA\dagger CB\dagger DD\dagger + F \dagger (1 - AA\dagger )CB\dagger DD\dagger + F \dagger CB\dagger (1 - DD\dagger ) - 1
2
F \dagger AWBB\dagger DD\dagger +
+F \dagger V (1 - BB\dagger )DD\dagger + (1 - F \dagger F )T.
Theorem 4.2 is proved.
5. Positive solutions of operator equation \bfitA \bfitX +\bfitX \ast \bfitA \ast = \bfitB . In this section, we provide an
approach to the study of the positive solutions to the operator equation (1.3) for adjointable operators
between Hilbert C\ast -modules.
Theorem 5.1. Let \scrX be a Hilbert \scrA -module, and A,B,C \in \scrL (\scrX ). Let A, B and BA\ast have
closed ranges and BA\ast be a positive operator, \mathrm{r}\mathrm{a}\mathrm{n}(B) = \mathrm{r}\mathrm{a}\mathrm{n}(BA\ast ) and (1 - A\dagger A)B\dagger = 0 . Then
the Eq. (1.3) has a positive solution X \in \scrL (\scrX )+ if and only if B is self-adjoint and (1 - AA\dagger )B(1 -
- AA\dagger ) = 0. In this case the general positive solution has the form
ISSN 1027-3190. Укр. мат. журн., 2021, т. 73, № 3
368 Z. NIAZI MOGHANI, M. MOHAMMADZADEH KARIZAKI, M. KHANEHGIR
X =
1
2
X0 +X0V (1 - A\dagger A) + (1 - A\dagger A)V \ast X0 + 2(1 - A\dagger A)V \ast X0V (1 - A\dagger A)+
+(1 - A\dagger A)W (1 - A\dagger A), (5.1)
where X0 = B\ast (BA\ast )\dagger B is a particular positive solution, V \in \scrL (\scrX )sa and W \in \scrL (\scrX )+ are
arbitrary.
Proof. Suppose that Eq. (1.3) has a solution X \in \scrL (\scrX )+. Obviously, B = B\ast . Also, we have
(1 - AA\dagger )B(1 - AA\dagger ) = (1 - AA\dagger )(AX +X\ast A\ast )(1 - AA\dagger ) =
= (1 - AA\dagger )AX(1 - AA\dagger ) + (1 - AA\dagger )X\ast A\ast (1 - AA\dagger ) = 0.
Conversely, assume that B is self-adjoint and (1 - AA\dagger )B(1 - AA\dagger ) = 0. We prove that the general
positive solution of Eq. (1.3) can be expressed as (5.1). For this purpose, take P1 = B\dagger B, P2 =
= 1 - B\dagger B and X1 = P1XP1, where X is as in (5.1). From the assumption (1 - A\dagger A)B\dagger = 0, we
have
X1 = P1XP1 = B\dagger BXB\dagger B =
1
2
B\ast (BA\ast )\dagger B,
and hence BA\ast \in \scrL (\scrX )+ implies that X1 \in \scrL (\scrX )+. Moreover, since \mathrm{r}\mathrm{a}\mathrm{n}(B) = \mathrm{r}\mathrm{a}\mathrm{n}(BA\ast ) we get
X1(2B
\dagger (BA\ast )(B\ast )\dagger )X1 =
1
2
B\ast (BA\ast )\dagger B(B\dagger (BA\ast )(B\ast )\dagger )B\ast (BA\ast )\dagger B =
=
1
2
B\ast (BA\ast )\dagger BB\dagger (BA\ast )BB\dagger (BA\ast )\dagger B =
1
2
B\ast (BA\ast )\dagger B,
thus, X1 is regular and X\dagger
1 = B\dagger (BA\ast )(B\ast )\dagger . According to Remark 2.2 we can write
X = X1 +X2 +X3 +X4 = P1XP1 + P1XP2 + P2XP1 + P2XP2,
where
X2 = P1XP2 = B\dagger BX(1 - B\dagger B) = B\ast (BA\ast )\dagger BV (1 - A\dagger A),
X3 = P2XP1 = (1 - B\dagger B)XB\dagger B = (1 - A\dagger A)V \ast B\ast (BA\ast )\dagger B,
X4 = P2XP2 = (1 - B\dagger B)X(1 - B\dagger B) = 2(1 - A\dagger A)V \ast B\ast (BA\ast )\dagger BV (1 - A\dagger A)+
+(1 - A\dagger A)W (1 - A\dagger A).
Evidently, X3 = X\ast
2 and X1X
\dagger
1X2 = X2. Furthermore, we have
X4 - X3X
\dagger
1X2 = 2(1 - A\dagger A)V \ast B\ast (BA\ast )\dagger BV (1 - A\dagger A) + (1 - A\dagger A)W (1 - A\dagger A) -
- ((1 - A\dagger A)V \ast B\ast (BA\ast )\dagger B)(B\dagger (BA\ast )(B\ast )\dagger )(2B\ast (BA\ast )\dagger BV (1 - A\dagger A)) =
= (1 - A\dagger A)W (1 - A\dagger A) \in \scrL (\scrX )+.
Applying Lemma 2.3, we derive that X is positive and any arbitrary positive solution to Eq. (1.3)
can be expressed as (5.1).
Theorem 5.1 is proved.
ISSN 1027-3190. Укр. мат. журн., 2021, т. 73, № 3
SOLUTIONS OF SYLVESTER EQUATION IN C\ast -MODULAR OPERATORS 369
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ISSN 1027-3190. Укр. мат. журн., 2021, т. 73, № 3
|
| id | umjimathkievua-article-152 |
| institution | Ukrains’kyi Matematychnyi Zhurnal |
| keywords_txt_mv | keywords |
| language | English |
| last_indexed | 2026-03-24T02:02:03Z |
| publishDate | 2021 |
| publisher | Institute of Mathematics, NAS of Ukraine |
| record_format | ojs |
| resource_txt_mv | umjimathkievua/5c/dfd42e7b5ac96eb995df34d66b43245c.pdf |
| spelling | umjimathkievua-article-1522025-03-31T08:48:21Z Solutions of Sylvester equation in $C^*$-modular operators Solutions of Sylvester equation in $C^*$-modular operators Solutions of Sylvester equation in $C^*$-modular operators Moghani, Z. Niazi Mohammadzadeh Karizaki, M. Khanehgir, M. Moghani, Z. Niazi Mohammadzadeh Karizaki, M. Khanehgir, M. Moghani, Z. Niazi Mohammadzadeh Karizaki, M. Khanehgir, M. Hilbert C ∗ -module Moore-Penrose inverse Operator equation Positive solution Sylvester equation Hilbert C ∗ -module Moore-Penrose inverse Operator equation Positive solution Sylvester equation UDC 517.9 We study the solvability of the Sylvester equation $AX + Y B = C$ and the operator equation $AXD + FY B = C$ in the general setting of the adjointable operators between Hilbert $C^*$ -modules. Based on the Moore – Penrose inverses of the associated operators, we propose necessary and sufficient conditions for the existence of solutions to these equations, and obtain the general expressions of the solutions in the solvable cases. We also provide an approach to the study of the positive solutions for a special case of Lyapunov equation. УДК 517.9 Розв’язки рiвняння Сiльвестра для  $C^*$-модульних операторiв Розглянуто розв’язнiсть рiвняння Сiльвестра $AX + Y B = C$ та операторного рiвняння $AXD + FY B = C$ при загальних умовах сумiжностi операторiв мiж гiльбертовими $C^*$- модулями. На основi обернених Мура – Пенроуза для зв’язаних операторiв отримано необхiднi та достатнi умови iснування розв’язкiв цих рiвнянь, а також загальнi вирази для розв’язкiв у випадку, коли вони iснують. Крiм того, запропоновано пiдхiд до вивчення додатних розв’язкiв у спецiальному випадку рiвняння Ляпунова. Institute of Mathematics, NAS of Ukraine 2021-03-11 Article Article application/pdf https://umj.imath.kiev.ua/index.php/umj/article/view/152 10.37863/umzh.v73i3.152 Ukrains’kyi Matematychnyi Zhurnal; Vol. 73 No. 3 (2021); 354 - 369 Український математичний журнал; Том 73 № 3 (2021); 354 - 369 1027-3190 en https://umj.imath.kiev.ua/index.php/umj/article/view/152/8979 Copyright (c) 2021 Mahnaz Khanehgir |
| spellingShingle | Moghani, Z. Niazi Mohammadzadeh Karizaki, M. Khanehgir, M. Moghani, Z. Niazi Mohammadzadeh Karizaki, M. Khanehgir, M. Moghani, Z. Niazi Mohammadzadeh Karizaki, M. Khanehgir, M. Solutions of Sylvester equation in $C^*$-modular operators |
| title | Solutions of Sylvester equation in $C^*$-modular operators |
| title_alt | Solutions of Sylvester equation in $C^*$-modular operators Solutions of Sylvester equation in $C^*$-modular operators |
| title_full | Solutions of Sylvester equation in $C^*$-modular operators |
| title_fullStr | Solutions of Sylvester equation in $C^*$-modular operators |
| title_full_unstemmed | Solutions of Sylvester equation in $C^*$-modular operators |
| title_short | Solutions of Sylvester equation in $C^*$-modular operators |
| title_sort | solutions of sylvester equation in $c^*$-modular operators |
| topic_facet | Hilbert C ∗ -module Moore-Penrose inverse Operator equation Positive solution Sylvester equation Hilbert C ∗ -module Moore-Penrose inverse Operator equation Positive solution Sylvester equation |
| url | https://umj.imath.kiev.ua/index.php/umj/article/view/152 |
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