Positive Solutions of Operator Equations AX = B , XC = D

: In this paper, using the technique of operator matrix, we consider the positive solution of the system of operator equations AX = B , XC = D in the framework of the Hilbert space; here, the ranges R ( A ) of A and R ( C ) of C are not necessarily closed. Firstly, we provide a new necessary and sufﬁcient condition for the existence of positive solutions of AX = B and also provide a representation of positive solutions, which generalize previous conclusions. Furthermore, using the above result, a condition of equivalence for the existence of common positive solutions of AX = B , XC = D is given, as well as the general forms of positive solutions


Introduction
The problem of solving operator equations has important applications in physics, control theory, and many other fields.It has been paid attention to by many scholars.For example, equations and AX = B, XC = D have been widely studied in matrix algebra [1], the operator space on the Hilbert space [2][3][4][5][6], and the adjointable operator space on Hilbert C * -modules [7][8][9][10].There is a classical result addressing the existence of solutions for Equation (1), which is the famous Douglas range inclusion theorem [4].Subsequent to this, many scholars have studied the existence and formulae of solutions for Equation (1) and common solutions for Equation (2).On occasion, specific requirements were imposed on these solutions, such as them being self-adjoint, positive, real positive, and so on.Dajić and Koliha [2,3] investigated positive solutions for Equation (1) and common positive solutions for Equation (2) in operator spaces where operators A and C have closed ranges.Qingxiang Xu [10] generalized the results to adjointable operators on Hilbert C *modules.The main tool used for closed-range operators is the Moore-Penrose generalized inverses.Unfortunately, this method does not work if the ranges of A and C are not closed.Liang and Deng [5] provided equivalent conditions for the existence of positive solutions for Equations (1) and ( 2), with no restrictions on the range.But determining the formulas for the common positive solution remains an open question.Recently, the authors of [9] considered the common positive solutions of system (2) for adjointable operators under certain conditions.
For the sake of the description, let us introduce some symbols.Throughout this note, H and K are complex separable Hilbert spaces.We denote the set of all bounded linear operators from H into K by B(H, K) and abbreviate B(H, K) to B(H) if K = H.For an operator A ∈ B(H, K), we denote the adjoint operator, and the absolute values of A by A * and |A|, respectively.Let R(A), R(A), and N(A) be the range, the closure of the range, and the null space of A, respectively.The Moore-Penrose inverse of A ∈ B(H, K) is defined as the operator A † ∈ B(K, H), satisfying the equations, An operator A has the (unique) Moore-Penrose inverse if and only if A is closed-range [11].An operator A ∈ B(H) is termed self-adjoint if A * = A, and positive if Ax, x ≥ 0 for all x ∈ H.We write A ≥ 0 if A is positive.By B + (H), we denote the set of all positive elements in B(H).We assume that M is a closed subspace of H, and the orthogonal projection onto M is denoted by P M .M ⊥ refers to the orthogonal complementary subspace of M. For an operator A ∈ B(H), we define P as the projection onto R(A * ) in the sequel.
In this paper, we discuss positive solutions for Equations ( 1) and ( 2) in the framework of the Hilbert space; the ranges of A and B may not be closed.Our main goal is to find the common positive solutions of system Equation (2).Firstly, we need to discuss the positive solutions for Equation (1).
In Section 2, using the operator matrix technique, we provide a new equivalent condition for the existence of positive solutions for Equation (1) by the range of reduced solutions without the condition for the range to be closed.Simultaneously, a specific representation of positive solutions is given.This representation is not as concise as in [5], which provides a method to study the common positive solutions of system (2).Moreover, we show that our result is equivalent to the related result in [2] when A and BA * have closed ranges.
In Section 3, we study common positive solutions for Equation (2).We use the result in Section 2; a necessary and sufficient condition for the existence of common positive solutions is given, and a formula for common positive solutions is obtained.
In Section 4, an example is provided to demonstrate that our results are valid.As shown by Example 1, Equation (1) has positive solutions, where R(A) and R(BA * ) are not closed.It also indicates that system (2) has common positive solutions, which are calculated by Theorem 2.

The Positive Solutions of AX = B
In this section, the positive solutions for Equation (1) are discussed.An equivalent condition for the existence was given by the operator inequality in [5]; simultaneously, a formula of positive solutions was provided by the reduced solution of (BA * ) 1 2 X = B.But this expression does not work for studying the common positive solution of system (2), to our knowledge.Here, we offer a new equivalent condition and a new formula of positive solutions, which can help us solve the relative question about the system of operator equations.

Lemma 1 ([4]). Let A, B ∈ B(H). The following three conditions are equivalent:
If one of these conditions holds, then there exists a unique solution G ∈ B(H) of equation AX = B, such that R(G) ⊆ R(A * ) and N(G) = N(B).This solution is called the Douglasreduced solution.If the operator A has a closed range and R(B) ⊆ R(A), then the Douglas-reduced solution is G = A † B.

Definition 1 ([12]
).An operator sequence {A n } converges to A ∈ B(H) in the strong operator topology if ||A n h − Ah|| → 0 as n → ∞ for all h in H.We denote by A = s.o.− lim n→∞ A n .

Lemma 2 ([6]
).Let A, B ∈ B(H) with R(B) ⊆ R(A) and A = U A |A| be the polar decomposition.Then the formula for the reduced solution G of Equation ( 1) is In this case, the general solution is of the form for all Y ∈ B(H).

Lemma 4 ([8]
).Let A, B ∈ B(H) be such that AX = B is solvable and G is the reduced solution.Then we have ), where G is the reduced solution of AX = B.In this case, the formula of positive solutions is for all Y in B + (H), where H is the reduced solution of (GP) Proof.Suppose that X is a solution of AX = B and G is the reduced solution.It follows from Lemma 2 that there is an operator Y ∈ B(H) satisfying With respect to the space decomposition , G and Y have the matrix forms as follows, and respectively.Then X is of the form '⇒' Assume that X is a positive operator.According to Lemma 3, G 11 ≥ 0 and 2 ).This shows GP ≥ 0 and R(G) ⊆ R((GP) 1 2 ).It is obvious that BA * ≥ 0, according to Lemma 4.
Next, we consider the general representation of the positive solution X. Recall X has matrix form (7) This shows that X can be represented by Formula (4).The proof is completed.
In fact, the above equivalent condition is consistent with the relative results in [2,5].Here, we will demonstrate this using Theorem 1.

Corollary 1 ([5]). Let A and B be operators in B(H).
Then AX = B has positive solutions if and only if there exists λ > 0, such that BB * ≤ λBA * .

It is natural to obtain that
Conversely, if BB * ≤ λBA * for some λ > 0, define a mapping T on R(A * ): Hence, T can be uniquely extended to R(A * ).Define T on R(A * ) ⊥ to be 0, T is a linear bounded operator an H and satisfies AT * = B.That is to say, AX = B is solvable.Suppose that G is the reduced solution, we have GG * ≤ λGP, since In particular, if the operators A and BA * have closed ranges, Theorem 1 can be transformed into the following form, which is Theorem 5.2 provided in [2].
2 ) when BA * ≥ 0. In this case, we have ).Therefore, A and B have the matrix forms as follows, and where A 11 is an invertible operator from R(A * ) onto R(A).Then A † and BA * can be represented by the following forms, BA * ≥ 0 implies BA * is a self-adjoint operator.From matrix form (13), we have B 21 A * 11 = 0. Hence, B 21 = 0.
If R(B) ⊆ R(BA * ), comparing the matrix forms (11) with (13), we have Moreover, the closedness of R(CA * ) implies that R(GP) is closed, according to Lemma 4, and then R(GP) = R((GP) Next, we prove that the general positive solution X is of form (9). According to Theorem 1, X has form (4), where (11) of B can be written as follows: Proof.'⇒': Combining Lemma 5 and Theorem 1, statements (i) and (ii) hold.Suppose that X is a common positive solution of system (2).Then there is a positive operator Y ∈ B(H), such that X has the form where G and H are, respectively, reduced solutions of AX = B and (GP) = N(H), we know that operators A, G, H, and C are of matrix forms, as follows: and We denote From Lemma 3, it is easy to obtain that Y 22 ≥ 0 since Y ≥ 0. Substituting the matrix forms (19), (20), and (23) into Formula (1); we see that X has the following matrix form From matrix forms (21) and (24), C * X * is of the form Recall that X is also a positive solution of C * X * = D * .This implies , for any Y 22 ∈ B + (H).

Conclusions
In this work, to find the common positive solutions for Equation (2) without the limit on the closeness of the range, we first characterize the positive solutions of Equation (1).Theorem 1 provides an equivalent condition for the existence of positive solutions for Equation (1) by the reduced solution.Meanwhile, the formula of positive solutions is given, which implies the matrix form of solutions with respect to the space decomposition . This result offers us a method to consider common positive solutions for Equation (2).By using operator-blocking techniques, the general formula of common positive solutions for Equation ( 2) is obtained in Theorem 2. Through Example 1, it is illustrated that Theorems 1 and 2 are useful for finding positive solutions.The approaches may also be valid for other operator equations, such as AXB = C, AX * + XA * = B, and so on.

1 2 .
Therefore, AX = B has positive solutions by Theorem 1.The proof is completed.

Corollary 2 (
[2]).Let A, B ∈ B(H) be such that A, BA * have closed ranges.Then AX = B has positive solutions if and only if R(B) ⊆ R(BA * ) and BA * ≥ 0. In this case, the positive solution has the formula X = B * (BA * ) † B + (I − P)Y(I − P), ∀Y ∈ B + (H).(9) Proof.Let A and BA * have closed ranges.According to Lemma 1 and Theorem 1, it is only to prove that R(B) ⊆ R(BA * ) if and only , where A 11 ∈ B(R(A * ), R(A)) is an injective operator with dense range.Moreover, R(D * ) ⊆ R(B * ) implies D has the matrix form