Introduction

The Sylvester matrix equation 𝐴𝑋 + 𝑋𝐵 = 𝐶, also sometimes called the continuous Sylvester equation, and the Stein matrix equation 𝑋 + 𝐴𝑋𝐵 = 𝐶, in turn, sometimes called the discrete Sylvester equation, as well as their special cases -the Lyapunov equations 𝐴 𝑇 𝑋 + 𝑋𝐴 = 𝐶 and 𝑋 − 𝐴𝑋𝐴 𝑇 = 𝐶, are well-studied and frequently encountered (for example, in the theory of differential equations) classes of matrix equations [1][2][3][4][5][6]. The conditions for the unique solvability of these equations have long been known; there are numerical algorithms for solving them, for example, the Bartels-Stewart and Golub-Nash-Van Loan algorithms.

An analogue of Sylvester's equation 𝐴𝑋 + 𝑋𝐵 = 𝐶 began to attract the attention of researchers relatively recently [7][8][9]. Although this equation is superficially very similar to Sylvester's equation, their natures are profoundly different. Let's give a simple example to illustrate this difference. If all matrices are square and 𝐴 = 𝐵, then Sylvester's equation has a unique solution 𝑋 -for any right-hand side 𝐶. For the same coefficients 𝐴 and 𝐵, the equation 𝐴𝑋 + 𝑋𝐵 = 𝐶 has a solution only if the matrix 𝐶 is symmetric. If this condition is met and 𝑋 satisfies this equation, then 𝑋 + 𝐾, where 𝐾 is an arbitrary skew-symmetric matrix, is also a solution.

Equation 𝐴 𝑇 𝑋 + 𝑋𝐴 = 𝐶 , as well as the equations 𝐴𝑋 + 𝑋𝐵 = 𝐶 we will generally call two member equations of Sylvester type. The relevance of studying this kind of equations is beyond doubt. Let us give several examples showing why the study of equations of Sylvester type is justified. Equation 𝐴𝑋 + 𝑋𝐵 = 𝐶 was first encountered by us in article [10][11][12], where it was studied under the additional assumption 𝐶 = 𝐶 𝑇 .

Solvability conditions and a description of the general solution were given in terms of generalized inverses for matrices 𝐴 and 𝐵 and their associated projectors. These conditions are not entirely constructive and are difficult to verify. Homogeneous equation 𝐴𝑋 + 𝑋𝐵 = 𝐶 was studied in article [2] in the special case 𝐵 = 𝐴. The authors were motivated by the fact that the set 𝐴𝑋 + 𝑋𝐵 ∈ 𝐶 𝑛×𝑛 is the tangent space (calculated at point 𝐴) of the orbit of matrix 𝐴 under the action of congruences. The codimension of this orbit is exactly the dimension of the space of solutions to the equation 𝐴𝑋 + 𝐴𝑋 = 0. The main result of work [2,3] was the establishment of the canonical structure of matrices in general position with respect to congruences. In a similar way, the same authors in [4] studied the equation 𝐴𝑋 + 𝐴𝑋 = 0. In the publication [4] the equations 𝐴𝑋 + 𝑋𝐵 = 𝐶 arise when constructing an algorithm for palindromic eigenvalue problems 𝐴𝑥 = 𝜆𝐴 𝑇 𝑥 .

In the process of reducing matrix A to antitriangular form, the need arises to solve these equations numerically. In article [5], the conditions for unique solvability and a numerical algorithm for solving the equation 𝐴𝑋 + 𝑋𝐵 = 𝐶 are formulated.

By analogy with equations of Sylvester type, equations of Stein type will be called equations 𝑋 + 𝐴𝑋𝐵 = 𝐶. The first of them was partially studied in publication [6]. The question naturally arises about the completion of this study. The topic of solvability of the other two equations, on the contrary, is explored exhaustively in this publication.

No less interesting is the problem of obtaining sufficient solvability conditions for the Sylvester matrix equation

𝐴 𝑇 𝐻 + 𝐻𝐴 + 𝐵 𝑇 𝐻𝐵 = −𝐶,

where 𝐶 is a positive definite matrix. The need to solve such equations is emphasized in [13][14].

It is noted in [9] that there are now several approaches to solving the Sylvester equation. The first is to reduce the matrix equation to a vector (linear algebraic equation of increased dimension) and then the condition for the solvability of this equation is expressed through the non-degeneracy of the corresponding matrix. The second approach uses the small parameter method. It is also possible to obtain a spectral sparsity criterion for an equation with mutually commutable matrices. An analytical solution of this equation is possible only for the case of the two-term Sylvester equation.

At the same time, the question of the existence of a positive definite solution in the general case remains open. The procedure for finding this solution is also of interest for another investigations [15][16]. The purpose of the article is to obtain sufficient conditions for the existence of a solution to the Sylvester matrix equation on a set of positive definite matrices. The article will also present an iterative procedure and algorithm for finding these solutions.

Main result

The approach is based on a modification of the contraction mapping method. If on a complete metric space 𝑀 the following operator is specified 𝐹[𝑥], 𝑥 ∈ 𝑀, that maps points 𝑥 ∈ 𝑀to points of the same space 𝐹[𝑥] ∈ 𝑀 and satisfies the contraction condition 𝜌(𝐹[𝑥], 𝐹[𝑦]) ≤ 𝛼𝜌(𝑥, 𝑦), 0 < 𝛼 < 1, where 𝜌(𝑥, 𝑦) is the metric of the space M, then the operator equation

𝑥 = 𝐹[𝑥]

has a unique solution 𝑥 * ∈ 𝑀, and it can be found by the method of successive iterations

𝑥 = lim 𝑛→∞ 𝑥 𝑛 , 𝑥 𝑛 = 𝐹[𝑥 𝑛−1 ], 𝑛 = 1,2, … , 𝑥 0 = 𝑥 0 .

Let 𝑥 0 -be an arbitrary point in space, H is some complete metric space with metric 𝜌(𝐻 1 , 𝐻 2 ). Let us fix the following operators: 𝐹, 𝐺: 𝐻 → 𝐻.

Consider the operator equation

𝐹[𝐻] = 𝐺[𝐻]

(1)

Let us show that to solve this equation we can use an iterative procedure based on the following implicit scheme

𝐹[𝐻 𝑠+1 ] = 𝐺[𝐻 𝑠 ], 𝐻 0 = 𝐻 0 , 𝑠 = 1,2, . . , ; 𝐻 0 ∈H (2)

Definition. We call method (2) converging to the solution of equation ( 1) if the sequence {𝑦 𝑠 }, 𝑦 𝑠 = 𝐹[𝐻 𝑠 ], at 𝑠 → ∞ converges. Let us prove a theorem whose conditions ensure the convergence of method (2) to the solution of equation ( 1).

Let us assume that the solution to equation ( 1

𝜌(𝐺[𝐻 1 ], 𝐺[𝐻 2 ]) < 𝑞 𝜌(𝐹[𝐻 1 ], 𝐹[𝐻 2 ]), 0 < 𝑞 < 1,(3)

and

𝜌(𝐺[𝐴], 𝐹[𝐴]) < (1 − 𝑞)𝑟 ,(4)

then operator equation ( 1) has a unique solution in the set 𝐻(𝑎, 𝑟) and the sequence {𝑦 𝑠 }, 𝑦 𝑠 = 𝐹[𝐻 𝑠 ] constructed according to scheme (2) converges to 𝑦 * = 𝐹[𝐻 * ] for any starting point 𝐻 0 𝜖𝐻(𝐴, 𝑟). For the error of method (2), the following estimate is valid:

𝜌(𝐹[𝐻 𝑘 ], 𝐹[𝐻 * ]) < 𝑞 𝑘 1−𝑞 𝜌(𝐺[𝐻 0 ], 𝐹[𝐻 0 ]).(5)

Proof. Let 𝐻 0 -be an arbitraryelement of H(A, r) . Let us prove that the sequence 𝐻 𝑘 built according to scheme (2) will not leave the set H(A, r). By condition 𝐻 0 ∈ 𝐻(𝐴, 𝑟). Let us assume that 𝐻 𝑘 ∈ 𝐻(𝐴, 𝑟)for some fixed k. Let us show that then 𝐻 𝑘+1 ∈ 𝐻(𝐴, 𝑟). Consider the equality:

𝐹[𝐻 𝑠+1 ] = 𝐺[𝐻 𝑠 ]

Let us subtract the 𝐹[𝐴] from both sides of the above mentioned equation, we get

𝐹[𝐻 𝑘+1 ] − 𝐹[𝐴] = 𝐺[𝐻 𝑘 ] − 𝐹[𝐴] = (𝐺[𝐻 𝑘 ] − 𝐺[𝐴]) + (𝐺[𝐴] − 𝐹[𝐴])

Then the inequality holds

𝜌(𝐹[𝐻 𝑘+1 ], 𝐹[𝐴]) < 𝜌(𝐺[𝐻 𝑘 ], 𝐺[𝐴]) + 𝜌(𝐺[𝐴], 𝐹[𝐴]).

Using the induction hypothesis, we have . And according to (6) we get: Let us prove the estimate of the error of method (2). Let k be fixed and 𝑝 → ∞. Then from ( 7) we obtain

𝜌(𝐺[𝐻 𝑘 ], 𝐺[𝐴]) < 𝑞𝜌(𝐹[𝐻𝜌(𝐹[𝐻 𝑘+𝑝 ], 𝐹[𝐻 𝑘 ]) < 𝑞 𝑘 1−𝑞 𝜌(𝐹[𝐻 1 ], 𝐹[𝐻 0 ]).𝜌(𝐹[𝐻 𝑘 ], 𝐹[𝐻 * ]) < 𝑞 𝑘 1−𝑞 𝜌(𝐹[𝐻 1 ], 𝐹[𝐻 0 ]), 𝑘 = 1,2, 𝐾.

Therefore the theorem is proven.

Existance of solution

Denote as H the space of symmetric matrices with the metric -𝜌(𝐻 1 , 𝐻 2 ) = |𝐻 1 − 𝐻 2 |. Let us write the Lagrange formula for the operators F and G. We have:

𝐹[𝐻 1 ] − 𝐹[𝐻 2 ] = 𝐻 2 (𝑥)(𝐻 1 − 𝐻 2 ), 𝐺[𝐻 1 ] − 𝐺[𝐻 2 ] = 𝐻 1 (𝑥)(𝐻 1 − 𝐻 2 ),

Where 𝐻 1 (.), 𝐻 2 (.) are the Gâteaux derivatives of the operators 𝐹 and 𝐺 at some midpoint. Assume that there is an inverse operator 𝐻 2 −1 . Then we get:

𝐺[𝐻 1 ] − 𝐺[𝐻 2 ] = 𝐻 1 (𝑥)𝐻 2 −1 (𝑥)(𝐹[𝐻 1 ] − 𝐹[𝐻 2 ]

).

Thus we have| 𝐺[𝐻 1 ] − 𝐺[𝐻 2 ]| ≤ |𝐻 1 (𝑥)𝐻 2 −1 (𝑥)||𝐹[𝐻 1 ] − 𝐹[𝐻 2 ]|(8)

It follows that condition (3) of Theorem 1 will be satisfied if

|𝐻 1 (𝑥)𝐻 2 −1 (𝑥)| ≤ 𝑞 < 1. (9)

This condition is not always convenient. Using Theorem 1 and inequality (9), we obtain the conditions for the solvability of the Sylvester matrix equation.

Let us define the operators F and G as follows:

𝐹[𝐻]=-𝐴 𝑇 𝐻 − HA, 𝐺[𝐻] = 𝐶 + 𝐵 𝑇 HB,

where 𝐴, 𝐵 are some matrices, 𝐶 is a positive definite matrix. Then the Sylvester matrix equation can be rewritten as:

𝐹[𝐻] = 𝐺[𝐻].(10)

Let us obtain constructive solvability conditions for this equation. Lemma 1. A necessary condition for the solvability of the Sylvester matrix equation on the set of positive definite matrices is that the matrix 𝐴 is Hurwitz.

Proof. Let 𝐻 0 is a positive definite solution to Sylvester's equation. Consider the expression −𝐴 𝑇 𝐻 0 − 𝐻 0 𝐴 = 𝐶 + 𝐵 𝑇 𝐻 0 𝐵.

Let's denote 𝐶 1 = 𝐶 + 𝐵 𝑇 𝐻 0 𝐵. Because 𝐶 and 𝐻 0 are positive definite, then 𝐶 1 will be a positive definite. Then −𝐴 𝑇 𝐻 0 − 𝐻 0 𝐴 = 𝐶 1 is a matrix Lyapunov equation and since 𝐻 0 is his solution, then 𝐴 is a Hurwitz matrix.

Iteration procedure

Lemma 2. Let matrices 𝐴, 𝐵 satisfy the following conditions:

1) 𝐴is Hurwitz matrix, ( 11)

2)|(𝐵 𝑇 ×B)(−𝐴 𝑇 ×I − I×A) −1 | ≤ 𝑞,(12)

where × denotes the Kronecker product, then there is a unique positive definite solution to the Sylvester matrix equation and it can be found using the iterative procedure −𝐴 𝑇 𝐻 𝑘+1 − 𝐻 𝑘+1 𝐴 = 𝐶 + 𝐵 𝑇 𝐻 𝑘 𝐵, where 𝐶 is an arbitrary positive definite matrix.

Proof. Let us apply inequality (9) to the operators defining the Sylvester equation. For operator 𝐺 we have:

𝛨 1 (𝜉) = 𝐵 𝑇 ×B.

Since 𝐴 is Hurwitz, then the operator 𝐻 2 −1 (𝜉) exists and can be written

𝛨 2 −1 (𝜉) = (−𝐴 𝑇 xI − IxA) −1 .

Consequently, we find that inequality (12) ensures that condition (3) of Theorem 1 is satisfied. Let us take as the working set the entire space of positive definite matrices, then 𝑟=∞. Thus, we obtain conditions for the existence and uniqueness of a solution to the Sylvester matrix equation.

Let us show that the solution obtained by the iterative procedure −𝐴 𝑇 𝐻 𝑘+1 − 𝐻 𝑘+1 𝐴 = 𝐶 + 𝐵 𝑇 𝐻 𝑘 𝐵 will be a positive definite matrix. Let us prove this by mathematical induction. Matrix 𝐻 0 positive definite -due to the choice of the starting point. Let 𝐻 𝑘 is a positive definite matrix. Let's show positive definiteness 𝐻 𝑘+1 . Consider the equation −𝐴 𝑇 𝐻 𝑘+1 − 𝐻 𝑘+1 𝐴 = 𝐶 𝑘 , where 𝐶 𝑘 = 𝐶 + 𝐵 𝑇 𝐻 𝑘 𝐵,𝑘 = 1,..., 𝑛,... Since 𝐴 is Hurwitz and 𝐶 𝑘 -is always positive definite, then𝐻 𝑘+1 , as a solution to the Lyapunov equation will be a positive definite matrix. The lemma is proven. where matrix 𝐴 is Hurwitz. We substitute the values of matrices 𝐴 and 𝐵 into condition (12), carry out calculations and find that all elements in the resulting matrix are less than 1 in absolute value, which indicates that the condition is met.

Application examples

In the second example