On the Low-Degree Solution of the Sylvester Matrix Polynomial Equation

We study the low-degree solution of the Sylvester matrix equation ( A 1 λ + A 0 ) X ( λ ) + Y ( λ )( B 1 λ + B 0 ) � C 0 , where A 1 λ + A 0 and B 1 λ + B 0 are regular. Using the substitution of parameter variables λ , we assume that the matrices A 0 and B 0 are invertible. Thus, we prove that if the equation is solvable, then it has a low-degree solution ( L ( λ ) , M ( λ )) , satisfying the degree conditions δ L ( λ ) < Ind ( A − 10 A 1 ) and δ M ( λ ) < Ind ( B 1 B − 10 )


Introduction
Dealing with the problems about regulation output in control theory leads to a generalized Sylvester matrix equation: where the matrices involved are the matrix polynomials (e.g., [1][2][3][4][5][6][7]). ere have been an extensive study and application of the generalized Sylvester matrix equation (e.g., [8][9][10][11]). is work investigates the bound of the low-degree solution of equation (1) with degree 1 matrix polynomials A and B.
We adopt the following terminology [12]. Let H(λ) � l i�0 H i λ i with H l ≠ 0, and we denote the degree of matrix polynomial H(λ) by δ(H) � l. If H(λ) � 0, we set the degree δ(H) � −∞. A matrix polynomial H(λ) is called regular if det H(λ) ≠ 0 and monic if H l � I is an identity matrix.
Wimmer used Jordan chains of polynomial to characterize the solvability of the generalized Sylvester equation in [13].
is condition about solvability extends results of Kučera [14] and Gohberg and Lerer [15]. In [16], Barnett studied the existence and uniqueness of the low-degree solutions. For monic matrices A, B, and δ(C) ≤ m + n −1, he proved that equation (1) has a unique solution satisfying δ(X) < n, δ(Y) < m if and only if the determinants of A(λ) and B(λ) are coprime. Feinstein and Bar-Ness [17] extended this result to the case with only A(λ) or B(λ) (not necessarily both) being regular. e solvability of matrix equation (1) was also studied in [18,19].
It is well known that there are polynomials u(λ) and v(λ), such that and δu < δg, δv < δf, where gcd(f(λ), g(λ)) is the monic greatest common factor of f(λ), g(λ). When we consider the case in Sylvester matrix polynomial equation (1), it is shown that for monic matrix polynomials A(λ) and B(λ), if equation (1) has solutions, then it has a solution (X, Y) satisfying δX < δB, δY < δA. However, the remark in [20] shows that the proposition is false when A(λ) and B(λ) are not monic matrix polynomials. As equation (1) with regular matrix polynomials A and B have not been developed fully about the degree, we will investigate the low-degree solution of where A 1 λ + A 0 and B 1 λ + B 0 are regular. We use the index of matrix to characterize the bound of low-degree solution.

Preliminary
To prove eorem 2, we first recall the division of matrix polynomials [12]. We restrict ourselves to the case when the dividend is a general matrix polynomial: and the divisor is a monic matrix polynomial: In this case, we have the following representation: where Q r (λ) is a matrix polynomial, which is called the right quotient, and R r (λ) is a matrix polynomial satisfying where Q r (λ) is the left quotient, and R l (λ) is the left remainder.
e following definitions about the Drazin inverse can be found in [21].
e smallest positive integer k for which holds is called the index of A and denoted by Ind A.
then X is called the Drazin inverse of A and denoted by A d .

Main Result
We start with the following theorem about the special Sylvester matrix equation. Ind(B). en, the equation has a solution if and only if the equation has a solution.
Proof. Suppose equation (12) holds. Multiplying B n−k on the right side of equation (12), we have is means equation (11) holds. On the other hand, by the property of the Drazin inverse of matrix, there exists a matrix B d satisfying Multiplying (B d ) n−k on the right side of equation (11), we have By equation (14), we have the following equation: us, there exists a matrix Y � X(B d ) n−k that satisfies e proof is completed.

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With the help of Lemma 1 and eorem 1, we can now prove the main result in this study. Theorem 2. Suppose A 0 , B 0 ∈ C r×r are invertible matrices. If the Sylvester matrix equation has solutions, then it has low-degree solution (L(λ), M(λ)) satisfying Proof. Suppose that the Sylvester matrix equation has a solution (X(λ), Y(λ)), where X(λ) � X n λ n + X n−1 λ n−1 + · · · + X 0 , en, By the division of matrix polynomials, we understand a representation in the form where Q(λ) � Cλ n − CBλ n−1 + · · · + (−1) n CB n . By substituting expression (25) into equation (23), it can be represented as By Lemma 1, equation (26) being solved is equivalent to there exists H, G ∈ C r×r satisfying i.e., By comparing the coefficients of λ on both sides of equation (28), we have H + G � 0. en, replacing G by −H, equation (27) can be reduced to AH − HB � (−1) n CB n . (29) By eorem 1, there exists a matrix D, such that (31) en, we have Similarly, equation (18) has a solution (L(λ), M(λ)) satisfying is completes the proof of eorem 2.

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We use the above theorem to calculate an example. is example also shows that the degree bound in eorem 1 is the lowest one.

Journal of Mathematics
where By eorem 2 and Ind −2 1 3 we assume δX � 1 and δY � 1, i.e., If we plug (36) into (34) and solve the equation, we obtain a solution Actually, there is no matrix polynomial with degree 0 satisfying equation (34).

Data Availability
No data, models, or codes were generated or used to support this study.

Conflicts of Interest
e authors declare that they have no conflicts of interest.