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

Authors: Yunbo Tian; Chao Xia
Journal: Journal of Mathematics
 (2021)

---

## Abstract

We study the low-degree solution of the Sylvester matrix equationA1λ+A0Xλ+YλB1λ+B0=C0, where A1λ+A0 and B1λ+B0 are regular. Using the substitution of parameter variables λ, we assume that the matrices A0 and B0 are invertible. Thus, we prove that if the equation is solvable, then it has a low-degree solution Lλ,Mλ, satisfying the degree conditions δLλ<IndA0−1A1andδMλ<IndB1B0−1.


---

## Body

## 1. Introduction

Dealing with the problems about regulation output in control theory leads to a generalized Sylvester matrix equation:(1)AX+YB=C,where the matrices involved are the matrix polynomials (e.g., [1–7]). There have been an extensive study and application of the generalized Sylvester matrix equation (e.g., [8–11]). This 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 Hλ=∑i=0lHiλi with Hl≠0, and we denote the degree of matrix polynomial Hλ by δH=l. If Hλ=0, we set the degree δH=−∞. A matrix polynomial Hλ is called regular if detHλ≠0 and monic if Hl=I is an identity matrix.Wimmer used Jordan chains of polynomial to characterize the solvability of the generalized Sylvester equation in [13]. This 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 δC≤m+n−1, he proved that equation (1) has a unique solution satisfying δX<n,δY<m if and only if the determinants of Aλ and Bλ are coprime. Feinstein and Bar-Ness [17] extended this result to the case with only Aλ or Bλ (not necessarily both) being regular. The solvability of matrix equation (1) was also studied in [18, 19].It is well known that there are polynomialsuλ and vλ, such that(2)uλfλ+vλgλ=gcdfλ,gλ,and δu<δg,δv<δf, where gcdfλ,gλ is the monic greatest common factor of fλ,gλ. When we consider the case in Sylvester matrix polynomial equation (1), it is shown that for monic matrix polynomials Aλ and Bλ, 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 Aλ and Bλ 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(3)A1λ+A0Xλ+YλB1λ+B0=C0,where A1λ+A0 and B1λ+B0 are regular. We use the index of matrix to characterize the bound of low-degree solution.


## 2. Preliminary

To prove Theorem2, we first recall the division of matrix polynomials [12]. We restrict ourselves to the case when the dividend is a general matrix polynomial:(4)Hλ=∑i=0lHiλi,and the divisor is a monic matrix polynomial:(5)Xλ=∑i=0m−1Xiλi+Iλm.In this case, we have the following representation:(6)Hλ=QrλXλ+Rrλ,where Qrλ is a matrix polynomial, which is called the right quotient, and Rrλ is a matrix polynomial satisfying δRrλ<m. The matrix polynomial Rrλ is called the right remainder on division of Hλ by Xλ. Similarly, we have(7)Hλ=XλQlλ+Rlλ,where Qrλ is the left quotient, and Rlλ is the left remainder.Lemma 1 (Lemma 3.1 in [20]).
Suppose that matrix polynomialsAλ and Bλ are monic, δA=m,δB=n,δC<m+n. If the matrix polynomial equation,(8)AλXλ+YλBλ=Cλ,is solvable, then it has a solution Xλ,Yλ satisfying δX<n, δY<m.The following definitions about the Drazin inverse can be found in [21].Definition 1.
The smallest positive integerk for which(9)rankAk=rankAk+1,holds is called the index of A and denoted by Ind A.Definition 2.
LetA be a r×r matrix, k=IndA. If X satisfies equations(10)AX=XA,Ak+1X=Ak,AX2=X,then X is called the Drazin inverse of A and denoted by Ad.


## 3. Main Result

We start with the following theorem about the special Sylvester matrix equation.Theorem 1.
SupposeA,B∈ℂr×r, n≥k=IndB. Then, the equation(11)AX−XB=CBn,has a solution if and only if the equation(12)AX−XB=CBk,has a solution.Proof.
Suppose equation (12) holds. Multiplying Bn−k on the right side of equation (12), we have(13)AXBn−k−XBn−kB=CBn.
This means equation (11) holds.
On the other hand, by the property of the Drazin inverse of matrix, there exists a matrixBd satisfying(14)BBd=BdB,BkBd=Bk−1.
MultiplyingBdn−k on the right side of equation (11), we have(15)AXBdn−k−XBBdn−k=CBnBdn−k.
By equation (14), we have the following equation:(16)AXBdn−k−XBdn−kB=CBk.
Thus, there exists a matrixY=XBdn−k that satisfies(17)AY−YB=CBk.
The proof is completed.With the help of Lemma1 and Theorem 1, we can now prove the main result in this study.Theorem 2.
SupposeA0,B0∈ℂr×r are invertible matrices. If the Sylvester matrix equation(18)A1λ+A0Xλ+YλB1λ+B0=C0,has solutions, then it has low-degree solution Lλ,Mλ satisfying(19)maxδLλ,δMλ≤minIndA0−1A1,IndB1B0−1−1.Proof.
Suppose that the Sylvester matrix equation(20)A1λ+A0Xλ+YλB1λ+B0=C0,has a solution Xλ,Yλ, where(21)Xλ=Xnλn+Xn−1λn−1+⋯+X0,Yλ=Ymλm+Ym−1λm−1+⋯+Y0.
Then,(22)A11λ+A0X1λ+Y1λB11λ+B0=C0.
Without loss of generality, we may assume thatn≥m. Multiplying both sides of equation (22) by λn+1, left side of equation (22) by A0−1, and right side of equation (22) by B0−1, we have(23)A+IλX˜λ+Y˜λB+Iλ=Cλn+1,where A=A0−1A1,B=B1B0−1,C=A0−1C0B0−1, and(24)X˜λ=XnB0−1+Xn−1B0−1λ+⋯+X0B0−1λn,Y˜λ=A0−1Ymλn−m+A0−1Ym−1λn−m+1+⋯+A0−1Y0λn.
By the division of matrix polynomials, we understand a representation in the form(25)Cλn+1=QλIλ+B+−1n+1CBn+1,where Qλ=Cλn−CBλn−1+⋯+−1nCBn.
By substituting expression (25) into equation (23), it can be represented as(26)A+IλX˜λ+Y˜λ−QλB+Iλ=−1nCBn.
By Lemma1, equation (26) being solved is equivalent to there exists H,G∈ℂr×r satisfying(27)A+IλH+GB+Iλ=−1nCBn,i.e.,(28)AH+Hλ+GB+Gλ=−1nCBn.
By comparing the coefficients ofλ on both sides of equation (28), we have H+G=0. Then, replacing G by −H, equation (27) can be reduced to(29)AH−HB=−1nCBn.
By Theorem1, there exists a matrix D, such that(30)AD−DB=−CBk,where k=IndB=IndB1B0−1.
Let(31)Lλ=−1k−1DB0λk−1,Mλ=A0−1kDλk−1+−1k−1CBk−1λk−1+⋯+−1λCB+C.
Then, we have(32)A1λ+A0Lλ+MλB1λ+B0=A0Aλ+ILλ+MλBλ+IB0=A0−1k−1ADλk+−1kDBλk+−1k−1CBkλk+CB0=−1k−1A0AD−DB+CBkB0λk+A0CB0=A0CB0=C0,δLλ≤k−1=IndB1B0−1−1,δMλ≤k−1=IndB1B0−1−1.
Similarly, equation (18) has a solution Lλ,Mλ satisfying(33)δLλ≤IndA0−1A1−1,δMλ≤IndA0−1A1−1.
This completes the proof of Theorem2.We use the above theorem to calculate an example. This example also shows that the degree bound in Theorem1 is the lowest one.Example 1.
Consider equation(34)−213−11110−2λ+I3Xλ+Yλ1−2010−11−21λ+I3=I3,where I3=100010001.
By Theorem2 and(35)Ind−213−11110−2=2,Ind1−2010−11−21=2,we assume δX=1 and δY=1, i.e.,(36)Xλ=X1λ+X0,Yλ=Y1λ+Y0.
If we plug (36) into (34) and solve the equation, we obtain a solution(37)Xλ=3−4−31−2−112−2λ+21−400002−2,Yλ=−1210100−10λ+−11201000−1.
Actually, there is no matrix polynomial with degree 0 satisfying equation (34).



---

*Source: 4612177-2021-07-31.xml*