# Serre’s Reduction and the Smith Forms of Multivariate Polynomial Matrices

Authors: Dongmei Li; Rui Liang
Journal: Mathematical Problems in Engineering
 (2020)

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## Abstract

The equivalence of systems plays a critical role in multidimensional systems, which are usually represented by the multivariate polynomial matrices. The Smith form of a matrix is one of the important research contents in polynomial matrices. This paper mainly investigates the Smith forms of some multivariate polynomial matrices. We have obtained several new results and criteria on the reduction of a given multivariate polynomial matrix to its Smith form. These criteria are easily checked by computing the minors of lower order of the given matrix.


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## Body

## 1. Introduction

The subject of multidimensionalnD systems is concerned with a mathematical framework for tackling a broad range of paradigms whose analysis or synthesis requires the use of functions and polynomials in several complex variables. Many physical systems, multiple-input multiple-output systems, data analysis procedures, and learning algorithms have a natural nD structure due to the presence of one spatial variable. So the theory of nD systems is widely applied in areas of image processing, linear multipass processes, geophysical exploration, iterative learning control systems, etc. [1–9]. The equivalence of systems is one important research problem in the nD system theory. It is often required to transform a given system into a simpler but equivalent form. As we all know, a multivariate nD polynomial matrix is often used to represent an nD system. So the equivalence problem of nD system is often transformed into the equivalence problem of nD polynomial matrices. For 1D systems, the equivalence problem has been solved [5, 7] by the quite mature theory of 1D polynomial matrix. For nD (n≥2) case, since the equivalence problem is equivalent to a highly difficult problem, the isomorphism problem of two finitely presented modules, there is no hope that the equivalence problem can be solved completely. There exist two primary problems on equivalence of multivariate nD polynomial matrices: one is to reduce an nD polynomial matrix to its Smith form. Kung et al. have obtained some interesting results about the equivalence of nD polynomial matrices to their Smith forms [5, 6, 10–13]. Furthermore, the Smith forms of some nD polynomial matrices can be computed by Maple [14]. The other is called Serre’s reduction problem. One of the motivations for doing Serre’s reduction for an nD polynomial matrix is to reduce an nD system to an equivalent system containing fewer equations and unknowns. Cluzeau et al. have studied Serre’s reduction and presented some new interesting results in [15, 16].The following problem, proposed by Serre in 1960s, plays an important role in the research problems ofnD systems. It is not only the problem of reducing a matrix to its Smith form, but also Serre’s reduction problem.Problem 1.
When is annD n≥2 polynomial matrix Fz equivalent to the matrix(1)Sz=Il−10l−1,101,l−1d,where d=detFz? Il−1 is the l−1×l−1 identity matrix and 0l,m is the l×m zero matrix.
For the real number fieldℝ, Lin et al. [12] have investigated Problem 1 for Fz∈ℝl×lz with detFz=z1−fz2,…,zn and proved that Fz is equivalent to its Smith form. Li et al. [17] have investigated Problem 1 for Fz∈ℝl×lz with detFz=z1−fz2,…,znq and obtained that Fz is equivalent to its Smith form with Fz satisfying some criteria. Cluzeau et al. [13–16] also studied it and gave some new interesting results.
In this paper, we will investigate Problem1 for the case of Fz∈Kl×lz and det Fz=z1−f1z2,…,znq1z2−f2z3,…,znq2, where q1,q2 are nonnegative integers, K is an arbitrary field (even a function field or a finite field). Then, we investigate the Smith forms of some rectangular polynomial matrices and consider the following problem.Problem 2.
LetFz∈Kl×mzl≤m with dFz=z1−fz2,…,znq1⋅z2−f2z3,…,znq2, where dFz is the greatest common divisor of the l×l minors of Fz, q1,q2 are nonnegative integers. When is Fz equivalent to its Smith form?
The paper is organized as follows. In Section2, we give some basic concepts on the equivalence of nD polynomial matrices. In Section 3, main results and some tractable criteria on equivalence of several kinds of polynomial matrices are proposed. In Section 4, an example is provided to illustrate the effectiveness of our constructive method.


## 2. Preliminaries

In the following,Kz=Kz1,z2,…,zn will denote the polynomial ring in n variables z1,z2,…,zn with coefficients in arbitrary field K, K¯ will be an algebraic closed field of K, K¯n will be the n dimensional vector space over K¯, Kl×mz will denote the set of l×m matrices with their entries in Kz, Kl×mz2,…,zn will denote the set of l×m matrices with their entries in Kz2,…,zn, the r×r identity matrix will be denoted by Ir and the r×t zero matrix will be denoted by 0r,t. A matrix over Kz with its determinant in K is said to be unimodular. Throughout the paper, the argument z is omitted whenever its omission does not cause confusion.Definition 1.
(see [18]).Let Fz∈Kl×mz be of full row (column) rank. Then Fz is said to be zero left prime (zero right prime) if the l×lm×m minors of Fz generate the unit ideal Kz.
IfFz∈Kl×mz is zero left prime (zero right prime), then Fz is called simply to be ZLP (ZRP). According to the Quillen-Suslin theorem [19], we have that Fz is ZLP if and only if there is a matrix Nz∈Km×mz such that Fz⋅Nz=Il0l×m−l. It is also equivalent to say that any ZLP (ZRP) matrix over Kz can be completed to an unimodular (invertible) matrix.Definition 2.
LetFz∈Kl×mz, l≤m, and Φi be polynomially defined as follows:(2)Φi=diF/di−1F,1≤i≤r,0,r<i≤l,where r is the rank of Fz, d0F≡1, diF is the greatest common divisor of the i×i minors of Fz, and Φi satisfies ΦiΦi+1, i=1,…,l. Then, we define the Smith form of Fz as(3)S=diagΦi0l×m−l.Definition 3.
(see [17]). Let T1z and T2z denote two matrices in Kl×mz, T1z, and T2z are said to be equivalent if there exist two invertible matrices Mz∈Kl×lz and Nz∈Km×mz such that T2z=MzT1zNz.


## 3. Main Results

In this section, the main results are presented. First, we give some well-known results and provide an answer to Problem1 for case of detFz=z1−f1z2,…,znz2−f2z3,…,zn in Subsection 3.1. Then we extend this result to more general case of detFz=z1−f1z2,…,znz2−f2z3,…,znq2 and present a complete answer to Problem 2 in Subsection 3.2.


### 3.1. Equivalent Theorem

In order to prove our main results, we first give several useful lemmas.Lemma 1.
(see [20]). Let f1z,…,fsz∈Kz, then f1z,…,fsz have no common zeros in K¯n (are zero coprime) if and only if there exist u1z,…,usz∈Kz such that(4)u1zf1z+⋯+uszfsz=1,i.e., f1z,…,fsz is a ZLP row vector in K1×sz, or f1z,…,fsz generate the unit ideal Kz.Lemma 2.
(see [20]). Let gz∈Kz1,z2,…,zn and fz2,…,zn∈Kz2,…,zn. Suppose that gfz2,…,zn,z2,…,zn=0, then z1−fz2,…,zn is a divisor of gz.Lemma 3.
(see [17]). Let Fz,F1z,F2z∈Kl×lz, Fz=F1z⋅F2z. If the l−1×l−1 minors of Fz have no common zeros in K¯n (generate Kz), then the l−1×l−1 minors of Fiz have no common zeros in K¯n (generate Kz) for i=1,2.Proof.
The proof is similar to that of Lemma 2.2 in [17], so it is omitted here.Lemma 4.
(see [18]). Let Q∈Kl−1×lz be of normal full rank, if the reduced minors of Q generate Kz, then there exists a ZLP matrix w∈Kl×1z such that Q⋅w=0l−1,1.The following result is presented in [20], for the convenience of the reader, we record it here.Lemma 5.
(see [20]). Let Fz=Fz1,z2,…,zn∈Kl×lz, fz2,…,zn∈Kz2,…,zn, if the l−1×l−1 minors of Fz generate Kz, then the l−1×l−1 minors of Ffz2,…,zn,z2,…,zn also generate Kz.Proof.
Assume that thel−1×l−1 minors of Fz are c1z1,…,zn,…,crz1,…,zn. If the l−1×l−1 minors of Fz generate Kz, then there exist u1z1,z2,…,zn,…,urz1,z2,…,zn∈Kz such that(5)u1z1,z2,…,znc1z1,z2,…,zn+⋯+urz1,z2,…,zncrz1,…,zn=1.
Substitutez1=fz2,…,zn, then(6)u1fz2,…,zn,z2,…,znc1fz2,…,zn,z2,…,zn+⋯+urfz2,…,zn,z2,…,zncrfz2,…,zn,z2,…,zn=1.
It is straightforward thatc1fz2,…,zn,z2,…,zncrfz2,…,zn,z2,…,zn are the l−1×l−1 minors of Ffz2,…,zn,z2,…,zn. Thus the l−1×l−1 minors of Ffz2,…,zn,z2,…,zn also generate Kz.Lemma 6.
LetFz∈Kl×lz, then detFz and the l−1×l−1 minors of Fz have no common zeros if and only if the l−1×l−1 minors of Fz have no common zeros.Proof.
Necessity. Assuming that the l−1×l−1 minors of Fz have a common zero point a0, by Laplace expansion, we can easily know that a0 is also a zero point of detFz, contradicting the fact that detFz and the l−1×l−1 minors of Fz have no common zeros. Thus, the l−1×l−1 minors of Fz have no common zeros.
Sufficiency. It is straightforward.
For the convenience, we first defineP1z,P2z,Pz as follows:(7)P1z=Il−10l−1,101,l−1z1−f1z2,…,zn,P2z=Il−10l−1,101,l−1z2−f2z3,…,zn,Pz=Il−10l−1,101,l−1z1−f1z2,…,znz2−f2z3,…,zn.
We have the following key conclusion which is very important to derive our main results.Lemma 7.
LetFz∈Kl×lz and Fz=P1zVzP2z, where Vz∈Kl×lz2,…,zn is unimodular. If all the l−1×l−1 minors of Fz generate Kz, then Fz is equivalent to its Smith form Pz.Proof.
We prove this by induction onl. When l=2, let(8)Vz=v11v12v21v22,then(9)Fz=v11v12⋅z2−f2z3,…,znv21⋅z1−f1z2,…,znv22⋅z1−f1z2,…,znz2−f2z3,…,zn.
Obviously,detFz=c⋅z1−f1z2,…,znz2−f2z3,…,zn,0≠c∈K. Note that Vz is unimodular, then v11,v12 have no common zeros. Assume that v11,v12z2−f2z3,…,zn have a zero α0=z20,z30,…,zn0, where z20=f2z30,…,zn0, then v11,v12⋅z2−f2z3,…,zn,v21⋅z1−f1z2,…,zn,v22⋅z1−f1z2,…,znz2−f2z3,…,zn has a common zero f1α0,z20,z30,…,zno. This is a contradiction. Thus, v11,v12⋅z2−f2z3,…,zn have no common zeros, and v11,v12⋅z2−f2z3,…,zn is a unimodular row. According to Quillen-Suslin Theorem, there exists a 2×2 unimodular matrix M1z∈K2×2z such that(10)v11,v12z2−f2z3,…,znM1z=1,0.
So(11)FzM1z=10p1zp2z,where p1z,p2z∈Kz. Then there exists the unimodular matrix M2z=10−p1z1 such that(12)M2zFzM1z=100p2z.
Note thatp2z=detM2zFzM1z=u⋅detFz=u′⋅z1−f1z2,…,znz2−f2z3,…,zn, 0≠u,u′∈K. Thus, Fz is equivalent to its Smith form(13)Pz=100z1−f1z2,…,znz2−f2z3,…,zn.
So the conclusion is true forl=2.
Assume that the conclusion is true forl−1, we investigate the case of l. Let Vz=V11V12V21V22, where V11∈Kl−1×l−1z2,…,zn, V12∈Kl−1×1z2,…,zn, V21∈K1×l−1z2,…,zn, V22∈K1×1z2,…,zn. Then(14)Fz=V11V12⋅z2−f2z3,…,znV21⋅z1−f1z2,…,znV22⋅z1−f1z2,…,znz2−f2z3,…,zn.
Letv11,…,v1,l−1,v1l be the first row of Vz. Since Vz is unimodular, using Laplace Theorem and expanding its first row, we obtain that v11a1z+⋯+v1,l−1al−1z+v1lalz=k, where a1z,…,al−1z,az are all the l−1×l−1 minors of Vz and 0≠k∈K. Combined with Lemma 1, we have that v11,…,v1,l−1,v1l have no common zeros. Assume that v11,…,v1,l−1,v1lz2−f2z3,…,zn have a common zero α0=z20,z30,…,zn0, then all the l−1×l−1 minors of Fz have a common zero f1α0,z20,z30,…,zn0. This is a contradiction. Thus, v11,…,v1,l−1,v1lz2−f2z3,…,zn is a unimodular row in Kz2,…,zn, by the Quillen-Suslin Theorem, there exists a unimodular matrix M1z such that(15)v11,…,v1,l−1,v1lz2−f2z3,…,znM1z=1,0,…,0.
So(16)FzM1z=101,l−1Q1zQ2z,where Q1z∈Kl−1×1z, Q2z∈Kl−1×l−1z. Then there exists the matrix(17)M2z=101,l−1−Q1zIl−1,such that(18)M2zFzM1z=101,l−10l−1,1Q2z.
SettingNz=M2zFzM1z, since all the l−1×l−1 minors of Fz have no common zeros, combined with Lemma 3, we have that all the l−1×l−1 minors of Nz have no common zeros. Note that an l−1×l−1 minor of Nz is just an l−2×l−2 minors of Q2z or detQ2z or 0, then detQ2z and the l−2×l−2 minors of Q2z have no common zeros. By Lemma 6, the l−2×l−2 minors of Q2z have no common zeros. By the inductive hypothesis, there exist two l−1×l−1 unimodular matrices N1z,N2z such that(19)N2zQ2zN1z=Il−20l−2,101,l−2z1−f1z2,…,znz2−f2z3,…,zn,then(20)101,l−10l−1,1N2zNz,w101,l−101,l−1N1z=101,l−10l−1,1N2zQ2zN1z=Il−10l−1,101,l−1z1−f1z2,…,znz2−f2z3,…,zn.
Thus,Nz is equivalent to the matrix Pz, combined with Fz is equivalent to the matrix Nz; then we obtain that Fz is equivalent to the matrix Pz, and Pz is the Smith form of Fz.
Now we are going to state one of our main results, which give partial answer to Problem1. We recall the notation(21)Pz=Il−10l−1,101,l−1z1−f1z2,…,znz2−f2z3,…,zn.Theorem 1.
LetFz∈Kl×lz with detFz=z1−f1z2,…,znz2−f2z3,…,zn. Fz is equivalent to its Smith form Pz if and only if all the l−1×l−1 minors of Fz generate Kz.Proof.
Sufficiency. Because detFz1,f2z3,…,zn,z3,…,zn=0, then the rank of Fz1,f2z3,…,zn,z3,…,zn≤l−1. Since all the l−1×l−1 minors of Fz generate Kz, by Lemma 5, the l−1×l−1 minors of Fz1,f2z3,…,zn,z3,…,zn also generate Kz, then rankFz1,f2z3,…,zn,z3,…,zn=l−1 for every z1,z3,…,zn∈K¯. By Lemma 4, there exists a ZRP column vector Y1z∈Kl×1z such that(22)Fz1,f2z3,…,zn,z3,…,znY1z≡0,…,0T.
By the Quillen-Suslin theorem, anl×l unimodular matrix U11z can be constructed such that Y1z is its last column. Then the elements of the last column of Fz1,f2z3,…,zn,z3,…,zn⋅U11z are zero polynomials. By Lemma 2, the last column of Fz⋅U11z have the common divisor z2−f2z3,…,zn, i.e.,(23)Fz⋅U11z=F1z⋅Il−10l−1,101,l−1z2−f2z3,…,zn,for some F1z∈Kl×lz. Let U1z=U11−1z, then we have(24)Fz=F1zP2zU1z.
Note thatU1z,U11z are unimodular, and detFz=z1−f1z2,…,znz2−f2z3,…,zn, then(25)detF1z=c⋅z1−f1z2,…,zn,0≠c∈K.
From Lemma3, we have that the l−1×l−1 minors of F1z also generate Kz. Note that detF1f1z2,…,zn,z2,z3,…,zn=0, combined with Lemma 5, we obtain that the l−1×l−1 minors of F1f1z2,…,zn,z2,z3,…,zn also generate Kz for z2,z3,…,zn∈K¯. Similarly, a unimodular matrix V11z∈Kl×lz2,…,zn can be constructed such that(26)F1zV11z=F2z⋅Il−10l−1,101,l−1z1−f1z2,…,zn,where F2z∈Kl×lz. Let V1z=V11−1z, then we have(27)F1z=F2zP1zV1z.
Note thatV1z∈Kl×lz2,…,zn is unimodular, and detF1z=c1⋅z1−f1z2,…,zn, then detF2z=c2,0≠c2∈K, F2z is unimodular, and(28)Fz=F2zP1zV1zP2zU1z.
By Lemma7, the matrix P1zV1zP2z is equivalent to(29)Pz=Il−10l−1,101,l−1z1−f1z2,…,znz2−f2z3,…,zn.
Combined withF2z,U1z are unimodular, we obtain that Fz is equivalent to its Smith form Pz.
Necessity. If Fz is equivalent to Pz, there exist two unimodular matrices Uz,Vz such that Fz=VzPzUz. Note that all the l−1×l−1 minors of Pz have no common zeros; combined with Lemma 3, we have that all the l−1×l−1 minors of Fz have no common zeros, i.e., all the l−1×l−1 minors of Fz generate Kz.
In the following, we will extend the above result to more general case.


### 3.2. Generalization

Lemma 8.
LetDz∈Kl×lz with Dz=Il−100hz, hz∈Kz, and Vz∈Kl×lz be a unimodular matrix. If all the l−1×l−1 minors of DszVzDtz generate Kz and s,t are positive integers, then DszVzDtz is equivalent to Ds+tz.Proof.
Let(30)Vz=V11zV12zV21zV22z,where V11z∈Kl−1×l−1z, V12z∈Kl−1×1z, V21z∈K1×l−1z, V22z∈K1×1z.
Then(31)Fz=DszVzDtz=V11zV12zhtzV21zhszV22zhs+tz,and detFz=c⋅hs+tz,0≠c∈K.
Next, we will proof thatV11zV12zhtz is a ZLP matrix. Note that Vz is unimodular, then V11zV12z is a ZLP matrix. Let c1z,c2z,…,clz denote all the l−1×l−1 minors of V11zV12z. Then all the l−1×l−1 minors of V11zV12zhtz are c1z,c2zhtz,…,clzhtz. We can prove that c1z,c2zhtz,…,clzhtz have no common zeros. Suppose that c1z,c2zhtz,…,clzhtz have a common zero α0 and combined with c1z,c2z,…,clz have no common zeros, so α0 is a zero of c1z and hz. Note that the elements of the last row of matrix Fz all have the factor hz and that an l−1×l−1 minor of Fz is just c1z or includes the factor hz, so the l−1×l−1 minors of Fz have the common zero α0, this is a contradiction. Thus, all the l−1×l−1 minors of V11zV12zhtz have no common zeros, and it is a ZLP matrix. According to Quillen-Suslin theorem, there exists an m×m unimodular matrix N1z such that(32)V11zV12zhtzN1z=Il−1z0l−1,1.
Then,(33)FzN1z=Il−1z0l−1,1H1zH2z,for some H1z,H2z∈Kz. Let(34)N2z=Il−1z0l−1,1−H1z1.
Then,(35)N2zFzN1z=Il−1z0l−1,101,l−1H2z.
Note thatN1z,N2z are unimodular, then H2z=u⋅detFz=u′⋅hs+tz,0≠u∈K,u′=u⋅c, thus Fz is equivalent to Ds+tz.
Now we investigate Problem1 for the case of q1=q2. Denote(36)Pqz=Il−10l−1,101,l−1z1−f1z2,…,znqz2−f2z3,…,znq.Theorem 2.
LetFz∈Kl×lz with detFz=z1−f1z2,..,znqz2−f2z3,…,znq, where q is a positive integer. Then Fz is equivalent to its Smith form Pqz if and only if all the l−1×l−1 minors of Fz generate Kz.Proof.
Sufficiency. Notice that detFz1,f2z3,…,zn,z3,…,zn=0, we have that rank Fz1,f2z3,…,zn,z3,…,zn≤l−1. Since all the l−1×l−1 minors of Fz generate the unit ideal Kz, according to Lemma 5, then the l−1×l−1 minors of Fz1,f2z3,…,zn,z3,…,zn also generate the unit ideal Kz. Hence, rankFz1,f2z3,…,zn,z3,…,zn=l−1 for every z1,z3,…,zn∈K¯, repeating the proof of Theorem 1, we have that(37)Fz=F1zP1zV1zP2zU1z,where V1z∈Kl×lz2,…,zn, U1z∈Kl×lz are unimodular, F1z∈Kl×lz with detF1z=z1−f1z2,…,znq−1z2−f2z3,…,znq−1. By Lemma 3, all the l−1×l−1 minors of P1zV1zP2z also generate Kz. According to Lemma 7, there exist unimodular matrices M1z,N1z∈Kl×lz such that P1zV1zP2z=M1zP1zP2zN1z=M1zPzN1z. Thus(38)Fz=F1zM1zPzN1zU1z.
Again by Lemma3, we obtain that all the l−1×l−1 minors of F1z generate Kz. If q≥2, iterating the preceding process, we obtain that(39)F1z=F2zM2zPzN2z,where F2z∈Kl×lz with detF2z=z1−f1z2,…,znq−2z2−f2z3,…,znq−2. Then(40)Fz=F2zM2zPzN2zM1zPzN1zU1z.
LetS2z=N2zM1z, we have that S2z is unimodular and all the l−1×l−1 minors of PzS2zPz generate Kz. By Lemma 8, PzS2zPz is equivalent to P2z, that is, there exist unimodular matrices L1z,T1z∈Kl×lz such that PzS2zPz=L1zP2zT1z. Furthermore,(41)Fz=F2zM2zL1zP2zT1zN1zU1z.
LetV2z=M2zL1z, U2z=T1zN1zU1z, then V2z,U2z are unimodular, and(42)Fz=F2zV2zP2zU2z.
Ifq≥3, iterating the same procedure successively, we obtain that(43)Fz=FqzVqzPqzUqz,where Vqz,Uqz are unimodular. Note that detFz=z1−f1z2,…,znqz2−f2z3,…,znq, then detFqz=c,c∈K; that is, Fqz is a unimodular matrix. So Fz is equivalent to its Smith form Pqz.
Necessity. It is straightforward that all the l−1×l−1 minors of Pqz generate Kz. According to Lemma 3, all the l−1×l−1 minors of Fz also generate Kz.
Next we investigate Problem1 for the case of Fz∈Kl×lz and det Fz=z1−f1z2,…,znq1z2−f2z3,…,znq2, where q1,q2 are nonnegative integers.Theorem 3.
LetFz∈Kl×lz with detFz=z1−f1z2,…,znq1z2−f2z3,…,znq2, where q1,q2 are nonnegative integers. If all the l−1×l−1 minors of Fz generate Kz, then Fz is equivalent to its Smith form(44)Qz=Il−10l−1,101,l−1z1−f1z2,…,znq1z2−f2z3,…,znq2.Proof.
(1) Ifq1=q2=0, then detFz=1, i.e., Fz is unimodular, so Fz is equivalent to Il.
(2) Ifq1=0 or q2=0, we have that detFz=z1−f1z2,…,znq1 or z2−f2z3,…,znq2. Without loss of generality, we assume that detFz=z1−f1z2,…,znq1. Because detFf1z2,…,zn,z2,…,zn=0, then the rank of Ff1z2,…,zn,z2,…,zn≤l−1. Since all the l−1×l−1 minors of Fz generate Kz, by Lemma 5, the l−1×l−1 minors of Ff1z2,…,zn,z2,…,zn also generate Kz; hence, rankFf1z2,…,zn,z2,…,zn=l−1 for every z2,z3,…,zn∈K¯n−1. By Lemma 4, we obtain a ZRP column vector Y1z∈Kl×1z such that(45)Ff1z1,…,zn,z2,…,znY1z≡0,…,0T.
According to the Quillen-Suslin Theorem, anl×l unimodular matrix U11zdetU11z=1 can be constructed such that Y1z is its last column. Then the last column of Ff1z1,…,zn,z2,…,zn⋅U11z are zero polynomials. By Lemma 2, the last column of Fz⋅U11z have the common divisor z1−f1z2,…,zn, i.e.,(46)Fz⋅U11z=F1z⋅Il−10l−1,101,l−1z1−f1z2,…,zn,for some F1z∈Kl×lz. Let U1z=U11−1z, then we obtain that(47)Fz=F1zP1zU1z,detF1z=z1−f1z2,…,znq1−1.
By Lemma3, all the l−1×l−1 minors of F1z also generate the unit ideal Kz. If q1≥2, iterating the preceding process, we obtain that(48)F1z=F2zP1zU2z.
Furthermore,(49)Fz=F2zP1zU2zP1zU1z.
By Lemma3, the l−1×l−1 minors of P1zU2zP1z generate the ideal Kz, according to Lemma 8, there exist unimodular matrices N1z,V1z such that P1zU2zP1z=N1zP12zV1z. Then(50)Fz=F2zN1zP12zV1zU1z.
Furthermore,detF2zN1z=z1−f1z2,…,znq1−2, combined with Lemma 3, the l−1×l−1 minors of F2zN1z also generate Kz.
Ifq1≥3, iterating the same procedure successively, we obtain that(51)Fz=Fq1zNq1−1zP1q1zVq1−1z,…,V1zU1z,with detFq1zNq1−1z=1. Thus, Fz is equivalent to its Smith form P1q1z=Qz.
(3) Ifq1,q2 are positive integers, then q1≥q2 or q2≥q1. Without loss of generality, we assume that q1≥q2. By Theorem 2, combined with the conclusion of the case (2) above, we have that Fz is equivalent to the matrix Pq2zVzP1q1−q2z. In the following, we prove that Pq2zVzP1q1−q2z is equivalent to the matrix Qz
Let(52)Vz=V11zV12zV21zV22z,where V11z∈Kl−1×l−1z, V12z∈Kl−1×1z, V21z∈K1×l−1z, and V22z∈K1×1z.
Then(53)Pq2zVzP1q1−q2z=V11zV12zp1q1−q2zV21zp1q2zp2q2zV22zp1q1zp2q2z,where p1z=z1−f1z2,…,zn, p2z=z2−f2z3,…,zn.
In fact, we can prove thatV11zV12zp1q1−q2z is a ZLP matrix. Note that Vz is unimodular, then V11zV12z is a ZLP matrix. Let c1z,c2z,…,clz denote all the l−1×l−1 minors of V11zV12z. Then all the l−1×l−1 minors of V11zV12zp1q1−q2z are c1z,c2zp1q1−q2z,…,clzp1q1−q2z. We can prove that c1z,c2zp1q1−q2z,…,clzp1q1−q2z have no common zeros. Suppose that c1z,c2zp1q1−q2z,…,clzp1q1−q2z have a common zero α0, note that c1z,c2z,…,clz have no common zeros, so c1z and p1z have the common zero α0. Moreover, the elements of the last row of matrix Pq2zVzP1q1−q2z have the common factor p1z, since an l−1×l−1 minors of Pq2zVzP1q1−q2z is just c1z or includes the factor p1z, so the l−1×l−1 minors of Pq2zVzP1q1−q2z have the common zero α0, this is a contradiction. Thus, all the l−1×l−1 minors of V11zV12zp1q1−q2z have no common zeros, and it is a ZLP matrix. By the Quillen-Suslin theorem, there exists an l×l unimodular matrix Nz such that(54)V11zV12zp1q1−q2zNz=Il−1z0l−1,1.
Then,(55)Pq2zVzP1q1−q2zNz=Il−1z0l−1,1H1zH2z,for some H1z,H2z∈Kz. There exists a unimodular matrix N1z such that(56)N1zPq2zVzP1q1−q2zNz=Il−1z0l−1,101,l−1H2z.
Note thatNz,N1z are unimodular, then H2z=u⋅detPq2zVzP1q1−q2z=u′⋅p1q1zp2q2z,0≠u∈K,u′=u⋅c; thus, Pq2zVzP1q1−q2z is equivalent to Qz. Note that Fz is equivalent to the matrix Pq2zVzP1q1−q2z, combined with the definition of the Smith form of a matrix, we obtain that Fz is equivalent to its Smith form Qz.
Theorem3 provides a positive answer to Problem 1 for the case of Fz∈Kl×lz and det Fz=z1−f1z2,…,znq1z2−f2z3,…,znq2. It also gives a sufficient condition to check this kind of matrices are equivalent to their Smith forms; in fact, this condition is also a necessary condition.Theorem 4.
LetFz∈Kl×lz with det Fz=z1−f1z2,…,znq1z2−f2z3,…,znq2, where q1,q2 are non-negative integers, then Fz is equivalent to its Smith form(57)Qz=Il−10l−1,101,l−1z1−f1z2,…,znq1z2−f2z3,…,znq2,if and only if all the l−1×l−1 minors of Fz generate Kz.Proof.
Sufficiency. From Theorem 3, it is straightforward.
Necessity. By computing, we can easily obtain that all the l−1×l−1 minors of Qz generate Kz. By Lemma 3, all the l−1×l−1 minors of Fz generate Kz.Remark 1.
In the theorem above,K is an arbitrary field. When K=ℝ is the real field and q2=0, Theorem 4 is same as Theorem 2.5 in Li et al. [17]. Furthermore, if q1=1 and q2=0, Theorem 4 is just Proposition 4 in Lin et al. [12]. So Theorem 4 extends the above two results.
In the following, we will investigate the Smith forms of some rectangular polynomial matrices and consider Problem2. Let dz denote the greatest common divisor (g.c.d) of the l×l minors of the matrix Fz, dl−1Fz denote the g.c.d of all the l−1×l−1 minors of Fz and denote(58)Bz=Il−10l−1,101,l−1dz.Theorem 5.
LetFz∈Kl×mz (l≤m) be of full row rank and suppose dz=z1−f1z2,…,znq1z2−f2z3,…,znq2, where q1,q2 are nonnegative integers. Then Fz is equivalent to its Smith form(59)Qz=Bz0l×m−l,if and only if all the l−1×l−1 minors of Fz generate Kz.Proof.
Sufficiency. According to Theorem 3.3 in [21], there exists Gz∈Kl×lz, F0z∈Kl×mz such that Fz=Gz⋅F0z, where detGz=dz, and F0z is a ZLP matrix. Combined with Theorem 5, we can obtain two unimodular matrices V1z,V2z∈Kl×lz such that Gz=V1z⋅Bz⋅V2z. Then(60)Fz=V1z⋅Bz⋅V2z⋅F0z.
It is obviously thatV2z⋅F0z is also ZLP. According to the Quillen-Suslin Theorem, we can construct an m×m unimodular matrix Uz such that V2z⋅F0z is its first l rows. Hence(61)Fz=V1z⋅Bz0l,m−l⋅Uz=V1z⋅Qz⋅Uz,and dl−1Qz=1. Since Qz=V1−1z⋅Fz⋅U−1z, combined with Lemma 3, we have that dl−1Fz=1, then the Smith form of Fz is Qz.
Necessity. If Fz is equivalent to its Smith form Qz=Bz0l,m−l, we see that all the l−1×l−1 minors of Qz generate Kz. Using Lemma 3, we obtain that all the l−1×l−1 minors of Fz generate Kz.Remark 2.
Letf1z,f2z,…,ftz∈Kz be nonzero polynomials, a necessary and sufficient condition for G=f1z,f2z,…,ftz generates that Kz is the reduced Gröbner basis of the ideal generated by G includes a unit in the field K. According to Theorems 4 and 5, we can check whether an nD polynomial matrix Fz is equivalent to its Smith form by using the existing Gröbner basis algorithms to the ideal generated by the lower minors of Fz. Thus, the conditions of Theorems 4 and 5 can be verified easily.


## 3.1. Equivalent Theorem

In order to prove our main results, we first give several useful lemmas.Lemma 1.
(see [20]). Let f1z,…,fsz∈Kz, then f1z,…,fsz have no common zeros in K¯n (are zero coprime) if and only if there exist u1z,…,usz∈Kz such that(4)u1zf1z+⋯+uszfsz=1,i.e., f1z,…,fsz is a ZLP row vector in K1×sz, or f1z,…,fsz generate the unit ideal Kz.Lemma 2.
(see [20]). Let gz∈Kz1,z2,…,zn and fz2,…,zn∈Kz2,…,zn. Suppose that gfz2,…,zn,z2,…,zn=0, then z1−fz2,…,zn is a divisor of gz.Lemma 3.
(see [17]). Let Fz,F1z,F2z∈Kl×lz, Fz=F1z⋅F2z. If the l−1×l−1 minors of Fz have no common zeros in K¯n (generate Kz), then the l−1×l−1 minors of Fiz have no common zeros in K¯n (generate Kz) for i=1,2.Proof.
The proof is similar to that of Lemma 2.2 in [17], so it is omitted here.Lemma 4.
(see [18]). Let Q∈Kl−1×lz be of normal full rank, if the reduced minors of Q generate Kz, then there exists a ZLP matrix w∈Kl×1z such that Q⋅w=0l−1,1.The following result is presented in [20], for the convenience of the reader, we record it here.Lemma 5.
(see [20]). Let Fz=Fz1,z2,…,zn∈Kl×lz, fz2,…,zn∈Kz2,…,zn, if the l−1×l−1 minors of Fz generate Kz, then the l−1×l−1 minors of Ffz2,…,zn,z2,…,zn also generate Kz.Proof.
Assume that thel−1×l−1 minors of Fz are c1z1,…,zn,…,crz1,…,zn. If the l−1×l−1 minors of Fz generate Kz, then there exist u1z1,z2,…,zn,…,urz1,z2,…,zn∈Kz such that(5)u1z1,z2,…,znc1z1,z2,…,zn+⋯+urz1,z2,…,zncrz1,…,zn=1.
Substitutez1=fz2,…,zn, then(6)u1fz2,…,zn,z2,…,znc1fz2,…,zn,z2,…,zn+⋯+urfz2,…,zn,z2,…,zncrfz2,…,zn,z2,…,zn=1.
It is straightforward thatc1fz2,…,zn,z2,…,zncrfz2,…,zn,z2,…,zn are the l−1×l−1 minors of Ffz2,…,zn,z2,…,zn. Thus the l−1×l−1 minors of Ffz2,…,zn,z2,…,zn also generate Kz.Lemma 6.
LetFz∈Kl×lz, then detFz and the l−1×l−1 minors of Fz have no common zeros if and only if the l−1×l−1 minors of Fz have no common zeros.Proof.
Necessity. Assuming that the l−1×l−1 minors of Fz have a common zero point a0, by Laplace expansion, we can easily know that a0 is also a zero point of detFz, contradicting the fact that detFz and the l−1×l−1 minors of Fz have no common zeros. Thus, the l−1×l−1 minors of Fz have no common zeros.
Sufficiency. It is straightforward.
For the convenience, we first defineP1z,P2z,Pz as follows:(7)P1z=Il−10l−1,101,l−1z1−f1z2,…,zn,P2z=Il−10l−1,101,l−1z2−f2z3,…,zn,Pz=Il−10l−1,101,l−1z1−f1z2,…,znz2−f2z3,…,zn.
We have the following key conclusion which is very important to derive our main results.Lemma 7.
LetFz∈Kl×lz and Fz=P1zVzP2z, where Vz∈Kl×lz2,…,zn is unimodular. If all the l−1×l−1 minors of Fz generate Kz, then Fz is equivalent to its Smith form Pz.Proof.
We prove this by induction onl. When l=2, let(8)Vz=v11v12v21v22,then(9)Fz=v11v12⋅z2−f2z3,…,znv21⋅z1−f1z2,…,znv22⋅z1−f1z2,…,znz2−f2z3,…,zn.
Obviously,detFz=c⋅z1−f1z2,…,znz2−f2z3,…,zn,0≠c∈K. Note that Vz is unimodular, then v11,v12 have no common zeros. Assume that v11,v12z2−f2z3,…,zn have a zero α0=z20,z30,…,zn0, where z20=f2z30,…,zn0, then v11,v12⋅z2−f2z3,…,zn,v21⋅z1−f1z2,…,zn,v22⋅z1−f1z2,…,znz2−f2z3,…,zn has a common zero f1α0,z20,z30,…,zno. This is a contradiction. Thus, v11,v12⋅z2−f2z3,…,zn have no common zeros, and v11,v12⋅z2−f2z3,…,zn is a unimodular row. According to Quillen-Suslin Theorem, there exists a 2×2 unimodular matrix M1z∈K2×2z such that(10)v11,v12z2−f2z3,…,znM1z=1,0.
So(11)FzM1z=10p1zp2z,where p1z,p2z∈Kz. Then there exists the unimodular matrix M2z=10−p1z1 such that(12)M2zFzM1z=100p2z.
Note thatp2z=detM2zFzM1z=u⋅detFz=u′⋅z1−f1z2,…,znz2−f2z3,…,zn, 0≠u,u′∈K. Thus, Fz is equivalent to its Smith form(13)Pz=100z1−f1z2,…,znz2−f2z3,…,zn.
So the conclusion is true forl=2.
Assume that the conclusion is true forl−1, we investigate the case of l. Let Vz=V11V12V21V22, where V11∈Kl−1×l−1z2,…,zn, V12∈Kl−1×1z2,…,zn, V21∈K1×l−1z2,…,zn, V22∈K1×1z2,…,zn. Then(14)Fz=V11V12⋅z2−f2z3,…,znV21⋅z1−f1z2,…,znV22⋅z1−f1z2,…,znz2−f2z3,…,zn.
Letv11,…,v1,l−1,v1l be the first row of Vz. Since Vz is unimodular, using Laplace Theorem and expanding its first row, we obtain that v11a1z+⋯+v1,l−1al−1z+v1lalz=k, where a1z,…,al−1z,az are all the l−1×l−1 minors of Vz and 0≠k∈K. Combined with Lemma 1, we have that v11,…,v1,l−1,v1l have no common zeros. Assume that v11,…,v1,l−1,v1lz2−f2z3,…,zn have a common zero α0=z20,z30,…,zn0, then all the l−1×l−1 minors of Fz have a common zero f1α0,z20,z30,…,zn0. This is a contradiction. Thus, v11,…,v1,l−1,v1lz2−f2z3,…,zn is a unimodular row in Kz2,…,zn, by the Quillen-Suslin Theorem, there exists a unimodular matrix M1z such that(15)v11,…,v1,l−1,v1lz2−f2z3,…,znM1z=1,0,…,0.
So(16)FzM1z=101,l−1Q1zQ2z,where Q1z∈Kl−1×1z, Q2z∈Kl−1×l−1z. Then there exists the matrix(17)M2z=101,l−1−Q1zIl−1,such that(18)M2zFzM1z=101,l−10l−1,1Q2z.
SettingNz=M2zFzM1z, since all the l−1×l−1 minors of Fz have no common zeros, combined with Lemma 3, we have that all the l−1×l−1 minors of Nz have no common zeros. Note that an l−1×l−1 minor of Nz is just an l−2×l−2 minors of Q2z or detQ2z or 0, then detQ2z and the l−2×l−2 minors of Q2z have no common zeros. By Lemma 6, the l−2×l−2 minors of Q2z have no common zeros. By the inductive hypothesis, there exist two l−1×l−1 unimodular matrices N1z,N2z such that(19)N2zQ2zN1z=Il−20l−2,101,l−2z1−f1z2,…,znz2−f2z3,…,zn,then(20)101,l−10l−1,1N2zNz,w101,l−101,l−1N1z=101,l−10l−1,1N2zQ2zN1z=Il−10l−1,101,l−1z1−f1z2,…,znz2−f2z3,…,zn.
Thus,Nz is equivalent to the matrix Pz, combined with Fz is equivalent to the matrix Nz; then we obtain that Fz is equivalent to the matrix Pz, and Pz is the Smith form of Fz.
Now we are going to state one of our main results, which give partial answer to Problem1. We recall the notation(21)Pz=Il−10l−1,101,l−1z1−f1z2,…,znz2−f2z3,…,zn.Theorem 1.
LetFz∈Kl×lz with detFz=z1−f1z2,…,znz2−f2z3,…,zn. Fz is equivalent to its Smith form Pz if and only if all the l−1×l−1 minors of Fz generate Kz.Proof.
Sufficiency. Because detFz1,f2z3,…,zn,z3,…,zn=0, then the rank of Fz1,f2z3,…,zn,z3,…,zn≤l−1. Since all the l−1×l−1 minors of Fz generate Kz, by Lemma 5, the l−1×l−1 minors of Fz1,f2z3,…,zn,z3,…,zn also generate Kz, then rankFz1,f2z3,…,zn,z3,…,zn=l−1 for every z1,z3,…,zn∈K¯. By Lemma 4, there exists a ZRP column vector Y1z∈Kl×1z such that(22)Fz1,f2z3,…,zn,z3,…,znY1z≡0,…,0T.
By the Quillen-Suslin theorem, anl×l unimodular matrix U11z can be constructed such that Y1z is its last column. Then the elements of the last column of Fz1,f2z3,…,zn,z3,…,zn⋅U11z are zero polynomials. By Lemma 2, the last column of Fz⋅U11z have the common divisor z2−f2z3,…,zn, i.e.,(23)Fz⋅U11z=F1z⋅Il−10l−1,101,l−1z2−f2z3,…,zn,for some F1z∈Kl×lz. Let U1z=U11−1z, then we have(24)Fz=F1zP2zU1z.
Note thatU1z,U11z are unimodular, and detFz=z1−f1z2,…,znz2−f2z3,…,zn, then(25)detF1z=c⋅z1−f1z2,…,zn,0≠c∈K.
From Lemma3, we have that the l−1×l−1 minors of F1z also generate Kz. Note that detF1f1z2,…,zn,z2,z3,…,zn=0, combined with Lemma 5, we obtain that the l−1×l−1 minors of F1f1z2,…,zn,z2,z3,…,zn also generate Kz for z2,z3,…,zn∈K¯. Similarly, a unimodular matrix V11z∈Kl×lz2,…,zn can be constructed such that(26)F1zV11z=F2z⋅Il−10l−1,101,l−1z1−f1z2,…,zn,where F2z∈Kl×lz. Let V1z=V11−1z, then we have(27)F1z=F2zP1zV1z.
Note thatV1z∈Kl×lz2,…,zn is unimodular, and detF1z=c1⋅z1−f1z2,…,zn, then detF2z=c2,0≠c2∈K, F2z is unimodular, and(28)Fz=F2zP1zV1zP2zU1z.
By Lemma7, the matrix P1zV1zP2z is equivalent to(29)Pz=Il−10l−1,101,l−1z1−f1z2,…,znz2−f2z3,…,zn.
Combined withF2z,U1z are unimodular, we obtain that Fz is equivalent to its Smith form Pz.
Necessity. If Fz is equivalent to Pz, there exist two unimodular matrices Uz,Vz such that Fz=VzPzUz. Note that all the l−1×l−1 minors of Pz have no common zeros; combined with Lemma 3, we have that all the l−1×l−1 minors of Fz have no common zeros, i.e., all the l−1×l−1 minors of Fz generate Kz.
In the following, we will extend the above result to more general case.


## 3.2. Generalization

Lemma 8.
LetDz∈Kl×lz with Dz=Il−100hz, hz∈Kz, and Vz∈Kl×lz be a unimodular matrix. If all the l−1×l−1 minors of DszVzDtz generate Kz and s,t are positive integers, then DszVzDtz is equivalent to Ds+tz.Proof.
Let(30)Vz=V11zV12zV21zV22z,where V11z∈Kl−1×l−1z, V12z∈Kl−1×1z, V21z∈K1×l−1z, V22z∈K1×1z.
Then(31)Fz=DszVzDtz=V11zV12zhtzV21zhszV22zhs+tz,and detFz=c⋅hs+tz,0≠c∈K.
Next, we will proof thatV11zV12zhtz is a ZLP matrix. Note that Vz is unimodular, then V11zV12z is a ZLP matrix. Let c1z,c2z,…,clz denote all the l−1×l−1 minors of V11zV12z. Then all the l−1×l−1 minors of V11zV12zhtz are c1z,c2zhtz,…,clzhtz. We can prove that c1z,c2zhtz,…,clzhtz have no common zeros. Suppose that c1z,c2zhtz,…,clzhtz have a common zero α0 and combined with c1z,c2z,…,clz have no common zeros, so α0 is a zero of c1z and hz. Note that the elements of the last row of matrix Fz all have the factor hz and that an l−1×l−1 minor of Fz is just c1z or includes the factor hz, so the l−1×l−1 minors of Fz have the common zero α0, this is a contradiction. Thus, all the l−1×l−1 minors of V11zV12zhtz have no common zeros, and it is a ZLP matrix. According to Quillen-Suslin theorem, there exists an m×m unimodular matrix N1z such that(32)V11zV12zhtzN1z=Il−1z0l−1,1.
Then,(33)FzN1z=Il−1z0l−1,1H1zH2z,for some H1z,H2z∈Kz. Let(34)N2z=Il−1z0l−1,1−H1z1.
Then,(35)N2zFzN1z=Il−1z0l−1,101,l−1H2z.
Note thatN1z,N2z are unimodular, then H2z=u⋅detFz=u′⋅hs+tz,0≠u∈K,u′=u⋅c, thus Fz is equivalent to Ds+tz.
Now we investigate Problem1 for the case of q1=q2. Denote(36)Pqz=Il−10l−1,101,l−1z1−f1z2,…,znqz2−f2z3,…,znq.Theorem 2.
LetFz∈Kl×lz with detFz=z1−f1z2,..,znqz2−f2z3,…,znq, where q is a positive integer. Then Fz is equivalent to its Smith form Pqz if and only if all the l−1×l−1 minors of Fz generate Kz.Proof.
Sufficiency. Notice that detFz1,f2z3,…,zn,z3,…,zn=0, we have that rank Fz1,f2z3,…,zn,z3,…,zn≤l−1. Since all the l−1×l−1 minors of Fz generate the unit ideal Kz, according to Lemma 5, then the l−1×l−1 minors of Fz1,f2z3,…,zn,z3,…,zn also generate the unit ideal Kz. Hence, rankFz1,f2z3,…,zn,z3,…,zn=l−1 for every z1,z3,…,zn∈K¯, repeating the proof of Theorem 1, we have that(37)Fz=F1zP1zV1zP2zU1z,where V1z∈Kl×lz2,…,zn, U1z∈Kl×lz are unimodular, F1z∈Kl×lz with detF1z=z1−f1z2,…,znq−1z2−f2z3,…,znq−1. By Lemma 3, all the l−1×l−1 minors of P1zV1zP2z also generate Kz. According to Lemma 7, there exist unimodular matrices M1z,N1z∈Kl×lz such that P1zV1zP2z=M1zP1zP2zN1z=M1zPzN1z. Thus(38)Fz=F1zM1zPzN1zU1z.
Again by Lemma3, we obtain that all the l−1×l−1 minors of F1z generate Kz. If q≥2, iterating the preceding process, we obtain that(39)F1z=F2zM2zPzN2z,where F2z∈Kl×lz with detF2z=z1−f1z2,…,znq−2z2−f2z3,…,znq−2. Then(40)Fz=F2zM2zPzN2zM1zPzN1zU1z.
LetS2z=N2zM1z, we have that S2z is unimodular and all the l−1×l−1 minors of PzS2zPz generate Kz. By Lemma 8, PzS2zPz is equivalent to P2z, that is, there exist unimodular matrices L1z,T1z∈Kl×lz such that PzS2zPz=L1zP2zT1z. Furthermore,(41)Fz=F2zM2zL1zP2zT1zN1zU1z.
LetV2z=M2zL1z, U2z=T1zN1zU1z, then V2z,U2z are unimodular, and(42)Fz=F2zV2zP2zU2z.
Ifq≥3, iterating the same procedure successively, we obtain that(43)Fz=FqzVqzPqzUqz,where Vqz,Uqz are unimodular. Note that detFz=z1−f1z2,…,znqz2−f2z3,…,znq, then detFqz=c,c∈K; that is, Fqz is a unimodular matrix. So Fz is equivalent to its Smith form Pqz.
Necessity. It is straightforward that all the l−1×l−1 minors of Pqz generate Kz. According to Lemma 3, all the l−1×l−1 minors of Fz also generate Kz.
Next we investigate Problem1 for the case of Fz∈Kl×lz and det Fz=z1−f1z2,…,znq1z2−f2z3,…,znq2, where q1,q2 are nonnegative integers.Theorem 3.
LetFz∈Kl×lz with detFz=z1−f1z2,…,znq1z2−f2z3,…,znq2, where q1,q2 are nonnegative integers. If all the l−1×l−1 minors of Fz generate Kz, then Fz is equivalent to its Smith form(44)Qz=Il−10l−1,101,l−1z1−f1z2,…,znq1z2−f2z3,…,znq2.Proof.
(1) Ifq1=q2=0, then detFz=1, i.e., Fz is unimodular, so Fz is equivalent to Il.
(2) Ifq1=0 or q2=0, we have that detFz=z1−f1z2,…,znq1 or z2−f2z3,…,znq2. Without loss of generality, we assume that detFz=z1−f1z2,…,znq1. Because detFf1z2,…,zn,z2,…,zn=0, then the rank of Ff1z2,…,zn,z2,…,zn≤l−1. Since all the l−1×l−1 minors of Fz generate Kz, by Lemma 5, the l−1×l−1 minors of Ff1z2,…,zn,z2,…,zn also generate Kz; hence, rankFf1z2,…,zn,z2,…,zn=l−1 for every z2,z3,…,zn∈K¯n−1. By Lemma 4, we obtain a ZRP column vector Y1z∈Kl×1z such that(45)Ff1z1,…,zn,z2,…,znY1z≡0,…,0T.
According to the Quillen-Suslin Theorem, anl×l unimodular matrix U11zdetU11z=1 can be constructed such that Y1z is its last column. Then the last column of Ff1z1,…,zn,z2,…,zn⋅U11z are zero polynomials. By Lemma 2, the last column of Fz⋅U11z have the common divisor z1−f1z2,…,zn, i.e.,(46)Fz⋅U11z=F1z⋅Il−10l−1,101,l−1z1−f1z2,…,zn,for some F1z∈Kl×lz. Let U1z=U11−1z, then we obtain that(47)Fz=F1zP1zU1z,detF1z=z1−f1z2,…,znq1−1.
By Lemma3, all the l−1×l−1 minors of F1z also generate the unit ideal Kz. If q1≥2, iterating the preceding process, we obtain that(48)F1z=F2zP1zU2z.
Furthermore,(49)Fz=F2zP1zU2zP1zU1z.
By Lemma3, the l−1×l−1 minors of P1zU2zP1z generate the ideal Kz, according to Lemma 8, there exist unimodular matrices N1z,V1z such that P1zU2zP1z=N1zP12zV1z. Then(50)Fz=F2zN1zP12zV1zU1z.
Furthermore,detF2zN1z=z1−f1z2,…,znq1−2, combined with Lemma 3, the l−1×l−1 minors of F2zN1z also generate Kz.
Ifq1≥3, iterating the same procedure successively, we obtain that(51)Fz=Fq1zNq1−1zP1q1zVq1−1z,…,V1zU1z,with detFq1zNq1−1z=1. Thus, Fz is equivalent to its Smith form P1q1z=Qz.
(3) Ifq1,q2 are positive integers, then q1≥q2 or q2≥q1. Without loss of generality, we assume that q1≥q2. By Theorem 2, combined with the conclusion of the case (2) above, we have that Fz is equivalent to the matrix Pq2zVzP1q1−q2z. In the following, we prove that Pq2zVzP1q1−q2z is equivalent to the matrix Qz
Let(52)Vz=V11zV12zV21zV22z,where V11z∈Kl−1×l−1z, V12z∈Kl−1×1z, V21z∈K1×l−1z, and V22z∈K1×1z.
Then(53)Pq2zVzP1q1−q2z=V11zV12zp1q1−q2zV21zp1q2zp2q2zV22zp1q1zp2q2z,where p1z=z1−f1z2,…,zn, p2z=z2−f2z3,…,zn.
In fact, we can prove thatV11zV12zp1q1−q2z is a ZLP matrix. Note that Vz is unimodular, then V11zV12z is a ZLP matrix. Let c1z,c2z,…,clz denote all the l−1×l−1 minors of V11zV12z. Then all the l−1×l−1 minors of V11zV12zp1q1−q2z are c1z,c2zp1q1−q2z,…,clzp1q1−q2z. We can prove that c1z,c2zp1q1−q2z,…,clzp1q1−q2z have no common zeros. Suppose that c1z,c2zp1q1−q2z,…,clzp1q1−q2z have a common zero α0, note that c1z,c2z,…,clz have no common zeros, so c1z and p1z have the common zero α0. Moreover, the elements of the last row of matrix Pq2zVzP1q1−q2z have the common factor p1z, since an l−1×l−1 minors of Pq2zVzP1q1−q2z is just c1z or includes the factor p1z, so the l−1×l−1 minors of Pq2zVzP1q1−q2z have the common zero α0, this is a contradiction. Thus, all the l−1×l−1 minors of V11zV12zp1q1−q2z have no common zeros, and it is a ZLP matrix. By the Quillen-Suslin theorem, there exists an l×l unimodular matrix Nz such that(54)V11zV12zp1q1−q2zNz=Il−1z0l−1,1.
Then,(55)Pq2zVzP1q1−q2zNz=Il−1z0l−1,1H1zH2z,for some H1z,H2z∈Kz. There exists a unimodular matrix N1z such that(56)N1zPq2zVzP1q1−q2zNz=Il−1z0l−1,101,l−1H2z.
Note thatNz,N1z are unimodular, then H2z=u⋅detPq2zVzP1q1−q2z=u′⋅p1q1zp2q2z,0≠u∈K,u′=u⋅c; thus, Pq2zVzP1q1−q2z is equivalent to Qz. Note that Fz is equivalent to the matrix Pq2zVzP1q1−q2z, combined with the definition of the Smith form of a matrix, we obtain that Fz is equivalent to its Smith form Qz.
Theorem3 provides a positive answer to Problem 1 for the case of Fz∈Kl×lz and det Fz=z1−f1z2,…,znq1z2−f2z3,…,znq2. It also gives a sufficient condition to check this kind of matrices are equivalent to their Smith forms; in fact, this condition is also a necessary condition.Theorem 4.
LetFz∈Kl×lz with det Fz=z1−f1z2,…,znq1z2−f2z3,…,znq2, where q1,q2 are non-negative integers, then Fz is equivalent to its Smith form(57)Qz=Il−10l−1,101,l−1z1−f1z2,…,znq1z2−f2z3,…,znq2,if and only if all the l−1×l−1 minors of Fz generate Kz.Proof.
Sufficiency. From Theorem 3, it is straightforward.
Necessity. By computing, we can easily obtain that all the l−1×l−1 minors of Qz generate Kz. By Lemma 3, all the l−1×l−1 minors of Fz generate Kz.Remark 1.
In the theorem above,K is an arbitrary field. When K=ℝ is the real field and q2=0, Theorem 4 is same as Theorem 2.5 in Li et al. [17]. Furthermore, if q1=1 and q2=0, Theorem 4 is just Proposition 4 in Lin et al. [12]. So Theorem 4 extends the above two results.
In the following, we will investigate the Smith forms of some rectangular polynomial matrices and consider Problem2. Let dz denote the greatest common divisor (g.c.d) of the l×l minors of the matrix Fz, dl−1Fz denote the g.c.d of all the l−1×l−1 minors of Fz and denote(58)Bz=Il−10l−1,101,l−1dz.Theorem 5.
LetFz∈Kl×mz (l≤m) be of full row rank and suppose dz=z1−f1z2,…,znq1z2−f2z3,…,znq2, where q1,q2 are nonnegative integers. Then Fz is equivalent to its Smith form(59)Qz=Bz0l×m−l,if and only if all the l−1×l−1 minors of Fz generate Kz.Proof.
Sufficiency. According to Theorem 3.3 in [21], there exists Gz∈Kl×lz, F0z∈Kl×mz such that Fz=Gz⋅F0z, where detGz=dz, and F0z is a ZLP matrix. Combined with Theorem 5, we can obtain two unimodular matrices V1z,V2z∈Kl×lz such that Gz=V1z⋅Bz⋅V2z. Then(60)Fz=V1z⋅Bz⋅V2z⋅F0z.
It is obviously thatV2z⋅F0z is also ZLP. According to the Quillen-Suslin Theorem, we can construct an m×m unimodular matrix Uz such that V2z⋅F0z is its first l rows. Hence(61)Fz=V1z⋅Bz0l,m−l⋅Uz=V1z⋅Qz⋅Uz,and dl−1Qz=1. Since Qz=V1−1z⋅Fz⋅U−1z, combined with Lemma 3, we have that dl−1Fz=1, then the Smith form of Fz is Qz.
Necessity. If Fz is equivalent to its Smith form Qz=Bz0l,m−l, we see that all the l−1×l−1 minors of Qz generate Kz. Using Lemma 3, we obtain that all the l−1×l−1 minors of Fz generate Kz.Remark 2.
Letf1z,f2z,…,ftz∈Kz be nonzero polynomials, a necessary and sufficient condition for G=f1z,f2z,…,ftz generates that Kz is the reduced Gröbner basis of the ideal generated by G includes a unit in the field K. According to Theorems 4 and 5, we can check whether an nD polynomial matrix Fz is equivalent to its Smith form by using the existing Gröbner basis algorithms to the ideal generated by the lower minors of Fz. Thus, the conditions of Theorems 4 and 5 can be verified easily.


## 4. Example and Algorithm

In this section, we first present a 2D example to illustrate our result and explain how to obtain the unimodular matrices associated with the equivalence of system matrices in the method. Then we give an algorithm to deal with the equivalence of the kind of matrices we discussed to their Smith forms.In many areas of engineering such as Circuits and Signals, the general 2D systems can be defined in terms of the generalized Rosenbrock system [8] as(62)Fz1,z2x=Uz1,z2u,y=Vz1,z2x+Wz1,z2u,where x∈Kn is the state vector, u∈Kl is the input vector, y∈Km is the output vector, Fz1,z2∈Kn×nz1,z2, Uz1,z2∈Kn×lz1,z2, Vz1,z2∈Km×nz1,z2, Wz1,z2∈Km×lz1,z2 are polynomial matrices, K is a field. The operators z1 and z2 may have various meanings depending on the type of system. For example, in delay-differential systems, z1 represents a differential operator and z2 a delay-operator. For 2D discrete systems, z1 and z2 represent horizontal and vertical shift operators, respectively. This system gives rise to the system matrix in the general form:(63)Tz1,z2=Fz1,z2Uz1,z2−Vz1,z2Wz1,z2.Example 1.
Consider a system matrix(64)Tz1,z2=Fz1,z2Uz1,z2−Vz1,z2Wz1,z2=z23−6z22z12−8z22z1−z22+4z2z12+2z2z1−z120z1z12−4−8z1+z2z1−2z22+z23z1+1z2z1−12z22−z1z2−6z2+4z110z1z200,
where(65)Fz1,z2=z23−6z22z12−8z22z1−z22+4z2z12+2z2z1−z120z12−4−8z1+z2z1−2z22+z23z1+1z1−12z22−z1z2−6z2+4z11,Uz1,z2=z1z20T,Vz1,z2=−z1−z20,Wz1,z2=0.
By computing,detFz=z1−z22z2−14, the 2×2 minors of Fz is as follows: c1z=z26−2z25+z24z1−8z23z1−4z23+6z22z14+8z22z13+z22z12−4z2z14−2z2z13+z14, c2z=z23z1+1, c3z=−z1+16z22z12+8z22z1+z22−4z2z12−2z2z1+z12, c4z=2z25−z24z1−6z24+4z23z1+6z22z13+2z22z12−7z22z1−z22−4z2z13+2z2z12+2z2z1+z13−z12, c5z=z13, c6z=−6z22z12−8z22z1−z22+4z2z12+2z2z1−z12, c7z=−z23z1+z23+2z22z12+2z22z1−2z22−z2z13−7z2z12+z2z1+4z13+8z12−4z1−4, c8z=1, c9z=z23−2z22z1−4z22+z2z12+8z2z1+6z2−4z12−12z1−4.
The reduced Gröbner basis of the ideal generated byc1z,…,c9z is 1. So the 2×2 minors of Fz1,z2 generate unit ideal Kz1,z2, by Theorem 4, Fz1,z2 is equivalent to its Smith form. Consider(66)Fz1,1=1−3z12−6z1−10z12−7z1−5z1+1z1−12z1−41.
Solving the equations,Fz1,1Y1=0, where Y1∈K3×1z1, we obtain Y1T=3z12+6z1+1,1,−3z13−3z12+2z1+5, where Y1T is the transpose of Y1. It then follows that(67)Fz1,13z12+6z1+11−3z13−3z12+2z1+5=000.
We complete3z12+6z1+1,1,−3z13−3z12+2z1+5T into a unimodular matrix U41∈K3×3z1 such that FU41=F1P2, where(68)F1=z2303z22z12+6z22z1+z22−3z2z12−2z2z1+z12z12−z1−1z22−z2+z1−1z1−1−12z2−z1−4,P2=10001000z2−1,U41=103z12+6z1+10010−1−3z13−3z12+2z1+5.
We then have(69)F=F1P2U4,where(70)U4=U41−1=1−3z12−6z1−100−3z13−3z12+2z1+5−1010.
We considerF1z1,z2 again(71)F1z1,1=10z12+4z1+1z12−z1−1z1−1z1−1−1−2−z1.
Repeating the procedure above forF1, we have that(72)F1=F2P2U3,where(73)F2=z230z22z12+4z22z1+z22−2z2z12−2z2z1+z12z12z1+1−z2z1−11−2,U3=10z12+4z1+10−1−z13−3z12+2z1−100−1.
Then,(74)F=F1P2U4=F2P2U3P2U4.
By Theorem 3.5 in [17], we have that P2U3P2 is equivalent to P22, and(75)P2U3P2=P22M2,where(76)M2=10z2−1z12+4z1+10−1−z2−1z13+3z12−2z1+100−1.
So,(77)F=F1P2U4=F2P2U3P2U4=F2P22M2U4.
ConsiderF2, factor it like above, we obtain that(78)F2=F3P2U2,where(79)F3=z2302z22z1+z22−z2z12−2z2z1+z12z12−z1−11z1−1−10,U2=102z1+10−12z12−z1+100−1.
Then,(80)F=F2P22M2U4=F3P2U2P22M2U4.
We considerF3 again(81)F3z2,z2=z230z23z22−z2−11z2−1−10.
Solving the equations,F3z2,z2Y2=0, where Y2∈K3×1z2, we obtain Y2T=1,z2−1,−1, where Y2T is the transpose of Y2. It then follows that(82)F3z2,z21z2−1−1=000.
We complete1,z2−1,−1T into a unimodular matrix V21∈K3×3z2 such that F3V21=F4P1, where(83)F4=z230z2−1z1−z2z12z1+1z1+1z1−111,P1=10001000z1−z2,V21=1010−1z2−100−1.
We obtain that(84)F3=F4P1V2,where(85)V2=V21−1=1010−11−z200−1.
Thus,(86)F=F3P2U2P22M2U4=F4P1V2P2U2P22M2U4.
Now we considerP1V2P2. By Lemma 7, we have that P1V2P2 is equivalent to P. And(87)P1V2P2=PM1,where(88)M1=10z2−10−1−z2−1200−1.
Thus,(89)F=F4P1V2P2U2P22M2U4=F4PM1U2P22M2U4.
Repeating the procedure above forF4, we obtain that F4=F5P2U1. Furthermore, we consider F5 and obtain F5=F6P1. So(90)F=F4PM1U2P22M2U4=F5P2U1PM1U2P22M2U4=F6P1P2U1PM1U2P22M2U4,where(91)F5=z2301−z2z12−z1−10z1−1−10,F6=z230−1z12−z1−10z1−1−10,U1=1000−1−100−1.
Setting(92)P=P1P210001000z1−z2z2−1,we obtain that(93)F=F6P1P2U1PM1U2P22M2U4=F6PU1PM1U2P22M2U4.
LetL1=M1U2, L2=M2U4, we have(94)F=F6PU1PL1P22L2.
Now we consider the matrixPU1P. From Theorem 2, we know that PU1P is equivalent to P2. And(95)PU1P=P2M3,where(96)M3=1000−1z2−z1z2−100−1.
Then we have(97)F=F6P2M3L1P22L2.
LetL3=M3L1, now consider the matrix P2L3P22. By Theorem 3, we have that P2L3P22 is equivalent to B, where(98)B=10001000z1−z22z2−14,P2L3P22=BL4,where(99)L4=10z2−122z1−z2+20−1z2−122z12−z2z1+z200−1.
Now(100)F=F6BL4L2.
LetL5=L4L2, then(101)L5=1z23−2z22z1−4z22+z2z12+8z2z1+6z2−4z12−12z1−400z23z1−z23−2z22z12−2z22z1+2z22+z2z13+7z2z12−z2z1+4z13−8z12+4z1+4−1010,we have(102)F=F6BL5.
By computing,detF6 = detF5 = 1.
ThusF is equivalent to the matrix(103)B=10001000z1−z22z2−14.
Now we can findX=01×3, Y=03×1, such that(104)F6−103×1X1FU−VW=BU1−V1W1L5Y01×31,where U1=F6U=z2z2z1−z2z24−z1T, V1=VL5−1=−z10−a1, a1=−z1z23+2z12z22+4z1z22−z2z13−8z2z12−6z2z1+z2+4z13+12z12+4z1, W1=0.
Thus, the system(105)Fz1,z2Uz1,z2−Vz1,z2Wz1,z2,is equivalent to the system(106)Bz1,z2U1z1,z2−V1z1,z2W1z1,z2,which is simpler.
With the help of Example1 and Theorem 3, we now can get the Algorithm 1.
The program of the function SyzygyModule which we use in algorithm can be found inhttps://faculty.math.illinois.edu/Macaulay2/doc/Macaulay2-1.15/share/doc/Macaulay2/MCMApproximations/html/_syzygy__Module.html.
Another function CompleteMatrix is available inhttp://wwwb.math.rwth-aachen.de/QuillenSuslin. Moreover, for a new algorithm for CompleteMatrix, see the package MatrixFactorization in http://www.mmrc.iss.ac.cn/dwang/software.html, which contained a ZLP algorithm. This algorithm can obtain a unimodular matrix whose inverse is the complete matrix of a given ZLP matrix.Algorithm 1: A matrix equivalence (ME) algorithm.
(i)
Step 1. Declare the ringKz=Kz1,…,zn over which the matrix is defined by declaring the indeterminates and the field of coefficients. Factor the determinant of F. Check that the determinant of F is the form zj−fjz2,…,znqjzk−fkz3,…,znqk. If yes, set q1=maxqj,qk, the polynomial it corresponds is z1−f1, q2=minqj,qk, the polynomial it corresponds is z2−f2, go to Step 2. Otherwise, return this method is not fit for F.
(ii)
Step 2. Compute the reduced Gröbner basisG of ideal generated by the lower minors of F. If G=1, go to Step 3; otherwise, return this method is not fit for F.
(iii)
Step 3. SetF0=F, P1=diag1,…,1,z1−f1, P2=diag1,…,1,z2−f2, P=P1P2, i=0 and t=q1−q2.
(iv)
Step 4. Substitutez1=f1 in Fi to obtain Fi¯. Compute a ZRP vector Yi∈Kz2,…,zn such that Fi¯Yi=0 by using the function SyzygyModule. Then compute a unimodular matrix U with Yi is its last column by using the function CompleteMatrix. Compute Fi+1 such that FiU=Fi+1P1. Compute U−1 and set Ui+1=U−1. Compute P1iUi+1P1=NiP1i+1Vi and obtain F=Fi+1NiP1i+1Vi, store Fi+1. L3=Fq1L1, i=i+1.
(v)
Step 5. Wheni≤t, go to Step 4. When t<i<q1, do Step 6; otherwise, compute Pq2Nq1P1t=L1Pq2P1tL2, let return F=L3Pq2P1tL2.
(vi)
Step 6. Substitutez1=f1 in Fi to obtain Fi¯, do procedure similar to the step 4. And obtain a ZRP vector Yi such that Fi¯Yi=0 and a unimodular matrix U with Yi is its last column. Then compute Fi+1 such that FiU=Fi+1P1, and compute U−1 and set Ui+1=U−1. Set Fi=Fi+1, substitute z2=f2 in Fi to obtain Fi¯. Compute a ZRP vector X∈Kz1,z3,…,zn such that Fi¯X=0 by using the function SyzygyModule. Then compute a unimodular matrix V with X is its last column by using the function CompleteMatrix. Compute Vi−t=V−1 and Fi+1 such that FiVi−t=Fi+1P2. Compute P1Ui+1P2=NiPMi and obtain F=Fi+1Pi−tNiP1tVi, where Vi=MiVi−t. Store Fi+1, i=i+1. Go to Step 5.


## 5. Conclusions

In this paper, we have investigated the equivalence problem of several kinds ofnD polynomials matrices over an arbitrary field, and have presented some interesting results. We have obtained some criteria for these matrices to equivalent to their Smith forms respectively. These criteria are easily checked by the existing Gröbner basis algorithm for the ideal generated by the minors of lower order of a given matrix. We also give an example to illustrate our method. All of these could provide useful information for engineers to reduce nD systems.



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*Source: 5430842-2020-05-13.xml*