# Some Further Results on the Reduction of Two-Dimensional Systems

Authors: Dongmei Li; Yingying Gui; Jinwang Liu; Man Wu
Journal: Complexity
 (2021)

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## Abstract

The reduction of two-dimensional systems plays an important role in the theory of systems, which is closely associated with the equivalence of the bivariate polynomial matrices. In this paper, the equivalence problems on several classes of bivariate polynomial matrices are investigated. Some new results on the equivalence of these matrices are obtained. These results are useful for reducing two-dimensional systems.


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## Body

## 1. Introduction

Multidimensional (nD) systems, especially two-dimensional (2D) systems, are widely used in the field of circuits, image, signal processing, control systems, etc. [1–5, 12]. And the theory of 2D systems has received increasing attention, among which the reduction of 2D systems is an important research content. Usually, a given system is desirable to reduce to an equivalent system with fewer equations or unknowns (named the equivalence of 2D systems). In this way, the characteristics of the original system can be studied in a better and simpler way. A 2D system can be represented with two types of dynamical elements, so 2D systems are often described by bivariate polynomial matrices. Hence, the equivalence problems of 2D systems are usually transformed into the equivalence problems of polynomial matrices [6–11].The equivalence problem of univariate polynomial matrices was solved by Rosenbrock in 1970 [6]. Using the Euclidean division property of the univariate polynomial ring, he proved that every univariate polynomial matrix in this ring is equivalent to the Smith form. The equivalence problem of the bivariate polynomial matrix is more complex. Lee and Zak gave an example of bivariate polynomial matrix z2−z1−1−z22z2, which is not equivalent to the Smith form [8]. Note that none of the bivariate polynomial rings is a Euclidean ring, and the Euclidean division property does not hold. So, many researchers study the equivalence of some special types of bivariate polynomial matrices. In 1986, Frost and Boudellioua proved that a full row rank bivariate polynomial matrix Tz,w is equivalent to the Smith form if and only if there exists a unimodular column vector U such that TU has a right inverse [9]. In 2012, Boudellioua wrote an algorithm to enable the decomposition and equivalence of some bivariate polynomial matrices to be realized in Maple [10]. In 2019, Li et al. presented some criteria on an l×l bivariate polynomial matrix Fz,w which is equivalent to the Smith form diagIl−1,detFz,w with detFz,w being an irreducible polynomial [11]. There are also some results on the equivalence of multivariate polynomial matrices in special cases [13, 14, 16, 18, 19], among which when an l×l polynomial matrix Fz which is equivalent to the Smith form diag Il−1,detFz is investigated, such as detFz=x1−fx2,…,xn [13], detFz=x1−fx2,…,xnq [14], and detFz=x1−f1x2,…,xnq1x2−f2x2,…,xnq2 [19], diag Il−1,detFz is a very important kind of matrices in the equivalence of multidimensional systems [9, 15, 17].In this paper,Kz,w denotes a bivariate polynomial ring with K being a field, and we consider arbitrary polynomial fz,w in Kz,w as a polynomial in w with coefficient in Kz, written as fz,w=∑i=0nαiwi, where αi=αiz∈Kz. Note that the coefficient ring Kz is a Euclidean ring, and combined with the Euclidean division property of Kz, we will investigate some classes of bivariate polynomial matrices with their entries in Kz,w. The following three problems are also considered.Problem 1.
LetFz,w∈Kl×lz,w, and detFz,w=hrz, where hz∈Kz is irreducible and r is a positive integer. When is Fz,w equivalent to(1)Il−rhz⋱hz?Problem 2.
LetFz,w∈Kl×lz with detFz,w=hq⋅rz, hz∈Kz be irreducible, and q,r be positive integers. When is Fz,w equivalent to(2)Il−rhqz⋱hqz?Problem 3.
LetFz,w∈Kl×mzl≤m with dlF=hq⋅rz, where dlF denotes the greatest common divisor of the l×l minor of Fz,w, hz∈Kz is irreducible, and q,r are positive integers. When is Fz,w equivalent to its Smith form?


## 2. Preliminaries

In the following,K is an arbitrary field, Kz is the univariate polynomial ring in variable z with coefficients in K, Kz,w is the bivariate polynomial ring in variables z,w whose coefficients are in K, K¯ is the algebraic closed field of K, 0m,n is the m×n zero matrix, and Im is the m×m identity matrix. For Fz,w∈Kl×mz,w, diF will be the greatest common divisor (g.c.d) of the i×i minors of Fz,w, i=1,…,l. Set Fz,w=fij∈Kl×mz,w, where fij∈Kz,w, hz∈Kz is irreducible, and Fz,w¯=fij¯∈Kl×mz,w, where fij¯ denotes fijmodhz. For convenience, the argument z,w is omitted whenever its omission does not cause confusion throughout this paper.Definition 1.
(see [16]). Let Fz,w∈Kl×mz,wl≤m with rank r and Φi be a polynomial defined as follows:(3)Φi=diFdi−1F,1≤i≤r,0,r<i≤l,where d0F≡1, diF is the g.c.d of the i×i minors of Fz,w, and Φi satisfies(4)Φ1|Φ2|⋯|Φr.
Then, the Smith form ofFz,w is given by(5)S=diagΦi0r,m−r0l−r,r0l−r,m−r.Definition 2.
(see [16]). Let Fz,w∈Kl×mz,w with full row rank; Fz,w is said to be zero left prime if the l×l minors of Fz,w have no common zero in K¯2.Definition 3.
LetUz,w∈Kl×lz,w; then, we say Uz,w to be a unimodular matrix if the determinant of Uz,w is a unit of K.Definition 4.
LetF1z,w,F2z,w∈Kl×mz,w; F1z,w is said to be equivalent to F2z,w if there are unimodular matrices Uz,w∈Kl×lz,w and Vz,w∈Km×mz,w such that(6)F1z,w=Uz,wF2z,wVz,w.Lemma 1 (see [16]).
SupposeFz,w, Qz,w∈Kl×lz,w; if Fz,w is equivalent to Qz,w, then diFz,w=diQz,w, for i=1,…,l.Lemma 2 (see [16]).
LetFz,w, F1z,w, F2z,w∈Kl×lz,w, and Fz,w=F1z,wF2z,w. If the l−r×l−rr≤l minors of Fz,w have no common zero in K¯2, then the l−r×l−r minors of Fi have no common zero in K¯2 for i=1,2.


## 3. Main Results

In this section, we investigate the three problems presented in Section1 and give the main results of this paper.In the following, forfz,w∈Kz,w, we consider it as an element in Kzw, written as fz,w=∑i=0nαiwi, where αi=αiz∈Kz. If hz∈Kz is irreducible, then fz,w¯ denotes fz,wmodhz and fz,w¯≡∑i=0n1βiwimodhz or 0, where βi=βiz∈Kz, degβiz<deghz. Hence, βn1z and hz are relatively prime in Kz; by Euclidean algorithm, there are xz,yz∈Kz such that xz⋅βn1z=1−yz⋅hz, i.e., xz⋅βn1z≡1modhz. Then, xz⋅fz,w¯≡f∗z,wmodhz, where f∗z,w is monic. In other words, xzfz,w¯ can be reduced to a monic polynomial f∗z,w by hz.Denote(7)Pz=Il−rhz⋱hz.First, we investigate Problem1.Theorem 1.
LetFz,w∈Kl×lz,w with detFz,w=hrz, where hz∈Kz is irreducible. If hz|dl−r+1Fz,w, then Fz,w is equivalent to the Smith form Pz.Proof.
Ifr rows of Fz,w are zero vectors mod hz, that is, r rows of Fz,w¯ are zero vectors, we obtain that(8)Fz,w=Uz,wIl−rhz⋱hzQz,w,and then Fz,w=Uz,wPzQz,w, where Uz,w is unimodular; by computing, detQz,w=1, so Qz,w is unimodular. Hence, Fz,w is equivalent to the Smith form Pz.
IfFz,w has no r rows which are the zero vectors mod hz, then Fz,w has r1 rows of zero vectors mod hz; in other words, Fz,w¯ has r1 rows of zero vectors, 0≤r1≤r−1. We premultiply and postmultiply Fz,w¯ by unimodular matrices U11 and V11 such that(9)U11Fz,w¯V11=fz,w¯f2z,w¯⋮fl−r1z,w¯X0r1,10r1,l−1,where fz,w¯≠0, degwfz,w¯≤degwfjz,w¯, or fjz,w¯=0,j=2,…,l−r1. Let fz,w¯=α0+α1w+⋯+αnwn, where degzαi<degzhz,i=0,…,n, and αn≠0. Note that αn and hz are relatively prime, so we can find xz,yz∈Kz such that xz⋅αn+yz⋅hz=1; then, xz⋅αn=1−yz⋅hz. We have xz⋅fz,w¯≡f′z,wmodhz, where f′z,w is monic. There are qjz,w,rjz,w∈Kz,w such that fjz,w¯=qjz,w⋅f′z,w+rjz,w, where degwrjz,w<degwf′z,w or rjz,w=0,j=2,…,l−r1. Therefore,(10)fjz,w¯≡qjz,w⋅xz⋅fz,w¯+rjz,w¯modhz,where degwrjz,w¯<degwfz,w¯ or rjz,w¯=0,j=2,…,l−r1.
Let(11)U12=101,l−1−xzq2z,w⋮−xzql−r1z,w0r1,l−r1Il−1.
Then,(12)U12U11Fz,w¯V11=fz,w¯r2z,w¯⋮rl−r1z,w¯X0r1,10r1,l−1modhz,where degwrjz,w¯<degwfz,w¯ or rjz,w¯=0,j=2,…,l−r1. If some of rjz,w¯ are not 0, j=2,…,l−r1, do some row transformations to the matrix U12U11Fz,w¯V11 such that the nonzero polynomial of the least degree in w among its first column is at position (1, 1). Repeating the previous steps, we obtain that(13)U1Fz,w¯V11=d¯11X0l−1,1F1z,w¯modhz,where U1=U1t⋯U12U11 is a unimodular matrix, the last r1 rows of F1z,w¯ are zero vectors, and d¯11≠0.
Ifr rows of F1z,w¯ are zero vectors mod hz, we can find two l−1×l−1 unimodular matrices U20′,V20′ such that(14)U20′F1z,w¯V20′=Bz,w0r,l−1modhz,where Bz,w∈Kl−r−1×l−1z,w.
LetU20=1U20′ and V20=1V20′; we have that(15)U20U1Fz,wV11V20=Il−rhz⋱hzQ1,where Q1∈Kl×lz,w and detQ1=1. Then, Fz,w=UPzV, where U=U1−1U20−1 and V=Q1V20−1V11−1 are unimodular matrices, so Fz,w is equivalent to the Smith form Pz.
IfF1z,w¯ has no r rows of zero vectors mod hz, then F1z,w¯ has r2 rows of zero vectors, r1≤r2≤r−2. Imitating the previous procedure to Fz,w¯, there are two l−2×l−2 unimodular matrices U21′,V21′ such that(16)U21′F1z,w¯V21′=d¯22X0l−2,1F2z,w¯modhz,where d¯22≠0 and the last r2 rows of F1z,w¯ are zero vectors.
LetU21=1U21′ and V21=1V21′; then,(17)U21U1Fz,w¯V11V21=d¯11∗d¯22X0l−2,2F2z,w¯modhz.
Repeating the procedure above successively, we obtain a series ofFiz,w¯,i=3,…,l−r−1. If there is some Fiz,w¯ which contains r rows of zero vectors mod hz, then the conclusion is straightforward. Otherwise, Fiz,w¯ has no r rows of zero vectors mod hz.
Then, we consider the case thatFiz,w¯ has no r rows of zero vectors mod hz, i=3,…,l−r−1. In this case, Fl−r−1z,w¯∈Kr+1×r+1z,w, and it has no r rows of zero vectors mod hz; there are r+1×r+1 unimodular matrices Ul−r,1′,Vl−r,1′ such that(18)Ul−r,1′Fl−r−1z,w¯Vl−r,1′=d¯l−r,l−rX0r,l−rFl−rz,w¯modhz,where d¯l−r,l−r≠0.
LetUl−r,1=Il−r−1Ul−r,1′ and Vl−r,1=Il−r−1Vl−r,1′; then,(19)Ul−r,1Ul−r−1,1⋯U21U1Fz,w¯V11V21⋯Vl−r−1,1Vl−r,1=d¯11∗⋯d¯22∗⋯⋱d¯l−r,l−rX0r,l−rFl−rz,w¯modhz.
Let(20)Az,w=d¯11∗…d¯22∗…⋱d¯l−r,l−rX0r,l−rFl−rz,w¯,Fl−rz,w¯=a11a12…a1ra21a22…a2r⋮⋮⋱⋮ar1ar2…arr,combined with hz|dl−r+1Fz,w; we have that dl−r+1Fz,w¯≡0modhz and dl−r+1Az,w≡0modhz. Considering the l−r+1×l−r+1 minors of Az,w, we see that d¯11d¯22⋯d¯l−r,l−r⋅ak,j≡0modhz for all k,j=1,…,r. Since d¯i,i≠0,i=1,…,l−r, ak,j≡0modhz, k,j=1,…,r. Hence, Fl−rz,w¯≡0r,rmodhz. Hence,(21)Ul−r,1Ul−r−1,1⋯U21U1Fz,wV11V21⋯Vl−r−1,1Vl−r,1=Az,w=Il−rhz⋱hzQ2,where Q2∈Kl×lz,w. Note that detFz,w=hrz and U1,Ui1,Vi1 are unimodular matrices, i=1,2,…,l−r; then, we obtain that detQ2=1 and Q2 is a unimodular matrix. Let U=Ul−r,1Ul−r−1,1⋯U1 and V=V11V21⋯Vl−r,1Q2−1; then, Fz,w=U−1PzV−1 with U−1,V−1 being unimodular matrices. Therefore, Fz,w is equivalent to the Smith form Pz.Remark 1.
Theorem1 provides a positive answer to Problem 1.Corollary 1.
LetFz,w∈Kl×lz,w and hz∈Kz be an irreducible polynomial. If hz|dl−r+1Fz,w and hrz|detFz,w, then Fz,w can be factorized as Tz,wPzQz,w, where Tz,w,Qz,w∈Kl×lz,w and Tz,w is unimodular.Theorem 2.
LetFz,w∈Kl×lz,w and hz∈Kz be an irreducible polynomial. Suppose detFz,w=hq⋅rz, r and q are positive integers. If hqz|dl−r+1Fz,w, then Fz,w can be factorized as(22)T1z,wPzT2z,wPz⋯Tqz,wPzTz,w,where Pz is defined as above and Tz,w,Tiz,w are unimodular matrices, i=1,2,…,q.Proof.
Whenq=1, by Theorem 1, Fz,w=T1z,wPzTz,w, where Tz,w,T1z,w are unimodular matrices, so the conclusion is true.
Whenq≥2, since hrz is a factor of detFz,w, according to Corollary 1, we see that Fz,w can be factorized as T1z,wPzF1z,w, where T1z,w is a unimodular matrix.
Then, we considerF1z,w. For detF1z,w=hq−1⋅rz, according to Lemma 1, we obtain(23)diFz,w=diPzF1z,w,i=1,…,l.
So,(24)dl−r+1Fz,w=dl−r+1PzF1z,w=dl−r+1Pzdl−r+1F1z,w=hzdl−r+1F1z,w.
Note thathqz|dl−r+1Fz,w; then, hq−1z|dl−r+1F1z,w. According to Corollary 1, F1z,w can be factorized as T2z,wPzF2z,w. Then,(25)Fz,w=T1z,wPzT2z,wPzF2z,w.
Imitating the procedure above successively, we can obtain that(26)Fz,w=T1z,wPzT2z,wPz⋯Tqz,wPzTz,w,where Tz,w,Tiz,w are unimodular matrices, i=1,2,…,q.Lemma 3.
LetPz be defined as above and Tz,w∈Kl×lz,w be unimodular. Suppose Fz,w=Ps1zTz,wPs2z. If the l−r×l−r minors of Fz,w have no common zero in K¯2, then Fz,w is equivalent to Ps1+s2z.Proof.
Let(27)Tz,w=T11T12T21T22,where T11∈Kl−r×l−rz,w, T12∈Kl−r×rz,w, T21∈Kr×l−rz,w, and T22∈Kr×rz,w.
Note that(28)Pz=Il−rhzIr.
Then,(29)Fz,w=Ps1zTz,wPs2z=T11T12hs2zT21hs1zT22hs1+s2z.
By computing,dl−r+1Fz,w=hs1+s2zdl−r+1Tz,w. Because Tz,w is unimodular and the l−r×l−r minors of Fz,w have no common zero,(30)d1Fz,w=d2Fz,w=⋯=dl−rFz,w=1,dl−r+kFz,w=hks1+s2z,1≤k≤r.
SinceT11,T12 is ZLP, the l−r×l−r minors of T11,T12 have no common zero. Set the l−r×l−r minors of T11,T12 to be e1,e2,…,et, where e1=detT11. Then, the l−r×l−r minors of T11,T12hs2z are e1,e2hm2z,…,ethmtz, where s2≤mi≤r⋅s2,i=2,…,t.
We prove thatT11,T12hs2z is ZLP.
Suppose thel−r×l−r minors of T11,T12hs2z have a common zero a0; then, a0 is a common zero of e1 and hz. Note that(31)Fz,w=T11T12hs2zT21hs1zT22hs1+s2z.
Then, thel−r×l−r minors of Fz,w have common zero a0. This is a contradiction. So, the l−r×l−r minors of T11,T12hs2z have no common zero. Hence, T11,T12hs2z is ZLP. By the Quillen–Suslin theorem, we can find a unimodular matrix M1∈Kl−r×l−rz,w that satisfies(32)T11,T12hzM1=Il−r0l−r,r.
Then,(33)Fz,wM1=Il−r0l−r,rN1N2,where N1∈Kr×l−rz,w and N2∈Kr×rz,w. Let(34)M2=Il−r0l−r,r−N1Irsuch that(35)M2Fz,wM1=Il−r0l−r,r0r,l−rN2.
Let(36)M3=Il−r0l−r,r0r,l−rN2.
We knowM3 is equivalent to Fz,w.
According to Lemma1, diM3=diFz,w,i=1,…,l. Hence,(37)d1M3=d2M3=⋯=dl−rM3=1,dl−r+kM3=hks1+s2z,1≤k≤r.
Notice that(38)M3=Il−r0l−r,r0r,l−rN2=Il−rbl−r+1,l−r+1bl−r+1,l−r+2⋯bl−r+1,lbl−r+2,l−r+1bl−r+2,l−r+2⋯bl−r+2,l⋮⋮⋱⋮bl,l−r+1bl,l−r+1⋯bl,l.
We have thathz divides every element of N2, so(39)M3=Il−rhs1+s2z⋱hs1+s2zM4.
Hence,M3=Ps1+s2z⋅M4, and detM4=1. Note that M2⋅Fz,w⋅M1=M3, so M2⋅Fz,w⋅M1=Ps1+s2z⋅M4; then, Fz,w=M2−1Ps1+s2z⋅M4M1−1, where M2−1 and M4M1−1 are unimodular matrices. Thus, Fz,w is equivalent to Ps1+s2z.Theorem 3.
LetFz,w∈Kl×lz,w and hz∈Kz be an irreducible polynomial. If detFz,w=hq⋅rz and hqz|dl−r+1Fz,w, where q and r are positive integers, then Fz,w is equivalent to the Smith form(40)Pqz=Il−rhqz⋱hqz,if and only if the l−r×l−r minors of Fz,w have no common zero in K¯2.Proof.
Sufficiency: by Theorem2, we have that(41)Fz,w=T1z,wPzT2z,wPz⋯Tqz,wPzTz,w,where(42)Pz=Il−rhz⋱hz.
Tz,w,Tiz,w are unimodular matrices, i=1,2,…,q.
We first prove thatPzT2z,wPz is equivalent to P2z. From the l−r×l−r minors of Fz,w which have no common zero in K¯2, we know that the l−r×l−r minors of PzT2z,wPz have no common zero in K¯2 by combining with Lemma 2. Then, according to Lemma 3, PzT2z,wPz is equivalent to P2z.
Repeating the procedure above, we obtain thatPzT2z,wPz⋯Tqz,wPz is equivalent to the matrix Pqz, so there exist U1z,w,V1z,w such that(43)PzT2z,wPz⋯Tqz,wPz=U1z,wPqzV1z,w,where U1z,w,V1z,w are unimodular matrices. Hence, we have that(44)Fz,w=Uz,wPqzVz,w,where Uz,w=T1z,wU1z,w and Vz,w=V1z,wTz,w are unimodular matrices. Therefore, Fz,w is equivalent to the Smith form Pqz.
Necessity: sinceFz,w is equivalent to Pqz, there exist unimodular matrices Uz,w and Vz,w such that Fz,w=Uz,wPqzVz,w. For the l−r×l−r minors of Pqz which have no common zero in K¯2, the l−r×l−r minors of Fz,w have no common zero in K¯2 by Lemma 2.Remark 2.
According to Theorem3, Problem 2 is solved, and a criterion for discriminating this class of bivariate polynomial matrices to be equivalent to the Smith form is also presented.Theorem 4.
LetFz,w∈Kl×mz,wl≤m be of full row rank and hz∈Kz be irreducible. Suppose dlF=hq⋅rz and hqz|dl−r+1F, where q and r are positive integers. Then, Fz,w is equivalent to the Smith form(45)Qz=Pqz0l×m−l,if and only if the l−r×l−r minors of Fz,w have no common zero in K¯2.Proof.
Sufficiency: by Theorem 3.3 in [17], there are Hz,w∈Kl×lz,w and F1z,w∈Kl×mz,w satisfying Fz,w=Hz,wF1z,w, where detHz,w=dF and F1z,w is ZLP. Then, the l−r×l−r minors of Hz,w have no common zero by using Lemma 2. Combined with hqz|dl−r+1H and Theorem 2, there are unimodular matrices U11z,w, U12z,w∈Kl×lz,w such that Hz,w=U11z,wPqzU12z,w. Then,(46)Fz,w=U11z,wPqzU12z,wF1z,w.
We know thatU12z,wF1z,w is also ZLP (U12z,w is unimodular). According to the Quillen-Suslin theorem, there exists an m×m unimodular matrix Uz,w which satisfies that U12z,wF1z,wUz,w=Il0l,m−l. Then,(47)Fz,w=U11z,wPqz0l,m−lUz,w=U1z,wQzUz,w.
According to thel−r×l−r minors of Fz,w which have no common zero and hqz|dl−r+1F, we see that Qz is the Smith form of Fz,w. Note that U1z,w and Uz,w are invertible matrices, so Fz,w is equivalent to the Smith form Qz.
Necessity: sinceFz,w is equivalent to the Smith form Qz, it is easy to see that the l−r×l−r minors of Qz have no common zero in K¯2. By Lemma 2, we have that the l−r×l−r minors of Fz have no common zero in K¯2.Remark 3.
A positive answer to Problem3 is presented in Theorem 4. And the equivalence of a rectangle matrix Fz,w and its Smith form is considered, which makes the result more general.


## 4. An Example

In this section, we use an example to illustrate our results and methods.Example 1.
Consider a3×3 2D polynomial matrix of ℝ3×3z,w:(48)Fz,w=a11a12a13a21a22a23a31a32a33,where(49)a11=z+1−w21−2w−2z2+2zw+2z2+z+12,a12=z+1−w2z−w+zw+1−zw+z+w−1z2+z+12,a13=wz+1−w2+zw+z2+1z2+z+12,a21=−w1−2w−2zz2+z+12,a22=−wz−w+zw+1−zz2+z+12,a23=−w2+zz2+z+12,a31=z1−2w−2z2+2z2+z+12,a32=zz−w+zzw+1−z+w−1z2+z+12,a33=zw+z2+1z2+z+12.By computing,d1Fz,w=1, d2Fz,w=z2+z+12, d3Fz,w=z2+z+14, and detFz,w=z2+z+14. We know that the 1×1 minors of Fz,w have no common zero in ℂ2. Combining z2+z+1∈Kz is irreducible and z2+z+12|d2Fz,w; by Theorem 3, we have that Fz,w is equivalent to the Smith form(50)P2z=1z2+z+12z2+z+12.LetFz,w¯=fij¯ denote Fz,w=fij mod z2+z+1; then,(51)Fz,w¯=z+1−w21−2wz+1−w2z−wwz+1−w22w2−ww2−zw−w2−2zw+z−zw−z−1zw,and no row of Fz,w¯ is zero vector mod z2+z+1. Note that the nonzero polynomial of the least degree in w among the first column is −2zw+z, so we postmultiply Fz,w¯ by a unimodular matrix U11=001010100 such that −2zw+z can be changed to the position (1, 1). Then,(52)U11Fz,w¯=−2zw+z−zw−z−1zw2w2−ww2−zw−w2z+1−w21−2wz+1−w2z−wwz+1−w2.We consider the element−2zw+z; −2z is the leading coefficient of −2zw+z. We know that −2z and z2+z+1 are relatively prime in ℝz, so we can find x1z=1/2z+1, y1z=1 such that x1z⋅−2z+y1z⋅z2+z+1=11/2z+1⋅−2z+z2+z+1=1; then, 1/2z+1⋅−2z=1−z2+z+1≡1modz2+z+1. So, x1z⋅−2zw+z=1/2z+1⋅−2zw+z≡w−1/2modz2+z+1; that is, −2zw+z can be reduced to a monic polynomial w−1/2 by mod z2+z+1.Then, we reduce other elements in the first column.For the element2w2−w, we can find q1z,w=2w and r1z,w=0 such that 2w2−w=w−1/2⋅q1z,w+r1z,w≡1/2z+1⋅−2zw+z⋅2w+0modz2+z+1. In reality, 2w2−w=1/2z+1⋅−2zw+z⋅2w−z2+z+1w−2w2.And for the elementz+1−w21−2w, we can find q2z,w=2w2−2z−2 and r2z,w=0 such that(53)z+1−w21−2w=w−12⋅q2z,w+r2z,w=w−12⋅2w2−2z−2+0≡12z+1⋅−2zw+z⋅2w2−2z−2+0modz2+z+1.Actually,z+1−w21−2w=1/2z+1⋅−2zw+z⋅2w2−2z−2−z2+z+1w2−z−11−2w.Let(54)U12=100−12z+12w10−12z+12w2−2z−201.We have(55)U12U11⋅Fz,w¯=−2zw+z−zw−z−1zw000000modz2+z+1,so(56)U12U11⋅Fz,w=1z2+z+1z2+z+1F1z,w,where(57)F1z,w=b11b12b13b21b22b23b31b32b33,b11=z1−2w−2z2+2z2+z+12,b12=zz−w+zzw+1−z+w−1z2+z+12,b13=zw+z2+1z2+z+12,b21=2w2−w+−2z+zw+w2z2+2z2+z+1,b22=w2−zw−zw+wzzw+1−z+w−1z2+z+1+zw+1−zz2+z+1,b23=−w2+z−wz+1z2+1z2+z+1,b31=1−2w−w2+1+z+−2z2−2zw−2+z+1−w2+1+z−2z2−2z2+z+1,b32=−w2+1+zz−w+zw+1−zw+z+w−1z2+z+1+z+1−w2+1+zzzw+1−z+w−1z2+z+1,b33=−w2+1+zw+z2+zw+1z2+z+1+z+1−w2+z+1z2+1z2+z+1.LetU1=U12U11; then,(58)Fz,w=U1−11z2+z+1z2+z+1F1z,w,F1z,w¯=−2zw+z−zw−z−1zw2w2−ww2−zw−w21−2w−w2+1+z−w2+z+1z−ww−w2+z+1.Now, considerF1z,w¯, for none of its rows are zero vectors mod z2+z+1; repeating the steps above, we can obtain that(59)U13⋅F1z,w¯=−2zw+z−zw−z−1zw000000modz2+z+1,where(60)U13=100−12z+12w10−12z+12w2−2z−201.Hence,(61)U13F1z,w=1z2+z+1z2+z+1F2z,w,where(62)F2z,w=c11c12c13c21c22c23c31c32c33,c11=z1−2w−2z2+2z2+z+12,c12=zz−w+zzw+1−z+w−1z2+z+12,c13=zw+z2+1z2+z+12,c21=2w2−w−2z+zw+w2z2+2z2+z+2,c22=w2+1−z−zw+wz2w+z−z2z2+z+2−w2−wz+1z2+z+2,c23=z−w2−wz+1z2+1z2+z+2,c31=1−2w−w2+1+z−2z2+2zw+2+z+1−w2+1+z−2z2−2z2+z+2,c32=−w2+1+zz−w+zw+1−zw+z+w−1+z2+z+2z+1−w2+1+zw−1+z2+z−w2+1+zzw+1−zz2+z+2,c33=−w2+1+zw+z2+zw+1+z2+z+2z+1−w2+z+1z2+1.So,(63)F1z,w=U13−11z2+z+1z2+z+1F2z,w.Combining(64)Fz,w=U1−1PzF1z,w,we have that(65)Fz,w=U1−1PzU13−1PzF2z,w.ConsiderPzU13−1Pz; combined with Lemma 3, we obtain(66)PzU13−1Pz=100wz+wz2+z+1z2+z+120−z+1−w2+1+zz2+z+10z2+z+12.Let(67)U14=100wz+wz2+z+110−z+1−w2+1+zz2+z+101.Then,(68)PzU13−1Pz=U14P2z.So,(69)Fz,w=U1−1U141z2+z+12z2+z+12F2z,w,and by computing, U1−1U14 and F2z,w are unimodular matrices. Therefore, Fz,w is equivalent to P2z.


## 5. Conclusions

In this paper, we have investigated the reduction of several kinds of bivariate polynomial matrices inKz,w, where K is an arbitrary field. Some new results on these matrices to be reduced to their Smith forms are presented. Furthermore, the conditions of these results are easily verified. An example is given to illustrate our method in the end of the article. All of these are useful for reducing 2D systems.



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