# Reduction to a Canonical Form of a Third-Order Polynomial Matrix with One Characteristic Root by means of Semiscalarly Equivalent Transformations

Authors: B. Z. Shavarovskii
Journal: Journal of Mathematics
 (2020)

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## Abstract

For the selected class of polynomial matrices of order three with one characteristic root with respect to the transformation of semiscalar equivalence, special triangular forms are established. The theorems of their uniqueness are proved. This gives reason to consider such canonical forms.


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## Body

## 1. Introduction

In [1], it is proved that the matrix Fx∈Mn,Cx of full rank by means of transformation:(1)Fx⟶PFxQx=Gx,where P∈GLn,C and Qx∈GLn,Cx can be reduced to the lower triangular form with invariant factors on the principal diagonal. Subdiagonal elements in a matrix of this form are ambiguously defined. The matrices Fx,Gx which are related by the transformation (1) are called semiscalarly equivalent [1]. In [2], the specified triangular form for polynomial 3×3 matrices with one characteristic root is a little simplified. The resulting matrix of a simplified triangular form is called a reduced. In [2], the invariants of the reduced matrix are established. In particular, the invariance of the location of zero subdiagonal elements is proved. In [3], the reduced matrix, if there are some zero elements under its principal diagonal, by means of transformations of the form (1) (i.e., by means of semiscalarly equivalent transformations) is reduced to such matrices, which are uniquely defined. This gives grounds to consider the obtained matrices canonical for the selected class of matrices. This article introduces canonical forms for reduced matrices with all nonzero subdiagonal elements.


## 2. Previous Information

Here are some definitions and notations that will be used in this article, which are known from [2, 3]. This applies to the definitions of the younger degree, of the younger term, of the younger coefficient, q-monomial, andq-coefficient of the polynomial and others. For example, monomial 4x2 and its degree 2 are, respectively, a younger term and younger degree of polynomial fx=−3x7+6x5−x4+4x2, and 4 is the younger coefficient of this polynomial. Monomial 6x5and its coefficient 6 are, respectively, a 5-monomial and 5-coefficient of polynomial fx.Let all the roots of the characteristic polynomialdetFx (= characteristic roots) of the matrix Fx be equal to each other; that is, the matrix Fx has only one (without taking into account the multiplicity) characteristic root. Without loss of generality, we assume that the only characteristic root is zero and the first invariant factor of matrix Fx is equal to one. With such assumptions, it is proved in [2] that, by means of semiscalarly equivalent transformations, the matrix Fx is reduced to the matrix of the form(2)Ax=100a1xxk10a3xa2xxk2,which satisfies the following conditions:(1)
dega1<k1, dega2,dega3<k2, a2x=xk1a2′x=0, a10=a30=a2′0=0.(2)
codega3≠codega1, codega3≠codega2′, if codega3<codega2.(3)
codega3≠2codega1+codega2′ and 2codega1-monomial in a1x is absent if codega3≥codega2.(4)
Younger coefficients ina1x and a2x are units.The matrixAx of the form (2) with conditions (1)–(4) in [2] is called the reduced matrix. Next, we consider the situation where the last two invariant multipliers of the matrix Ax do not coincide, that is, k1<k2. The case k1=k2 was considered in [4]. The notation Ax≈Bx means that the matrices Ax and Bx are semiscalarly equivalent. It should be noted that the problem of classification with respect to semiscalar equivalence of matrices of the second order is solved in the article [5]. Thus, this article discusses other situations that differ from [4, 5]. In [2], it is proved that, in case Ax≈Bx, we can choose the left transformation matrix in the transition from Ax to the reduced matrix(3)Bx=100b1xxk10b3xb2xxk2,of the lower triangular form. We will then apply semiscalarly equivalent transforms Ax⟶SAxRx=Bx to the matrix Ax to obtain a reduced matrix Bx of the form (3) with predefined properties. Let us show that, by a given reduced matrix Ax of form (2) and a matrix(4)S=1s12s1301s23001,we can find the matrix Bx and the right transformation matrix Rx so that Ax≈Bx=SAxRx. Using the method of uncertain coefficients for given elements a1x,a2x,a3x and s12,s13,s23 of matrices Ax and S, respectively, with congruence(5)a1x+s23a3x−b1x1+s12a1x+s13a3x≡0modxk1,we find b1x∈Cx, degb1<k1. We denote such elements by rxuv,u,v=1,2: (6)r11x=1+s12a1x+s13a3x,r12x=s12xk1+s13a2x,r21x=a1x+s23a3x−b1xr11xxk1∈Cx,r22x=1+s23a2′x−s12b1x−s13b1xa2′x.Herea2′x=a2x/xk1∈Cx. We form the matrix rxuv12 and consider the congruence(7)b3xb2xruvx12≡a3xa2xmodxk2,with the unknown b2x,b3x. Since the free member of a matrix polynomial rxuv12 is a unit matrix, we can use the method of uncertain coefficients to solve this congruence and find b2x,b3x∈Cx, degb2,degb3<k2. We can check that b2′x=b2x/xk1∈Cx. In addition to the above, we also denote(8)r13x=s13xk2,r23x=s23xk2−k1−b1xr13x,r31x=a3x−b3xr11x−b2xr21xxk2∈Cx,r32x=a2x−b3xr12x−b2xr22xxk2∈Cx,r33x=1−b3xr13x−b2xr23xxk2∈Cx.By the aboverijxi,j=1,2,3, and from the congruences (5) and (7) bix, we construct rijx13 and a matrix Bx of the form (3), respectively. We can be convinced of equality SAx=Bxrijx13. This means that rijx13 is reversible and its inverted matrix together with the matrix S reduces Ax to Bx. If the matrix S (4) in the transition from Ax to Bx has one of the following views: (9)10001s23001,1s120010001,or10s13010001,then we will say that transformations of type I, transformations of type II, or transformations of type III, respectively, are applied to the matrix Ax. We shall use the following notation for matrices Ax of form (2) and Bx of form (3):(10)δAx:=deta1x1a3xa2′x,δBx:=detb1x1b3xb2′x.


## 3. The Main Results

Theorem 1.
Suppose that in the reduced matrixAx of the form (2), we have a1x,a2x,a3x≠0, a2x=xk1a2′x, q1:=codega1, q2:=codega2′, q3:=codega3, nj:=q3,j=1q2+q3,j=3, mj:=qj+q3, j=1,3. Then, the matrix Ax is semiscalarly equivalent to the reduced matrix Bx of the form (3), where elements b1x,b2x,b3x≠0 satisfy one of the following conditions:(1)
2q3-monomial is absent in b3x, if q3<q1 and q3<q2.(2)
2q3- and q1+q3-monomials are absent in b3x if q3<q1 and q3>q2.(3)
Ifq3>q1 and q3<q2, then in the first of polynomials bjx, j=1,3, which satisfies condition nj<kj, nj-monomial is absent, and in the first of these polynomials, which satisfies condition mj<kj, mj-monomial is absent.The matrixBx is uniquely defined.Proof.
Existence.(1)
If2q3≥k2, then Ax is the desired matrix. Otherwise, we denote by d0 and d1, respectively, the lower coefficient and the 2q3-coefficient of the polynomial a3x and apply to Ax transformations of the type III. In the left transformation matrix (see (9)), we put s13=d1/d02. The elements bix, i=1,2,3, of the obtained in this way matrix Bx satisfy the congruence:(11)a1x−b1x−s13a3xb1x≡0modxk1,(12)a2x−b2x−s13a2xδBx≡0modxk2,(13)a3x+δBx+s13a3xδBx−a1xb2′x≡0modxk2.First, we obtain from (11) and (12) that the lower terms in b1x,b2x are identical with the lower terms in b1x,b2x, respectively. Further note that the younger terms in δBx and b3x coincide, the lower degrees of the last two additions in the left-hand side (13) exceed codegb3=q3, and inequality codega3xδBx=2q3<codega1xb2′x holds. Therefore, by comparing the 2q3-coefficients in both parts of (13), we obtain zero for such 2q3-coefficient in b3x. So, there Bx is a desired matrix.(2)
If in matrixAx has 2q3≥k2, then everything is proven—this matrix is the desired one. Otherwise, we will apply to it the transformation mentioned in Section 1. To show the absence of the 2q3-monomial in b3x, one must take into account that codega1x−b1x≥q1+q3 (see (11)).Therefore, in (13), we have codegb2′xa1xb1x>2q3. The remaining considerations are the same as in paragraph 1. In order to not introduce new notations, we further assume that there is no 2q3-monomial in the element a3x of original matrix Ax. If q1+q3≥k2, then everything is proven—the matrix Ax is the desired one. Otherwise to Ax, we apply transformation of the type II. At the same time, in the left transformation matrix (see (9)), we put s12=d2/d0, where d0 and d2 are, respectively, the lower coefficient and q1+q3-coefficient of the polynomial a3x. The elements bix, i=1,2,3, of the reduced matrix Bx obtained in this way satisfy the congruence:(14)a1x−b1x−s12a1xb1x≡0modxk1,(15)a2x−b2x+s12ΔBx≡0modxk2,(16)a3x+δBx+s12a1xδBx−a1xb2′x≡0modxk2.It can be seen from (14) and (15) that the younger terms in b1x,b2x are the same as the lower terms in a1x,a2x, respectively (their coefficients are equal to one).Let us write (16) as follows:(17)a3x−b3x+s12a1xb1xb2′x−b3x−−a1x−b1xb2′x≡0modxk2.From (14), we have codega1x−b1x≥2q1. Because(18)codegb2′xa1xb1x=2q1+q2,codegb2′xa1x−b1x≥2q1+q2,andcodega1xb3x=q1+q3<2q1+q2, then by comparing the q1+q3-coefficients in both parts of (17), we find that b3x contains no q1+q3-monomial. And because 2q3<q1+q3, then in b3x, as in a3x, there is no 2q3-monomial.(3)
Suppose that conditionsq3>q1 and q3<q2 are satisfied in matrix Ax.(1)
Ifq3≥k1 and 2q3≥k2, then all is proved—matrix Ax is the desired one.(2)
Letq3≥k1 and 2q3<k2. Since q3<codega2, then q3 (as well as q1) is invariant (see Proposition 6 [2]). We apply to Ax the transformation specified in Section 1. As a result, we obtain the matrix Bx in the form (3). Its elements satisfy the congruences (11)–(13). Since q1+q3>k1, then from (11), we have a1x=b1x. It can be seen from (12) that a2x=b2x. Now we can represent (13) as(19)a3x−b3x−s13a3xb1xb2′x−b3x≡0modxk2.From the last congruence, we havecodegb3=q3. Therefore, Bx is a reduced matrix. If we take into account q1+q2>q3, then by comparing the 2q3-coefficients at both times (19), we will conclude that, in b3x, there is no 2q3-monomial. If q2+q3≥k2, then everything is proved-matrix Bx is the desired one. Otherwise, we take another step. In order not to introduce new notations, we will assume that element a3x of matrix Ax does not contain 2q3-monomial. We apply to Ax transformations of the type I. In the left transformation matrix (see (9)), we put s23=d2/d0, where d0 and d2 are, respectively, the lower coefficient and the q1+q3 coefficient of polynomial a3x. The elements bix, i=1,2,3 of the resulting matrix Bx satisfy the congruence:(20)a1x−b1x+s23a3x≡0modxk1,(21)a2x−b2x−s23a2′xb2x≡0modxk2,(22)a3x−b3x+b1x−a1xb2′x−s23b2′xa3x≡0modxk2.From (20), we have that a1x=b1x, and from (21), it follows codega2=codegb2. Then, from (22), we get(23)a3x−b3x−s23b2′xa3x≡0modxk2,where we can getcodegb3=q3. Comparing the coefficients in both parts of (23), we conclude that there is no the monomial of degree q1+q3 in polynomial b3x. At the same time, in b3x, as in a3x, there is no 2q3-monomial.(3)
Now suppose that, in matrixAx, we have q3<k1 and 2q3≥k2. Apply to Ax transformations of the type I. In the left transformation matrix (see (9)), we put s23=−d3/d0, where d0 and d3 are, respectively, the lower coefficient of the polynomial a3x and the q3-coefficient of the polynomial a1x. As a result, we obtain a matrix Bx of the form (3) whose elements bix, i=1,2,3 satisfy the congruences (20)–(22). From (20), we have that codegb1=q1 and q3-monomial in b1x is absent. From (21), codegb2=codega2, and from (22), we have codegb3=q3. It also follows from (20)–(22) that the lower coefficients in aix and bix, i=1,2,3 coincide. That is, Bx is a reduced matrix. If q1+q3≥k1, then everything is already proven. Then, Bx is the desired matrix. Otherwise, in order to not introduce new notations, we consider the q3-coefficient in the element a1x of the matrix Ax null. Denote by d0 and d4, respectively, the lower polynomial coefficient of the polynomial a3xand q1+q3-coefficient of the polynomial a1x. We perform over the matrix Ax transformation of the type III. For this, we put s13=d4/d0 in the left transformation matrix (see (9)). The elements of the resulting matrix Bx satisfy the congruences (11)–(13). From (21), we obtain that codegb1=q1, the lower coefficient of the polynomial b1x is 1, and its q3- and q1+q3-coefficients are zero. From (12), it is seen that the lower coefficient in b2x, as in a2x, is equal to 1. Therefore, the matrix Bx has the necessary properties.(4)
Letq3<k1 and 2q3<k2. We can assume that the q3-coefficient in a1x of matrix Ax is zero. If this is not the case, then to Ax, we will apply transformation of the type I described in Section 3. If q1+q3<k1, then to Ax, we apply transformation of the type III described in Section 3. Then, the resulting matrix will be zero q1+q3-coefficient and will remain zero q3-coefficient of the polynomial in position 2,1. If q1+q3≥k1, then from the matrix Ax by means of transformations of the type III referred to in item 1, we go to the redundant matrix Bx, in which 2q3-monomial of polynomial b3x is absent. Then, q3-factor in b1x will also remain zero. This proves the first part of the theorem (existence).


### 3.1. Uniqueness of the Matrix in Theorem 1

(1)
Suppose that, for the reducible matricesAx, Bxof forms (2) and (3), condition 1 of theorem holds, and, in addition, we have Ax≈Bx. Then, the left transformative matrix S in the equality SAxRx=Bx can be chosen in the form (9) (see Corollary 1 and Remark 1 [2]) and elements aix and bix, i=1,2,3, of these matrices satisfy the congruence(24)a3x+δBx−s13a3xδBx−a1xb2′x≡0modxk2.We havecodegδBxa3x=2q3<q1+q2. If 2q3≥k2, then from (24), a3x−b3x≡0modxk2 follows. Otherwise, in (24), we have s13=0 since 2q3-monomials in a3x and b3x are absent. In any case, Ax=Bx.(2)
Suppose that the reduced matricesAx, Bx of the forms (2) and (3) satisfy condition 2 of theorems and Ax≈Bx. Then, in the left transformative matrix S (4), in the transition from Ax to Bx, we have s23=0 (see Corollary 1 and Remark 1 [2]) and the elements aix and bix of the matrices Ax and Bx satisfy the congruences:(25)a1x−b1x1+s12a1x+s13a3x≡0modxk1,a2x−b2x+s12ΔBx+s13a2′xΔBx≡0modxk2,a3x+δBx+δBxs12a1x+s13a3x−a1xb2′x≡0modxk2.From (25), we can write(26)a3x−b3x+δBxs12a1x+s13a3x+b2′xb1x−a1x≡0modxk2.From (25), we have codega1x−b1x≥q1+q3. It is easy to see that(27)codegδBxa3x=2q3<q1+q3==codegδBxa1x<codegb2′xb1x−a1x.If2q3≥k2, then from (26), we have a3x−b3x≡0modxk2; hence, it follows a3x=b3x.Since2q3<codegΔA<codega2′xΔAx, then from (26), a2x−b2x≡0modxk2 follows, whence a2x=b2x. From (25), taking into account 2q3<q3+q1<codega1x2, we get a1x−b1x≡0modxk1 from where a1x=b1x. So, we have Ax=Bx.If2q3<k2, then from (25), we get s13=0. If q1+q3≥k2, then taking into account q1+q3<codegδA+k1 and q1+q3<2q1 from (25) and (26), we have ajx−bjx≡0modxkj, j=1,2, and a3x−b3x≡0modxk2. Therefore, Ax,Bx coincide. If q1+q3<k2, then from (26), we obtain s12=0. Hence, in this case, the matrices Ax,Bx also coincide.(3)
Suppose that the reduced matricesAx,Bx of the forms (2) and (3) satisfy condition 3 of theorems and Ax≈Bx. Then, for the elements of these matrices, we can write the congruences:(28)a1x−b1x+s23a3x−s13a3xb1x≡0modxk1,a2x−b2x−s23a2′xb2x+s13a2xδBx≡0modxk2,a3x+δBx−a3xs23b2′x+s13δBx−a1xb2′x≡0modxk2.Ifq3≥k1, then q1+q3>k1, and from (28), we get a1x=b1x. Then, (28) will take the form(29)a3x−b3x−a3xs23b2′x+s13δBx≡0modxk2.Obviously,codega3xδBx=2q3. If 2q3≥k2, then (29) implies a3x=b3x since codega3xb2′x>codega3xδBx.Then, from (28), we get a2x=b2x since(30)codegb2xa2′x>codegb2′xa3x,codegΔBxa2′x>codegδBxa3x.If2q3<k2, then (29) implies s13=0. If, moreover, codega3xb2′x<k2, then from (29), it yields s23=0 and all is proved. If codega3xb2′x≥k2, then all the same from (28) and (29), we have a3x=b3x and a2x=b2x, respectively.Ifq3<k1, then from (28), we get s23=0. If in addition q1+q3<k1, then from (28), it follows also s13=0 and all is proved. If q1+q3≥k1, then a1x=b1x, and again from (28), we go to (29). It follows from this that s13=0, if codega3xδBx<k2. And if codega3xδBx≥k2, then immediately from (28) and (29), we have a3x=b3x and a2x=b2x, respectively. Theorem is proved.Suppose that, in the reduced matricesAx,Bx of the forms (2) and (3), we have a1x,a2x,a3x,b1x,b2x,b3x≠0. Let us keep the notation given in theorem:(31)q1:=codega1,q2:=codega2′,q3:=codega3,a2′x=a2xxk1∈Cx,b2′x=b2xxk1∈Cx.We define polynomials:(32)a11x:≡a1x2modxk1,a22x:≡a2′x2modxk2−k1,a32x:≡a3xa2′xmodxk2,a04x:≡δAxmodxk2−k1.a14x:≡a1xδAxmodxk2,a34x:≡a3xδAxmodxk2.From the coefficients of each of the polynomialsa1x, a3x, and a11x, we form, respectively, columns a¯1, a¯03, and a¯11 of height k1−q1. In the first place, in these columns, we put q1-coefficients, and below in order of increasing degrees, we place the rest of their coefficients, up to degree k1−1 inclusive. We denote by a¯2, a¯22, and a¯04, the columns of height k2−k1−q2, constructed from the coefficients of polynomials a2x, a22x, and a04x, respectively. In the first place in each of these columns, we put q2-coefficients. Below we place the rest of their coefficients (including zero) up to the degree k2−k1−1. Similarly, from the coefficients of polynomials a3x, a32x, a34x, and a14x, we form columns a¯3, a¯32, a¯34, and a¯14 and height k2−q3. Here, we also put in the first place q3-coefficients, and then, in the order of increasing degrees, we place all other coefficients. In the last places, there will be k2−1-coefficients. For Ax, by the columns formed, we construct the matrices of the following form:(33)KA=a¯1a¯2a¯3K0A,K0A=K1AK2AK3A,(34)K1A=−a¯030¯a¯11,K2A=a¯220¯−a¯04,K3A=a¯32−a¯34−a¯14.In complete analogy forBx, we construct matrices of the following form:(35)KB=b¯1b¯2b¯3K0B,K0B=K1BK2BK3B,(36)K1B=−b¯030¯b¯11,K2B=b¯220¯−b¯04,K3B=b¯32−b¯34−b¯14.Obviously, in these matrices, each row consists of monomial coefficients of the same degrees.Theorem 2.
Let in the reduced matrixAx of the form (2), we have a1x,a2x,a3x≠0, q3>q1, q3>q2 and n1:=q1+q3, n2:=q2+codegδA+k1. Then, Ax≈Bx, where in the reduced matrix Bx of the form (3), all elements b1x,b2x,b3x are nonzero, polynomial b3x does not contain n1-monomial if n1<k1, and polynomial δBx does not contain n2−k1-monomial if n2<k2.In addition, one of the following conditions is true:(1)
Inb1x, n1-monomial is absent, if n1<k1 and n2<k2.(2)
Inb2x, codegδA+k1- and n2-monomials are absent, if n1≥k1 and n2<k2.(3)
Inb1x, q3- and n1-monomials are absent, if n1<k1 and n2≥k2.(4)
In the first column of the matrixKB (35), the coefficients of the polynomials b1x,b2x,b3x are zero elements that correspond to the maximum system of the first linearly independent rows of the submatrix K0B, if n1≥k1 and n2≥k2.The matrixBx is uniquely defined.Proof.
Existence. Letn1<k1.
We apply toAx transformation of the type II with the left transformation matrix of the form (9). At the same time, we put s12=d1/d0, where d0 is the younger coefficient and d1 is the n1-coefficient in a3x. The elements bix, i=1,2,3, of the thus obtained reduced matrix Bx satisfy the congruences (14)–(16). We write (16) in the form(37)a3x−b3x−s12a1xb3x−b2xr21x≡0modxk2,where r21x=a1x−b1x−s12a1xb1x/xk1∈Cx. Comparing the n1-coefficients in both parts of the last congruence, we have that b3x does not contain n1-monomial. We further assume that element a3x of the matrix Ax does not contain n1-monomial (if n1<k1).
Letn2<k2. Denote by c0 and c1, respectively, the junior and n2−k1-coefficients of the polynomial δAx. Apply to Ax transformations of the type I with the left transformation matrix of the form (9), while putting s23=c1/c0.
The elementsbix, i=1,2,3, of the thus obtained reduced matrix Bx satisfy the congruences (20)–(22) (with the one listed here s23). From (21) and (22), we obtain(38)δAx−δBx−s23b2′xδAx≡0modxk2−k1.
If we compare then2−k1-coefficients in both parts of the last congruence, we will conclude that δBx does not contain n2-monomial:(1)
Suppose that, in elementa3x of matrix Ax, there is no monomial of degree n1<k1, and in polynomial δAx, there is no monomial of degree n2<k2. Denote by d0 and c1, respectively, the lower coefficient in a3x and n1-coefficient in a1x. With the help transformation of the type III, we pass from Ax to the reduced matrix Bx. In the left transformation matrix (see (9)), we put s13=c1/d0. The elements of the resulting matrix Bx satisfy the congruences (11)–(13) (with the one specified here s13). From (11), we get that, in element b1x, n1-monomial is missing.We write (13) in the form(39)a3x−b3x−s13a3xb3x−b2xr21x≡0modxk2,wherer21x=a1x−b1x−s13a3xb1x/xk1∈Cx. Since 2q3>n1, then, as seen from the last congruence, in b3x, as in a3x, there is no n1-monomial. Also in δBx, as in δAx, there is no n2−k1-monomial. This is evident from the congruence(40)δAx−δBx+s13δAxδBx≡0modxk2−k1,which is recorded on the basis of (12) and (13) since codegδAxδBx>n2−k1. This proves the existence of matrix Bx with condition (1) specified in theorem.(2)
Suppose that conditionsn1≥k1, n2<k2, are satisfied in matrix Ax, and n2−k1-monomial is absent in polynomial δAx. We denote by c0 and d2, respectively, the lower coefficient in δAx and the codegδA+k1-coefficient in a2x. Let us do over matrix Ax transformation of the type II. To do this we put s12=−d2/c0 in the left transformation matrix (see (9)). We obtain a reduced matrix Bx whose elements satisfy the congruences of the form (14)–(16) (with s12 indicated here). Taking into account that the lower coefficients in δAx and δBx coincide, then from (15) we find that codegδA+k1-monomial is absent in b2x. From (15) and (16), we have δAx−δBx≡0modxk2−k1. It follows that, in δBx, as in δAx, there are no monomials of degree n2−k1.Next, we consider the absence ofcodegδA+k1-monomial in element a2x of the matrix Ax. Denote by c0 and d3, respectively, the lower coefficient in δAx and n2-coefficient in a2x. Above the matrix Ax, we carry out the transformation of the type III. Here, we put s13=d3/c0 in the left transformation matrix (see (9)). The elements of the obtained reduced matrix Bx satisfy the congruences of the form (11)–(13) (with s13 indicated here). It can be seen from (12) that n2-monomial is absent in b2x. Also codegδA+k1-coefficient in b2x will remain zero since codegδA<n2−k1. As can be seen from (40), in δBx, as in δAx, n2−k1-monomial is absent since codegδAxδBx>n2−k1.The existence of the required matrixBx with condition (2) is proved.(3)
Letn2≥k2 for Ax and in a3x be absent monomial of degree n1<k1. In the first step, we apply to the matrix Ax transformation of the type I with the left transformative matrix (see (9)), in which s23=−c2/d0, where d0 and c2 are, respectively, the lower coefficient in the a3x and the q3-coefficient in a1x. As a result, we obtain a reduced matrix Bxof the form (3) whose elements satisfy the conditions of the form (20)–(22) (with s23 selected here). From (20), it is seen that, in b1x, the q3-monomial is absent. From (20) and (22), it can be written as(41)a3x−b3x−b2xr21x≡0modxk2,wherer21x=a1x−b1x+s23a3x/xk1∈Cx. From the last congruence, it can be seen that n1-monomial is absent in b3x as in a3x.Let alreadya1x in Ax not contain q3-monomial. Denote by d0 and c3, respectively, the lower coefficient in a3x and the n1-coefficient in a1x and let s13=c3/d0. In the second step, with the help of the transformation of the type III with the specified s13 in the left transformative matrix (see (9)), we pass from Ax to some reduced matrix Bx of the form (3). For elements of the matrix Bx, conditions (11)–(13) (with the specified here s13) are satisfied. From (11), it follows that, in b1x, there is no n1-monomial. In addition, b1x does not contain q3-monomial. On the basis of (11) and (13), we can write the congruence of the form (41) in which r21x=a1x−b1x−s13a3xb1x/xk1∈Cx. It shows that, in b3x, in comparison with a3x, the zero coefficient of n1-monomial is preserved. This proves the existence for the matrix Ax a semiscalarly equivalent reduced matrix Bx with condition 3.(4)
Suppose that conditionsn1≥k1, n2≥k2, are satisfied in the reduced matrix Ax. If K0A=0 in KA (33), then the desired matrix is Ax and everything is already proven. Otherwise, in the first step, we fix in the matrix K0A the first nonzero row u¯1=d11d12d13 and the corresponding row d1d11d12d13 in KA. Let u¯1 consist of h1-coefficients and be the l1-rd row in K0A. We find an arbitrary solution x10x20x30t of the equation(42)d11d12d13x1x2x3t=d1.
We apply toAx a semiscalarly equivalent transformation with the left transformative matrix S of the form (4). At the same time, in S, we put s23=x10, s13=x20, and s12=x30. The elements bix, i=1,2,3, of the obtained reduced matrix Bx of the form (3) satisfy the congruence:(43)a1x−b1x+s23a3x−s12a1xb1x≡0modxk1,a2′x−b2′x−s23a2′xb2′x+s12δBx≡0modxk2−k1,a3x−b3x−s23a3xb2′x+s12a1x+s13a3xδBx≡0modxk2.
Depending on which of the matricesK1A, K2A, or K3A (see (34)) row u¯1 belongs, let us consider the congruence (43), respectively. By comparing the h1-coefficients in both parts of that congruence, we conclude that the l1-th element of the first column of matrix KB (35) is zero. In addition, all rows in KB, which precede the l1-th, coincide with the corresponding rows of the matrix KA.
IfrankK0A=1, then everything is already proven. Matrix Bx is the desired one. Otherwise, we assume that the l1-th element of the first column of the matrix KA is zero. In the second step, we fix in K0A the first linearly independent of u¯1 row u¯2=d21d22d23, as well as the corresponding to it row d2d21d22d23 in KA and the degree h2 of monomials, the coefficients of which form these rows. Also let u¯2 be the l2-th row in K0Al2>l1.
We find some solutiony10y20y30t of the equation(44)d11d12d13d21d22d23y1y2y3t=0d2t.
We apply toAx a semiscalarly equivalent transformation with the left transformative matrix S of the form (4), putting s23=y10, s13=y20, and s12=y30. We obtain a reduced matrix Bx of the form (3).
Again, as in the previous step, we consider one of the congruences (43) depending on which of the matrices K1A, K2A, or K3A (see (34)) contains row u¯2. In both parts of this congruence, we compare the coefficients of the h2-monomials and conclude that the l2-th element of the first column of the matrix KB (34) is equal to zero. Also from this and the previous congruences, we get that every row preceding the l2-th in KA coincides with the corresponding row in KB. If rankK0A=2, then everything is already proven. Then, matrix Bx is the desired one. Otherwise, in order to not introduce new designations, we assume that the first column of matrix KA has zero l1-th and l2-th elements. In matrix K0A, we fix the l3-th row, which is the first linearly independent of u¯1, u¯2 (l3>l2>l1). Let this be line u¯3=d31d32d33. To him, KA corresponds to d3d31d32d33. Also let h3 be the exponent that corresponds to these rows. We find the (unique) solution z10z20z30t of the equation(45)d11d12d13d21d22d23d31d32d33z1z2z3t=00d3t.
We apply toAx a semiscalarly equivalent transformation with the left transformation matrix S of the form (4) putting s23=z10, s13=z20, and s12=z30. We obtain the matrix Bx. The above considerations show that Bx is the desired matrix.


### 3.2. Uniqueness of the Matrix in Theorem 2

Suppose that, for the reduced matricesAx, Bx of forms (2) and (3), we have Ax≈Bx. Suppose also that elements a3x, b3x of these matrices do not contain n1-monomials if n1<k1, and in polynomials δAx, δBx, there are no n2−k1-monomials if n2<k2. Let us first show that the matrix S in the transition from Ax to Bx can be selected in the form(46)S=10s1301s23001,if n1<k1, or in the form(47)S=1s12s13010001,if n2<k2.Indeed, the elements of the matricesAx, Bx satisfy the congruence(48)a3x−b3x−s12a1x+s13a3xb3x−b2xr21x≡0modxk2.If we compare the coefficients of the monomers of degreen1<k1 in both parts of this congruence, we get s12=0. Also, from equivalence Ax≈Bx, it is easy to get congruence(49)δAx−δBx+s13δAxδBx−s23δAxb2′x≡0modxk2−k1.If we compare the coefficients of the monomers of degreen2<k2 in both parts of the last congruence, then we come to s23=0:(1)
In casen1<k1, n2<k2, the transition matrix S from Ax to Bx has the form (46) and (47) simultaneously. Therefore, we have s12=s23=0. Elements a1x, b1x in Ax, Bx satisfy (11). From here, we get s13=0. For this reason, matrices 1 and 2 coincide.(2)
Sincen2<k2, then matrix S of the transition from Ax to Bx has the form (47), and the elements a2x, b2x in Ax, Bx satisfy (25). In a2x, b2x, there are no codegδA+k1- and n2-monomials, so from (25), we get s12=s13=0. So, Ax=Bx.(3)
Ifn1<k1, then the matrix S of the transition from Ax to Bx has the form (46). Elements a1x, b1x in Ax, Bx satisfy the congruence (28). From it, we have s23=s13=0, since in a1x, as in b1x, there are no q3-and n1-monomials. Therefore, in this case, Ax, Bx coincide.(4)
Suppose that matrixBx satisfies condition 4, that is, in KB, the elements of the first column corresponding to the maximum system of the first linearly independent rows of the submatrix K0B are zero. Suppose that matrix Ax also has the same property, and in addition, condition Ax≈Bx holds. Then, the elements aix, bix, i=1,2,3, of these matrices satisfy the congruences (43). If in KA, we have K0A=0, then(50)minq3,q12≥k1,minq12,codegδA≥k2−k1,minq2+q3,codega1xδAx≥k2.Therefore, as can be seen from (43), aix=bix, i=1,2,3.If inKA, we have K0A≠0, and l1 is the number of the first nonzero row u¯1 in K0A, then the first l1 elements in the first column of the matrix KA coincide with the corresponding elements in the matrix KB; moreover, l1-th elements are zero. Therefore, in K0A, the first l1+1 rows coincide with the corresponding rows of the matrix K0B. In addition, from congruences (43), we have u¯1s23s13s12t=0. If the next after u¯1 row v¯ in K0A (or in K0B) is linearly dependent on u¯1, then(51)v¯s23s13s12t=0.Then, from (43), we obtain that the first l1+1 elements in the first column of the matrix KA coincide with the corresponding elements in KB. If u¯1 and v¯ are linearly independent, then (51) is still satisfied since in this case, the l1+1-th elements in the first columns of matrices KA and KB are zero. Then, the l1+2 th row in K0A coincides with the corresponding row of the matrix K0B. We think of this row in the same way as it was done above with row v¯. Let u¯2 be the first linearly independent of row u¯1 and l2 be its number in K0A. Then, this row coincides with the l2-th row in K0B, and the first l2 elements of the first column in KA coincide with the corresponding elements in KB, with l2th elements being zero. Then from (43), we have u¯2s23s13s12t=0. If w¯ is the l2+1-th row in K0A, then the corresponding l2+1-th row in K0B is also w¯. If w¯ is linearly dependent on the system u¯1, u¯2, then(52)w¯s23s13s12t=0,and the l2+1-th elements in the first columns of matrices KA,KB coincide. Otherwise, these elements also coincide because they are null. Continuing our considerations, we show that, in KA, KB, the first columns coincide, or at some steps, we will get s12=s13=s23=0. In each case, Ax=Bx. Theorem is proved.Example 1.
MatricesAx=100x3x40x6+x4+x2x7x8, Bx=100x3x40x2x7x8, and Cx=x4−x2x2−1−x0x30x4x2x5+x2 are semiscalarly equivalent. In this case, Ax is a reduced, and Bx is a canonical matrix for Cx.


## 3.1. Uniqueness of the Matrix in Theorem 1

(1)
Suppose that, for the reducible matricesAx, Bxof forms (2) and (3), condition 1 of theorem holds, and, in addition, we have Ax≈Bx. Then, the left transformative matrix S in the equality SAxRx=Bx can be chosen in the form (9) (see Corollary 1 and Remark 1 [2]) and elements aix and bix, i=1,2,3, of these matrices satisfy the congruence(24)a3x+δBx−s13a3xδBx−a1xb2′x≡0modxk2.We havecodegδBxa3x=2q3<q1+q2. If 2q3≥k2, then from (24), a3x−b3x≡0modxk2 follows. Otherwise, in (24), we have s13=0 since 2q3-monomials in a3x and b3x are absent. In any case, Ax=Bx.(2)
Suppose that the reduced matricesAx, Bx of the forms (2) and (3) satisfy condition 2 of theorems and Ax≈Bx. Then, in the left transformative matrix S (4), in the transition from Ax to Bx, we have s23=0 (see Corollary 1 and Remark 1 [2]) and the elements aix and bix of the matrices Ax and Bx satisfy the congruences:(25)a1x−b1x1+s12a1x+s13a3x≡0modxk1,a2x−b2x+s12ΔBx+s13a2′xΔBx≡0modxk2,a3x+δBx+δBxs12a1x+s13a3x−a1xb2′x≡0modxk2.From (25), we can write(26)a3x−b3x+δBxs12a1x+s13a3x+b2′xb1x−a1x≡0modxk2.From (25), we have codega1x−b1x≥q1+q3. It is easy to see that(27)codegδBxa3x=2q3<q1+q3==codegδBxa1x<codegb2′xb1x−a1x.If2q3≥k2, then from (26), we have a3x−b3x≡0modxk2; hence, it follows a3x=b3x.Since2q3<codegΔA<codega2′xΔAx, then from (26), a2x−b2x≡0modxk2 follows, whence a2x=b2x. From (25), taking into account 2q3<q3+q1<codega1x2, we get a1x−b1x≡0modxk1 from where a1x=b1x. So, we have Ax=Bx.If2q3<k2, then from (25), we get s13=0. If q1+q3≥k2, then taking into account q1+q3<codegδA+k1 and q1+q3<2q1 from (25) and (26), we have ajx−bjx≡0modxkj, j=1,2, and a3x−b3x≡0modxk2. Therefore, Ax,Bx coincide. If q1+q3<k2, then from (26), we obtain s12=0. Hence, in this case, the matrices Ax,Bx also coincide.(3)
Suppose that the reduced matricesAx,Bx of the forms (2) and (3) satisfy condition 3 of theorems and Ax≈Bx. Then, for the elements of these matrices, we can write the congruences:(28)a1x−b1x+s23a3x−s13a3xb1x≡0modxk1,a2x−b2x−s23a2′xb2x+s13a2xδBx≡0modxk2,a3x+δBx−a3xs23b2′x+s13δBx−a1xb2′x≡0modxk2.Ifq3≥k1, then q1+q3>k1, and from (28), we get a1x=b1x. Then, (28) will take the form(29)a3x−b3x−a3xs23b2′x+s13δBx≡0modxk2.Obviously,codega3xδBx=2q3. If 2q3≥k2, then (29) implies a3x=b3x since codega3xb2′x>codega3xδBx.Then, from (28), we get a2x=b2x since(30)codegb2xa2′x>codegb2′xa3x,codegΔBxa2′x>codegδBxa3x.If2q3<k2, then (29) implies s13=0. If, moreover, codega3xb2′x<k2, then from (29), it yields s23=0 and all is proved. If codega3xb2′x≥k2, then all the same from (28) and (29), we have a3x=b3x and a2x=b2x, respectively.Ifq3<k1, then from (28), we get s23=0. If in addition q1+q3<k1, then from (28), it follows also s13=0 and all is proved. If q1+q3≥k1, then a1x=b1x, and again from (28), we go to (29). It follows from this that s13=0, if codega3xδBx<k2. And if codega3xδBx≥k2, then immediately from (28) and (29), we have a3x=b3x and a2x=b2x, respectively. Theorem is proved.Suppose that, in the reduced matricesAx,Bx of the forms (2) and (3), we have a1x,a2x,a3x,b1x,b2x,b3x≠0. Let us keep the notation given in theorem:(31)q1:=codega1,q2:=codega2′,q3:=codega3,a2′x=a2xxk1∈Cx,b2′x=b2xxk1∈Cx.We define polynomials:(32)a11x:≡a1x2modxk1,a22x:≡a2′x2modxk2−k1,a32x:≡a3xa2′xmodxk2,a04x:≡δAxmodxk2−k1.a14x:≡a1xδAxmodxk2,a34x:≡a3xδAxmodxk2.From the coefficients of each of the polynomialsa1x, a3x, and a11x, we form, respectively, columns a¯1, a¯03, and a¯11 of height k1−q1. In the first place, in these columns, we put q1-coefficients, and below in order of increasing degrees, we place the rest of their coefficients, up to degree k1−1 inclusive. We denote by a¯2, a¯22, and a¯04, the columns of height k2−k1−q2, constructed from the coefficients of polynomials a2x, a22x, and a04x, respectively. In the first place in each of these columns, we put q2-coefficients. Below we place the rest of their coefficients (including zero) up to the degree k2−k1−1. Similarly, from the coefficients of polynomials a3x, a32x, a34x, and a14x, we form columns a¯3, a¯32, a¯34, and a¯14 and height k2−q3. Here, we also put in the first place q3-coefficients, and then, in the order of increasing degrees, we place all other coefficients. In the last places, there will be k2−1-coefficients. For Ax, by the columns formed, we construct the matrices of the following form:(33)KA=a¯1a¯2a¯3K0A,K0A=K1AK2AK3A,(34)K1A=−a¯030¯a¯11,K2A=a¯220¯−a¯04,K3A=a¯32−a¯34−a¯14.In complete analogy forBx, we construct matrices of the following form:(35)KB=b¯1b¯2b¯3K0B,K0B=K1BK2BK3B,(36)K1B=−b¯030¯b¯11,K2B=b¯220¯−b¯04,K3B=b¯32−b¯34−b¯14.Obviously, in these matrices, each row consists of monomial coefficients of the same degrees.Theorem 2.
Let in the reduced matrixAx of the form (2), we have a1x,a2x,a3x≠0, q3>q1, q3>q2 and n1:=q1+q3, n2:=q2+codegδA+k1. Then, Ax≈Bx, where in the reduced matrix Bx of the form (3), all elements b1x,b2x,b3x are nonzero, polynomial b3x does not contain n1-monomial if n1<k1, and polynomial δBx does not contain n2−k1-monomial if n2<k2.In addition, one of the following conditions is true:(1)
Inb1x, n1-monomial is absent, if n1<k1 and n2<k2.(2)
Inb2x, codegδA+k1- and n2-monomials are absent, if n1≥k1 and n2<k2.(3)
Inb1x, q3- and n1-monomials are absent, if n1<k1 and n2≥k2.(4)
In the first column of the matrixKB (35), the coefficients of the polynomials b1x,b2x,b3x are zero elements that correspond to the maximum system of the first linearly independent rows of the submatrix K0B, if n1≥k1 and n2≥k2.The matrixBx is uniquely defined.Proof.
Existence. Letn1<k1.
We apply toAx transformation of the type II with the left transformation matrix of the form (9). At the same time, we put s12=d1/d0, where d0 is the younger coefficient and d1 is the n1-coefficient in a3x. The elements bix, i=1,2,3, of the thus obtained reduced matrix Bx satisfy the congruences (14)–(16). We write (16) in the form(37)a3x−b3x−s12a1xb3x−b2xr21x≡0modxk2,where r21x=a1x−b1x−s12a1xb1x/xk1∈Cx. Comparing the n1-coefficients in both parts of the last congruence, we have that b3x does not contain n1-monomial. We further assume that element a3x of the matrix Ax does not contain n1-monomial (if n1<k1).
Letn2<k2. Denote by c0 and c1, respectively, the junior and n2−k1-coefficients of the polynomial δAx. Apply to Ax transformations of the type I with the left transformation matrix of the form (9), while putting s23=c1/c0.
The elementsbix, i=1,2,3, of the thus obtained reduced matrix Bx satisfy the congruences (20)–(22) (with the one listed here s23). From (21) and (22), we obtain(38)δAx−δBx−s23b2′xδAx≡0modxk2−k1.
If we compare then2−k1-coefficients in both parts of the last congruence, we will conclude that δBx does not contain n2-monomial:(1)
Suppose that, in elementa3x of matrix Ax, there is no monomial of degree n1<k1, and in polynomial δAx, there is no monomial of degree n2<k2. Denote by d0 and c1, respectively, the lower coefficient in a3x and n1-coefficient in a1x. With the help transformation of the type III, we pass from Ax to the reduced matrix Bx. In the left transformation matrix (see (9)), we put s13=c1/d0. The elements of the resulting matrix Bx satisfy the congruences (11)–(13) (with the one specified here s13). From (11), we get that, in element b1x, n1-monomial is missing.We write (13) in the form(39)a3x−b3x−s13a3xb3x−b2xr21x≡0modxk2,wherer21x=a1x−b1x−s13a3xb1x/xk1∈Cx. Since 2q3>n1, then, as seen from the last congruence, in b3x, as in a3x, there is no n1-monomial. Also in δBx, as in δAx, there is no n2−k1-monomial. This is evident from the congruence(40)δAx−δBx+s13δAxδBx≡0modxk2−k1,which is recorded on the basis of (12) and (13) since codegδAxδBx>n2−k1. This proves the existence of matrix Bx with condition (1) specified in theorem.(2)
Suppose that conditionsn1≥k1, n2<k2, are satisfied in matrix Ax, and n2−k1-monomial is absent in polynomial δAx. We denote by c0 and d2, respectively, the lower coefficient in δAx and the codegδA+k1-coefficient in a2x. Let us do over matrix Ax transformation of the type II. To do this we put s12=−d2/c0 in the left transformation matrix (see (9)). We obtain a reduced matrix Bx whose elements satisfy the congruences of the form (14)–(16) (with s12 indicated here). Taking into account that the lower coefficients in δAx and δBx coincide, then from (15) we find that codegδA+k1-monomial is absent in b2x. From (15) and (16), we have δAx−δBx≡0modxk2−k1. It follows that, in δBx, as in δAx, there are no monomials of degree n2−k1.Next, we consider the absence ofcodegδA+k1-monomial in element a2x of the matrix Ax. Denote by c0 and d3, respectively, the lower coefficient in δAx and n2-coefficient in a2x. Above the matrix Ax, we carry out the transformation of the type III. Here, we put s13=d3/c0 in the left transformation matrix (see (9)). The elements of the obtained reduced matrix Bx satisfy the congruences of the form (11)–(13) (with s13 indicated here). It can be seen from (12) that n2-monomial is absent in b2x. Also codegδA+k1-coefficient in b2x will remain zero since codegδA<n2−k1. As can be seen from (40), in δBx, as in δAx, n2−k1-monomial is absent since codegδAxδBx>n2−k1.The existence of the required matrixBx with condition (2) is proved.(3)
Letn2≥k2 for Ax and in a3x be absent monomial of degree n1<k1. In the first step, we apply to the matrix Ax transformation of the type I with the left transformative matrix (see (9)), in which s23=−c2/d0, where d0 and c2 are, respectively, the lower coefficient in the a3x and the q3-coefficient in a1x. As a result, we obtain a reduced matrix Bxof the form (3) whose elements satisfy the conditions of the form (20)–(22) (with s23 selected here). From (20), it is seen that, in b1x, the q3-monomial is absent. From (20) and (22), it can be written as(41)a3x−b3x−b2xr21x≡0modxk2,wherer21x=a1x−b1x+s23a3x/xk1∈Cx. From the last congruence, it can be seen that n1-monomial is absent in b3x as in a3x.Let alreadya1x in Ax not contain q3-monomial. Denote by d0 and c3, respectively, the lower coefficient in a3x and the n1-coefficient in a1x and let s13=c3/d0. In the second step, with the help of the transformation of the type III with the specified s13 in the left transformative matrix (see (9)), we pass from Ax to some reduced matrix Bx of the form (3). For elements of the matrix Bx, conditions (11)–(13) (with the specified here s13) are satisfied. From (11), it follows that, in b1x, there is no n1-monomial. In addition, b1x does not contain q3-monomial. On the basis of (11) and (13), we can write the congruence of the form (41) in which r21x=a1x−b1x−s13a3xb1x/xk1∈Cx. It shows that, in b3x, in comparison with a3x, the zero coefficient of n1-monomial is preserved. This proves the existence for the matrix Ax a semiscalarly equivalent reduced matrix Bx with condition 3.(4)
Suppose that conditionsn1≥k1, n2≥k2, are satisfied in the reduced matrix Ax. If K0A=0 in KA (33), then the desired matrix is Ax and everything is already proven. Otherwise, in the first step, we fix in the matrix K0A the first nonzero row u¯1=d11d12d13 and the corresponding row d1d11d12d13 in KA. Let u¯1 consist of h1-coefficients and be the l1-rd row in K0A. We find an arbitrary solution x10x20x30t of the equation(42)d11d12d13x1x2x3t=d1.
We apply toAx a semiscalarly equivalent transformation with the left transformative matrix S of the form (4). At the same time, in S, we put s23=x10, s13=x20, and s12=x30. The elements bix, i=1,2,3, of the obtained reduced matrix Bx of the form (3) satisfy the congruence:(43)a1x−b1x+s23a3x−s12a1xb1x≡0modxk1,a2′x−b2′x−s23a2′xb2′x+s12δBx≡0modxk2−k1,a3x−b3x−s23a3xb2′x+s12a1x+s13a3xδBx≡0modxk2.
Depending on which of the matricesK1A, K2A, or K3A (see (34)) row u¯1 belongs, let us consider the congruence (43), respectively. By comparing the h1-coefficients in both parts of that congruence, we conclude that the l1-th element of the first column of matrix KB (35) is zero. In addition, all rows in KB, which precede the l1-th, coincide with the corresponding rows of the matrix KA.
IfrankK0A=1, then everything is already proven. Matrix Bx is the desired one. Otherwise, we assume that the l1-th element of the first column of the matrix KA is zero. In the second step, we fix in K0A the first linearly independent of u¯1 row u¯2=d21d22d23, as well as the corresponding to it row d2d21d22d23 in KA and the degree h2 of monomials, the coefficients of which form these rows. Also let u¯2 be the l2-th row in K0Al2>l1.
We find some solutiony10y20y30t of the equation(44)d11d12d13d21d22d23y1y2y3t=0d2t.
We apply toAx a semiscalarly equivalent transformation with the left transformative matrix S of the form (4), putting s23=y10, s13=y20, and s12=y30. We obtain a reduced matrix Bx of the form (3).
Again, as in the previous step, we consider one of the congruences (43) depending on which of the matrices K1A, K2A, or K3A (see (34)) contains row u¯2. In both parts of this congruence, we compare the coefficients of the h2-monomials and conclude that the l2-th element of the first column of the matrix KB (34) is equal to zero. Also from this and the previous congruences, we get that every row preceding the l2-th in KA coincides with the corresponding row in KB. If rankK0A=2, then everything is already proven. Then, matrix Bx is the desired one. Otherwise, in order to not introduce new designations, we assume that the first column of matrix KA has zero l1-th and l2-th elements. In matrix K0A, we fix the l3-th row, which is the first linearly independent of u¯1, u¯2 (l3>l2>l1). Let this be line u¯3=d31d32d33. To him, KA corresponds to d3d31d32d33. Also let h3 be the exponent that corresponds to these rows. We find the (unique) solution z10z20z30t of the equation(45)d11d12d13d21d22d23d31d32d33z1z2z3t=00d3t.
We apply toAx a semiscalarly equivalent transformation with the left transformation matrix S of the form (4) putting s23=z10, s13=z20, and s12=z30. We obtain the matrix Bx. The above considerations show that Bx is the desired matrix.


## 3.2. Uniqueness of the Matrix in Theorem 2

Suppose that, for the reduced matricesAx, Bx of forms (2) and (3), we have Ax≈Bx. Suppose also that elements a3x, b3x of these matrices do not contain n1-monomials if n1<k1, and in polynomials δAx, δBx, there are no n2−k1-monomials if n2<k2. Let us first show that the matrix S in the transition from Ax to Bx can be selected in the form(46)S=10s1301s23001,if n1<k1, or in the form(47)S=1s12s13010001,if n2<k2.Indeed, the elements of the matricesAx, Bx satisfy the congruence(48)a3x−b3x−s12a1x+s13a3xb3x−b2xr21x≡0modxk2.If we compare the coefficients of the monomers of degreen1<k1 in both parts of this congruence, we get s12=0. Also, from equivalence Ax≈Bx, it is easy to get congruence(49)δAx−δBx+s13δAxδBx−s23δAxb2′x≡0modxk2−k1.If we compare the coefficients of the monomers of degreen2<k2 in both parts of the last congruence, then we come to s23=0:(1)
In casen1<k1, n2<k2, the transition matrix S from Ax to Bx has the form (46) and (47) simultaneously. Therefore, we have s12=s23=0. Elements a1x, b1x in Ax, Bx satisfy (11). From here, we get s13=0. For this reason, matrices 1 and 2 coincide.(2)
Sincen2<k2, then matrix S of the transition from Ax to Bx has the form (47), and the elements a2x, b2x in Ax, Bx satisfy (25). In a2x, b2x, there are no codegδA+k1- and n2-monomials, so from (25), we get s12=s13=0. So, Ax=Bx.(3)
Ifn1<k1, then the matrix S of the transition from Ax to Bx has the form (46). Elements a1x, b1x in Ax, Bx satisfy the congruence (28). From it, we have s23=s13=0, since in a1x, as in b1x, there are no q3-and n1-monomials. Therefore, in this case, Ax, Bx coincide.(4)
Suppose that matrixBx satisfies condition 4, that is, in KB, the elements of the first column corresponding to the maximum system of the first linearly independent rows of the submatrix K0B are zero. Suppose that matrix Ax also has the same property, and in addition, condition Ax≈Bx holds. Then, the elements aix, bix, i=1,2,3, of these matrices satisfy the congruences (43). If in KA, we have K0A=0, then(50)minq3,q12≥k1,minq12,codegδA≥k2−k1,minq2+q3,codega1xδAx≥k2.Therefore, as can be seen from (43), aix=bix, i=1,2,3.If inKA, we have K0A≠0, and l1 is the number of the first nonzero row u¯1 in K0A, then the first l1 elements in the first column of the matrix KA coincide with the corresponding elements in the matrix KB; moreover, l1-th elements are zero. Therefore, in K0A, the first l1+1 rows coincide with the corresponding rows of the matrix K0B. In addition, from congruences (43), we have u¯1s23s13s12t=0. If the next after u¯1 row v¯ in K0A (or in K0B) is linearly dependent on u¯1, then(51)v¯s23s13s12t=0.Then, from (43), we obtain that the first l1+1 elements in the first column of the matrix KA coincide with the corresponding elements in KB. If u¯1 and v¯ are linearly independent, then (51) is still satisfied since in this case, the l1+1-th elements in the first columns of matrices KA and KB are zero. Then, the l1+2 th row in K0A coincides with the corresponding row of the matrix K0B. We think of this row in the same way as it was done above with row v¯. Let u¯2 be the first linearly independent of row u¯1 and l2 be its number in K0A. Then, this row coincides with the l2-th row in K0B, and the first l2 elements of the first column in KA coincide with the corresponding elements in KB, with l2th elements being zero. Then from (43), we have u¯2s23s13s12t=0. If w¯ is the l2+1-th row in K0A, then the corresponding l2+1-th row in K0B is also w¯. If w¯ is linearly dependent on the system u¯1, u¯2, then(52)w¯s23s13s12t=0,and the l2+1-th elements in the first columns of matrices KA,KB coincide. Otherwise, these elements also coincide because they are null. Continuing our considerations, we show that, in KA, KB, the first columns coincide, or at some steps, we will get s12=s13=s23=0. In each case, Ax=Bx. Theorem is proved.Example 1.
MatricesAx=100x3x40x6+x4+x2x7x8, Bx=100x3x40x2x7x8, and Cx=x4−x2x2−1−x0x30x4x2x5+x2 are semiscalarly equivalent. In this case, Ax is a reduced, and Bx is a canonical matrix for Cx.


## 4. Conclusion

The matricesBx, whose existence is established in Theorems 1 and 2, can be considered canonical in the class of semiscalarly equivalent matrices. The method of their construction follows from the proof of the first parts of these theorems. This completes the study of semiscalar equivalence of third-order polynomial matrices with one characteristic root, started in the previous works of the author.The results obtained in this article, as well as the results of the works cited here, are applicable to the study of the simultaneous similarity of sets of numerical matrices. In this context, the works of [6–9] should be noted. These results also have utility in solving Sylvester-type matrix equations over polynomial rings. Such equations often arise in applied problems.



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*Source: 7960987-2020-10-28.xml*