# Conditions of Semiscalar Equivalence of One Class 3×3 Matrices of Simple Structure

Authors: B. Z. Shavarovskii
Journal: Journal of Mathematics
 (2022)

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## Abstract

This paper solves the problem of classifying matrices over a ring of polynomials up to semiscalarly equivalence transformations. For the selected set of polynomial matrices of simple structure, the so-called oriented by characteristic roots reduced matrix is constructed. The latter, in addition to the triangular form and the presence of invariant factors on the main diagonal, has some predefined properties. Invariants and conditions of semiscalar equivalence are established for such matrices.


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## Body

## 1. Introduction

LetC be the field of complex numbers and Cx the ring of polynomials in an indeterminate x over C. Let Mn,C, Mn,Cx denote the algebras of n-by-n matrices over C, Cx, respectively, and GLn,C, GLn,Cx their corresponding groups of units (general linear groups). Two matrices Fx, Gx of Mn,Cx are said to be semiscalarly equivalent when there exists a pair P,Qx∈GLn,C×GLn,Cx such that (see [1, 2])(1)PFxQx=Gx.This is an equivalence relation; its equivalence classesPFxQx are called semiscalarly equivalence classes. Therefore, there is a problem of classification of polynomial matrices up to semiscalar equivalence. If a matrix Fx has a full rank, then in the class PFxQx, there is a lower triangular matrix in which its invariant factors stand on the main diagonal (see [1] Theorem 1 §1 Section 4). A similar result was also obtained in [3, 4]. The specified triangular matrix is defined by the class PFxQx ambiguously; therefore, it cannot be used at once for the establishment of belonging of any matrix to this class. Thus, there is a need to specify in class PFxQx such a matrix, which is determined with a lesser degree of ambiguity and for which we can find a system of invariants. To do this, it is necessary to clarify the mentioned triangular form of the matrix, that is, to establish such a matrix, which, in addition to the triangular form and the presence on the main diagonal of its invariant factors, has some predetermined properties. The problem of classifying of polynomial matrices of the same size up to semiscalar equivalence contains the matrix pair problem, which is the problem of classifying a pair of square matrices of the same size up to similarity transformations. Donovan and Freislich [5] call a matrix problem “wild” if it contains the matrix pair problem and “tame” otherwise. A certain characterization of time and wild problems is given by Drozd [6]. Thus, the problem of classifying polynomial matrices up to semiscalar equivalence is wild.We investigate the problem of determining when two matrices are semiscalarly equivalent. This study aims to construct a simpler form in the class of semiscalarly equivalent matrices and to find its invariants. In this paper, we limit ourselves to the case when a matrixFx∈M3,Cx has a simple structure, a complete rank, and a unit first invariant factor. The latter condition is not significant. More precisely, all the obtained results can be applied to a wider class of matrices. It should also be noted that this article is a continuation of the research initiated by the author [7]. Much useful information on the issues discussed in this article can be found in the monograph [8]. In particular, this monograph also investigates other types of equivalences of matrices and finite sets of matrices over different rings. Some generalizations of the concept of semiscalar equivalence in the case of matrices over quadratic rings and their application to the solution of matrix equations can be found in recent work [9].


## 2. Preliminary Information

Suppose that in the classPFxQx (see (1)), we have a matrix(2)Ax=100a1xϕ1x0a3xa2xϕ2x,where dega1x<degϕ1x, dega2x,dega3x<degϕ2x, ϕ1x divides a2x and ϕ2x, that is, 1, ϕ1x, and ϕ2x are invariant factors. Since Ax has a simple structure, polynomials ϕ1x, ϕ2x do not have multiple roots. We assume that degϕ1x,degϕ2x/ϕ1x>1. Otherwise, we have a1x≡0 or a2x≡0. Then, the task is slightly simplified and is the subject of another study. Denote by M1 and M2 the sets of roots of polynomials ϕ1x and ϕ12x=ϕ2x/ϕ1x, respectively. The notation ax|bx means ax divides bx, and ax,bx is the greatest common divisor of polynomials ax and bx. The corresponding notation will be used in the case of three polynomials. Let also be a2′x≔a2x/ϕ1x for element a2x of matrix Ax. In the future, we will denote the zero element of the main field and zero rows and columns as 0. We can assume that in the matrix Ax for some fixed root, α0∈M1 is satisfied the condition(3)a1α0=a3α0=0.This is easily achieved by adding some multiples with the numerical coefficients of the first row of a matrixAx to its other rows.Proposition 1.
The greatest common divisora1x,a3x,ϕ1x for the class PFxQx (see (1)) does not depend on the choice of the matrix Ax with condition (3) with a fixed characteristic root α0∈M1.Proof.
Suppose, in addition toAx, to the class PFxQx belongs some other matrix(4)Bx=100b1xϕ1x0b3xb2xϕ2x,where degb1x<degϕ1x, degb2x,degb3x<degϕ2x, b1α0=b3α0=0.
Then, equality(5)sij13100a1xϕ1x0a3xa2xϕ2x=100b1xϕ1x0b3xb2xϕ2xrijx13,is satisfied for some sij13∈GL3,C, rijx13∈GL3,Cx. If we put x=α0 in the latter equality, then we have s21=s31=0, so s11≠0. Also from (5), you can write(6)s22a1x+s23a3x=b1xr11x+ϕ1xr21x,(7)s32a1x+s33a3x=b3xr11x+b2xr21x+ϕ2xr31x.
Sinceϕ1x|r12x,r13x, then r11x,ϕ1x=1. Therefore, from (6) and (7), we have that a1x,a3x,ϕ1x|b1x,b3x. If we take into account the symmetry of semiscalar equivalence, then we have that b1x,b3x,ϕ1x|a1x,a3x. Therefore, a1x,a3x,ϕ1x=b1x,b3x,ϕ1x.
Ifa1x,a3x,ϕ1x=ϕ1x, then a1x≡0. This case is the subject of labor research [7]. In this article, we consider the situation when a1x,a3x,ϕ1x≠ϕ1x. Then, there is a root α1∈M1 such that a1α1a3α1≠0.


## 3. Reduction to a Special Triangular Form

Proposition 2.
If for matrixAx, (2) of class PFxQx condition (3) and a1x,a3x,ϕ1x≠ϕ1x are satisfied, then in this class, there is such a matrix Bx (4) that b1α0=b3α0=0 and b1α1≠0, b3α1=0, for some α1∈M1.Proof.
If inAx, for a fixed α1∈M1, we have a1α1≠0, then the desired matrix is Bx achieved by adding some multiple (with a numerical coefficient) of the second row of the matrix Ax to its third row. If a1α1=0, a3α1≠0, then first we choose s23≠0 so that 1+s23a2′β≠0 for every β∈M2. Then, from the congruences(8)a1x+s23a3x−b1x≡0modϕ1x,a2′x−b2′x1+s23a2′x≡0modϕ12x,a3x−b2′xa1x+s23a3x−b1x−b3x≡0modϕ2x,consistently find b1x, b2′x, b3x of degrees degb1x<degϕ1x, degb2′x<degϕ12x, degb3x<degϕ2x. Next, from elements s23, b1x, b2x=b2′xϕ1x, and b3x, we construct a matrix Bx of the form (4) and matrices(9)sij13=10001s23001,rijx13,where(10)r11x=1,r12x=r13x=0,r21x=a1x+s23a3x−b1xϕ1x,r22x=1+s23a2′x,r23x=s23ϕ12x,r31x=a3x+b2′xa1x+s23a3x−b1x−b3xϕ2x,r32x=a2′x−b2′x1+s23a2′xϕ12x,r33x=1−s23b2′x.
It is easy to see thatr21x,r31x,r32x∈Cx and rijx13∈GL3,Cx. By direct verification, we establish that the matrices Bx, sij13, and rijx13 together with the original matrix Ax satisfy equality (5). This means that Bx belongs to the class PFxQx and also b1α1≠0, b1α0=b3α0=0. By adding to the third row of the matrix Bx its second row multiplied by −b3α1/b1α1, we obtain the matrix we need with zero value of the (3, 1)-element at x=α1.
We can further assume thatb1α1=1 for the matrix Bx referred to in Proposition 2.Proposition 3.
If the elements of matricesAx (2), Bx (4) of class PFxQx satisfy the conditions:(11)a1α0=b1α0=a3α0=b3α0=a3α1=b3α1=0,a1α1=b1α1=1,for some α0,α1∈M1, then a3x,ϕ1x=b3x,ϕ1x, and the matrix sij13 in relation (5) has an upper triangular form.Proof.
The proof of Proposition1 shows that in the matrix sij13 from equality (5), there is s21=s31=0. If we put x=α1 in (7), we have s32=0. Then, from (7), it is seen that b3x,ϕ1x|a3x since s33≠0. Given the symmetry of the relation of semiscalar equivalence, we can obtain that a3x,ϕ1x|b3x. Therefore, a3x,ϕ1x=b3x,ϕ1x.Proposition 4.
If for elements of matricesAx (2), Bx (4) of class PFxQx are true conditions (8) for some roots α0,α1∈M1, and a3x,ϕ1x=ϕ1x, then the following conditions are also satisfied:(i)
a1x,ϕ1x=b1x,ϕ1x, a1x−1,ϕ1x=b1x−1,ϕ1x;(ii)
Equivalencea1αi=a1αj⇔b1αi=b1αj is performed for arbitrary αi,αj∈M1;(iii)
IfIijA≔a1αj−1a1αi/a1αi−1a1αj, IijB≔b1αj−1b1αi/b1αi−1b1αj then IijA=IijB for such αi,αj∈M1 that a1αi,a1αj≠0, a1αi,a1αj≠1.Proof.
If we take into account thata3x,ϕ1x=ϕ1x, then from (6), we can write(12)s22a1x−b1xs11+s12a1x≡0modϕ1x,where do we get the first part of the condition (i) Putting in (12) x=α1,x=αi,x=αj, we have, respectively,(13)s22−s11−s12=0,(14)s22a1αi−b1αis11+s12a1αi=0,(15)s22a1αj−b1αjs11+s12a1αj=0.
Sinces11,s22≠0, then from (13), (14), we obtain that a1αi=1⇔b1αi=1 for an arbitrary αi∈M1. Also from (14), (15) follows such equivalence a1αi=a1αj⇔b1αi=b1αj. This fully proves conditions (i) and (ii).
EqualityIijA=IijB based on the proved (i), (ii) is obvious for those αi,αj∈M1 that a1αi=a1αj. In the case of a1αi≠a1αj, this equality follows from (13)–(15), if we take into account s11,s22≠0.Corollary 1.
Suppose that the conditions of Proposition4 for matrices Ax (2), Bx (4) are satisfied, and let(16)JijA≔a1αj−1a1αia1αj−a1αi,JijB≔b1αj−1b1αib1αj−b1αi.
Then,JijA=JijB for such αi,αj∈M1 that a1αi,a1αj≠0, a1αi,a1αj≠1, a1αi≠a1αj.Proof.
It is easy to see thatJijA=IijA/IijA−1, JijB=IijB/IijB−1. Under Proposition 4, we have IijA≠1. Therefore, JijA=JijB.Proposition 5.
Suppose that for matrixAx (2), the conditions of Proposition 4 hold. Let also α2 and αj be a fixed and an arbitrary root of the set M1, respectively, for which both values a1α2,a1αj are different from 0 and 1, and a1α2≠a1αj. Then, in the class PFxQx, there is a matrix Bx (4) with the conditions of Proposition 4 and an arbitrarily fixed value b1α2≠0 other than 1 and J2jA (16).Proof.
Let us choose for an unknown polynomialb1x at x=α2 some value b1α2≠0 other than 1 and J2jA. Consider the equation(17)b1αj−1b1α2b1α2−1b1αj=I2jA,with the unknown b1αj for each element αj specified in Proposition, where I2jA is defined in (9). Let us write the latter in this form(18)b1αjb1α21−I2jA+I2jA=b1α2.
Sinceb1α2≠I2jA/I2jA−1=J2jA (see the proof of Corollary 1), it is solvable for each element αj stated in Proposition. Let us now construct a polynomial b1x of degree <degϕ1x by its values b1αi for each root αi∈M1 as follows:(19)b1αi=0,ifa1αi=0,1,ifa1αi=1,b1α2,ifa1αi=a1α2,b1αj,ifa1αi=a1αj.
Consider the equation(20)1−11a1α2−1b1α2−11a1αj−1b1αj−11x1x2x3=0,relative to x1x2x3T. If we take into account the equality I2jA=I2jB for every αj specified in Proposition, and the inequality a1α2≠b1α2, then the rank of the matrix of this equation is 2, and it has a solution 1s22s12T with nonzero all its components. Based on this solution, we write the congruence(21)s22a1x−b1x1+s12a1x≡0modϕ1x,and consider the equation(22)b3xb2xrijx12≡a3xa2xmodϕ2x,with unknown b3xb2x, where(23)rijx12=1+s12a1xs12ϕ1xs22a1x−b1x1+s12a1xϕ1xs22−s12b1x∈M2,Cx.
From (22), we can find polynomials b2xb3x, of degrees <degϕ2x, since detrijx12=s22≠0. From the found b1x, b2xb3x, and 1s22s12T, we construct a matrix B (x) of the form (4) and a matrix(24)sij13=1s1200s220001.
We also construct a matrixrijx13, in which the upper left 2×2-block coincides with the above matrix rijx12 and, in addition, r13x=r23x≡0, r33x=1,(25)r31x=a3x−b3x1+s12a1x−b2xr21xϕ2x∈Cx,r32x=a2x−b3xr12x−b2xr22xϕ2x∈Cx.
It is easy to see that the matrixAx together with the constructed Bx, non-singular sij13 and invertible rijx13 satisfy equality (5). In addition, the element b1x of the matrix Bx has a preselected value b1α2. Therefore, Bx is the desired matrix.Proposition 6.
Suppose that for matricesAx (2), Bx (4) with the conditions of Proposition 4, we have 0≠a1α2=b1α2≠1 for some α2∈M1 and(26)a2′x≔a2xϕ1x,b2′x≔b2xϕ1x,δAx≔deta1x1a3xa2′x,δBx≔detb1x1b3xb2′x.
Then, in these matricesa1x=b1x, a3x,ϕ2x=b3x,ϕ2x, a2′x,ϕ12x=b2′x,ϕ12x, δAx,ϕ12x=δBx,ϕ12x and in the upper triangular matrix sij13 from relation (5), we have s12=0, s11=s22.Proof.
For the elementsa1x,b1x of the matrices Ax,Bx, the congruence (12) is performed. Putting in it x=α1 and x=α2, we obtain, respectively, equality (13) and(27)s22a1α2−b1α2s11+s12a1α2=0.
From them, it immediately follows thats12=0, s11=s22, because the values a1α2=b1α2 are different from 0 and 1. Then, if we return to (12), we obtain a1x=b1x. If we take into account what has already been proved, then from (5), we can write the congruences as follows:(28)a3xs33+s13δBx−s23b2′x≡s11b3xmodϕ2x,(29)a2′xs33+s13δBx−s23b2′x≡s22b2′xmodϕ12x,from which it follows(30)δAxs33+s13xδBx−s23b2′x≡s11δBxmodϕ12x.
From (28), (29), and (30), it is easy to see that a3x,ϕ2x=b3x,ϕ2x, a2′x,ϕ12x=b2′x,ϕ12x and δAx,ϕ12x=δBx,ϕ12x, respectively.Corollary 2.
For matricesAx (2), Bx (4) of class PFxQx with the conditions of Proposition 6, we have the following equivalences:(31)a2x≡0⇔b2x≡0,a3x≡0⇔b3x≡0,δAαi≡0modϕ12x⇔δBαi≡0modϕ12x.Proposition 7.
Suppose that for the elements of matricesAx (2), Bx (4), the conditions of Proposition 4 are satisfied, and let a2′x,b2′x, δAx,δBx are defined in Proposition 6. Then equivalences(32)a2′βi=δAβi⇔b2′βi=δBβi,a2′βiδAβi−1=a2′βjδAβj−1⇔b2′βiδBβi−1=b2′βjδBβj−1,are fulfilled for such βi,βj∈M2 that δAβi,δAβj≠0.Proof.
MatricesAx, Bx satisfy equation (5), where the matrix sij13 according to Proposition 3 has an upper triangular form. From (5), except (12), we obtain the congruences as follows:(33)a3xs33+s13δBx−s23b2′x+s11δBx≡a1xs22b2′x−s12δBxmodϕ2x.(34)a2′xs33+s13δBx−s23b2′x≡s22b2′x−s12δBxmodϕ12x.
If we exclude from (33) and (34) the members containing s13, s23, s33, we get(35)s11a2′xδBx−s22b2′xδAx+s12δAxδBx≡0modϕ12x.
From (12) for x=α1, we have (13), and from (35) for x=βi and x=βj, respectively, we obtain(36)s11a2′βiδAβi−1−s22b2′βiδBβi−1+s12=0,(37)s11a2′βjδAβj−1−s22b2′βjδBβj−1+s12=0.
From (13) and (36) follows the first of the required equivalences, and from (36) and (37)-the second.Proposition 8.
Suppose that in the matrixAx (2) with the conditions of Proposition 4, element a1x of the set M1 acquires only the values 0 and 1. Let also be a2′β0≠δAβ0≠0 for some β0∈M2, where a2′x, δAx are defined in Proposition 6. Then, in the class PFxQx, there is a matrix Bx of the form (4) with the conditions of Proposition 4, in which b1x=a1x and b2′β0=−δBβ0≠0.Proof.
Denotec≔a2′β0δAβ0−1, s12≔−0,5c+1, s22≔−0,5c−1 and construct matrices(38)sij13=1s1200s220001,rijx12=1+s12a1xs12ϕ1xs22a1x−a1x1+s12a1xϕ1xs22−s12a1x∈M2,Cx.
Obviously,s22=1+s12≠0 and rijx12∈GL2,Cx. We also construct a matrix B (x) of the form (4), in which b1x=a1x and elements b2x,b3x of degrees <degϕ2x are defined from the congruence(39)b3xb2xrijx12≡a3xa2xmodϕ2x.
Based on the elements of matrixrijx12 and elements r13x=r23x≡0,(40)r31x=a2x−b3xr12x−b2xr22xϕ2x∈Cx,r32x=a3x−b3xr11x−b2xr21xϕ2x∈Cx,r33x=1,we construct matrix rijx13. Obviously, rijx13∈GL3,Cx. We make sure that the constructed matrices sij13, Bx, and rijx13 together with Ax satisfy equality (5). Therefore, Bx, like Ax, belongs to class PFxQx. The elements of the matrices Ax, Bx satisfy the congruence (35), where s11=1, and s12, s22 are defined above. If we put x=β0 in this congruence, we get b2′β0=−δBβ0.Proposition 9.
Suppose that for matricesAx (2), Bx (4) of the class, PFxQx conditions of Proposition 8 are satisfied, and we have a2′β0=−δAβ0, b2′β0=−δBβ0 for some β0∈M2. Then, the conclusion of Proposition 6 is true.Proof.
Congruences (13) and (35) are performed for the elements of matrices Ax, Bx. If in these congruences to put accordingly x=α1 and x=β0, then from the received, we will receive that s12=0 and s11=s22. The rest of the proof is the same as in the proof of Proposition 6.Corollary 3.
For matricesAx (2), Bx (4) with the conditions of Proposition 9, we have a3x≡0⇔b3x≡0 and ϕ12x|δAx⇔ϕ12x|δBx.


## 4. Main Results

Definition 1.
MatrixBx of Proposition 4, in which b1αi=0 or b1αi=1 for each αi∈M1 and δBβj=0 or δBβj=b2′βj≠0 for each βj∈M2, is called oriented by the characteristic roots α0,α1∈M1 reduced matrix. Matrix Bx of Proposition 5, for which(41)bα2=−1,ifJ2j≠−1for eachαj∈M1,maxJ2j+1,in opposite case,is called oriented by the characteristic roots α0,α1,α2∈M1 reduced matrix. Matrix Bx of Proposition 8 is called oriented by the characteristic roots α0,α1∈M1, β0∈M2 reduced matrix.Theorem 1.
Suppose that in the oriented by the same characteristic rootsα0,α1,α2∈M1 or α0,α1∈M1, β0∈M2 reduced matrices Ax (2), Bx (4), we have a2x,a3x≠0, and(42)cβi≔a2′βi,ifa2′βi≠0,δAβi,ifa2′βi=0,δAβi≠0,0,ifa2′βi=δAβi=0,dβi≔b2′βi,ifb2′βi≠0,δBβi,ifb2′βi=0,δBβi≠0,0,ifb2′βi=δBβi=0,where βi∈M2 (notation a2′x,b2′x,δAx,δBx see Proposition 6). Matrices Ax, Bx are semiscalarly equivalent if and only if their elements a1x,b1x coincide, and the rest satisfy the following conditions:(i)
a2′x,ϕ12x=b2′x,ϕ12x, a3x,ϕ12x=b3x,ϕ12x, δAx,ϕ12x=δBx,ϕ12x;(ii)
for arbitrary such rootsβi∈M2 that a2′βi≠0, we have(43)δAβia2′βi−1=δBβib2′βi−1,a3βia2′βi−1=b3βib2′βi−1.(iii)
for arbitrary rootsβi,βj∈M2, we have(44)a2′βi−a2′βjδAβi−δAβj=0⇔b2′βi−b2′βjδBβi−δBβj=0.(iv)
for arbitrary rootsβi,βj,βl∈M2 such that cβi,cβj,cβl≠0 and rows a2′βiδAβi, a2′βjδAβja2′βlδAβl are pairwise different, there is a number k≠0 such that (45)kcβicβjcβldeta2′βiδAβi1a2′βjδAβj1a2′βlδAβl1=1dβidβjdβldetb2′βiδBβi1b2′βjδBβj1b2′βlδBβl1.(46)kcβicβjdetcβi1cβj1=kdβidβjdetdβi1dβj1,ifdeta2′βiδAβia2′βjδAβj=0.Proof.
(i)
Necessity(ii)
For semiscalarly equivalent matricesAx, Bx the equality a1x=b1x, condition (i) and condition (ii) for zero values a3βi, δAβi follow from Propositions 6 or 9. For the elements of matrices Ax, Bx, the congruences (33) and (34) are satisfied, from which (35) follows. If we consider that s12=0 and s11=s22 (see Propositions 6 or 9), then from (35), we have(47)a2′xδBx−b2′xδAx≡0modϕ12x.From the latter equivalence fora2′βi≠0 (b2′βi≠0) and nonzero values δAβi, δBβj, it follows the first equality from (43). The second of them follows from the first, since δAx=a1xa2′x−a3x, δBx=b1xb2′x−b3x, a1x=b1x.(iii)
Equivalence (43) for zero values a2′βi,a2′βj and δAβi,δAβj follows from (i). If a2′βi=a2′βj=0 and a3βi=δAβi=δAβj=a3βj≠0, then from (33) for x=βi,x=βj, we have(48)s33b3βi−1−s11a3βi−1−s13=0,s33b3βj−1−s11a3βj−1−s13=0,whenceb3βi=b3βj. Since in this case δBβi=−b3βi, δBβj=−b3βj, so δBβi=δBβj. This proves the equivalence (44) for zero values a2′βi,a2′βj and nonzero values δAβi,δAβj. Consider now the case where a2′βi=a2′βj≠0 and δAβi=δAβj. If we take into account (43), then from (34), where s12=0, we have(49)s33b2′βi−1−s22a2′βi−1+s13δAβia2′βi−1−s23=0,s33b2′βj−1−s22a2′βj−1+s13δAβja2′βj−1−s23=0,whenceb2′βi=b2′βj. Then, it is easy to trace the chain of implications(50)a2′βiδAβi=a2′βjδAβj⇒⇒a2′βi1δAβia2′βi−1=a2′βj1δAβja2′βj−1⇒⇒1δBβib2′βi−1=1δBβjb2′βj−1⇒b2′βiδBβi=b2′βjδBβj.(iv)
Suppose we have three rootsβi,βj,βl∈M2, for which cβi,cβj,cβl≠0 (respectively, dβi,dβj,dβl≠0). For elements of matrices Ax, Bx the congruences (28), (29) hold. From (28) and (29) follows the congruence (30). From (29) or (30), depending on whether cβn=a2′βn or cβn=δAβn, for x=βn, n=i,j,l, we obtain(51)s33cβi−kdβi+s13cβiδBβi−s23cβib2′βi=0,s33cβj−kdβj+s13cβjδBβj−s23cβjb2′βj=0,s33cβl−kdβl+s13cβlδBβl−s23cβlb2′βl=0,where k=s11s33−1≠0. Since s33≠0, then(52)detcβi−kdβicβiδBβicβib2′βicβj−kdβjcβjδBβjcβjb2′βjcβl−kdβlcβlδBβlcβlb2′βl=0
If we take into account thatcβnδBβn=dβnδAβn and cβnb2′βn=dβna2′βn (see (43)), then from (57), we have equality(53)detcβicβiδBβicβib2′βicβjcβjδBβjcβjb2′βjcβlcβlδBβlcβlb2′βl=kdetdβidβiδAβidβia2′βidβjdβjδAβjdβja′βjdβldβlδAβldβla′βl,from which it follows equality (45). From condition(54)deta2′βiδAβia2′βjδAβj=0,follows either cβi=a2′βi≠0, cβj=a2′βj≠0, or a2′βi=a2′βj=0, cβi=δAβi≠0, cβj=δAβj≠0. Then(55)detb2′βiδBβib2′βjδBβj=0,and cβidβi=cβib2′βi, cβjdβj=cβjb2′βj, or cβidβi=cβiδBβi, cβjdβj=cβjδBβj. Since s33≠0, then in each case, from system (51), we have(56)detcβi−kdβicβidβicβj−kdβjcβjdβj=0,where it follows (43).
Sufficiency. Letβ1,…,βm are all those elements from the set M2 for which cβ1,…,cβm≠0, and all rows a2′β1δAβ1,…,a2′βmδAβm are pairwise different. Under the conditions of theorem (see (iii)), the rows b2′β1δBβ1,…,b2′βmδBβm are also pairwise different, and each has a nonzero first element. Consider the equation(57)cβ1−kdβ1cβ1δBβ1−cβ1b2′β1⋮⋮⋮cβm−kdβmcβmδBβm−cβmb2′βmx1x2x3=0,with the unknown x1x2x3T. If we take into account that cβiδBβi=dβiδAβi and cβib2′βi=dβia2′βi, 1,…,m, (see condition (ii)), then from (45), it follows that in the matrix of equation (57), every three rows are linearly dependent. Hence, this equation has a nonzero solution. If for some βi,βj∈M2, 1≤i,j≤m, we have(58)deta2′βiδAβia2′βjδAβj=0,then from conditions (i), (ii) can be obtained(59)detb2′βiδBβib2′βjδBβj=0.
Then, it follows from (46) that in the matrix of equation (57), the rows for the same βi,βj are linearly dependent. Therefore, this equation has a solution x10x20x30T with a nonzero first component x10. Based on this solution, taking into account conditions (i)–(iii), we can write the congruences(60)a2′xs33+s13δBx−s23b2′x−s22b2′x≡0modx−β1…x−βm,(61)δAxs33+s13δBx−s23b2′x−s11δBx≡0modx−β1…x−βm.where s33=x10, s11=s22=kx10, s13=x20, s23=x30. It should be borne in mind that s33+s13δBβi−s23b2′βi≠0 for each βi=β1,…,βm. It is clear that from (60) and (61), it is possible to pass to (29), (30). If we take into account that ϕ1x,ϕ12x=1 and a1x=b1x, then by adding to the congruence (30) multiplied by −a1x the congruence (29), we proceed to the congruence (28) with the same above-mentioned coefficients s33,s11,s13,s23. Based on the elements of matrices Ax, Bx and elements s33,s11,s22=s11,s13,s23, we construct matrices(62)sij13=s110s130s22s2300s33,rijx13,where(63)r11x=s11+s13a3x,r12x=s13a2x,r13x=s13ϕ2x,r21x=a3xs23−s13b1xϕ1x∈Cx,r22x=s22+a2′xs23−s13b1x,r23x=ϕ12xs23−s13b1x,r33x=s33+s13δBx−s23b2′x,r31x=a3xs33+s13δBx−s23b2′x−s11b3xϕ2x∈Cx,r32x=a2′xs33+s13δBx−s23b2′x−s22b2′xϕ12x∈Cx.
It is easy to verify the reversibility of matrixrijx13 and the truth of equality (5) for constructed sij13, rijx13 and the original matrices Ax, Bx. The theorem is proved.Theorem 2.
Suppose that in the oriented by the same characteristic rootsα0,α1∈M1 the reduced matrices Ax (2), Bx (4) (see Definition 1) we have\(64)M21≔βp:βp∈M2,δAβp≠0,M22≔βq:βq∈M2,δAβq=0,where δAx is defined in Proposition 6. Matrices Ax, Bx are semiscalarly equivalent if and only if their elements a1x,b1x coincide, and the rest satisfy the following conditions:(i)
δAx,ϕ12x=δBx,ϕ12x, a2′x,ϕ12x=b2′x,ϕ12x, a3x,ϕ12x=b3x,ϕ12x;(ii)
for arbitrary rootsβi,βj∈M21 or βi,βj∈M22, we have a2′βi=a2′βj⇔b2′βi=b2′βj;(iii)
for arbitrary rootsβi,βj,βl∈M21 or βi,βj,βl∈M22 such that a2′βi,a2′βj,a2′βl are nonzero and pairwise different, we have(65)a2′βla2′βj−a2′βia2′βja2′βl−a2′βi=b2′βlb2′βj−b2′βib2′βjb2′βl−b2′βi.Proof.
Necessity. According to Proposition4, the elements a1x,b1x of matrices Ax, Bx on the set M1 acquire the same values (0 and 1). Since dega1x,degb1x<M1, then a1x=b1x.(i)
MatricesAx, Bx satisfy equality (5), in which the matrix sij13 according to Proposition 3 has an upper triangular form. For the elements of these matrices from (5) follow the congruences (33), (34). If we exclude s12 from them, we can obtain a congruence (30). From the latter, given the symmetry of the relation of semiscalar equivalence, it follows δAx,ϕ12x=δBx,ϕ12x. Therefore, δBβi≠0 for every βi∈M21. This means that b2′βi≠0 for the same βi∈M21. Also from (34), for x=βi∈M22, we have(66)a2′βjs33−s23b2′βj−s22b2′βj=0.whencea2′βj=0⇔b2′βj=0. Therefore, a2′x,ϕ12x=b2′x,ϕ12x. If we put βi∈M21 in (33) and take into account that b2′βi=δBβi=b1βib2′βi−b3βi, s22=s11+s12, we can write(67)a3βis33+s13−s23b2′βi−s11b3βi=0.If you putx=βj∈M22 and take into account b1βib2′βi=b3βi, then from (33), we have(68)a3βis33−s23b3βi−s22b3βi=0.From the obtained equalities (67), (68), it follows that a3βi=0⇔b3βi=0, a3βj=0⇔b3βj=0, whence a3x,ϕ12x=b3x,ϕ12x.(ii)
The required equivalence for zero valuesa2′βi,a2′βj follows from (i), because then the values b2′βi,b2′βj are also zero. Let now a2′βi,a2′βj≠0. From (34) for βi,βj∈M21, we have(69)a2′βis33+s13−s23b2′βi−s11b2′βi=0,a2′βjs33+s13−s23b2′βj−s11b2′βj=0.and forβi,βj∈M22, we have(70)a2′βis33−s23b2′βi−s22b2′βi=0.and (66). From each of the obtained pairs of equalities, it follows that b2′βi=b2′βj.(iii)
Letβi,βj,βl∈M21 or βi,βj,βl∈M22, and all values a2′βi,a2′βj,a2′βl are nonzero and pairwise different. Under conditions (i), (ii), b2′βi,b2′βj,b2′βl are also nonzero and pairwise different. If we put x=βi,x=βj,x=βl in (66), then in each of the cases βi,βj,βl∈M21 or βi,βj,βl∈M22, we get systems(71)s11a2′βi−1−s33b2′βi−1−s13+s23=0,s11a2′βj−1−s33b2′βj−1−s13+s23=0,s11a2′βl−1−s33b2′βl−1−s13+s23=0,or(72)s22a2′βi−1−s33b2′βi−1+s23=0,s22a2′βj−1−s33b2′βj−1+s23=0,s22a2′βl−1−s33b2′βl−1+s23=0,respectively. Since s11,s22,s33≠0, then from each of the obtained systems, we have(73)deta2′βj−1−a2′βi−1b2′βj−1−b2′βi−1a2′βl−1−a2′βi−1b2′βl−1−b2′βi−1=0,whence it follows (65).
Sufficiency. Let in matricesAx,Bx the elements a1x,b1x coincide, and for the rest conditions (i)–(iii) are fulfilled. If a2x≡0, then ϕ12x|δAx (see definition) and a3x≡0. Then, from (i), it follows that b2x=b3x≡0. In this case, Ax=Bx and everything is proved. Now let a2x≠0 and β1,…,βm is the maximum subset of the set M2 for which all values a2′β1,…,a2′βm are nonzero and a2′βi≠a2′βj if βi,βj∈M21 or βi,βj∈M22, 1≤i,j≤m. Obviously, m≥1. Under conditions (i), (ii), we have b2′β1,…,b2′βm≠0 and b2′βi≠b2′βj. Let also β1,…,βr⊂M21, βr+1,…,βm⊂M22, 0≤r≤m. Consider the equation(74)a2′β1−10−b2′β1−1−11⋮⋮⋮⋮⋮a2′βr−10−b2′βr−1−110a2′βr+1−1−b2′βr+1−101⋮⋮⋮⋮⋮0a2′βm−1−b2′βm−101x1x2x3x4x5=0,relative to the unknown x1x2x3x4x5T. The rank of the matrix of this equation does not exceed 4, because, as follows from (iii), the rank of each of the selected submatrices is not more than 2. This means that the equation has a nonzero solution. Moreover, it is easy to see that there is a solution x10x20x30x40x50T, where x10,x20,x30≠0. Based on this solution, the congruences can be written:(75)a2′xs33+s13−s23b2′x−s11b2′x≡0modx−β1…x−βr,(76)a2′xs33−s23b2′x−s22b2′x≡0modx−βr+1…x−βm,where s11=x10, s22=x20, s33=x30, s13=x40, s23=x50. Since a2′βi=δAβi, b2′βi=δBβi, for every βi∈β1,…,βr (see definition and condition (i)), (75) can be written as(77)a2′xs33+s13δBx−s23b2′x+s12δBx−s22b2′x≡0modx−β1…x−βr,where s12=s22−s11. From the congruences (76) and (77), we can write(78)a2′xs33+s13δBx−s23b2′x+s12δBx−s22b2′x≡0modx−β1…x−βm,since all β1,…,βm are different and(79)δAx≡δBx≡0modx−βr+1…x−βm.
Forβi,βj∈M21 or βi,βj∈M22 under conditions (i), (ii), the following implications are true:(80)δAβi=δAβj⇐a2′βi=a2′βj⇒b2′βi=b2′βj⇒δBβi=δBβj.
Therefore, the congruence (78) is performed modulo ϕ12x, that is, the congruence (34) is true. Since a2′βi=δAβi, b2′βi=δBβi for every βi∈M21 and δAβj=δBβj=0 for every βj∈M22, we can move from (75) to congruence(81)δAxs33+s13xδBx−s23b2′x≡s11δBxmodx−β1…x−βm,and from it to the congruence (30), if we take into account (80). If we add to the congruence (30), the congruence (34) multiplied by −a1x=−b1x and take into account that a3x,ϕ1x=b3x,ϕ1x=ϕ1x, ϕ1x,ϕ12x=1, then we can obtain the congruence (33). According to the above sij, i,j=1,2,3, i≤j, and elements of matrices Ax,Bx, we construct matrices(82)sij13=s11s12s130s22s2300s33,rijx13,where(83)r11x=s11+s13a3x,r12x=s12ϕ1x+s13a2x,r13x=s13ϕ2x,r21x=s22a1x+s23a3x−b1xs11+s12a1x+s13a3xϕ1x,r22x=s22+s23a2′x−b1xs12+s13a2′x,r23x=ϕ12xs23−s13b1x,r33x=s33−s13b3x−b2′xs23−s13b1x,r31x=a3xs33+s13δBx−s23b2′x+s11δBx+a1xs12δBx−s22b2′xϕ2x,r32x=a2′xs33+s13δBx−s23b2′x+s12δBx−s22b2′xϕ12x.
The first of these matrices is not singular. In the second matrixr21x∈Cx, since ϕ1x|a3x, s12=s22−s11 and the equal polynomials a1x,b1x on the set M1 acquire only two values (0 and 1). We also have r31x,r32x∈Cx, considering (33) and (34). We can see that rijx13∈GL3,Cx, and moreover, both of these matrices sij13, rijx13 together with Ax, Bx satisfy equality (5). This means that Ax, Bx are semiscalarly equivalent. This completes the proof of the theorem.


## 5. Conclusion

In the class of semiscalarly equivalent polynomial matrices of simple structure, the so-called oriented by characteristic roots reduced matrix is established. It has a lower triangular form, invariant factors on the main diagonal, and some other properties. The left transformation matrix in the transition from one oriented by characteristic roots reduced matrix to another semiscalarly equivalent oriented (by the same characteristic roots) reduced matrix has an upper triangular form (Propositions3, 6, 9). Theorems 1 and 2 give the necessary and sufficient conditions of the semiscalarly equivalence of oriented by the same characteristic roots reduced matrices. Proof of all statements is constructive. Based on them, we can specify the method of constructing the left and right transformation matrices in the transition from an arbitrary triangular matrix with invariant factors on the main diagonal to the oriented by characteristic roots reduced matrix (Propositions 2, 5, 8). This makes it possible to apply the obtained results to the solution of linear two-sided matrix equations of the type XAx−BxYx=0 with unknown nonsingular matrix X over C and invertible Yx over Cx. Matrix equations of this type are often found in mathematical physics, other sections of mathematics, and applied problems.



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