# Oriented by Characteristic Roots Reduced Matrices in the Class of Semiscalarly Equivalent

Authors: B. Z. Shavarovskii
Journal: Journal of Mathematics
 (2021)

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## Abstract

A set of polynomial3×3-matrices of simple structure has been singled out, for which a so-called oriented by certain characteristic roots reduced matrix is established in the class of semiscalarly equivalent. The invariants of such reduced matrices and the conditions of their semiscalar equivalence are indicated. The obtained results will also be applied to the problem of similarity of sets of numerical matrices.


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## Body

## 1. Introduction

By Theorem1 (see Section 4 in [1] and [2]), matrix Fx∈Mn,Cx of full rank by transformation(1)Fx⟶PFxQx=Gx,where P∈GLn,C and Qx∈GLn,Cx are reduced to the lower triangular form with invariant factors on the main diagonal. We consider the case when Fx is a 3×3-matrix of simple structure and its first invariant factor is equal to one.The latter condition does not limit the generality in the sense that everything stated in this article can be extended to3×3-matrices, which after reduction (i.e., division of all their elements) by the first invariant factor become matrices of simple structure. By the definition of a matrix of simple structure, all its elementary divisors are linear. This concept is taken from [1], where it is introduced for polynomial matrices of arbitrary degree by analogy with the concept of numerical matrices (above the field) of simple structure with [3]. Note that the numerical matrices of a simple structure in [4] are called simple and in [5] are nondefective. Next, we will use the terminology from [1, 2], namely, the transformation (1) and the matrices Fx and Gx will be called semiscalarly equivalent (abbreviation: ssk.e., notation: Fx≈Gx). Our task is to construct in the class PFxQxa lower triangular matrix with invariant factors on the main diagonal and some predefined properties, to find a complete system of its invariants and to establish on this basis the conditions of ssk.e. for two such matrices. This problem for 2×2-matrices in the general case is solved by the author in [6] and for 3×3-matrices with one characteristic root in [7].


## 2. Preliminary Information

In classPFxQx, there is a matrix of the form(2)Ax=100a1xϕ1x0a3xa2xϕ2x,where dega1x<degϕ1x, dega2x,dega3x<degϕ2x, and ϕ1x divides a2x and ϕ2x. Let us denote ϕ12x:=ϕ2x/ϕ1x and a2′x:=a2x/ϕ1x. We assume that degϕ1x,degϕ12x>1. Otherwise, Ax in class PFxQx can be chosen so that a1x≡0 or a2x≡0. In these cases, the task of finding the conditions ssk.e. matrices are greatly simplified. Hereinafter, the notation M1 and M2 will mean the sets of roots of polynomials ϕ1x and ϕ12x, respectively. Recall that element α∈M1∪M2 is called the characteristic root of the matrix Ax and M1∪M2 for class PFxQx is defined unambiguously (i.e., does not depend on the choice of matrix Ax). Obviously, Ax in class PFxQx can be chosen so that, for some α0∈M1, the following condition(3)a1α0=a3α0=0is satisfied. In the future, this condition will be mandatory for Ax and for all triangular matrices of class PFxQx. Hereinafter, record ax,bx will mean the greatest common divisor of polynomials ax and bx. We will use the corresponding notation for the greatest common divisor of three polynomials.Proposition 1.
The greatest common divisora1x,a3x,ϕ1x for class PFxQx does not depend on the choice of the matrix Ax with a fixed characteristic root α0∈M1.Proof.
Let the matrix(4)Bx=100b1xϕ1x0b3xb2xϕ2x,where degb1x<degϕ1x, degb2x,degb3x<degϕ2x, and b1α0=b3α0=0, belongs to class PFxQx.
Then, for some matricessij13∈GL3,C and rijx13∈GL3,Cx, the equality(5)sij13100a1xϕ1x0a3xa2xϕ2x=100b1xϕ1x0b3xb2xϕ2xrijx13holds. From the latter, we can write(6)r11x=s11+s12a1x+s13a3x,(7)s21+s22a1x+s23a3x=b1xr11x+ϕ1xr21x,(8)s31+s32a1x+s33a3x=b3xr11x+b2xr21x+ϕ2xr31x.
If we put in (7) and (8) x=α0, we get s21=0 and s31=0, respectively. Since ϕ1x divides r12x and r13x, then r11x,ϕ1x=1. Therefore, from (7) and (8), we have that a1x,a3x,ϕ1x divides b1x and b3x. Given that ssk.e. has the property of symmetry (moreover, ssk.e. is the ratio of equivalence), we can write that b1x,b3x,ϕ1x divides a1x and a3x. Therefore, a1x,a3x,ϕ1x=b1x,b3x,ϕ1x. The proposition is proved.
In this study, we will limit ourselves to the case whena1x,a3x,ϕ1x=ϕ1x. Then, a1x≡0 and ϕ1x divides a3x. Considering the opposite case, when a1x,a3x,ϕ1x≠ϕ1x, is the subject of another study.


## 3. Invariants of Triangular Matrices

Proposition 2.
Ifa1x,a3x,ϕ1x=ϕ1x, then a3x,ϕ2x for class PFxQx does not depend on the choice of matrix Ax.Proof.
LetBx (4) be some other matrix of class PFxQx for which a condition similar to (3) holds. Since a1x=b1x≡0, then, from (6) and (7), we have r11x=s11+s13a3x and r21x=s23a3x/ϕ1x. Then, (8), where s31=01, can be written as(9)s33a3x=b3xs11+s13a3x+s23b2xa3xϕ1x+ϕ2xr31x.
Sinces11≠0, the latter shows that a3x,ϕ2x divides b3x. Therefore, given the symmetry of the ratio nsk.e., we have that b3x,ϕ2x divides a3x and finally a3x,ϕ2x=b3x,ϕ2x. The proposition is proved.
We divide the setM2 of roots of polynomial ϕ12x into subsets according to the following principle: we assign elements α1,α2∈M2 to the same subset if condition a3α1a2′α1=a3α2a2′α2 holds for Ax. Thus, we get some partition of the set M2:(10)M2=K1∪…∪Km.Proposition 3.
Partition (10) of the set M2 for class PFxQx does not depend on the choice of matrix Ax if a1x,a3x,ϕ1x=ϕ1x.Proof.
Suppose that, in addition toAx, the matrix Bx (4) with a condition similar to (3) also belongs to class PFxQx. Then, for some sij13∈GL3,C and rijx13∈GL3,Cx, equation (5) holds, from which we have (9) and(11)s32+s33a2′x=b3xs12+s13a2′x+b2′xs22+s23a2′x+ϕ12xr32x,where b2′x:=b2x/ϕ1x. Let α1andα2 belong to one of the subsets of (10), for example, to K1. If x=α1 and x=α2 is put alternately and the obtained equations are subtracted in (9), then we get b3α1−b3α2s11+s13a3α1+b2′α1−b2′α2s23a3α1=0.
Similarly, from (11), we obtain b3α1−b3α2s12+s13a2′α1+b2′α1−b22′α2s22+s23a2′α1=0.
The last two equations are written in matrix formb3α1−b3α2b2′α1−b2′α2s11+s13a3α1s12+s13a2′α1s23a3α1s22+s23a2′α1=0.
It is easy to see thatdets11+s13a3α1s12+s13a2′α1s23a3α1s22+s23a2′α1=detrijα112.
Since inrijx13, for α1, we have r13α1=r23α1=0, then rijα112≠0. Therefore, b3α1b2′α1=b3α2b2′α2. The proposition is proved.
If, in (10), m=1, then the elements a2x and a3x in Ax are determined to the nearest multiplier, and then, Ax can be chosen so that a2x≡0. In this case, Ax can be considered canonical in class PFxQx. Therefore, we consider the case when m>1.


## 4. Reduction to a Special Triangular Form

Proposition 4.
Suppose that, forAx (2), the conditions hold: a1x,a3x,ϕ1x=ϕ1x, in (10), m>1 and a3β1 or a3β2 are nonzero for β1∈K1 and β2∈K2. Then, in the class PFxQx, there is a matrix Bx (4), in which b3β1≠b3β2 and b2β1=b2β2=0.Proof.
We can assume thata2β1≡0. Otherwise, this is achieved by adding the second row of matrix Ax multiplied by some number to its third row. If a3β1=a3β2≠0 and a2β2≠0, then, in the first step from Ax, we move on to such a matrix Bx (4) in class PFxQx that b3β1≠b3β2. To do this, choose c1≠0 so that 1+c1a2′βi≠0 for every βi∈M2. Then, from the congruence a2′x−b2′x1+c1a2′x≡0modϕ12x, we find b2′x, degb2′x<degϕ12x. Obviously, b2′β1=0 and b2′β2≠0. Next, from the congruence a3x−b3x−c1b2′xa3x≡0modϕ2x, we find b3x, degb3x<degϕ2x. It is easy to see that b3β1=a3β1 and b3β2≠a3β2, since c1b2′β2a3β2≠0. Based on the above selected c1 and obtained b2′x and b3x, we construct matrices sij13=10001c1001 and Bx (4), where b2x=b2′xϕ1x, b1x≡0, and rijx13=100c1a3x/ϕ1x1+c1a2′xc1ϕ12xa3x1−c1b2′x−b3x/ϕ2xa2′x1−c1b2′x−b2′x/ϕ12x1−c1b2′xrijx13∈GL3,Cx. We make sure that the constructed matrices sij13, Bx, rijx13 together with the original matrix Ax satisfy equality (5). In order not to enter new notations, we assume that conditions a3β1≠a3β2 and a2β1=0 are satisfied for Ax. In the second step in class PFxQx from Ax, we pass to matrix Bx with the properties specified in proposition. To do this, first, from equation a3β2−a3β1c2=a2′β2, we find c2≠0 and then, from congruence, b2′x+c2a3x−a3β1−a2′x≡0modϕ21x, we find b2′x, degb2′x<degϕ12x. According to the found c2, b2′x, we construct matrices sij13=1c200100c21 and Bx (4), where b1x≡0, b2x=b2′xϕ1x, b3x=a3x, and rijx13=1c2ϕ1x00100a2′x−c2a3x−a3β1−b2′x/ϕ12x1∈GL3,Cx.
For the original matrixAx and constructed sij13, Bx, and rijx13, we check the truth of equality (5). Since b2′β1=b2′β2=0, then Bx is the required matrix in class PFxQx. The proposition is proved.Proposition 5.
If in matricesAx (2) and Bx (4) of class PFxQx the conditions of Proposition 4 are satisfied and, for β1,β2∈M2, we have a3β1≠a3β2, b3β1≠b3β2, and a2β1=a2β2=b2β1=b2β2=0, then the matrix sij13 in (5) is the upper triangular, where s12=0. In addition, a2′x,ϕ12x=b2′x,ϕ12x.Proof.
If we put in (11) x=β1 and x=β2, we have s32+s12b3β1=0,s32+s12b3β2=0, where we immediately get s32=s12=0 because b3β1≠b3β2. Then, (11) will take the form(12)s33a2′x=s13a2′xb3x+b2′xs22+s23a2′x+ϕ12xr32x.
From (12), it is seen that a2′x,ϕ12x=b2′x,ϕ12x, since s22,s33≠0. The proposition is proved.
Denote byN1 and N2 the subsets of the set M2, for elements βk∈N1 and γl∈N2 of which we have a3βk≠0 and a2γl≠0, respectively. Propositions 2 and 5 imply the following.Corollary 1.
SubsetsN1=βk,βk∈M2,a3βk≠0 and N2=γl,γl∈M2,a2γl≠0 for class PFxQx for fixed β1,β2∈M2 do not depend on the choice of the matrix Ax, which satisfies the conditions of Proposition 5.Since we have imposed the condition that, in (10), m>1, then N1 is not an empty set. If N2 is empty, then a2x≡0. In this case, the problem of ssk.e. for 3×3-matrices is reduced to a similar problem for 2×2-matrices. The latter, as mentioned in Section 1, is resolved in [6]. Therefore, in the future, we assume that N2 is not an empty set.LetAx satisfy the conditions of Proposition 5. It is clear that one of the values a3β1 and a3β2 is nonzero. Let a3β1≠0. We can also assume that, in Ax, one of the following conditions is fulfilled: a3β3=a2′β3=1 for some β3∈N1∩N2 if the intersection N1∩N2 is not empty or a3β3=a2′β3=1 for some β3∈N2 otherwise. Each of these conditions for Ax is achieved by multiplying the first two rows and the first two columns by the corresponding numerical factors. Such a matrix Ax is called oriented by characteristic rootsβ1,β2,β3reduced matrix.Proposition 6.
If the intersectionN1∩N2 for oriented by the same characteristic roots β1,β2,β3 reduced matrices Ax (2) and Bx (4) of class PFxQx is not empty, then s11=s22 in the matrix sij13 of equation (5).Proof.
According to Proposition5, the matrix sij13 in question is an upper triangular, and s12=0. Equations (9) and (12) hold for elements of matrices Ax and Bx. From them, we can get a congruence:(13)s11a2′xb3x−s22b2′xa3x≡0modϕ12x.
If we put, in (13), x=β3, we immediately get s11−s22=0. The proposition is proved.


## 5. Conditions of Semiscalar Equivalence of Reduced Matrices

Theorem 1.
LetAx (2) and Bx (4) be oriented by the same characteristic roots β1,β2,β3 reduced matrices and ΩAx:=deta3xa2′x1a3β101111, ΩBx:=detb3xb2′x1b3β101111.
If, for the setsN1andN2defined in the corollary, the intersectionN1∩N2is not empty, then the matricesAx and Bxare ssk.e. if and only if the following conditions are met:(i)
a2′α/a3α=b2′α/b3αfor eachα∈N1∩N2.(ii)
ValuesΩAβ, ΩBβare simultaneously equal to zero or are nonzero for eachβ∈N1∪N2.(iii)
IfΩAβi≠0for someβi∈N1orΩAγj≠0for someγj∈N2, then(14)ΩAβka3βiΩAβia3βk=ΩBβkb3βiΩBβib3βk,ΩAγla3βiΩAβia2′γl=ΩBγlb3βiΩBβib2′γl,or(15)ΩAβka2′γjΩAγja3βk=ΩBβkb2′γjΩBγjb3βk,ΩAγla2′γjΩAγja2′γl=ΩBγlb2′γjΩBγjb2′γl,for eachβk∈N1 and γl∈N2.
If the intersectionN1∩N2is empty, then the matricesAx and Bxare ssk.e. if and only if the following conditions are met:(1)
Valuesa3βk−1 and b3βk−1, as well as valuesa2′γl−1 and b2′γl−1, are simultaneously equal to zero or are nonzero for eachβk∈N1 and γl∈N2.(2)
Ifa3βi≠1for someβi∈N1, then(16)a3βk−1a3βia3βi−1a3βk=b3βk−1b3βib3βi−1b3βk,for everyβk∈N1, and ifa2′γj≠1for someγj∈N2, then(17)a2′γl−1a2′γja2′γj−1a2′γl=b2′γl−1b2′γjb2′γj−1b2′γl,for eachγl∈N2.Proof (necessity).
LetAx≈Bx. Then, for the elements of matrices Ax and Bx are equations (9) and (12), in which according to Proposition 6 in the absence of intersection of sets N1 and N2 we have s11=s22. Therefore, from (13), we immediately have (i). If in equations (9) and (12) put x=β1, x=β3, x=βk, and x=γl, where βk and γl mean an arbitrary element of N1 and N2, respectively, the result can be written in the matrix form as follows:(18)1a3β1−1b3β1101−1111a3βk−1b3βk1b2′βkb3βk1a2′γl−1b2′γla3γla2′γl1s11s33s13s23=0.
Hence, we have(19)1a3β1−1b3β1101−1a3β1−1−1b3β101ΩAβka3βka3β1−ΩBβkb3βkb3β100ΩAγla2′γla3β1−ΩBγlb2′γlb3β100s11s33s13s23=0.
From the last equality (ii), since s11,s33≠0. If ΩAβi≠0 for some βi∈N1, then, from equality, ΩAβi/a3βia3β1ΩBβi/b3βib3β1ΩAβk/a3βka3β1ΩBβk/b3βkb3β1ΩAγl/a2′γla3β1ΩBγl/b2′γlb3β1s11s33=0.
For arbitraryβk∈N1 and γl∈N2, we have each equality with (14), since s11,s33≠0. If ΩAγj≠0 for some γj∈N2, then, for any βk∈N1 and γl∈N2, similarly we have each of equation (15).
Now, let the setsN1,N2 not intersect and βk∈N1 and γl∈N2. If we put x=β1, x=βk and x=β3, x=γl, respectively, in (9) and (12), we obtain equations that can be written as(20)10−1101a3βk0−1b3βk1001−10101a2′γl−1b2′γl01s11s22s33s13s23=0.
Hence, we have1−1/a3βk0−1−1/b3βk01−1/a2′γl−1−1/b2′γls11s22s33=0.
Sinces11,s22,s33≠0, then, from the last equality, we obtain (j). If a3βi≠1, for some βi∈N1, then, from equality,(21)1−1a3βi0−1−1b3βi1−1a3βk0−1−1b3βks11s22s33=0.
We get (16) for each βk∈N1. Similarly, if a2′γj≠1, for some γj∈N2, then, from equality,(22)01−1a2′γj−1−1b2′γj01−1a2′γl−1−1b2′γls11s22s33=0.
We obtain (17) for every γl∈N2. The necessity is proved.


### 5.1. Sufficiency

Let the setsN1,N2 intersect. Consider (19) as an equation with the unknown s11s33s13s23T. We mean that βk and γl acquire all values from sets N1 and N2, respectively. Therefore, the matrix on the left side of this equation may have more rows than four. Under the conditions of the theorem, the rank of this matrix is equal to 2 if ΩAβk=ΩAγl=0 for every βk∈N1 and γl∈N2. Otherwise, when ΩAβi≠0 for some βi∈N1 or ΩAγj≠0 for some γj∈N2, the rank of the matrix is 3. In each case, this equation has a solution s110s330s130s230T with nonzero first two components s110,s330. If we take into account (i), then, from (19), we can proceed to the equivalent equation (18). Since, as already mentioned, βk and γl acquire all values from N1 and N2, respectively; based on the solution s110s330s130s230T of equation (18), we can write such a system of congruences:(23)s330a3x−b3xs110+s130a3x−s230a3xb2′x≡0modϕ12x,s330a2′x−b2′xs220+s230a2′x−s130a2′xb3x≡0modϕ12x,where s220=s110 and s110,s330≠0. The first of the congruences is performed modulo ϕ2x=ϕ1xϕ12x because ϕ1x,ϕ12x=1 and ϕ1x divides a3x and b3x. Also, we can pass on module ϕ2xand in the second congruence after its multiplication by ϕ1x. So, we obtain s330a3x−b3xs110+s130a3x−s230a3xb2′x≡0modϕ2x,s330a2x−b2xs220+s230a2′x−s130a2xb3x≡0modϕ2x.From the elements of solutions110s330s130s230T, we construct a matrix sij13=s1100s1300s220s23000s330, where, as before, s220=s110, and the matrix rijx13, where r11x=s110+s130a3x, r12x=s130a2x, r13x=s130ϕ2x, r21x=s230a3x/ϕ1x∈Cx, r22x=s220+s230a2′x, r23x=s230ϕ12x, r31x=s330a3x−b3xs110+s130a3x−s230a3xb2′x/ϕ2x∈Cx, r31x=s330a2x−b2xs220+s230a2′x−s130a2xb3x/ϕ2x∈Cx, and r33x=s330−s130b3x−s230b2′x.For the constructed matricessij13 and rijx13and for the original Ax and Bx, we are convinced of the truth of equality (5). This means that Ax≈Bx.Now, let set thatN1 and N2 do not intersect. Consider (21) and (22) together as a system of two equations with the unknown s11s22s33T. Note that, in this system βk and γl, acquire all values from N1 and N2, respectively. Under the conditions (1) and (2) of the theorem, the rank of the matrix of this system does not exceed 2. It is also easy to see that this system has a solution of s110s220s330T with nonzero of all three of its components, i.e., s110,s220,s330≠0. If we consider (20) as an equation with the unknown s11s22s33s13s23T and put in this equation s110s220s330T instead of s11s22s33T, we can unambiguously find s13=s130 and s23=s230. This will find the solution s110s220s330s130s230T of equation (20) with nonzero of first three components. Based on this solution, we can write a system of congruences (23). Then, by the same considerations as in the previous case, we come to the conclusion that Ax≈Bx. The theorem is proved.


## 5.1. Sufficiency

Let the setsN1,N2 intersect. Consider (19) as an equation with the unknown s11s33s13s23T. We mean that βk and γl acquire all values from sets N1 and N2, respectively. Therefore, the matrix on the left side of this equation may have more rows than four. Under the conditions of the theorem, the rank of this matrix is equal to 2 if ΩAβk=ΩAγl=0 for every βk∈N1 and γl∈N2. Otherwise, when ΩAβi≠0 for some βi∈N1 or ΩAγj≠0 for some γj∈N2, the rank of the matrix is 3. In each case, this equation has a solution s110s330s130s230T with nonzero first two components s110,s330. If we take into account (i), then, from (19), we can proceed to the equivalent equation (18). Since, as already mentioned, βk and γl acquire all values from N1 and N2, respectively; based on the solution s110s330s130s230T of equation (18), we can write such a system of congruences:(23)s330a3x−b3xs110+s130a3x−s230a3xb2′x≡0modϕ12x,s330a2′x−b2′xs220+s230a2′x−s130a2′xb3x≡0modϕ12x,where s220=s110 and s110,s330≠0. The first of the congruences is performed modulo ϕ2x=ϕ1xϕ12x because ϕ1x,ϕ12x=1 and ϕ1x divides a3x and b3x. Also, we can pass on module ϕ2xand in the second congruence after its multiplication by ϕ1x. So, we obtain s330a3x−b3xs110+s130a3x−s230a3xb2′x≡0modϕ2x,s330a2x−b2xs220+s230a2′x−s130a2xb3x≡0modϕ2x.From the elements of solutions110s330s130s230T, we construct a matrix sij13=s1100s1300s220s23000s330, where, as before, s220=s110, and the matrix rijx13, where r11x=s110+s130a3x, r12x=s130a2x, r13x=s130ϕ2x, r21x=s230a3x/ϕ1x∈Cx, r22x=s220+s230a2′x, r23x=s230ϕ12x, r31x=s330a3x−b3xs110+s130a3x−s230a3xb2′x/ϕ2x∈Cx, r31x=s330a2x−b2xs220+s230a2′x−s130a2xb3x/ϕ2x∈Cx, and r33x=s330−s130b3x−s230b2′x.For the constructed matricessij13 and rijx13and for the original Ax and Bx, we are convinced of the truth of equality (5). This means that Ax≈Bx.Now, let set thatN1 and N2 do not intersect. Consider (21) and (22) together as a system of two equations with the unknown s11s22s33T. Note that, in this system βk and γl, acquire all values from N1 and N2, respectively. Under the conditions (1) and (2) of the theorem, the rank of the matrix of this system does not exceed 2. It is also easy to see that this system has a solution of s110s220s330T with nonzero of all three of its components, i.e., s110,s220,s330≠0. If we consider (20) as an equation with the unknown s11s22s33s13s23T and put in this equation s110s220s330T instead of s11s22s33T, we can unambiguously find s13=s130 and s23=s230. This will find the solution s110s220s330s130s230T of equation (20) with nonzero of first three components. Based on this solution, we can write a system of congruences (23). Then, by the same considerations as in the previous case, we come to the conclusion that Ax≈Bx. The theorem is proved.


## 6. Conclusions

In this paper, a special form is established for the selected set of matrices of simple structure, which is achieved with the help of ssk.e. transformations and is called the oriented by characteristic roots reduced matrix (Proposition4). The matrix of this special shape is lower triangular and has invariant factors on the main diagonal and some predefined properties. Propositions 1–3 and 5 and Corollary indicate the invariants of the reduced matrix. Propositions 5 and 6 set out the form of the left transform matrix, which together with some right transform (polynomial) matrix translates one reduced matrix into another within class PFxQx. This matrix has an upper triangular form with a zero element in position 1,2. In some cases, it has the same first two diagonal elements. The theorem specifies the necessary and sufficient conditions under which the reduced matrices are ssk.e. Thus, Propositions 1–3 and 5, corollary, and theorem indicate a complete system of invariants of the reduced matrix with respect to ssk.e. It is clear that theorem can be applied to establish ssk.e for arbitrary matrices Fx and Gx (not reduced) if they belong to the set selected in this paper. Namely, Fx and Gx are ssk.e. if and only if their oriented by the same characteristic roots reduced matrices Ax and Bxsatisfy the conditions of theorem. Also, as already mentioned in Section 1, everything described in this article can be extended to 3×3-matrices, which after division by the first invariant factor become matrices of simple structure. The proved theorem can be applied to the problem of classification of finite sets of numerical matrices with accuracy to similarity. In particular, let sets F and G of numerical (from the field) 3×3-matrices be given. On the matrices of these sets, as on the coefficients, we construct matrix polynomials Fx and Gx with single higher coefficients. Suppose that, after division by the first invariant factor of the polynomial matrices Fx and Gx obtained in this way, we have particles F′x and G′x of simple structure. Then, for the similarity of sets F and G, it is necessary and sufficient to satisfy the conditions of theorem for consolidated matrices Ax and Bx such that Ax≈F′x and Bx≈G′x.



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*Source: 5592756-2021-04-19.xml*