# Toeplitz Matrices in the Problem of Semiscalar Equivalence of Second-Order Polynomial Matrices

Authors: B. Z. Shavarovskii
Journal: International Journal of Analysis
 (2017)

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## Abstract

We consider the problem of determining whether two polynomial matrices can be transformed to one another by left multiplying with some nonsingular numerical matrix and right multiplying by some invertible polynomial matrix. Thus the equivalence relation arises. This equivalence relation is known as semiscalar equivalence. Large difficulties in this problem arise already for 2-by-2 matrices. In this paper the semiscalar equivalence of polynomial matrices of second order is investigated. In particular, necessary and sufficient conditions are found for two matrices of second order being semiscalarly equivalent. The main result is stated in terms of determinants of Toeplitz matrices.


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## Body

## 1. Introduction

LetC be a field of complex numbers and C[x] the ring of polynomials in an indeterminate x over C. Let M(n,C) and M(n,C[x]) denote the algebras of n×n matrices over C and C[x], respectively, and GL(n,C), GL(n,C[x]) their corresponding groups of units. Given two matrices F(x),G(x)∈M(n,C[x]), the question of determining whether there exist the matrices P∈GL(n,C), Q(x)∈GL(n,C[x]), such that (1)Fx=PGxQx,has attracted much attention for many years. This proved to be a bigger problem than originally anticipated. Large difficulties in this problem arise already for elements of M(2,C[x]). The matrices F(x),G(x)∈M(n,C[x]) are called semiscalarly equivalent, if the equality (1) is satisfied for some nonsingular matrix P∈M(n,C) and for some invertible matrix Q(x)∈M(n,C[x]) [1] (see also [2]). Due to this fact the problem of finding the conditions under which the matrices are semiscalarly equivalent is of current interest. In this paper the indicated problem for the matrices of second order is solved. Toeplitz matrices plays an important role in the conditions under which two matrices of second order can be transformed to one another by semiscalar equivalent transformation. In spite of Toeplitz matrices forming a special matrix class, many classical problems related Laurent series, moment’s problem, orthogonal polynomials, and others are reduced to them. A more serious interest to Toeplitz matrices has, to a large extent, the following explanation. Any matrix has connection with Toeplitz matrices in the sense that every matrix can be represented as a sum of the products of Toeplitz matrices. Much applied problems of electrodynamics, geophysics, acoustics, and automatic control require investigation of Toeplitz matrices. Also, there is a correspondence between complex functions and Fourier series and the latter is closely related to some sequence of Toeplitz matrices. The monographs [3–5] present plenty of material about the research of Toeplitz and Hankel matrices. The articles [6–8] refer to modern problems of these matrices. The results of this paper may be applied in the solving of the matrix equations, which are utilized in many problems of engineering.


## 2. Preliminaries

LetF(x),G(x)∈M(2,C[x]). If matrices F(x) and G(x) are semiscalarly equivalent then it is necessary for them to satisfy the following condition: det⁡F(x)=cdet⁡G(x), c∈C, c≠0. If the matrices F(x) and G(x) are of full rank, then according to [1] (see also [2]), they are semiscalarly equivalent to the lower triangular matrices A(x) and B(x), respectively, with the invariant multipliers of the main diagonal. Similar results are published in [9, 10]. We may assume, without loss of generality, that first invariant multipliers of matrices A(x) and B(x) are identities. Therefore, these matrices can be considered in the form (2)Ax=10axδx,Bx=10bxδx,deg⁡ax,deg⁡bx<deg⁡δx.Denote by a(t)(α) and b(t)(α) the values at x=α of the t-th derivatives of a(x) and b(x), respectively, in the matrices A(x) and B(x). The determinant A(x)=δ(x) is called the characteristic polynomial and its roots are called the characteristic roots of the matrix A(x) (resp., for matrix B(x)). Let us denote by M the set of characteristic roots of matrix A(x) of the form (2). Now consider a partition (3)M=⋃v=1uMvof the set M into subsets Mv such that α,β∈Mv if a(α)=a(β). Evidently, each two different subsets Mv of the partition (3) are disjoint.The following two assertions are valid.Proposition 1.
The partition (3) of the set M of characteristic roots of matrix A(x) of the form (2) is invariant for the class {CA(x)Q(x)} of semiscalarly equivalent matrices.Proof.
Let the matricesA(x), B(x) of the form (2) be semiscalarly equivalent; that is, the equality(4)s11s12s21s2210axδx=10bxδxr11xr12xr21xr22x,where s11s12s21s22∈GL(2,C), r11(x)r12(x)r21(x)r22(x)∈GL(2,C[x]), holds. We deduce from (4) the relations (5)s11+s12ax=r11x,(6)s12δx=r12x,(7)s21+s22ax=bxr11x+δxr21x.Setting x=α and x=β, α≠β, α,β∈M, we obtain the relations(8)s21+s22aα=bαr11α,s21+s22aβ=bβr11β.
Ifa(α)=a(β), then left sides of the resulting relations are equal. Therefore, from the equality of right sides, taking into account that r11(α)=r11(β)≠0 (see (5), (6)), we have b(α)=b(β). The notion of semiscalar equivalence is a symmetric relation. Then b(α)=b(β) by the similar argument yields a(α)=a(β). This completes the proof.Proposition 2.
Letα be the characteristic root of the multiplicity n of matrix A(x) of the form (2). Let also m be the lowest (nonzero) order of the nonzero derivative a(m)(α)≠0 of the entry a(x) of this matrix at x=α. Then the number m (as n) is an invariant for the class {CA(x)Q(x)} of semiscalarly equivalent matrices, if m<n.Proof.
Let the matricesA(x), B(x) of the form (2) be semiscalarly equivalent and m′ be the lowest (nonzero) order of the nonzero derivative b(m)(α)≠0 of the entry b(x) of matrix B(x) at x=α. Suppose that m<m′. From relations (5) and (7), we obtain(9)s21+s22ax-s11bx-s12axbx=δxr21x.
Substitutingx=α into (9), we find(10)s21+s22aα-s11bα-s12aαbα=0.
Takingm-the derivative of (9) at x=α, we obtain (11)s22amα-s12amαbα=0.
Dividing both sides of obtained equality bya(m)(α)≠0 and the substituting in (10) yields (12)s22-s12bα=0,s21-s11bα=0.
But this is impossible, since matrixsij12 is nonsingular. Therefore, m≥m′. Inasmuch the semiscalar equivalence is a symmetric relation, we have m=m′. The Proposition is proved.Next it is assumed thatu≥3 in the partition (3). The simplest cases are when u=1 or u=2 may be found in some other articles of the author (see, for example, [11, 12]). We may take a(α)=b(α)=0, for entries a(x),b(x) of A(x),B(x), if α∈M1. By Proposition 2 for the matrices A(x) and B(x) being semiscalarly equivalent, it is necessary that a(β),b(β)≠0, when β∈M, β∉M1.We use the following notations:s=deg⁡δ(x) is a degree of characteristic polynomial of the matrices A(x),B(x) of the form (2); ni is the multiplicity of characteristic root αi of these matrices; mi is the lowest (nonzero) order of the nonzero derivative a(m)(αi)≠0 of a(x). If mi<ni, then by Proposition 2 the semiscalar equivalence of A(x) and B(x) implies that b(mi)(αi)≠0, b(v)(αi)=0, 0<v<mi, for the entry b(x) of B(x). For entries a(x),b(x) of A(x),B(x) and for their arbitrary characteristic root αi we denote by ai,ait,bi,bit the following quantities: (13)ai=aαi,ait=ami+tαimi+t!,bi=bαi,bit=bmi+tαimi+t!,t=0,1,…,s-mi-1.The following cases are possible:The Case  1.
m
i
≥
n
i for every root αi∈M.The Case  2. There is a root αj∈M such that mj<nj and 2mi≥ni for every root αi∈M.The Case  3. There is a root αj∈M such that 2mj<nj.Let us now consider each of them separately.


## 3. The Case  1

Let us now formulate the following Theorem based on the defined notation and the assumption.Theorem 3.
Let the partition of the setM of characteristic roots of matrix A(x) (2) be of the form (3) and αv∈Mv, v=1,…,u, u≥3. Also suppose the entry a(x) of the matrix A(x) satisfies the condition a(α1)=0 and mi≥ni for each root αi∈M. The matrix A(x) in the class {CA(x)Q(x)} of semiscalarly equivalent matrices is determined up to constant factor c≠0 of the row (14)a2-a3a2a3⋯a2-aua2au.Proof.
Let the entryb(x) of the matrix B(x) of the form (2) satisfy the condition b1=0 (see notations (13)). If the matrices A(x) and B(x) are semiscalarly equivalent, then the equality (4) implies s21=0 and(15)s22ax-s11bx-s12axbx≡0modδx,where s11,s22≠0. Substituting x=α2,α3,…,αu in the congruence (15), we obtain the system of equalities(16)s22a2-s11b2-s12a2b2=0,s22a3-s11b3-s12a3b3=0,⋮s22au-s11bu-s12aubu=0,where a2,…,au,b2,…,bu≠0. By excluding s12 in the system (16), we find that (17)ca2-a3a2a3⋯a2-aua2au=b2-b3b2b3⋯b2-bub2bu,where c=s11/s22.
Conversely, the quality (17), after some transformations, can be written in the form (18)a2-cb2a2b2=a3-cb3a3b3,…,a2-cb2a2b2=au-cbuaubu.
Introducing the notations(19)s11=c,s22=1,s12=a2-cb2a2b2,
based on the equalities (18), we obtain the system (16). Since mi≥ni for each root αi∈M then every term of degree 1,…,ni-1 in the binomial decomposition of the entries a(x),b(x) of the matrices A(x),B(x) vanishes in the powers of x-αi. This and the system (16) mean that congruence (15) is valid. This implies that (20)s22ax-s11bx-s12axbxδx∈Cx.
Denotingr21(x)=(s22ax-s11bx-s12axbx)/δ(x) and r11(x)=s11+s12a(x), r12(x)=s12δ(x), r22(x)=s22-s12b(x), it is easy to make sure that the equality (4) holds true, where s21=0, s11,s22≠0 (see (19)), and r11(x)r12(x)r21(x)r22(x)=s11s22. This means that the matrices A(x) and B(x) are semiscalarly equivalent. The Theorem is proved.


## 4. The Case  2

Further we shall retain the earlier introduced notations (see, in particular, (13)).Theorem 4.
Let the partition of the setM of characteristic roots of matrices A(x), B(x) (2) be of the form (3) and αv∈Mv, v=1,…,u, u≥3. Let the entries a(x), b(x) of this matrices satisfy the condition a(α1)=b(α1)=0. Let also 2mi≥ni for every root αi∈M and mj<nj for some roots αj∈M. Matrices A(x), B(x) are semiscalarly equivalent if and only if there exists a number c≠0; the following conditions hold:(1)
ap0ap1⋯ap,np-mp-1=cbp0bp1⋯bp,np-mp-1 for every root αp∈M1 such that mp<np;(2)
aq0/aq2aq1/aq2⋯aq,nq-mq-1/aq2=1/cbq0/bq2bq1/bq2⋯bq,nq-mq-1/bq2 for every root αq∈M such that αq∉M1 and mq<nq;(3)
a2-a3/a2a3…a2-au/a2au=1/cb2-b3/b2b3…b2-bu/b2bu.Proof.
Necessity. If the matrices A(x) and B(x) are semiscalarly equivalent, then from congruence (15) we obtain the system of equalities(21)s22ap0-s11bp0=0,s22ap1-s11bp1=0,⋮s22ap,np-mp-1-s11bp,np-mp-1=0,for every root αp∈M1 such that mp<np. The condition s11,s22≠0 implies the condition (1) of Theorem, for c=s11/s22.
From the congruence (15) we can write the system of equations (22)s22aq-s11bq-s12aqbq=0,s22aq0-s11bq0-s12aq0bq+bq0aq=0,s22aq1-s11bq1-s12aq1bq+bq1aq=0,⋮s22aq,nq-mq-1-s11bq,nq-mq-1-s12aq,nq-mq-1bq+bq,nq-mq-1aq=0for every root αq∈M such that αq∉M1 and mq<nq. It is understood that αq,bq,s11,s22≠0. From first equation of system (22) we find that (23)s12=s11aq-s22bqaqbq.
Substituting it into second and every succeeding equalities we obtain(24)s11aq0aq2=s22bq0bq2,s11aq1aq2=s22bq1bq2⋮s11aq,nq-mq-1aq2=s22bq,nq-mq-1bq2.
The condition (2) of the Theorem is proved.
As in the proof of Theorem3, from congruence (15) by virtue of the substitution x=αv, v=1,…,u, we can easily obtain the equalities (16). By excluding s12, we arrive at condition (3) of Theorem. The necessity of the conditions (1)–(3) of Theorem is proved.
Sufficiency. Suppose that the conditions of Theorem are satisfied. If we introduce the notations s11=c,s22=1, then the condition (1) denotes the equalities (21) being satisfied for every root αp∈M1 such that mp<np. From this it follows immediately the congruence (25)s22ax-s11bx-s12axbx≡0modx-αlnl,where s12 is arbitrary number, for every root αl∈M1 (but not necessarily ml<nl).
From the condition (3) of Theorem follows the equalities (18). If we introduce the notations (19), then from (18) we can obtain the system (16). This means that the congruence (26)s22ax-s11bx-s12axbx≡0modx-αjis valid for every root αj∈M such that αj∉M1.
Using the notations (19) we can proceed from condition (2) of Theorem to the system of equations (22). Last system is equivalent to the congruence(27)s22ax-s11bx-s12axbx≡0modx-αqnqfor every root αq∈M such that αq∉M1 and mq<nq. When it is considered the congruence (26), then from (27) we actually have(28)s22ax-s11bx-s12axbx≡0modx-αknkfor every root αk∉M1 (but not necessarily mk<nk). Combining (25) with (28), we obtain the congruence (15). We complete the proof of Theorem in a way analogous to the end of the proof of Theorem 3.


## 5. Auxiliary Statements

In the following studies we need to use Lemmas5 and 6 that we prove in the current section.Lemma 5.
The matrix equation(29)a0b00a1b10⋮⋮⋮am-1bm-10ambma0b0am+1bm+1a0b1+a1b0⋮⋮⋮akbk∑t=0k-matbk-m-tx1x2x3=0-,over C, where a0,b0≠0, has nonzero solutions if and only if the following conditions are satisfied:(30)a1a2⋯av-1ava0a1⋯av-2av-1⋱⋱⋮⋮⋱a1a20a0a1=cvb1b2⋯bv-1bvb0b1⋯bv-2bv-1⋱⋱⋮⋮⋱b1b20b0b1,where v=1,…,m-1,m+1,…,k, c=a0/b0. If the conditions of Lemma are satisfied then for every nonzero solution x10x20x30T of (29) x10,x20≠0.Proof.
Necessity. Let (29) have the solution x10x20x30T≠0-. Then (31)a0x10+b0x20=0,a1x10+b1x20=0,⋮am-1x10+bm-1x20=0,(32)amx10+bmx20+a0b0x30=0,am+1x10+bm+11x20+a0b1+a1b1x30=0,⋮akx10+bkx20+x30∑t=0k-matbk-m-t=0.
We assume thatx10=0 or x20=0. Then, from first equalities of (31) and (32) we obtain x10=x20=x30=0. This contradicts the assumption that solution x10x20x30T is nonzero. So, we have x10,x20≠0. From equalities (31) we obtain the conditions (30), where c=-x20/x10=a0/b0, for v=1,…,m-1.
We conclude from (32), if am=cbm, then x30=0 and ah=cbh for h=m+1,…,k. From this it follows that equality (30) holds true for v=m+1,…,k. For this reason in the following assume that am≠cbm, where c=-x20/x10=a0/b0. From the first and second equalities (32) by excluding x30, we obtain (33)a0am+1-amcb1+a1=c2b0bm+1-cbmcb1+a1.
Ifm=1 then a0a2-a12=c2(b0b2-b12). This means that conditions (30) are satisfied for v=m+1=2. If m>1, then a1=cb1, and by multiplication of the both sides by a0m-1=cm-1b0m-1 in (33) we find (34)a0mam+1-2a0m-1a1am=cb0mbm+1-2b0m-1b1bm.
Denote byAuw, Buw the submatrices obtained, respectively, from matrices (35)a1a2…amam+1a0a1…am-1am⋱⋱⋮⋮⋱a1a20a0a1,b1b2…bmbm+1b0b1…bm-1bm⋱⋱⋮⋮⋱b1b20b0b1by obliterating of two last columns and u-th and w-th rows. Also denote by Δm+1(A), Δm+1(B) the determinants of matrices (35), respectively. Decompose them by the minors of order two that are contained in the last two columns. Because Auw=Buw=0 (for u≠m+1), we have (36)Δm+1A=-1m+1amam+1a0a1A1,m+1-am-1ama0a1A2,m+1+⋯+a1a2a0a1Am,m+1,(37)Δm+1B=-1m+1bmbm+1b0b1B1,m+1-bm-1bmb0b1B2,m+1+⋯+b1b2b0b1Bm,m+1.
Sincea0a1⋯am-1=cb0b1⋯bm-1 each summand of expression in parenthesis for Δm+1(A), possibly except the first two, differs from the corresponding summand for Δm+1(B) by the multiplier cm+1. From this fact and from the equality (34) for v=m+1 follows equality (30).
Denote byΔv(A) and Δv(B) the determinants in the left and right sides of equality (30), respectively. Suppose by induction Δr(A)=crΔr(B) for all r such that m<r<k. For the sake of determinacy we assume that r>2m. In the case when r≤2m the proof is not different in principle. From the first r-m+2 equalities (32), by excluding x30 and by sufficiently evident transformations, we arrive at the system(38)am+1-a0b0-1am∑u=01aub1-u-a0mΔr-mA=cr+1bm+1-a0b0-1bm∑u=01aub1-u-b0mΔr-mB,am+2-a0b0-1am∑u=02aub2-u-a0m+1Δr-m-1A=cr+1bm+2-a0b0-1bm∑u=02aub2-u-b0m+1Δr-m-1B,⋮ar-m+1-a0b0-1am∑u=0r-2m+1aubr-2m-u+1-ar-m+1+a0b0-1am∑u=0r-2m+1aubr-2m-u+1-a0r-mΔmA=cr+1br-m+1-a0b0-1bm∑u=0r-2m+1aubr-2m-u+1-br-m+1+a0b0-1bm∑u=0r-2m+1aubr-2m-u+1-b0r-mΔmB,⋮ar-a0b0-1am∑u=0r-maubr-m-u-a0r-1Δ1A=cr+1br-a0b0-1bm∑u=0r-maubr-m-u-b0r-1Δ1B,ar+1-a0b0-1am∑u=0r-m+1aubr-m-u+1-a0r=cr+1br+1-a0b0-1bm∑u=0r-m+1aubr-m-u+1-b0r,where c=a0/b0.
If we add left sides of equalities (38) and separate right sides, we obtain(39)-a0rar+1+-a0r-1arΔ1A+⋯+-a0r-mar-m+1ΔmA+⋯+-a0mam+1Δr-mA+-a0m-1amΔr-m+1A-a0b0-1amb1Δr-mA-a0m+1+b2Δr-m-1A-a0m+2+⋯+bj,r-mjδj1A-aj0r+a0b0-1amb1Δr-mA-a0m+1+b2Δr-m-1A-a0m+2+⋯+br-mΔ1A-a0r+br-m+1-a0r+1-ar-m+1-a0b0-1am∑u=0r-2m+1aubr-2m-u+1-a0r-mΔmA=cr+1-b0rbr+1+-b0r-1brΔ1B+⋯+-b0r-mbr-m+1ΔmB+⋯+-b0mbm+1Δr-mB+-b0m-1bmΔr-m+1B-a0b0-1bma1Δr-mB-b0m+1+a2Δr-m-1B-b0m+2+⋯+ar-mΔ1B-b0r+a0b0-1bma1Δr-mB-b0m+1+a2Δr-m-1B-b0m+2+⋯+ar-mΔ1B-b0r+ar-m+1-b0r+1-br-m+1-a0b0-1bm∑u=0r-2m+1aubr-2m-u+1-bj0r-mjΔmB.
Gathering similar terms in both sides of obtained equalities, we obtain(40)-a0rar+1+-a0r-1arΔ1A+⋯+-a0r-mar-m+1ΔmA+⋯+-a0mam+1Δr-mA+-a0m-1amΔr-m+1A+a0b0-1ambr-m+1-a0r+1-ar-m+1-a0b0-1am∑u=1r-2m+1aubr-2m-u+1-a0r-mΔmA=cr+1-b0rbr+1+-b0r-1brΔ1B+⋯+-b0r-mbr-m+1ΔmB+⋯+-b0mbm+1Δr-mB+-b0m-1bmΔr-m+1B+a0b0-1bmar-m+1-b0r+1-br-m+1-a0b0-1bm∑u=1r-2m+1aubr-2m-u+1-b0r-mΔmB.
It follows from (32) that(41)ar-m+1+a0b0-1am∑u=0r-2m+1aubr-2m-u+1=cbr-m+1+a0b0-1bm∑u=0r-2m+1aubr-2m-u+1.
From this relation it is easy to say that the following equality holds:(42)a0b0-1ambr-m+1-a0r+1-ar-m+1-a0b0-1am∑u=0r-2m+1aubr-2m-u+1-a0r-mΔmA=cr+1a0b0-1bmar-m+1-b0r+1-br-m+1-a0b0-1bm∑u=0r-2m+1aubr-2m-u+1-b0r-mΔmB;
From (31) and the induction hypothesis, we can write (43)-a0m-2am-1Δr-m+2A+⋯+-a0a2Δr-1A+a1ΔrA=cr+1-b0m-2bm-1Δr-m+2B+⋯+-bj0bj2δj,r-1B+b1ΔrB.
Comparing (40), (42), and (43), we obtain equality (44)-a0rar+1+-a0r-1arΔ1A+⋯+-a0a2Δr-1A+a1ΔrA=cr+1-b0rbr+1+-b0r-1brΔ1B+⋯+-b0b2Δr-1B+b1ΔrB,that is, Δr+1(A)=cr+1Δr+1(B), where c=a0/b0. The necessity of conditions of Lemma is proved.
Sufficiency. Consider the equalities (31) and (32) as one system of equations in three indeterminate x10,x20,x30. In conditions (30) c=a0/b0. This means that x10=1, x20=-c satisfy the first equation in the system (31). From condition (30) with v=1 it follows that x10=1, x20=-c satisfy the second equation in system (31). Next, third and all following equalities in system (31) for x10=1, x20=-c can be recurrently obtained from conditions (30) with v=2,…,m-1. Evidently, the values (45)x10=1,x20=-c,x30=cbm-ama0b0-1satisfy the first equation (32). We compute both determinants in equality (30) with v=m+1. After the annihilation of equal summands on both sides of obtained equality and after division by a0m=cmb0m using the of simple transformations, we can obtain the following relation: (46)am+1-cbm+1+cbm-ama0b0-1a0b1+a1b1=0.
This means that (45) satisfies the second equation of system (32).
Assume by induction that (45) satisfies first r-m+1 equations of system (32), that is, (47)am-cbm+cbm-ama0b0-1a0b0=0,am+1-cbm+1+cbm-ama0b0-1∑u=01aub1-u=0,⋮ar-cbr+cbm-ama0b0-1∑u=0r-maubr-m-u=0.
While proceeding we may assumer>2m. In the opposite case the proof is completely analogous. Taking into account the conditions (30) and the assumption of induction we can write the equalities (42), (43), and (44). From these equalities we obtain the relation (40). This relation implies the equality (39). It is evident that from the second and all following equalities of (47) we find that first r-m equalities of (38) are valid. The first r-m equalities of (38) along with relation (39) yield the last equality of (38). Dividing by (-a0)r=cr(-b0)r and after some simplifications this equality can be written in the form(48)ar+1-cbr+1+cbm-ama0b0-1∑u=0r-m+1aubr-m-u+1=0.
This means that (45) is the solution of (r-m+2)-th equation of system (32).
We have inductively proved the existence of the nonzero solution (45) of systems (31) and (32). Thus, matrix equation (29) has nonzero solution. Lemma is proved.Lemma 6.
The matrix equation(49)ababa0b0ab0+a0ba1b1ab1+a1b⋮⋮⋮am-1bm-1abm-1+am-1bambmabm+a0b0+ambam+1bm+1abm+1+a0b1+a1b0+am+1b⋮⋮⋮akbkabk+∑t+0k-matbk-m-t+akbx1x2x3=0-,over C, where a,b,a0,b0≠0, k≥m, m≥1, has nonzero solutions if and only if the following conditions are satisfied:(50)am+v-1a0-10⋯0a0a1⋯ava⋱⋱⋱⋱⋱⋮⋱⋱⋱⋱⋱a1⋱⋱⋱⋱a0⋱⋱⋱0⋱⋱⋮0a0︸m+v=bm+v-1b0-10⋯0b0b1⋯bvb⋱⋱⋱⋱⋱⋮⋱⋱⋱⋱⋱b1⋱⋱⋱⋱b0⋱⋱⋱0⋱⋱⋮0b0︸m+v,v=0,1,…,k.
If the conditions of Lemma are satisfied then every nonzero solutionx10x20x30T of the equation (49) has x10,x20≠0.Proof.
Necessity. The equality (50) for v=0 holds true trivially. Let (49) have nonzero solution. Then the matrix of this equation has rank less than 3. The two first rows of the matrix are linearly independent. Then, each of the following rows of the matrix linearly depends on these first two rows. Because (51)ababa0b0ab0+a0bavbvabv+avb=0,v=1,…,m-1,we have av/a0=bv/b0. This implies the equality (50) for v=1,…,m-1. From the equality (52)ababa0b0ab0+a0bambmabm+a0b0+amb=0we have (53)ab02a0-a0bmab+amb0ab-a02b0b=0,where, after division of both sides by aba0b0, we obtain equality (50) for v=m. Assume by induction that equality (50) holds true for all v=m,m+1,…,r,m≤r<k. Consider the equality(54)ababa0b0ab0+a0bar+1br+1abr+1+∑t=0r+1-matbr+1-m-t+ar+1b=0.
After the calculation of the determinant on the left, we obtain the equation(55)ab0∑t=0r+1-matbr+1-m-t-a0br+1ab+ar+1b0ab-a0b∑t=0r+1-matbr+1-m-t=0.
We can divide byaba0b0 both sides to obtain the equality(56)ar+1a0-1-br+1b0-1-ab0-1∑t=0r+1-matbr+1-m-t+a0b-1∑t=0r+1-matbr+1-m-t=0.
Let us denote byΔv(a) and Δv(b) the determinants in the left and right sides of the equality (50), respectively. Considering the inductive assumption, we can write the equalities (57)-1m-1a0a-1ara0-1Δr+1-ma-brb0-1Δr+1-mb=0,-1ma1a-1ar-1a0-1Δr-ma-br-1b0-1Δr-mb=0,⋮-1r-1ar-ma-1ama0-1Δ1a-bmb0-1Δ1b=0,-1rar+1-ma-1am-1a0-1Δ0a-bm-1b0-1Δ0b=0,(58)-1m-1b0b-1ara0-1Δr+1-ma-brb0-1Δr+1-mb=0,-1mb1b-1ar-1a0-1Δr-ma-br-1b0-1Δr-mb=0,⋮-1r-1br-mb-1ama0-1Δ1a-bmb0-1Δ1b=0,-1rbr+1-mb-1am-1a0-1Δ0a-bm-1b0-1Δ0b=0.
Decompose the determinantsΔr+1(a) and Δr+1(b) for entries of their last columns. Then the left and right sides of equality (50) for v=r+1 can be written in the form(59)am+ra0-1Δr+1a=am+ra0-1-1m+rar+1am+r+-1rar+1-marΔ0a+-1r-1ar-mar-1Δ1a+⋯+-1m+1a2am+1Δr-m-1a+-1ma1amΔr-ma+-1m-1a0am-1Δr-m+1a,(60)bm+rb0-1Δr+1b=bm+rb0-1-1m+rbr+1bm+r+-1rbr+1-mbrΔ0b+-1r-1br-mar-1Δ1b+⋯+-1m+1b2bm+1Δr-m-1b+-1mb1bmΔr-mb+-1m-1b0bm-1Δr-m+1b.
It is easy to make sure that the following identity is true:(61)b-1b0ara0-1Δr-m+1a-1m-1+b1ar-1a0-1Δr-ma-1m+⋯+br-mama0-1Δ1a-1r-1+br-m+1am-1a0-1Δ0a-1r+a-1-a0brb0-1Δr-m+1b-1m-1-a1br-1b0-1Δr-mb-1m-⋯-ar-mbmb0-1Δ1b-1r-1-ar-m+1bm-1b0-1Δ0b-1r+-1m+ra0b-1∑t=0r+1-matbr+1-m-t--1m+rab0-1∑t=0r+1-matbr+1-m-t=0.
We add the left sides of equalities (57) and (58) and multiply by (-1)m+r the left side of the equality (56) and separately add right sides of the these equalities. Taking into account the expression (59), (60) for (am+ra0)-1Δr+1(a), (bm+rb0)-1Δr+1(b) and the identity (61), we obtain (am+ra0)-1Δr+1(a)-(bm+rb0)-1Δr+1(b)=0. This means that the equality (50) for v=r+1 is true. The necessity of the conditions of Lemma is proved.
Sufficiency. Let the conditions (50) for v=1,…,m-1 be satisfied. Then the submatrix formed by first m+1 rows of the matrix of (49) has rank less than 3. In fact, this rank is equal to 2, because (as has been stated above) the first two rows of this matrix are linearly independent. The equality (50) for v=m implies that (m+2)-th row of the matrix of equation (49) linearly depends on its first two rows. Our inductive assumption is the following. Let each of the first r+2, m≤r<k, rows of the matrix of (49) linearly depend on its first two rows. We now reverse the order of arguments, as compared to the proof of the necessity, passing from relation (50) for v=r+1 to relation (56). This relation implies that the minor of order 3 of the matrix of (49) that is contained in 1-st, 2-nd, and (r+3)-th rows is equal to zero. This means that indicated rows are linearly dependent. The above argument inductively proves that the matrix of equation (49) has the rank 2. From this it follows that (49) has the nonzero solution. The rest of the Lemma will be proved by contradiction. Let x10x20x30T be the nonzero solution of the equation (49), where x10=0 or x20=0. Hence we have the equality (62)babb0ab0+a0bx20x30=0-
or(63)aaba0ab0+a0bx10x30=0-.
Sincea,b,a0,b0≠0, the determinants of the 2×2 matrices of these equalities are nonzero. Hence, x20x30T=0- or x10x30T=0-; that is, x10x20x30T=0- contrary to the our assumption. Therefore in the nonzero solution x10x20x30T of (49) necessarily x10,x20≠0. Lemma is proved.


## 6. The Case  3

The notations are the same as in the Cases  1 and  2 (in particular, see (13)).Theorem 7.
Let the partition of the setM of characteristic roots of matrices A(x), B(x) (2) be of the form (3), where u≥3. Let the entries a(x), b(x) of these matrices satisfy the condition a(α1)=b(α1)=0 for root α1∈M1. Let also 2mj<nj for some root αj∈M. Matrices A(x), B(x) are semiscalarly equivalent if and only if there exists a number c≠0 such that the following conditions hold:(1)
ap0=cbp0 and (64)ap1ap2…ap,sp-1ap,spap0ap1…ap,sp-2ap,sp-1⋱⋱⋮⋮⋱ap1ai20ap0ap1=cspbp1bp2…bp,sp-1bp,spbp0bp1…bp,sp-2bp,sp-1⋱⋱⋮⋮⋱bp1bi20bp0bp1,sp=1,…,mp-1,mp+1,…,np-mp-1, for every root αp∈M1 such that mp<np;(2)
(65)ak,mkak02-al,mlal02=1cbk,mkbk02-bl,mlbl02for every pair of the roots αk,αl∈M1 such that 2mk<nk, 2ml<nl;(3)
(66)1aqmq+sq+10⋯0aq0aq1⋯aq,sqaq⋱⋱⋱⋱⋱⋮⋱⋱⋱⋱⋱aq1⋱⋱⋱⋱aq0⋱⋱⋱0⋱⋱⋮0aq0︸mq+sq=1cbqmq+sq+10⋯0bq0bq1⋯bq,sqbq⋱⋱⋱⋱⋱⋮⋱⋱⋱⋱⋱bq1⋱⋱⋱⋱bq0⋱⋱⋱0⋱⋱⋮0bq0︸mq+sqfor every root αq∈M such that αq∉M1 and mq<nq;(4)
(67)ar-aharah=1cbr-bhbrbhfor every pair of the roots αr,αh∈M such that αr,αh∉M1;(5)
(68)ap,mpap02+1aq=1cbp,mpbp02+1bqfor every pair of the roots αp,αq∈M such that αp∈M1, 2mp<np and αq∉M1.Proof.
Necessity. As we already know, from semiscalar equivalence of the matrices A(x), B(x) it follows that the entries a(x), b(x) of these matrices satisfy the congruence (15), where s11,s22≠0. Taking into account that(69)ax=ap+∑t=0s-mp-1aptx-αpmp+t,ap0≠0,(70)bx=bp+∑t=0s-mp-1bptx-αpmp+t,bp0≠0,we compare the coefficients of equal degrees of binomial x-αp on both sides of the congruence (15). Then we obtain(71)s22apt-s11bpt=0,t=0,1,…,lp-mp-1,lp=min2mp,np.
From this it follows thatap0=cbp0 and equality (64), where c=s11/s22, for all roots αp∈M1 such that 2mp≥np.
Taking into account the rootsαp∈M1 such that 2mp<np and comparing the coefficients of equal degrees of binomial x-αp on both sides of the congruence (15), we obtain the system of equalities, which in the matrix form can be written(72)ap0bp00ap1bp10⋮⋮⋮ap,mp-1bp,mp-10ap,mpbp,mpap0bp0ap,mp+1bp,mp+1ap0bp1+ap1bp0⋮⋮⋮ap,np-mp-1bp,np-mp-1∑t=0np-2mp-1aptbp,np-2mp-1-ts22-s11-s12=0-.
Sinces11,s22≠0, by Lemma 5 from the obtained equality we have that the condition (1) of Theorem is completely satisfied.
Letαk,αl∈M1 be the arbitrary pair of the roots such that 2mk<nk, 2ml<nl. For the coefficients ak,mk,al,ml,bk,mk,bl,ml of the compositions of types (69) and (70) of the entries a(x), b(x) into degrees of binomials x-αk, x-αl from the congruence (15) we can write the relations(73)s22ak,mk-s11bk,mk-s12ak0bk0=0,s22al,ml-s11bl,ml-s12al0bl0=0.
By exclusion from these equalitiess12 and considering that ak0=cbk0, al0=cbl0, we have equality (65). This proves the condition (2) of Theorem.
From the congruence (15) for the coefficients of the decomposition of types (69) and (70) of the entries a(x), b(x) into degrees of binomial x-αq for root αq∈M such that αq∉M1 and mq<nq can be written the system(74)s22aq-s11bq-s12aqbq=0,s22aq0-s11bq0-s12aq0bq+bq0aq=0,s22aq1-s11bq1-s12aq1bq+bq1aq=0,⋮s22aq,lq-mq-1-s11bq,lq-mq-1-s12aq,lq-mq-1bq+bq,lq-mq-1aq=0,where lq=min(2mq,nq). Excluding from this system s12, we have(75)s22aqbqbqt-s11bqaqaqt=0,or(76)aqtaq2=1cbqtbq2,t=0,1,…,lq-mq-1,where c=s11/s22. This implies the equality (66) for the roots αq∈M such that αq∉M1, mq<nq, and 2mq≥nq.
If2mq<nq, then from congruence (15) the system of equalities can be obtained, which can be written in the matrix form(77)FGHs22-s11-s12=0-,where (78)F=aqaq0aq1⋮aq,mq-1aq,mqaq,mq+1⋮aq,nq-mq-1,G=bqbq0bq1⋮bq,mq-1bq,mqbq,mq+1⋮bq,nq-mq-1,(79)H=aqbqaq0bq+aqbq0aq1bq+aqbq1⋮aq,mq-1bq+aqbq,mq-1aq,mqbq+aq0bq0+aqbq,mqaq,mq+1bq+aq0bq1+aq1bq0+aqbq,mq+1⋮aq,nq-mq-1bq+∑t+0nq-2mq-1aqtbq,nq-2mq-1-t+aqbq,nq-mq-1.
Sinces11,s22≠0, by Lemma 6 from the obtained equality, we have(80)1aqmq+sq-1aq00⋯0aq0aq1⋯aq,sqaq⋱⋱⋱⋱⋱⋮⋱⋱⋱⋱⋱aq1⋱⋱⋱⋱aq0⋱⋱⋱0⋱⋱⋮0ap0︸mq+sq=1bqmq+sq-1bq00⋯0bq0bq1⋯bq,sqbq⋱⋱⋱⋱⋱⋮⋱⋱⋱⋱⋱bq1⋱⋱⋱⋱bq0⋱⋱⋱0⋱⋱⋮0bp0︸mq+sq,where sq=0,1,…,nq-mq-1. Here for t=0,1,…,mq-1 the equalities (76) hold. Now we multiply the left side of the equality (80) by the left side of the equality (76) at t=0 and carry out analogous operation with the right sides. As a result we obtain the equality (66). This proves the condition (3) of Theorem entirely.
For the arbitrary pair of rootsαr∈Mr, αh∈Mh such that r,h≠1, r≠h, the congruence (15) implies(81)s22ar-s11br-s12arbr=0,s22ah-s11bh-s12ahbh=0.
Excludings12 it follows the relation (67). The condition (4) of Theorem is proved.
Letαp,αq∈M be the pair of the roots such that αp∈M1, 2mp<np and αq∉M1. For the coefficients ap0,ap,mp,bp0,bp,mp of the decomposition (69) and (70) and values αq=a(αq),bq=b(αq) from the congruence (15) the system of equalities(82)s22ap,mp-s11bp,mp-s12ap0bp0=0,s22aq-s11bq-s12aqbq=0can be obtained. From these equalities we exclude s12. Considering ap0=cbp0, we have relation (68). The necessity of the conditions of Theorem is completely proved.
Sufficiency. Let the conditions (1)–(5) of Theorem be satisfied. Consider the matrix equations(83)ap0bp00ap1bp10⋮⋮⋮ap,mp-1bp,mp-10x1x2x3=0-,(84)ap,mpbp,mpap0bp0ap,mp+1sp,mp+1ap0bp1+ap1bp0⋮⋮⋮ap,np-mp-1bp,np-mp-1∑t=0np-2mp-1aptbp,np-2mp-1-tx1x2x3=0-for the indeterminate vector-column x1x3x3T. They are written using the coefficients of the decomposition (69) and (70) for some root αp∈M1 such that mp<np. If 2mp≥np, then the condition (1) implies that (83) has nonzero solution 1-cs12T. Moreover this solution does not depend on the choice of the root αp∈M1 such that mp<np. If 2mi<ni, then by Lemma 5 both these equations (83) and (84) have the common nonzero solution. Evidently, those first rows of the matrices of these equations are linearly independent and any other of their rows are linear combination of these rows. Since(85)ap0bp00ap,mpbp,mpap0bp01-ccbp,mp-ap,mpap0bp0-1=0-,this means that (86)x1x3x3T=1-ccbp,mp-ap,mpap0bp0-1Tis the common nonzero solution of (83) and (84).
From the condition (2) we deduce the relation(87)cbk,mk-ak,mkak0bk0-1=cbl,ml-al,mlal0bl0-1for arbitrary pair of the roots αk,αl∈M1 such that 2mk<nk, 2ml<nl. This means that the solution (86) does not depend on the choice of the root αp∈M satisfying the condition (1). Hence it can be written the congruence(88)ax-cbx-c1axbx≡0modx-αpnp,where (89)c1=ap,mp-cbp,mpap0bp0-1for every root αp∈M1. Clearly, here are included also the roots αi∈M1 for which mi≥ni (if such exist).
Let the equality (66) hold true for some root αq∈M such that αq∉M1 and mq<nq. The fulfillment of these equalities for sq=0 means that caq0/aq2=bq0/bq2. Note that the relations (80) follow from the equality (66) for sq=1,…,nq-mq-1. Therefore by Lemma 6 the equation(90)FGHx1x2x3=0-,where F, G, and H are defined as in (78) and (79), has nonzero solution. Since the first two rows of the matrix of (90) are linearly independent and(91)aqbqaqbqaq0bq0aq0bq+aqbq01-ccbq-aqaqbq-1=0-,then the column(92)x1x3x3T=1-ccbq-aqaqbq-1Tis the solution of (90).
From the condition (4) it follows that the relation (93)cbr-ararbr-1=cbh-ahahbh-1holds true for every pair of the roots αr,αh∈M such that αr,αh∉M1. For this reason the solution (92) does not depend on the choice of the root αq∈M such that αq∉M1. This means that the congruence(94)ax-cbx-c2axbx≡0modx-αqnq,where (95)c2=aq-cbqaqbq-1,holds for every root αq∉M1, including the roots αj∉M1 such that mj≥nj, if such exist.
From the condition (5) of Theorem can be easily obtained the relation(96)ap,mp-cbp,mpap0bp0-1=aq-cbqaqbq-1for every pair of the roots αp,αq∈M such that αp∈M1, 2mp<np, and αq∉M1. This means that the coefficients c1 and c2 in the congruences (88) and (94) coincide. This makes it possible to write the congruence (15), where s11=c,s22=1,s12=c1=c2.
The conclusion of the proof of Theorem can be fulfilled in the same way as of the Theorem3. The Theorem is proved.



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*Source: 6701078-2017-10-26.xml*