# Canonical Form of Reduced 3-by-3 Matrix with One Characteristic Root and with Some Zero Subdiagonal Elements

Authors: B. Z. Shavarovskii
Journal: Journal of Mathematics
 (2019)

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## Abstract

A canonical form for a reduced matrix of order 3 with one characteristic root and with some zero subdiagonal elements is constructed. Thus, the problem of classification with respect to semiscalar equivalence of a selected set of polynomial matrices is solved.


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## Body

## 1. Introduction

Let a matrixF(x)∈M(n,C[x]) have a unit first invariant factor and only one (without taking into account multiplicity) characteristic root. Without loss of generality, we assume that this uniquely characteristic root is zero. Consider the transformation F(x)→PF(x)Q(x)=G(x), where P∈GL(n,C), Q(x)∈GL(n,C[x]). In accordance with [1] (see also [2]) matrices F(x), G(x) are called semiscalarly equivalent (abbreviation: ss.e.; notation: F(x)≈G(x)). In [3], the author proved that in the class {PF(x)Q(x)}, where F(x)∈M(3,C[x]), there exists a matrix(1)Ax=100a1xxk10a3xa2xxk2,which has the following properties:(i)
dega1(x)<k1, dega2(x),dega3(x)<k2, a2(x)=xk1a2′(x), a1(0)=a2′(0)=a3(0)=0 (see Proposition 1 [3]);(ii)
codega3(x)≠codega1(x),codega2′(x), if codega3(x)<codega2(x) (see Propositions 4, 5 [3]);(iii)
codega3(x)≠2codega1(x)+codega2′(x) and in a1(x) the monomial of the degree 2codega1(x) is absent, if codega3(x)≥codega2(x) (see Propositions 6, 7 [3]).Herecodeg denotesthe junior degree of polynomial.Junior degreeof polynomial f(x)∈C[x], f(x)≠0, is the least degree of the monomial (of nonzero coefficient) of this polynomial; notation codegf. The monomial of degree codegf and its coefficient are called thejunior term andjunior coefficient, respectively. Denote by symbol +∞ the junior degree of the polynomial f(x)≡0. If both elements a1(x), a2(x) of the matrix A(x) are nonzero, then we may take their junior coefficients to be identity elements. In the opposite case, we may take the junior coefficients of the nonzero subdiagonal elements of the matrix A(x) to be one. Such matrix A(x) in [3] is called thereduced matrix. The purpose of this paper is to construct the canonical form of the matrix F(x)∈M(3,C) in the class {PF(x)Q(x)} of ss.e. matrices. For this purpose we base our work on the reduced matrix A(x) of the form (1). This article is a continuation of the work [3]. The case a1(x)=a2(x)=a3(x)≡0 is trivial. Then the Smith form is canonical for the matrix F(x) in the class {PF(x)Q(x)}. If k1=k2, then a2(x)≡0. This case is considered in the author’s paper [4]. For this reason in the sequel we shall take k1≠k2. In this paper we consider the case, when some of the elements a1(x), a2(x), a3(x) of the matrix A(x) are equal to zero and at least one of them is different from zero. Recall that the zero equality of some subdiagonal elements of the matrix A(x) is an invariant (see Proposition 2 [3]). A case in which all subdiagonal elements of the matrix A(x) are nonzero will be the subject of another study. It should be noted that the problem of ss.e. matrices of the second order is solved in the article [5]. Some aspects of this problem for the matrices of the arbitrary order are considered in the papers [6–9]. We also add that the work close to [1, 2] is [10, 11].


## 2. The Canonical Form of a Reduced Matrix with Two Zero Subdiagonal Elements

To abbreviate in the sequel the expressions “monomial of degreeq” and “coefficient of the monomial of degree q” of some polynomial we will use “q-monomial” and “q-coefficient,” respectively. Further, by the symbol 0 we often denote, besides the zero element of the field C, the rectangular zero matrix if it is clear from the context what we have in mind. By 0- we denote zero column of the necessary height.Theorem 1.
Let in the reduced matrixA(x) of the form (1) the conditions ai(x)≠0, for some index i from set {1,2,3} and aj(x)≡0 for the rest j∈{1,2,3},j≠i, be fulfilled. Then A(x)≈B(x), where in the reduced matrix(2)Bx=100b1xxk10b3xb2xxk2,the element bi(x)≠0 does not contain ni-monomial,(3)ni=2codegai,i=1,3,2codega2′+k1,i=2,a2′(x)=a2(x)/xk1, bj(x)≡0.The matrixB(x)is uniquely defined.Proof.
Existence. Let the conditions of the Theorem hold true. If(4)ni≥k1,i=1,k2,i=2,3,then in ai(x) the ni-monomial is absent, since(5)degaix<k1,i=1,k2,i=2,3.Suppose that(6)ni<k1,i=1,k2,i=2,3.Denote ni-coefficient of the polynomial ai(x) by c. We will apply to the matrix A(x) a semiscalarly equivalent transformation (ss.e.t.) with left transforming matrix of one of the forms:(7)S=1s120010001,if i=1;(8)S=10001s23001,if i=2, or(9)S=10s13010001,if i=3. In doing so every time instead of s12, s23, or s13 in the transforming matrix S we put c, that is,(10)c=s12,i=1,s23,i=2,s13,i=3.In article [3], for the given matrices A(x) of the form (1) and S one of the form (7), (8), or (9) is a method for finding the matrix B(x) of the form (2) such that A(x)≈B(x). In doing so, we can find also the right transforming matrix. Depending on the form (7), (8), or (9) of the matrix S we shall say that the ss.e.t.-I, ss.e.t.-II, or ss.e.t.-III is applied to matrix A(x), respectively.
Ifi=1 and ss.e.t.-I is applied to matrix A(x), then the elements a1(x)≠0,a2(x),a3(x)≡0 and b1(x),b2(x),b3(x) of the matrices A(x) and B(x) satisfy the following congruences:(11)b1x-a1x1-s12b1x≡0modxk1,(12)b2x-a2x-s12a1xa2x-xk1a3x≡0modxk2,(13)b3x-a3x+s12a3xb1x-a2xr12x≡0modxk2,where r21(x)=b1(x)-a1(x)(1-s12b1(x))/xk1∈C[x]. From (12) and (13) we have that b2(x)=b3(x)≡0. From (11) we obtain that b1(x)≠0 and in b1(x) the n1-monomial is absent.
Ifi=2, then the matrix B(x) is obtained by means of ss.e.t-II. Then the elements of A(x) and B(x) fulfill the following congruences:(14)a1x+s23a3x-b1x≡0modxk1,(15)a2x-b2x1+s23a2′x≡0modxk2,(16)a3x-b3x-b2xr21x≡0modxk2,where r21(x)=a1(x)+s23a3(x)-b1(x))/xk1∈C[x]. From (14) and (16) it follows at once that b1(x)≡0, r21(x)≡0 and b3(x)≡0. From (15) we have that b2(x)≠0 and moreover b2(x) does not contain the n2-monomial.
Leti=3 and we pass from A(x) to B(x) with the help of ss.e.t.-III. From ss.e. of the matrices A(x) and B(x) it follows that their elements satisfy the congruences:(17)b1x+s13a3xb1x≡0modxk1,(18)b2x≡0modxk2,(19)a3x-b3x-s13a3xb3x-b2xr21x≡0modxk2,where r21(x)=-b1(x)(1+s13a3(x))/xk1∈C[x]. From (17) and (18) we obtain b1(x)=b2(x)≡0, r21(x)≡0. From (19) we have that b3(x)≠0 and in b3(x) the n2-monomial is absent. Evidently, the obtained matrix B(x) is reduced in each under consideration case to i=1,2,3. The first part of the theorem is proved.
Observe that the equalitybj(x)≡0, j≠i, each time follows from Corollaries 2, 3 [3].
Uniqueness. It suffices to prove for some of the cases: i=1, i=2, or i=3. The proof in two other cases is analogous. Assume in the reduced matrices A(x) and B(x) of the forms (1) and (2) we have ai(x),bi(x)≠0, aj(x)=bj(x)≡0, i≠j and in ai(x),bi(x) the ni-monomial is absent. If i=3, then from A(x)≈B(x) we obtain the congruence (19), where r21(x)≡0. Comparing the n3-coefficients in both sides of this congruence, we obtain s13=0. From this it follows that a3(x)=b3(x), because dega3,degb3<k2. The theorem is proved completely.


## 3. The Canonical Form of a Reduced Matrix with One Zero Subdiagonal Element

Let us now consider the ss.e.t. with the left transforming matrix of one of the next forms:(20)S=1s12001s23001,(21)S=1s12s13010001,(22)S=10s1301s23001Such transformations are called ss.e.t-I-II, ss.e.t-I-III, and ss.e.t-II-III, respectively.Leta1(x)=b1(x)≡0, a2(x),a3(x),b2(x),b3(x)≠0, a2(x)=xk1a2′(x), b2(x)=xk1b2′(x) in the reduced matricesA(x) and B(x) of the forms (1) and (2). We determine the polynomials a32(x) and a33(x) as follows:(23)a32x:≡a3xa2′xmodxk2,a33x:≡a3x2modxk2.Next, we form the columns a-3, a-32, and a-33 from the coefficients of the polynomials a3(x), a32(x), and a33(x), respectively. In doing so we place on the first positions in these columns the coefficients of the monomials of degree codega3. Below we arrange every succeeding coefficients in the order of growth of the degrees of the monomials, up to (k2-1)-coefficient. At the same time we do not omit zero coefficients. Take into consideration the polynomials(24)a22x:≡a2′x2modxk2-k1,a23x:≡a2′xa3xmodxk2-k1.With the coefficients of the polynomials a2′(x), a3(x), a22(x), and a23(x) we form the columns a-2, a-30, a-22 and a-23, respectively, of the height k2-codega2. Here we place on the first positions the coefficients of monomials of degree codega2-k1. Further we put all coefficients of the monomials of higher degrees up to degree k2-k1-1. Let us build for A(x) the matrices:(25)MA=a-3a-2M0A,M0A=M3AM2A,M3A=a-32a-330-,M2A=a-22a-23a-30Evidently, each column in these matrices is composed by the coefficients of the monomials of the same degrees. By complete analogy for B(x) we build the matrices:(26)MB=b-3b-2M0B,M0B=M3BM2B,M3B=b-32b-330-,M2B=b-22b-23b-30.Theorem 2.
Let A ( x ) be the reduced matrix of the form (1),in whicha1(x)≡0, a2(x),a3(x)≠0, a2(x)=xk1a2′(x),andq2:=codega2′, q3:=codega3.ThenA(x)≈B(x),where in the reduced matrixB(x)of the form (2) b1(x)≡0and the elementsb2(x),b3(x)≠0satisfy one of the next conditions.(1)
Inb3(x) the (2q3)-monomial is absent, if q3<q2, q3<k1.(2)
Inb3(x) the (2q3)- and (q2+q3)-monomials are absent, if q3<q2, q3≥k1.(3)
Inb3(x) and b2′(x), where b2′(x)=b2(x)/xk1∈C[x], respectively, the (2q3)- and q3-monomials are absent, if q3>q2, q3<k1.(4)
In the first column of the matrixMB (26) with the coefficients of the polynomials b3(x), b2′(x) are zero elements, which correspond to the maximal system of the first linearly independent rows of the submatrix M0B, if q3>q2, q3≥k1.
The matrixB(x) is uniquely defined.Proof.
Existence. (1) Let q3<q2, q3<k1, and c1 identify (2q3)-coefficient of the polynomial a3(x) in the matrix A(x). Let us apply to the matrix A(x) ss.e.t.-III. In so doing, we set s13=c1 in the left transforming matrix S (see (9)). In the obtained reduced matrix B(x) the element b3(x) satisfies congruence (19), where r21(x)≡0. From (19) we deduce that for b3(x) the condition (1) of the theorem is fulfilled.
(2) Letq3<q2, q3≥k1. We may take that the element a3(x) of the matrix A(x) does not contain (2q3)-monomial. In the opposite case we apply to the matrix A(x) ss.e.t.-III, as has been shown above in p. (1). Let us denote (q2+q3)-coefficient of the polynomial a3(x) by c2 and let us apply to the matrix A(x) ss.e.t.-II. At that, we set in left transforming matrix (see (8)) s23=c2. For the elements b2(x), b3(x) of the obtained reduced matrix B(x) the congruence (16) holds true, where a1(x)=b1(x)≡0. For this reason it may be written in the form(27)a3x-b3x-s23a3xb2′x≡0modxk2.From the above it follows that in b3(x) the (q2+q3)-coefficient is zero. Since q3<q2, then 2q3<q2+q3. Therefore, from the last congruence we see that in b3(x), as in a3(x), the (2q3)-monomial is absent.
(3) Letq3>q2, q3<k1. We may take that in a3(x) the (2q3)-monomial is absent. We can always do this as has been shown above in p. (1). Let us denote by c3 the q3-coefficient in a2′(x). We subject A(x) to action of ss.e.t-I. In this case we set in the left transforming matrix (see (7)) s12=c3. As a result we obtain the reduced matrix B(x) of the form (2), in which b3(x)=a3(x) and the congruence(28)a2′x≡b2′x+s12a3xmodxk2-k1.holds true. The last congruence implies that in b2′(x) the q3-monomial is absent. Then B(x) is the matrix to be found.
(4) Letq3>q2, q3≥k1. If M0A=0 in the matrix MA (see (25)), then everything is proved; that is A(x) is the matrix to be found. If the matrix M0A is nonzero, but its block M3A is zero, then the element a3(x) in A(x) has the required property. In this case the second column of the matrix M0A is zero. Let the first nonzero row d110d13 of the matrix M0A and the corresponding row d1d110d13, d1≠0, of the matrix MA be formed of m1-coefficients. Let us find some solution x10x30t of the equation d11d13x10x30t=d1 and use ss.e.t.-I-II to A(x). Herewith we set in left transforming matrix S of the form (20) s12=x30, s23=x10. The divisor b2′(x) of the element b2(x), b2(x)=xk1b2′(x), of the obtained reduced matrix B(x) satisfies the congruence(29)a2′x-s23a2′xb2′x-s12a3x-b2′x≡0modxk2-k1.From the last one it follows that in b2′(x) the m1-monomial is absent and b2′(x)≡a2′(x)(modxm1).
IfrankM0A=1, then everything is proved; that is B(x) is the matrix to be found. In the opposite case we take that already for A(x) the polynomial a2′(x) does not contain m1-monomial. Let the row d210d23 of the matrix M0A be the first linearly independent row of d110d13 and d2d210d23 is the respective row of the matrix MA. Let these rows be formed of the m2-coefficients. We find the (unique) solution y10y30t of the equation(30)d11d13d21d23y1y3=0d2.By ss.e.t-I-II we pass from A(x) to the reduced matrix B(x). Here in the matrix S (see (20)) we set s12=y30, s23=y10. The divisor b2′(x) of the element b2(x) in B(x), as is seen from the congruence (29), does not contain m1- and m2-coefficients.
If the matrixM3A is nonzero, then the row u-1=101 of (q2+q3)-coefficients is its first nonzero row. At the firststep_ we apply to A(x) ss.e.t-II. In addition we set in the left transforming matrix S of the form (8) in place of s23 the coefficient of the monomial of degree q2+q3 in the polynomial a3(x) (see p. (2)). In the obtained reduced matrix the polynomial in the position (3,1) does not contain the (q2+q3)-monomial. If 2q3<k2, then we consider the row of the form u-2=∗10 formed of the (2q3)-coefficients in the matrix, analogous to M3A. This row is the first linearly independent of u-1. We do the secondstep_. This is ss.e.t-III of the obtained matrix after the first step. For this purpose we set (2q3)-coefficient of the polynomial in the position (3,1) of the obtained matrix after the first step instead s13 in the left transforming matrix of the form (9) (see p. (1)). In the obtained matrix after the second step the polynomial in position (3,1) does not contain (q2+q3)- and (2q3)-monomials. In order not to introduce the new notations, we will consider that in the matrix A(x) its element a3(x) possesses this property. If q3<k2-k1, then in the submatrix M2A of the matrix M0A the row u-3=∗01 of the q3-coefficients is the first linearly independent row with the collection u-1, u-2. Then we make the thirdstep_. It is ss.e.t.-I of the matrix A(x). For this purpose in the left transforming matrix of the form (7) instead of s12 we set q3-coefficient in a2′(x) (see p. (3)). After this step we obtain the matrix in which the element in the position (3,1) is not changing and the polynomial in the position (3,2) does not contain (q3+k1)-monomial. We obtain the required matrix.
If after the first step it turns out that2q3≥k2, that is, rankM3A=1 but q3<k2-k1, then we immediately take the third step. Next, if q2+q3<k2-k1, then the row ∗1∗ of the (q2+q3)-coefficients is the first linearly independent row with the collection u-1, u-3 in the submatrix, analogous to M0A. Then we take the fourthstep_. It is ss.e.t.-III with a left transforming matrix (see (9)), in which instead of s13 is (q2+q3+k1)-coefficients of the polynomial in the position (3,2) of the matrix obtained after the third step. After that, the element in the position (3,1) of the resulting matrix will not change, and in position (3,2) the desired element will stand. The obtained matrix is the required one.
Uniqueness. (1) Let the matrices A(x), B(x) of the form (1), (2) satisfy the condition (1) and A(x)≈B(x). Then the congruence (19), where r(x)21≡0, and the congruence(31)a2′x≡b2′x+s13a2′xb3x≡0modxk2-k1are fulfilled (see Corollary 1 and Remark [3]). If 2q3≥k2, then q2+q3≥k2-k1 and from (19) we have a3(x)=b3(x), and from (31) we will obtain a2′(x)=b2′(x). If 2q3<k2, then comparing (2q3)-coefficients in both sides (19), we get s13=0. Therefore, in each case A(x) and B(x) coincide.
(2) Let condition (2) for the matricesA(x) and B(x) of the form (1) and (2) be fulfilled. Then on the basis of Corollary 1 and Remark [3] on A(x)≈B(x) we can write the congruences:(32)a2′x≡b2′x+s23a2′xb2′x+s13a2′xb3x≡0modxk2-k1,(33)a3x≡b3x+s23a3xb2′x+s13a3xb3xmodxk2.It should be noted that 2q3<q2+q3<codega2+q3<codega2+q2. If 2q3≥k2, then q2+q3≥k2-k1. Then from (32) and (33) we have a2′(x)=b2′(x) and a3(x)=b3(x), respectively. In the case of 2q3<k2 by comparison, in both parts (33) of the (2q3)-coefficients we arrive at s13=0. If q2+q3≥k2, then from (32) and (33), as before, we have a2′(x)=b2′(x) and a3(x)=b3(x), respectively. If q2+q3<k2, then by comparing the (q2+q3)-coefficients in both parts (33), we have s23=0. Thus, in this case A(x)=B(x) too.
(3) For the reduced matricesA(x), B(x) of the form (1), (2) such that A(x)≈B(x), in the case of satisfying the condition (3) we have congruence (19) where r(x)21≡0, and congruence(34)a2′x≡b2′x+s12b3x+s13a2′xb3xmodxk2-k1.First let us note that 2q3<q3+k1<codega2+q3. If 2q3≥k2, then q3≥k2-k1. Then from (34) and (19) we have a2′(x)=b2′(x) and a3(x)=b3(x), respectively. If 2q3<k2, then by comparing the (2q3)-coefficients from (19) we have s13=0 and a3(x)=b3(x). If further q3≥k2-k1, then from (34) a2′(x)=b2′(x) immediately follows. In the case q3<k2-k1 we compare q3-coefficients in both parts (34) and come to s12=0. So A(x)=B(x).
(4) Suppose that for the reduced matricesA(x), B(x) of the form (1), (2) which satisfy condition (4) we have A(x)≈B(x). For their elements, taking into account Remark [3], we can write the congruence (33) and congruence(35)a2′x≡b2′x+s12b3x+s13a2′xb3x+s23a2′xb2′xmodxk2-k1.If q2+q3≥k2 and m:=min(q3,2q2)≥k2-k1, then 2q3≥k2 and q2+q3≥k2-k1. Then the matrices M0A, M0B are zero and, as can be seen, from (33) and (35), we have a3(x)=b3(x) and a2′(x)=b2′(x). Let q2+q3≥k2 but m<k2-k1. Then, in M0A, M0B the submatrices M3A, M3B and the second columns are zero, and from (33), (35) we have a3(x)=b3(x), a2′(x)≡b2′(x)(modxm), respectively. Thus, in the columns a-2, b-2 of the matrices MA, MB the corresponding first l=m-q2 elements coincide and the first l rows in the matrices M2A, M2B are zero. The first nonzero rows in these matrices are their (l+1)-th rows. They are equal among themselves. We will denote them by u110u13. The corresponding elements in the columns a-2, b-2 are zero (see (8)). Therefore, we have a2′(x)≡b2′(x)(modxm+1). This means that the following (l+2)-th rows in M2A, M2B coincide. We will denote them by u210u23. From (35) it is clear that u11u13s23s12t=0. If u11u13, u21u23 are linearly independent, then the corresponding elements in a-2, b-2 are zero. On this basis, from (35) we have u21u23s23s12t=0. Therefore, s23=s12=0 and all is proved. If u11u13, u21u23 are linearly dependent, then u21u23s23s12t=0, and this, as can be seen from (35), means that in the columns a-2, b-2 their (l+2)-th elements coincide. Therefore, the following (l+3)-th rows in the matrices M2A, M2B coincide. Let us denote them by u310u33 and consider the cases where the rows u11u13,u31u33 are linearly independent and when they are linearly dependent. In the first case, as above, we will get s23=s12=0. Then everything is proved. In the second case we obtain that the (l+3)-th elements in a-2, b-2 coincide. Therefore, the (l+4)-th rows in M2A, M2B coincide. Continuing similar consideration, at some step we will get s23=s12=0. This on the basis of (35) means a2′(x)≡b2′(x)(modxk2-k1), that is a2′(x)=b2′(x), or we will come to a-2=b-2. In any case we will have A(x)=B(x).
Letq2+q3<k2. Then in the columns a-3, b-3, as is seen from (33), the first l1=q2 elements coincide, and in the matrices M3A, M3B the first l1 rows are zero. The first nonzero row in each of the matrices M3A, M3B is their (l1+1)-th row, which has the form u-1=100. Since (l1+1)-th elements in the columns a-3, b-3 are zero, then as is clear from (33), s23=0. This means, that the corresponding first l2=q3, l2>l1, elements in the columns a-3, b-3 coincide. If 2q3<k2, then the first linearly independent with the row u-1 in each of the matrices M3A, M3B is the same (l2+1)-th row u-2=∗10. Since (l2+1)-th elements in a-3, b-3 are zero, then from (33) we have s13=0. That means that a3(x)=b3(x). As can be seen from (35), in the columns a-2, b-2 the corresponding first l3=q3-q2 elements coincide. Then in the case q3≥k2-k1 we have a2(x)=b2(x); that is, everything has already been proved. If q3<k2-k1, then in each of the matrices M2A, M2B, the first l3 rows are linearly dependent on the collection u-1, u-2, and their (l3+1)-th row u-3=∗01 is the first linearly independent of u-1, u-2 in these matrices. Since in the columns a-2, b-2 their (l3+1)-th elements are zero, then from (35) we obtain s12=0. Then a-2=b-2, a2(x)=b2(x) and everything is proved.
If2q3≥k2, then from (33) we have a3(x)=b3(x). Recall that in (33), (35) s23=0. Also s12=0, if q3<k2-k1. Therefore, if q2+q3≥k2-k1, then everything has been proved. Otherwise, if q2+q3<k2-k1, then in the columns a-2, b-2 the corresponding first l2=q3 elements coincide and in each of the matrices M2A, M2B their (l2+1)-th row ∗1∗ is first linearly independent of the collection u-1, u-3. Since (l2+1)-th elements in a-2, b-2 are zero, then from (35) we get s13=0. That is, every time A(x)=B(x). The theorem is proved.Let in reduced matricesA(x),B(x) of the form (1), (2) a2(x)=b2(x)≡0, a1(x),a3(x),b1(x),b3(x)≠0. Define the polynomials a31(x) and a33(x) as follows:(36)a31x:≡a3xa1xmodxk2,a33x:≡a3x2modxk2.We will construct columns a-3, a-31, and a-33 of height k2-codega3 from the coefficients of the polynomials a3(x), a31(x), and a33(x), respectively. In these columns, in the first place, we put the coefficients of the monomials of degree codega3. Then, in the order of increasing powers of monomials, we place the rest of the coefficients, together with zero ones up to (k2-1)-coefficients. We create columns a-1, a-03 and a-13a-11 of height k1-codega1 from the coefficients of polynomials a1(x), a3(x) and from the coefficients of polynomials:(37)a13x:≡a1xa3xmodxk1,a11x:≡a1x2modxk1,respectively. Here, in the first place, we put the coefficients of the monomials of degree codega1, and then in the order of increasing powers of monomials we place the remaining coefficients up to the monomials of degree k1-1 inclusive. For A(x) we construct matrices of the form(38)NA=a-3a-1N0A,N0A=N3AN1A,N3A=0-a-33a-31,N1A=a-03a-13a-11.Obviously, each row in these matrices consists of the coefficient of the monomials of the same degree.Quite similarly forB(x) we construct matrices:(39)NB=b-3b-1N0B,N0B=N3BN1B,N3B=0-b-33b-31,N1B=b-03b-13b-11.Theorem 3.
Let in the reduced matrix A ( x ) of the form (1) a2(x)≡0, a1(x),a3(x)≠0andq1:=codega1, q3:=codega3.ThenA(x)≈B(x),where in thereduced matrixB(x)of the form (2) b2(x)≡0and the elementsb1(x),b3(x)≠0satisfy one of the following conditions.(1)
Inb3(x) the (2q3)-monomial is absent, if q3<q1, q3<k2-k1.(2)
Inb3(x) the (2q3)- and the (q1+q3)-monomials are absent, if q3<q1, q3≥k2-k1.(3)
Inb1(x) the q3-monomial is absent and in the first of the polynomials bix,i=1,3, for which qi+q3<ki, the (qi+q3)-monomial is absent, if q3>q1, q3<k2-k1.(4)
In the first column of the matrixNB (39) with the coefficients of the polynomials b3(x), b1(x) there are zero elements corresponding to the maximal system of the first linearly independent rows of the submatrix N0B, if q3>q1, q3≥k2-k1.
The matrixB(x) is uniquely defined.Proof.
Existence. (1) The proof is completely analogous to the proof of condition (1) in Theorem 2.
(2) We can assume that in the matrixA(x) element a3(x) does not contain (2q3)-monomial. Otherwise, we act as in p. (1) of Theorem 2. If q1+q3≥k2, then everything is proved. Otherwise, we denote the (nonzero) coefficient of the (q1+q3)-monomial in a3(x) by c1 and apply to A(x) ss.e.t.-I. In this case, in the left transforming matrix (see (7)) we put s12=c1. Elements b1(x), b3(x) of the resulting reduced matrix B(x) satisfy the congruence(40)a3x≡b3x+s12a1xb3xmodxk2,from which it is seen that in b3(x), there is no (q1+q3)-monomial. Since q3<q1, then in b3(x), as in a3(x), there is no (2q3)-monomial.
(3) Letq3>q1 and q3<k2-k1. If q3≥k1, then in a1(x) there is no q3-monomial, since dega1<k1. Otherwise, we denote the (nonzero) coefficient of the q3-monomial in a1(x) by c2 and apply to A(x) ss.e.t.-II. Thus, in the left transforming matrix of the form (8) we put s23=-c2. In the resulting reduced matrix B(x) element b1(x) satisfy the congruence (14) and b3(x)=a3(x). From (14) it is clear that in b1(x) there is no q3-monomial. In order not to introduce new notation, we assume that already in A(x) the polynomial a1(x) does not contain a q3-monomial. If qi+q3≥ki,i=1,3, then everything has already been proved. Then the matrix A(x) is the desired one. Otherwise, we find the first of two values i=1, i=3, for what qi+q3<ki. Denote by c3 the coefficient of the (qi+q3)-monomial in the polynomial ai(x) for this value i (fixed now). Apply to A(x) ss.e.t.-III. Here in the left transforming matrix of the form (9) we put s13=c3. As a result in the reduced matrix B(x) the element bi(x) (for the above-defined index i) satisfies the congruence(41)aix≡bix+s13a3xbixmodxki.From this congruence it follows that in bi(x) there is no (qi+q3)-monomial. If i=1, then from the same congruence it is also evident that in b1(x), as in a1(x), there is no q3-monomial. The same thing is proved by the congruence(42)a1x≡b1x+s13a3xb1xmodxk1,if i=3.
(4) IfN0A=0, then everything has already been proved. Then the matrix A(x) is the desired one. If N0A≠0, but N3A=0, then the element a3(x) in A(x) satisfies condition (4). This element will not change with all subsequent ss.e.t. Then q1+q3≥k2. This means that in N1A the second column is zero. Let d110d13 be the first nonzero row of the matrix N1A and d1d110d13 is the corresponding row of the matrix NA. Let these rows be composed of the coefficients of the monomials of degree m1. We find some solution x10x30t of the equation -d11d13x1x3t=d1 and apply to A(x) ss.e.t.-I-II. In this case, we put s12=x30, s23=x10, in the left transforming matrix S (see (20)). The element b3(x) of the resulting reduced matrix B(x) coincides with a3(x), and the element b1(x) satisfies the congruence(43)a1x+s23a3x≡b1x+s12a1xb1xmodxk1.It follows from this that in b1(x) there is no m1-monomial and a1(x)≡b1(x)(modxm1). If rankN0A=1, then all subsequent rows in N1A are linearly dependent on d110d13, then the matrix B(x) is to be sought. In order not to introduce new notations, we assume that the element a1(x) in A(x) does not contain m1-monomial. Let d210d23 be the first linearly independent row from the row d110d13 of the matrix N0A and let d2d210d23 be the corresponding row of the matrix NA. Let these rows be composed of the coefficients of the monomials of degree m2>m1. From equation(44)-d11d13-d21d31y1y3=0d2we find its (unique) solution y10y30t≠0. Using ss.e.t.-I-II we pass from A(x) to the reduced matrix B(x). Here in the left transforming matrix S of the form (20) we assume s12=y30, s23=y10. The element b1(x) in B(x), as is shown in (43), does not contain m1- and m2-monomials and b3(x)=a3(x). Therefore, B(x) is the desired matrix.
Next let us consider the situation whenN3A≠0. Then, n1:=q1+q3<k2 and the row u-1=001 of the n1-coefficients is the first nonzero row of the matrix N3A. We do the firststep_. It is ss.e.t.-I of the matrix A(x), in which we put the n1-coefficient of polynomial a3(x) instead of s12 in the left transforming matrix (see (7)). In the obtained reduced matrix B(x), element b3(x) does not contain a n1-monomial (see p. (2)). If n2:=2q3<k2, then the first linearly independent row with row u-1 in N3B is a row u-2=01∗ of n2-coefficients, n2>n1. We do the secondstep_. This is ss.e.t.-III of the matrix B(x), in which in the left transforming matrix of the form (9) we assume the n2-coefficient of polynomial b3(x) instead of s13. In the matrix obtained after the second step, the element in position (3,1) does not contain n1- and n2-monomials. In order not to introduce new notations for the matrices resulting from transformations, we assume that the element a3(x) in A(x) has this property. If n3:=q3<k1, then in the matrix N0A the row u-3=10∗ of the n3-coefficients is the first linearly independent row with the collection u-1, u-2. To A(x), apply ss.e.t.-II. In this case, in place of s23 in the left transforming matrix (see (8)) we put the n3-coefficient of the polynomial a1(x) taken with the opposite sign. This will be the thirdstep_. In the resulting matrix, the element in position (2,1) does not contain a n3-coefficient. Since the element in position (3,1) does not change at the same time, such a matrix is the desired one.
If after the first step it turns out thatn2≥k2, but n3<k1, then we do the third step. Further, if n4:=q1+q3<k1, then the row ∗1∗ of the n4-coefficients in the matrix analogous to N0A is the first linearly independent with the collection u-1, u-3. We do the fourthstep_. It will be the ss.e.t-III of the matrix obtained after the third step. In this case, in the left transforming matrix (see (9)) instead of s13, we put the n4-coefficient of the polynomial in the position (2,1). As a result, the element in the position (3,1) will not change, and (2,1)-element will be free from n3- and n4-monomials. The resulting matrix is the required one.
Uniqueness. (1) Let A(x)≈B(x) and the elements of the reduced matrices A(x), B(x) of the form (1), (2) satisfy condition (1) of the theorem. Then, taking into account Corollary 1 and Remark [3], we can write the congruences:(45)a1x-b1x-s12a1xb1x-s13a3xb1x≡0modxk1,(46)a3x-b3x-s12a1xb3x-s13a3xb3x≡0modxk2.Since s12xk1b3(x)≡0(modxk2), then s12=0. If 2q3≥k2, then q1+q3>k2. Then from (45) and (46) we have a1(x)=b1(x) and a3(x)=b3(x), respectively. Otherwise, when 2q3<k2, comparing the (2q3)-coefficients in both parts of the congruence (46), we arrive at s13=0. Therefore, in each case, A(x)=B(x).
(2) Let condition (2) of the theorem be satisfied for the reduced matricesA(x), B(x) of the form (1), (2) and A(x)≈B(x). Then we can write (45) and (46). If 2q3≥k2, then q1+q3>k2 and 2q1≥k1. Then from (45) and (46) we have a1(x)=b1(x) and a3(x)=b3(x), respectively. Otherwise, when 2q3<k2, from (46) we obtain s13=0, since q1+q3>2q3 and in a3(x), b3(x) there are no (2q3)-monomials. If q1+q3≥k2, then again from (45) and (46) we obtain the coincidence a1(x), a3(x) from b1(x), b3(x), respectively. If q1+q3<k2, then comparing the (q1+q3)-coefficients in both parts (46) we obtain s12=0. It means A(x)=B(x).
(3) Let the condition (3) of the theorem be satisfied in the reduced ss.e. matricesA(x), B(x) of the form (1), (2). Then for the elements of these matrices we have the fulfillment of the congruence (46), where s12=0, and the congruence(47)a1x+s23a3x-b1x-s13a3xb1x≡0modxk1.If q3<k1, then from (47) we have s23=0. In any case, the second term in (47) can be omitted. Then (46) (where s12=0) and (47) can be presented in the form(48)aix-bix-s13a3xbix≡0modxki,i=1,3If qi+q3≥ki for each i=1,3, then immediately we have ai(x)=bi(x). Otherwise, if we compare the (qi+q3)-coefficients in both parts (48) for the first of two values of the index i such that qi+q3<ki, then we obtain s13=0. In any case A(x)=B(x).
(4) Let the reduced matricesA(x),B(x) of the form (1), (2) be ss.e. and let them satisfy condition (4) of the theorem. Then for their elements it is possible to write the congruence (46) and congruence(49)a1x+s23a3x-b1x-s12a1xb1x-s13a3xb1x≡0modxk1.If q1+q3≥k2 and m=min(q3,2q1)≥k1, then in matrices NA,NB we have N0A=N0B=0 and, as can be seen from (46), (49), a3(x)=b3(x),a1(x)=b1(x).
Ifq1+q3≥k2 and m<k1, then in matrices N0A,N0B we have N3A=N3B=0 and the second columns in N1A,N1B are zero. In this case, from (46) and (49) we have a3(x)=b3(x) and a1(x)≡b1(x)(modxm), respectively. Therefore, in subcolumns a-1,b-1 of the matrices NA,NB the first l=m-q1 elements coincide and the corresponding l rows in the matrices N1A,N1B are zero. In these matrices their (l+1)-th rows are the first nonzero rows. They are the same. We will denote them by v-1=v110v13. The elements in a-1,b-1 corresponding to these rows are zero (see (8)). So we really have a1(x)≡b1(x)(modxm+1). This means that the (l+2)-th rows in the matrices N1A,N1B coincide. We will denote them by v-2=v210v23. From (49) it is clear that v11v13s23s13t=0. If v-1,v-2 are linearly independent, then in a-1,b-1 there are zero elements corresponding to v-2. Therefore, on the basis of (49) we have v21v23s23s13t=0. Hence s23=s12=0 and everything is proved. If v-1,v-2 are linearly dependent, then v21v23s23s13t=0, and this on the basis of (49) means that the (l+2)-th elements in a-1,b-1 coincide. Therefore, the following (l+3)-th rows in N1A,N1B coincide. Denote them as v-3=v310v33 and again consider two situations: when v-1,v-3 are linearly independent and when they are linearly dependent. In the first case, we will have that s23=s12=0. Then the proof is finished. In the second case we have that the (l+3)-th elements in a-1,b-1 coincide. This means that in N1A,N1B the (l+4)-th rows coincide. Continuing similarly to our considerations, at some point we will obtain s23=s12=0 or eventually we will have a-1=b-1. In any case A(x)=B(x).
Let us now considerq1+q3<k2. Then in columns a-3,b-3, as can be seen from (46), their first l1=q1 corresponding elements coincide, and in matrices N3A,N3B their first l1 rows are zero. The first nonzero row in these matrices N3A,N3B is a row u-1=001. Since the (l1+1)-th elements in a-3,b-3 are zero (see (8)), then from (46) s12=0 follows. This means that a3(x)≡b3(x)(modx2l2), where l2=q3. If 2l2<k2, then in a-3,b-3 the first l2 corresponding elements coincide. Therefore, in N3A,N3B the first l2 corresponding rows coincide, and each of them is linearly dependent with u-1. The first linearly independent row with u-1 in each of the matrix N3A,N3B is their (l2+1)-th row u-2=01∗. Since (l2+1)-th corresponding elements in a-3,b-3 are zero, it follows from (46) that s13=0 and a3(x)=b3(x). If q3≥k1, then a1(x)=b1(x) and all is proved. In another case, as can be seen from (49), in a-1,b-1, the first l3=q3-q1 corresponding elements coincide, and in each of the matrix N1A,N1B each of their first l3 rows is linearly dependent on the collection u-1,u-2. The first linearly independent row with the collection u-1,u-2 in each of the matrices N1A,N1B is one and the same (l3+1)-th row u-3=10∗. Since (l3+1)-th elements in a-1,b-1 are zero, then from (49) s23=0 follows. Then a1(x)=b1(x).
Let2l2≥k2, and, as before, q1+q3<k2. Then from (46), where s12=0, we have at once a3(x)=b3(x). If q1+q3≥k1, then from (49) a1(x)=b1(x) follows. If q1+q3<k1, then in a-1,b-1 the corresponding first l2 elements and in N1A,N1B the corresponding first l2 rows coincide. Each of the these rows is linearly dependent on the collection u-1, u-3. In each of the matrices N1A,N1B their (l2+1)-th rows ∗1∗ are the first linearly independent rows with the collection u-1,u-3. Since (l2+1)-th elements in a-1,b-1 are zero, then from (49), where s23=0, we will have s13=0. In any case A(x)=B(x). The theorem is proved.Theorem 4.
Let the elements of the reduced matrixA(x) of the form (1) satisfy conditions: a1(x),a2(x)≠0, a3(x)≡0, a2(x)=xk1a2′(x), and q1:=codega1, q2:=codega2′. Then, A(x)≈B(x), where the elements of the reduced matrix B(x) of the form (2) satisfy conditions: b1(x),b2(x)≠0, b3(x)≡0, b2(x)=xk1b2′(x), and in b2(x) there are no (2q2+k1)- and (2q2+q1+k1)-monomials. The matrix B(x) is uniquely determined.Proof.
Existence. If 2q2-k1≥k2, then everything has been already proved. Then B(x)=A(x). Otherwise, apply to A(x) ss.e.t.-II. In this case, instead of s23 in the left transforming matrix (see (8)) we assume the (2q2+k1)-coefficient in a2(x). Then we obtain a matrix B(x) of the form (2), which will also be reduced, and its elements b1(x)=a1(x), b2(x)≠0, b3(x)≡0, satisfy the congruence (15). From the latter it can be seen that in b2(x) there is no (2q2+k1)-monomial and codega2=codegb2. If in the obtained matrix B(x) we have 2q2+q1+k1≥k2, then everything has been already proved. Then the matrix B(x) is the desired one. Otherwise apply to B(x) ss.e.t.-III. To do this, in the left transforming matrix (see (9)) instead of s13 we put the (2q2+q1+k1)-coefficient in b2(x), taken with opposite sign. The resulting matrix will be the sought one.
Uniqueness. Let A(x)≈B(x), where A(x), B(x) are reduced matrices of the form (1), (2), in which a3(x)=b3(x)=0, a1(x),a2(x),b1(x),b2(x)≠0, and in each of a2(x), b2(x) there are no (2q2+k1)- and (2q2+q1+k1)-monomials. For elements of the matrices A(x), B(x) a congruence (11) and congruences(50)b2′xa1x-b1x-s12a1xb1x≡0modxk2,(51)a2x-b2x-s23a2′xb2x+s12b1xb2x+s13b1xb2xa2′x≡0modxk2.are fulfilled. We recall that according to the definition of the reduced matrix in a1(x),b1(x) there are no (2q1)-monomials. If 2q1<k1, then from (11) we obtain s12=0. The same is true from (50), if 2q1≥k1, but 2q1+q2<k2. Then a1(x)=b1(x). If 2q2+k1<k2, then from (51) we have s23=0. Similarly, if 2q2+q1+k1<k2, then from (51) we obtain s13=0. Then a2(x)=b2(x). Note that from 2q1+q2≥k2 follows q1+q2+k1≥k2. Therefore, regardless of whether 2q2+k1≥k2 or 2q2+q1+k1≥k2, the coincidence a2(x) and b2(x) will be obtained from (51). The theorem is proved.


## 4. Conclusions

The matrixB(x), established by each of Theorems 1–4, can be considered canonical for the class {PF(x)Q(x)} of ss.e. matrices. It can be applied to the classification of sets of numerical matrices (over the field C) with respect to simultaneous similarity. In this context, the work of [12–15] should be noted. From the proof of Theorems 1–4 we can construct an algorithm for finding the transforming matrices (left nonsingular numerical and right invertible polynomial) that reduce A(x) to the canonical matrix B(x).



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