Hindawi International Journal of Analysis Volume 2017, Article ID 6701078, 14 pages https://doi.org/10.1155/2017/6701078
Research Article Toeplitz Matrices in the Problem of Semiscalar Equivalence of Second-Order Polynomial Matrices B. Z. Shavarovskii Department of Algebra, Pidstryhach Institute for Applied Problems of Mechanics and Mathematics, National Academy of Sciences of Ukraine, Lviv 79060, Ukraine Correspondence should be addressed to B. Z. Shavarovskii;
[email protected] Received 21 June 2017; Accepted 24 September 2017; Published 26 October 2017 Academic Editor: Shwetabh Srivastava Copyright © 2017 B. Z. Shavarovskii. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. 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.
1. Introduction Let C 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 𝐹 (𝑥) = 𝑃𝐺 (𝑥) 𝑄 (𝑥) ,
(1)
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 Let 𝐹(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) = 𝑐 det 𝐺(𝑥), 𝑐 ∈ C, 𝑐 ≠ 0. If the matrices 𝐹(𝑥) and 𝐺(𝑥) are of full rank, then
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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 1 0 , 𝐴 (𝑥) = 𝑎 (𝑥) 𝛿 (𝑥) 1 0 (2) , 𝐵 (𝑥) = 𝑏 (𝑥) 𝛿 (𝑥) deg 𝑎 (𝑥) , deg 𝑏 (𝑥) < deg 𝛿 (𝑥) . Denote by 𝑎(𝑡) (𝛼) and 𝑏(𝑡) (𝛼) the values at 𝑥 = 𝛼 of the 𝑡th 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)
V=1
of 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 matrices 𝐴(𝑥), 𝐵(𝑥) of the form (2) be semiscalarly equivalent; that is, the equality 𝑠 𝑠 1 0 11 12 𝑠 𝑠 𝑎 (𝑥) 𝛿 (𝑥) 21 22 (4) 1 0 𝑟11 (𝑥) 𝑟12 (𝑥) , = 𝑏 (𝑥) 𝛿 (𝑥) 𝑟21 (𝑥) 𝑟22 (𝑥) 𝑟11 (𝑥) 𝑟12 (𝑥) 𝑠11 𝑠12 where 𝑠21 𝑠22 ∈ 𝐺𝐿(2, C), 𝑟21 (𝑥) 𝑟22 (𝑥) ∈ 𝐺𝐿(2, C[𝑥]), holds. We deduce from (4) the relations 𝑠11 + 𝑠12 𝑎 (𝑥) = 𝑟11 (𝑥) ,
(5)
𝑠12 𝛿 (𝑥) = 𝑟12 (𝑥) ,
(6)
𝑠21 + 𝑠22 𝑎 (𝑥) = 𝑏 (𝑥) 𝑟11 (𝑥) + 𝛿 (𝑥) 𝑟21 (𝑥) .
(7)
Setting 𝑥 = 𝛼 and 𝑥 = 𝛽, 𝛼 ≠ 𝛽, 𝛼, 𝛽 ∈ M, we obtain the relations 𝑠21 + 𝑠22 𝑎 (𝛼) = 𝑏 (𝛼) 𝑟11 (𝛼) , 𝑠21 + 𝑠22 𝑎 (𝛽) = 𝑏 (𝛽) 𝑟11 (𝛽) .
Proposition 2. Let 𝛼 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 matrices 𝐴(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 𝑠21 + 𝑠22 𝑎 (𝑥) − 𝑠11 𝑏 (𝑥) − 𝑠12 𝑎 (𝑥) 𝑏 (𝑥) = 𝛿 (𝑥) 𝑟21 (𝑥) .
(9)
Substituting 𝑥 = 𝛼 into (9), we find
𝑢
M = ⋃ MV
If 𝑎(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.
(8)
𝑠21 + 𝑠22 𝑎 (𝛼) − 𝑠11 𝑏 (𝛼) − 𝑠12 𝑎 (𝛼) 𝑏 (𝛼) = 0.
(10)
Taking 𝑚-the derivative of (9) at 𝑥 = 𝛼, we obtain 𝑠22 𝑎(𝑚) (𝛼) − 𝑠12 𝑎(𝑚) (𝛼) 𝑏 (𝛼) = 0.
(11)
Dividing both sides of obtained equality by 𝑎(𝑚) (𝛼) ≠ 0 and the substituting in (10) yields 𝑠22 − 𝑠12 𝑏 (𝛼) = 0, 𝑠21 − 𝑠11 𝑏 (𝛼) = 0.
(12)
But this is impossible, since matrix ‖𝑠𝑖𝑗 ‖21 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 that 𝑢 ≥ 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
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their arbitrary characteristic root 𝛼𝑖 we denote by 𝑎𝑖 , 𝑎𝑖𝑡 , 𝑏𝑖 , 𝑏𝑖𝑡 the following quantities:
𝑎2 − 𝑎3 𝑎 − 𝑎𝑢 𝑏2 − 𝑏3 𝑏2 − 𝑏𝑢 = ⋅⋅⋅ 2 ⋅ ⋅ ⋅ 𝑐 , 𝑎2 𝑎𝑢 𝑏2 𝑏3 𝑎2 𝑎3 𝑏2 𝑏𝑢
𝑎𝑖 = 𝑎 (𝛼𝑖 ) , 𝑎𝑖𝑡 =
𝑎(𝑚𝑖 +𝑡) (𝛼𝑖 ) , (𝑚𝑖 + 𝑡)!
(13)
𝑏𝑖 = 𝑏 (𝛼𝑖 ) , 𝑏𝑖𝑡 =
where 𝑎2 , . . . , 𝑎𝑢 , 𝑏2 , . . . , 𝑏𝑢 ≠ 0. By excluding 𝑠12 in the system (16), we find that
where 𝑐 = 𝑠11 /𝑠22 . Conversely, the quality (17), after some transformations, can be written in the form 𝑎2 − 𝑐𝑏2 𝑎3 − 𝑐𝑏3 𝑎 − 𝑐𝑏2 𝑎𝑢 − 𝑐𝑏𝑢 = ,..., 2 = . 𝑎2 𝑏2 𝑎3 𝑏3 𝑎2 𝑏2 𝑎𝑢 𝑏𝑢
(𝑚𝑖 +𝑡)
(𝛼𝑖 ) 𝑏 , (𝑚𝑖 + 𝑡)!
(17)
(18)
Introducing the notations 𝑠11 = 𝑐,
𝑡 = 0, 1, . . . , 𝑠 − 𝑚𝑖 − 1. The following cases are possible:
𝑠22 = 1,
The Case 1. 𝑚𝑖 ≥ 𝑛𝑖 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
𝑠12 =
(19)
𝑎2 − 𝑐𝑏2 , 𝑎2 𝑏2
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
Let us now formulate the following Theorem based on the defined notation and the assumption.
𝑠22 𝑎 (𝑥) − 𝑠11 𝑏 (𝑥) − 𝑠12 𝑎 (𝑥) 𝑏 (𝑥) ∈ C [𝑥] . 𝛿 (𝑥)
Theorem 3. Let the partition of the set M 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
Denoting 𝑟21 (𝑥) = (𝑠22 𝑎(𝑥) − 𝑠11 𝑏(𝑥) − 𝑠12 𝑎(𝑥)𝑏(𝑥))/𝛿(𝑥) and 𝑟11 (𝑥) = 𝑠11 + 𝑠12 𝑎(𝑥), 𝑟12 (𝑥) = 𝑠12 𝛿(𝑥), 𝑟22 (𝑥) = 𝑠22 − 𝑠12 𝑏(𝑥), it is easy to make sure that the equality (4) holds true, 𝑟12 (𝑥) where 𝑠21 = 0, 𝑠11 , 𝑠22 ≠ 0 (see (19)), and 𝑟𝑟1121 (𝑥) (𝑥) 𝑟22 (𝑥) = 𝑠11 𝑠22 . This means that the matrices 𝐴(𝑥) and 𝐵(𝑥) are semiscalarly equivalent. The Theorem is proved.
𝑎2 − 𝑎3 𝑎 − 𝑎𝑢 . ⋅⋅⋅ 2 𝑎2 𝑎𝑢 𝑎2 𝑎3
4. The Case 2 (14)
Proof. Let the entry 𝑏(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 𝑠22 𝑎 (𝑥) − 𝑠11 𝑏 (𝑥) − 𝑠12 𝑎 (𝑥) 𝑏 (𝑥) ≡ 0
(mod 𝛿 (𝑥)) , (15)
where 𝑠11 , 𝑠22 ≠ 0. Substituting 𝑥 = 𝛼2 , 𝛼3 , . . . , 𝛼𝑢 in the congruence (15), we obtain the system of equalities 𝑠22 𝑎2 − 𝑠11 𝑏2 − 𝑠12 𝑎2 𝑏2 = 0, 𝑠22 𝑎3 − 𝑠11 𝑏3 − 𝑠12 𝑎3 𝑏3 = 0, .. . 𝑠22 𝑎𝑢 − 𝑠11 𝑏𝑢 − 𝑠12 𝑎𝑢 𝑏𝑢 = 0,
(20)
(16)
Further we shall retain the earlier introduced notations (see, in particular, (13)). Theorem 4. Let the partition of the set M 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) 𝑎𝑝0 𝑎𝑝1 ⋅ ⋅ ⋅ 𝑎𝑝,𝑛𝑝 −𝑚𝑝 −1 = 𝑐 𝑏𝑝0 𝑏𝑝1 ⋅ ⋅ ⋅ 𝑏𝑝,𝑛𝑝 −𝑚𝑝 −1 for every root 𝛼𝑝 ∈ M1 such that 𝑚𝑝 < 𝑛𝑝 ; 2 2 2 (2) 𝑎𝑞0 /𝑎𝑞 𝑎𝑞1 /𝑎𝑞 ⋅ ⋅ ⋅ 𝑎𝑞,𝑛𝑞 −𝑚𝑞 −1 /𝑎𝑞 = (1/ 2 2 2 𝑐) 𝑏𝑞0 /𝑏𝑞 𝑏𝑞1 /𝑏𝑞 ⋅ ⋅ ⋅ 𝑏𝑞,𝑛𝑞 −𝑚𝑞 −1 /𝑏𝑞 for every root 𝛼𝑞 ∈ M such that 𝛼𝑞 ∉ M1 and 𝑚𝑞 < 𝑛𝑞 ; = (1/ (3) (𝑎2 − 𝑎3 )/𝑎2 𝑎3 . . . (𝑎2 − 𝑎𝑢 )/𝑎2 𝑎𝑢 (𝑏 − 𝑏 )/𝑏 𝑏 . . . (𝑏 − 𝑏 )/𝑏 𝑏 𝑐) 2 3 2 3 . 2 𝑢 2 𝑢
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Proof. Necessity. If the matrices 𝐴(𝑥) and 𝐵(𝑥) are semiscalarly equivalent, then from congruence (15) we obtain the system of equalities
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 𝑠22 𝑎 (𝑥) − 𝑠11 𝑏 (𝑥) − 𝑠12 𝑎 (𝑥) 𝑏 (𝑥) ≡ 0
𝑠22 𝑎𝑝0 − 𝑠11 𝑏𝑝0 = 0,
𝑛
(mod (𝑥 − 𝛼𝑙 ) 𝑙 ) ,
𝑠22 𝑎𝑝1 − 𝑠11 𝑏𝑝1 = 0, (21)
.. . 𝑠22 𝑎𝑝,𝑛𝑝 −𝑚𝑝 −1 − 𝑠11 𝑏𝑝,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 𝑠22 𝑎𝑞 − 𝑠11 𝑏𝑞 − 𝑠12 𝑎𝑞 𝑏𝑞 = 0, 𝑠22 𝑎𝑞0 − 𝑠11 𝑏𝑞0 − 𝑠12 (𝑎𝑞0 𝑏𝑞 + 𝑏𝑞0 𝑎𝑞 ) = 0,
− 𝑠12 (𝑎𝑞,𝑛𝑞 −𝑚𝑞 −1 𝑏𝑞 + 𝑏𝑞,𝑛𝑞 −𝑚𝑞 −1 𝑎𝑞 ) = 0
𝑠11 𝑎𝑞 − 𝑠22 𝑏𝑞 𝑎𝑞 𝑏𝑞
.
(23)
Substituting it into second and every succeeding equalities we obtain
𝑠11
𝑎𝑞2 𝑎𝑞1 𝑎𝑞2
= 𝑠22 = 𝑠22
𝑏𝑞0 𝑏𝑞2
𝑎𝑞,𝑛𝑞 −𝑚𝑞 −1 𝑎𝑞2
= 𝑠22
𝑛𝑞
(mod (𝑥 − 𝛼𝑞 ) )
(27)
𝑏𝑞1 𝑏𝑞2
(24)
𝑏𝑞,𝑛𝑞 −𝑚𝑞 −1 𝑏𝑞2
𝑛
(mod (𝑥 − 𝛼𝑘 ) 𝑘 )
(28)
for 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 Lemmas 5 and 6 that we prove in the current section.
,
.. . 𝑠11
(26)
is 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
𝑠22 𝑎 (𝑥) − 𝑠11 𝑏 (𝑥) − 𝑠12 𝑎 (𝑥) 𝑏 (𝑥) ≡ 0
for 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
𝑎𝑞0
(mod (𝑥 − 𝛼𝑗 ))
for every root 𝛼𝑞 ∈ M such that 𝛼𝑞 ∉ M1 and 𝑚𝑞 < 𝑛𝑞 . When it is considered the congruence (26), then from (27) we actually have
𝑠22 𝑎𝑞,𝑛𝑞 −𝑚𝑞 −1 − 𝑠11 𝑏𝑞,𝑛𝑞 −𝑚𝑞 −1
𝑠11
𝑠22 𝑎 (𝑥) − 𝑠11 𝑏 (𝑥) − 𝑠12 𝑎 (𝑥) 𝑏 (𝑥) ≡ 0
(22)
.. .
𝑠12 =
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
𝑠22 𝑎 (𝑥) − 𝑠11 𝑏 (𝑥) − 𝑠12 𝑎 (𝑥) 𝑏 (𝑥) ≡ 0
𝑠22 𝑎𝑞1 − 𝑠11 𝑏𝑞1 − 𝑠12 (𝑎𝑞1 𝑏𝑞 + 𝑏𝑞1 𝑎𝑞 ) = 0,
(25)
.
The condition (2) of the Theorem is proved. As in the proof of Theorem 3, 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.
Lemma 5. The matrix equation 𝑎 𝑏0 0 0 𝑎 𝑏 0 1 1 . . . .. . . . . 𝑥 𝑎 0 1 𝑚−1 𝑏𝑚−1 𝑥 𝑎 𝑏 𝑎 𝑏 2 = 0, 𝑚 𝑚 0 0 𝑥 𝑎 𝑏 𝑎 𝑏 + 𝑎 𝑏 𝑚+1 𝑚+1 0 1 1 0 3 . . . . .. .. . 𝑘−𝑚 𝑎 𝑏𝑘 ∑ 𝑎𝑡 𝑏𝑘−𝑚−𝑡 𝑘 𝑡=0
(29)
International Journal of Analysis
5
over C, where 𝑎0 , 𝑏0 ≠ 0, has nonzero solutions if and only if the following conditions are satisfied: 𝑎 𝑎 1 2 𝑎 𝑎 0 1 d 0
⋅ ⋅ ⋅ 𝑎V−1 ⋅ ⋅ ⋅ 𝑎V−2 d
.. .
d 𝑎1 𝑎0
𝑏 𝑏 𝑎V 1 2 𝑏 𝑏 𝑎V−1 0 1 .. = 𝑐V d . 𝑎2 0 𝑎1
⋅ ⋅ ⋅ 𝑏V−1 ⋅ ⋅ ⋅ 𝑏V−2 d
.. .
d 𝑏1 𝑏0
𝑏V 𝑏V−1 .. , . 𝑏2 𝑏1
Denote by 𝐴 𝑢𝑤 , 𝐵𝑢𝑤 the submatrices obtained, respectively, from matrices
(30)
where V = 1, . . . , 𝑚 − 1, 𝑚 + 1, . . . , 𝑘, 𝑐 = 𝑎0 /𝑏0 . If the conditions of Lemma are satisfied then for every nonzero T solution 𝑥10 𝑥20 𝑥30 of (29) 𝑥10 , 𝑥20 ≠ 0. Proof. T Necessity. Let (29) have the solution 𝑥10 𝑥20 𝑥30 ≠ 0. Then 𝑎0 𝑥10 + 𝑏0 𝑥20 = 0, 𝑎1 𝑥10 + 𝑏1 𝑥20 = 0, (31)
.. . 𝑎𝑚−1 𝑥10 + 𝑏𝑚−1 𝑥20 = 0, 𝑎𝑚 𝑥10 + 𝑏𝑚 𝑥20 + 𝑎0 𝑏0 𝑥30 = 0, 𝑎𝑚+1 𝑥10 + 𝑏𝑚+11 𝑥20 + (𝑎0 𝑏1 + 𝑎1 𝑏1 ) 𝑥30 = 0, .. .
(32) 𝑘−𝑚
𝑎𝑘 𝑥10 + 𝑏𝑘 𝑥20 + 𝑥30 ∑ 𝑎𝑡 𝑏𝑘−𝑚−𝑡 = 0. 𝑡=0
We assume that 𝑥10 = 0 or 𝑥20 = 0. Then, from first equalities of (31) and (32) we obtain 𝑥10 = 𝑥20 = 𝑥30 = 0. T This contradicts the assumption that solution 𝑥10 𝑥20 𝑥30 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 𝑎0 𝑎𝑚+1 − 𝑎𝑚 (𝑐𝑏1 + 𝑎1 ) = 𝑐2 𝑏0 𝑏𝑚+1 − 𝑐𝑏𝑚 (𝑐𝑏1 + 𝑎1 ) .
(33)
If 𝑚 = 1 then 𝑎0 𝑎2 − 𝑎12 = 𝑐2 (𝑏0 𝑏2 − 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−1 𝑏0𝑚−1 in (33) we find 𝑎0𝑚 𝑎𝑚+1 − 2𝑎0𝑚−1 𝑎1 𝑎𝑚 = 𝑐 (𝑏0𝑚 𝑏𝑚+1 − 2𝑏0𝑚−1 𝑏1 𝑏𝑚 ) .
(34)
𝑎 𝑎 1 2 𝑎 𝑎 0 1 d 0
. . . 𝑎𝑚
𝑏 𝑏 1 2 𝑏 𝑏 0 1 d 0
. . . 𝑏𝑚
. . . 𝑎𝑚−1 d
.. .
d
𝑎1 𝑎0
. . . 𝑏𝑚−1 d
.. .
d
𝑏1 𝑏0
𝑎𝑚+1 𝑎𝑚 .. , . 𝑎2 𝑎1
(35)
𝑏𝑚+1 𝑏𝑚 .. . 𝑏2 𝑏1
by obliterating of two last columns and 𝑢-th and 𝑤-th rows. Also denote by Δ 𝑚+1 (𝐴), Δ 𝑚+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 |𝐴 𝑢𝑤 | = |𝐵𝑢𝑤 | = 0 (for 𝑢 ≠ 𝑚 + 1), we have 𝑎 𝑎 𝑚 𝑚+1 𝐴 Δ 𝑚+1 (𝐴) = (−1)𝑚+1 ( 𝑎0 𝑎1 1,𝑚+1 𝑎 𝑎 𝑎 𝑚−1 𝑎𝑚 𝐴 2,𝑚+1 + ⋅ ⋅ ⋅ + 1 2 𝐴 𝑚,𝑚+1 ) , − 𝑎0 𝑎1 𝑎0 𝑎1 𝑏 𝑏 𝑚+1 𝑚 𝑚+1 ( Δ 𝑚+1 (𝐵) = (−1) 𝐵 𝑏0 𝑏1 1,𝑚+1 𝑏 𝑏 𝑏 𝑚−1 𝑏𝑚 1 2 − + ⋅ ⋅ ⋅ + ) . 𝐵 𝐵 𝑏0 𝑏1 2,𝑚+1 𝑏0 𝑏1 𝑚,𝑚+1
(36)
(37)
Since 𝑎0 𝑎1 ⋅ ⋅ ⋅ 𝑎𝑚−1 = 𝑐 𝑏0 𝑏1 ⋅ ⋅ ⋅ 𝑏𝑚−1 each summand of expression in parenthesis for Δ 𝑚+1 (𝐴), possibly except the first two, differs from the corresponding summand for Δ 𝑚+1 (𝐵) by the multiplier 𝑐𝑚+1 . From this fact and from the equality (34) for V = 𝑚 + 1 follows equality (30). Denote by Δ V (𝐴) and Δ V (𝐵) the determinants in the left and right sides of equality (30), respectively. Suppose by induction Δ 𝑟 (𝐴) = 𝑐𝑟 Δ 𝑟 (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 −1
1
𝑚
(𝑎𝑚+1 − (𝑎0 𝑏0 ) 𝑎𝑚 ∑ 𝑎𝑢 𝑏1−𝑢 ) (−𝑎0 ) Δ 𝑟−𝑚 (𝐴) 𝑢=0
−1
1
𝑚
= 𝑐𝑟+1 (𝑏𝑚+1 − (𝑎0 𝑏0 ) 𝑏𝑚 ∑ 𝑎𝑢 𝑏1−𝑢 ) (−𝑏0 ) 𝑢=0
⋅ Δ 𝑟−𝑚 (𝐵) ,
6
International Journal of Analysis 2
−1
𝑚+1
(𝑎𝑚+2 − (𝑎0 𝑏0 ) 𝑎𝑚 ∑ 𝑎𝑢 𝑏2−𝑢 ) (−𝑎0 ) 𝑢=0
𝑟+1
=𝑐
𝑚+2
𝑟
+ ⋅ ⋅ ⋅ + 𝑏𝑗,𝑟−𝑚𝑗 𝛿𝑗1 (𝐴) (−𝑎𝑗0 ) )
−1
2
−1
⋅ (−𝑎0 )
Δ 𝑟−𝑚−1 (𝐴)
(𝑏𝑚+2 − (𝑎0 𝑏0 ) 𝑏𝑚 ∑ 𝑎𝑢 𝑏2−𝑢 ) (−𝑏0 )
𝑚+1
+ (𝑎0 𝑏0 ) 𝑎𝑚 (𝑏1 Δ 𝑟−𝑚 (𝐴) (−𝑎0 )
𝑚+1
𝑚+2
+ 𝑏2 Δ 𝑟−𝑚−1 (𝐴) (−𝑎0 )
𝑢=0
+ ⋅ ⋅ ⋅ + 𝑏𝑟−𝑚 Δ 1 (𝐴)
⋅ Δ 𝑟−𝑚−1 (𝐵) , 𝑟
𝑟−2𝑚+1
−1
−1
) − (𝑎𝑟−𝑚+1 − (𝑎0 𝑏0 )
𝑟−2𝑚+1
(𝑎𝑟−𝑚+1 − (𝑎0 𝑏0 ) 𝑎𝑚 ∑ 𝑎𝑢 𝑏𝑟−2𝑚−𝑢+1 − 𝑎𝑟−𝑚+1
𝑟−𝑚
⋅ 𝑎𝑚 ∑ 𝑎𝑢 𝑏𝑟−2𝑚−𝑢+1 ) (−𝑎0 )
𝑢=0
𝑢=0
𝑟−2𝑚+1
−1
𝑟+1
⋅ (−𝑎0 ) + 𝑏𝑟−𝑚+1 (−𝑎0 )
.. .
𝑟−𝑚
+ (𝑎0 𝑏0 ) 𝑎𝑚 ∑ 𝑎𝑢 𝑏𝑟−2𝑚−𝑢+1 ) (−𝑎0 )
= 𝑐𝑟+1 ((−𝑏0 ) 𝑏𝑟+1 + (−𝑏0 )
⋅ Δ 𝑚 (𝐴) = 𝑐𝑟+1 (𝑏𝑟−𝑚+1
+ (−𝑏0 )
𝑟
𝑢=0
𝑟−𝑚
− (𝑎0 𝑏0 ) 𝑏𝑚 ∑ 𝑎𝑢 𝑏𝑟−2𝑚−𝑢+1 − 𝑏𝑟−𝑚+1 𝑟−2𝑚+1
𝑟−𝑚
𝑢=0
𝑏𝑚 Δ 𝑟−𝑚+1 (𝐵) 𝑚+1
𝑚+2
Δ 𝑚 (𝐵) ,
+ 𝑎2 Δ 𝑟−𝑚−1 (𝐵) (−𝑏0 )
+ ⋅⋅⋅
𝑟
+ 𝑎𝑟−𝑚 Δ 1 (𝐵) (−𝑏0 ) ) + (𝑎0 𝑏0 )
.. .
−1
𝑚+1
𝑟−𝑚
𝑟−1
(𝑎𝑟 − (𝑎0 𝑏0 ) 𝑎𝑚 ∑ 𝑎𝑢 𝑏𝑟−𝑚−𝑢 ) (−𝑎0 ) 𝑢=0
⋅ 𝑏𝑚 (𝑎1 Δ 𝑟−𝑚 (𝐵) (−𝑏0 )
Δ 1 (𝐴)
𝑚+2
+ 𝑎2 Δ 𝑟−𝑚−1 (𝐵) (−𝑏0 )
𝑟−𝑚
−1
𝑟
⋅ Δ 1 (𝐵) , 𝑟−𝑚+1
𝑟+1
+ 𝑎𝑟−𝑚 Δ 1 (𝐵) (−𝑏0 ) + 𝑎𝑟−𝑚+1 (−𝑏0 )
𝑢=0
−1
) − (𝑏𝑟−𝑚+1
𝑟−2𝑚+1
𝑟−𝑚𝑗
− (𝑎0 𝑏0 ) 𝑏𝑚 ∑ 𝑎𝑢 𝑏𝑟−2𝑚−𝑢+1 ) (−𝑏𝑗0 )
𝑟
(𝑎𝑟+1 − (𝑎0 𝑏0 ) 𝑎𝑚 ∑ 𝑎𝑢 𝑏𝑟−𝑚−𝑢+1 ) (−𝑎0 )
𝑢=0
𝑢=0
𝑟−𝑚+1
−1
+ ⋅⋅⋅
𝑟−1
= 𝑐𝑟+1 (𝑏𝑟 − (𝑎0 𝑏0 ) 𝑏𝑚 ∑ 𝑎𝑢 𝑏𝑟−𝑚−𝑢 ) (−𝑏0 )
−1
𝑚−1
−1
+ (𝑎0 𝑏0 ) 𝑏𝑚 ∑ 𝑎𝑢 𝑏𝑟−2𝑚−𝑢+1 ) (−𝑏0 )
−1
𝑚
− (𝑎0 𝑏0 ) 𝑏𝑚 (𝑎1 Δ 𝑟−𝑚 (𝐵) (−𝑏0 )
𝑢=0
−1
𝑏𝑟 Δ 1 (𝐵) + ⋅ ⋅ ⋅
𝑏𝑟−𝑚+1 Δ 𝑚 (𝐵) + ⋅ ⋅ ⋅ + (−𝑏0 )
⋅ 𝑏𝑚+1 Δ 𝑟−𝑚 (𝐵) + (−𝑏0 )
𝑟−2𝑚+1
−1
𝑟−1
Δ 𝑚 (𝐴)
⋅ Δ 𝑚 (𝐵)) .
= 𝑐𝑟+1 (𝑏𝑟+1 − (𝑎0 𝑏0 ) 𝑏𝑚 ∑ 𝑎𝑢 𝑏𝑟−𝑚−𝑢+1 ) 𝑢=0
(39)
𝑟
⋅ (−𝑏0 ) ,
(38) where 𝑐 = 𝑎0 /𝑏0 . If we add left sides of equalities (38) and separate right sides, we obtain 𝑟
𝑟−1
(−𝑎0 ) 𝑎𝑟+1 + (−𝑎0 )
𝑟−𝑚
𝑎𝑟 Δ 1 (𝐴) + ⋅ ⋅ ⋅ + (−𝑎0 ) 𝑚
⋅ 𝑎𝑟−𝑚+1 Δ 𝑚 (𝐴) + ⋅ ⋅ ⋅ + (−𝑎0 ) 𝑎𝑚+1 Δ 𝑟−𝑚 (𝐴) 𝑚−1
+ (−𝑎0 )
Gathering similar terms in both sides of obtained equalities, we obtain 𝑟
𝑟−1
(−𝑎0 ) 𝑎𝑟+1 + (−𝑎0 )
𝑟−𝑚
𝑎𝑟 Δ 1 (𝐴) + ⋅ ⋅ ⋅ + (−𝑎0 ) 𝑚
⋅ 𝑎𝑟−𝑚+1 Δ 𝑚 (𝐴) + ⋅ ⋅ ⋅ + (−𝑎0 ) 𝑎𝑚+1 Δ 𝑟−𝑚 (𝐴) 𝑚−1
+ (−𝑎0 )
−1
𝑎𝑚 Δ 𝑟−𝑚+1 (𝐴) + (𝑎0 𝑏0 )
⋅ 𝑎𝑚 𝑏𝑟−𝑚+1 (−𝑎0 )
𝑟+1
−1
− (𝑎𝑟−𝑚+1 − (𝑎0 𝑏0 )
−1
𝑎𝑚 Δ 𝑟−𝑚+1 (𝐴) − (𝑎0 𝑏0 ) 𝑚+1
⋅ 𝑎𝑚 (𝑏1 Δ 𝑟−𝑚 (𝐴) (−𝑎0 )
+ 𝑏2 Δ 𝑟−𝑚−1 (𝐴)
𝑟−2𝑚+1
𝑟−𝑚
⋅ 𝑎𝑚 ∑ 𝑎𝑢 𝑏𝑟−2𝑚−𝑢+1 ) (−𝑎0 ) 𝑢=1
Δ 𝑚 (𝐴)
International Journal of Analysis
7
𝑟
= 𝑐𝑟+1 ((−𝑏0 ) 𝑏𝑟+1 + (−𝑏0 ) 𝑟−𝑚
+ (−𝑏0 )
𝑟−1
that is, Δ 𝑟+1 (𝐴) = 𝑐𝑟+1 Δ 𝑟+1 (𝐵), where 𝑐 = 𝑎0 /𝑏0 . The necessity of conditions of Lemma is proved.
𝑏𝑟 Δ 1 (𝐵) + ⋅ ⋅ ⋅
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
𝑚
𝑏𝑟−𝑚+1 Δ 𝑚 (𝐵) + ⋅ ⋅ ⋅ + (−𝑏0 ) 𝑚−1
⋅ 𝑏𝑚+1 Δ 𝑟−𝑚 (𝐵) + (−𝑏0 ) −1
+ (𝑎0 𝑏0 ) 𝑏𝑚 𝑎𝑟−𝑚+1 (−𝑏0 )
𝑏𝑚 Δ 𝑟−𝑚+1 (𝐵)
𝑟+1
𝑟−2𝑚+1
−1
− (𝑏𝑟−𝑚+1 − (𝑎0 𝑏0 ) 𝑏𝑚 ∑ 𝑎𝑢 𝑏𝑟−2𝑚−𝑢+1 ) 𝑢=1
𝑟−𝑚
⋅ (−𝑏0 )
𝑥10 = 1,
Δ 𝑚 (𝐵)) .
𝑥20 = −𝑐, (40)
−1
𝑥30 = (𝑐𝑏𝑚 − 𝑎𝑚 ) (𝑎0 𝑏0 )
It follows from (32) that 𝑟−2𝑚+1
−1
𝑎𝑟−𝑚+1 + (𝑎0 𝑏0 ) 𝑎𝑚 ∑ 𝑎𝑢 𝑏𝑟−2𝑚−𝑢+1 𝑢=0
(41)
𝑟−2𝑚+1
−1
= 𝑐 (𝑏𝑟−𝑚+1 + (𝑎0 𝑏0 ) 𝑏𝑚 ∑ 𝑎𝑢 𝑏𝑟−2𝑚−𝑢+1 ) . 𝑢=0
From this relation it is easy to say that the following equality holds: −1
𝑟+1
(𝑎0 𝑏0 ) 𝑎𝑚 𝑏𝑟−𝑚+1 (−𝑎0 )
−1
− (𝑎𝑟−𝑚+1 − (𝑎0 𝑏0 ) 𝑟−𝑚
⋅ 𝑎𝑚 ∑ 𝑎𝑢 𝑏𝑟−2𝑚−𝑢+1 ) (−𝑎0 ) 𝑢=0
−1
𝑟+1
(42)
𝑎𝑟 − 𝑐𝑏𝑟 + (𝑐𝑏𝑚 − 𝑎𝑚 ) (𝑎0 𝑏0 )
𝑢=0
Δ 𝑚 (𝐵)) ;
From (31) and the induction hypothesis, we can write 𝑎𝑚−1 Δ 𝑟−𝑚+2 (𝐴) + ⋅ ⋅ ⋅ + (−𝑎0 ) 𝑎2 Δ 𝑟−1 (𝐴) 𝑚−2
+ 𝑎1 Δ 𝑟 (𝐴) = 𝑐𝑟+1 ((−𝑏0 )
𝑏𝑚−1 Δ 𝑟−𝑚+2 (𝐵) + ⋅ ⋅ ⋅ (43)
+ (−𝑏𝑗0 ) 𝑏𝑗2 𝛿𝑗,𝑟−1 (𝐵) + 𝑏1 Δ 𝑟 (𝐵)) . Comparing (40), (42), and (43), we obtain equality 𝑟
(−𝑎0 ) 𝑎𝑟+1 + (−𝑎0 )
𝑟−1
𝑎𝑟 Δ 1 (𝐴) + ⋅ ⋅ ⋅ + (−𝑎0 )
𝑟−1
+ (−𝑏0 )
𝑏𝑟 Δ 1 (𝐵) + ⋅ ⋅ ⋅ + (−𝑏0 ) 𝑏2 Δ 𝑟−1 (𝐵)
+ 𝑏1 Δ 𝑟 (𝐵)) ,
𝑟−𝑚
−1
∑ 𝑎𝑢 𝑏𝑟−𝑚−𝑢 = 0.
𝑢=0
While proceeding we may assume 𝑟 > 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 −1
(44)
𝑢=0
(47)
𝑎𝑟+1 − 𝑐𝑏𝑟+1 + (𝑐𝑏𝑚 − 𝑎𝑚 ) (𝑎0 𝑏0 )
𝑟
⋅ 𝑎2 Δ 𝑟−1 (𝐴) + 𝑎1 Δ 𝑟 (𝐴) = 𝑐𝑟+1 ((−𝑏0 ) 𝑏𝑟+1
1
∑ 𝑎𝑢 𝑏1−𝑢 = 0,
.. .
𝑟−2𝑚+1
−1
− (𝑏𝑟−𝑚+1 − (𝑎0 𝑏0 ) 𝑏𝑚 ∑ 𝑎𝑢 𝑏𝑟−2𝑚−𝑢+1 )
𝑚−2
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,
𝑎𝑚+1 − 𝑐𝑏𝑚+1 + (𝑐𝑏𝑚 − 𝑎𝑚 ) (𝑎0 𝑏0 ) −1
(−𝑎0 )
−1 𝑎𝑚+1 − 𝑐𝑏𝑚+1 + (𝑐𝑏𝑚 − 𝑎𝑚 ) (𝑎0 𝑏0 ) (𝑎0 𝑏1 + 𝑎1 𝑏1 ) = 0. (46)
𝑎𝑚 − 𝑐𝑏𝑚 + (𝑐𝑏𝑚 − 𝑎𝑚 ) (𝑎0 𝑏0 ) 𝑎0 𝑏0 = 0,
Δ 𝑚 (𝐴)
= 𝑐𝑟+1 ((𝑎0 𝑏0 ) 𝑏𝑚 𝑎𝑟−𝑚+1 (−𝑏0 )
𝑟−𝑚
satisfy 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 = 𝑐𝑚 𝑏0𝑚 using the of simple transformations, we can obtain the following relation:
−1
𝑟−2𝑚+1
⋅ (−𝑏0 )
(45)
𝑟−𝑚+1
∑ 𝑎𝑢 𝑏𝑟−𝑚−𝑢+1
𝑢=0
(48)
= 0. This means that (45) is the solution of (𝑟 − 𝑚 + 2)-th equation of system (32).
8
International Journal of Analysis
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.
linearly independent. Then, each of the following rows of the matrix linearly depends on these first two rows. Because 𝑎 𝑏 𝑎𝑏 𝑎 𝑏 𝑎𝑏 + 𝑎 𝑏 = 0, 0 0 0 0 𝑎 𝑏 𝑎𝑏 + 𝑎 𝑏 V V V V
Lemma 6. The matrix equation 𝑎 𝑏 𝑎𝑏 𝑎 𝑏0 𝑎𝑏0 + 𝑎0 𝑏 0 𝑎 𝑏1 𝑎𝑏1 + 𝑎1 𝑏 1 . .. .. .. . . 𝑥1 𝑎 𝑎𝑏𝑚−1 + 𝑎𝑚−1 𝑏 𝑚−1 𝑏𝑚−1 𝑥2 = 0, (49) 𝑎 𝑎𝑏𝑚 + 𝑎0 𝑏0 + 𝑎𝑚 𝑏 𝑚 𝑏𝑚 𝑥3 𝑎 𝑚+1 𝑏𝑚+1 𝑎𝑏𝑚+1 + 𝑎0 𝑏1 + 𝑎1 𝑏0 + 𝑎𝑚+1 𝑏 . .. .. . . . . 𝑘−𝑚 𝑎 𝑏 𝑎𝑏 + 𝑎 𝑏 + 𝑎 𝑏 ∑ 𝑘 𝑘 𝑘 𝑡 𝑘−𝑚−𝑡 𝑘 𝑡+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: 0 ⋅ ⋅ ⋅ 0 𝑎 𝑎 ⋅ ⋅ ⋅ 𝑎 0 1 V .. 𝑎 d d d d d . d d d d d 𝑎 1 −1 d d d d 𝑎0 (𝑎𝑚+V−1 𝑎0 ) d d d 0 .. d d . 0 𝑎 0 ⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟ (50)
𝑚+V
V = 0, 1, . . . , 𝑘. If the conditions of Lemma are satisfied then every nonzero T solution 𝑥10 𝑥20 𝑥30 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
(51)
we have 𝑎V /𝑎0 = 𝑏V /𝑏0 . This implies the equality (50) for V = 1, . . . , 𝑚 − 1. From the equality 𝑎 𝑏 𝑎𝑏 𝑎 𝑏 = 0 𝑎𝑏0 + 𝑎0 𝑏 0 0 𝑎 𝑏 𝑎𝑏 + 𝑎 𝑏 + 𝑎 𝑏 𝑚 𝑚 𝑚 0 0 𝑚
(52)
𝑎𝑏02 𝑎0 − 𝑎0 𝑏𝑚 𝑎𝑏 + 𝑎𝑚 𝑏0 𝑎𝑏 − 𝑎02 𝑏0 𝑏 = 0,
(53)
we have
where, after division of both sides by 𝑎𝑏𝑎0 𝑏0 , 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 𝑎 𝑏 𝑎𝑏 𝑎 𝑏 𝑎𝑏 + 𝑎 𝑏 0 0 0 0 = 0. 𝑟+1−𝑚 𝑎𝑟+1 𝑏𝑟+1 𝑎𝑏𝑟+1 + ∑ 𝑎𝑡 𝑏𝑟+1−𝑚−𝑡 + 𝑎𝑟+1 𝑏 𝑡=0
𝑚+V
0 ⋅ ⋅ ⋅ 0 𝑏 𝑏 ⋅ ⋅ ⋅ 𝑏 0 1 V . . 𝑏 d d d d d . d d d d d 𝑏 1 −1 d d d d 𝑏0 , = (𝑏𝑚+V−1 𝑏0 ) d d d 0 .. d d . ⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟ 0 𝑏 0
V = 1, . . . , 𝑚 − 1,
(54)
After the calculation of the determinant on the left, we obtain the equation 𝑟+1−𝑚
𝑎𝑏0 ∑ 𝑎𝑡 𝑏𝑟+1−𝑚−𝑡 − 𝑎0 𝑏𝑟+1 𝑎𝑏 + 𝑎𝑟+1 𝑏0 𝑎𝑏 𝑡=0
(55)
𝑟+1−𝑚
− 𝑎0 𝑏 ∑ 𝑎𝑡 𝑏𝑟+1−𝑚−𝑡 = 0. 𝑡=0
We can divide by 𝑎𝑏𝑎0 𝑏0 both sides to obtain the equality −1
𝑎𝑟+1 𝑎0−1 − 𝑏𝑟+1 𝑏0−1 − (𝑎𝑏0 ) + (𝑎0 𝑏)
−1
𝑟+1−𝑚
𝑟+1−𝑚
∑ 𝑎𝑡 𝑏𝑟+1−𝑚−𝑡
𝑡=0
(56)
∑ 𝑎𝑡 𝑏𝑟+1−𝑚−𝑡 = 0.
𝑡=0
Let us denote by Δ V (𝑎) and Δ V (𝑏) the determinants in the left and right sides of the equality (50), respectively.
International Journal of Analysis
9
Considering the inductive assumption, we can write the equalities −1
(−1)𝑚−1 𝑎0 𝑎−1 ((𝑎𝑟 𝑎0 ) Δ 𝑟+1−𝑚 (𝑎)
𝑟−1
(−1) 𝑎1 𝑎 ((𝑎 𝑟−1
− (𝑏
+ (−1)𝑟 𝑏𝑟+1−𝑚 𝑏𝑟 Δ 0 (𝑏) + (−1)𝑟−1 𝑏𝑟−𝑚 𝑎𝑟−1 Δ 1 (𝑏) (60)
+ (−1)𝑚 𝑏1 𝑏𝑚 Δ 𝑟−𝑚 (𝑏)
−1
−1
−1
+ ⋅ ⋅ ⋅ + (−1)𝑚+1 𝑏2 𝑏𝑚+1 Δ 𝑟−𝑚−1 (𝑏)
− (𝑏𝑟 𝑏0 ) Δ 𝑟+1−𝑚 (𝑏)) = 0, 𝑚
−1
(𝑏𝑚+𝑟 𝑏0 ) Δ 𝑟+1 (𝑏) = (𝑏𝑚+𝑟 𝑏0 ) ((−1)𝑚+𝑟 𝑏𝑟+1 𝑏𝑚+𝑟
+ (−1)𝑚−1 𝑏0 𝑏𝑚−1 Δ 𝑟−𝑚+1 (𝑏)) .
−1
𝑎0 ) Δ 𝑟−𝑚 (𝑎)
It is easy to make sure that the following identity is true:
−1
𝑏0 ) Δ 𝑟−𝑚 (𝑏)) = 0,
−1
𝑏−1 (𝑏0 (𝑎𝑟 𝑎0 ) Δ 𝑟−𝑚+1 (𝑎) (−1)𝑚−1
.. .
(57)
−1
−1
(−1)𝑟−1 𝑎𝑟−𝑚 𝑎−1 ((𝑎𝑚 𝑎0 ) Δ 1 (𝑎) − (𝑏𝑚 𝑏0 ) Δ 1 (𝑏))
−1
+ 𝑏1 (𝑎𝑟−1 𝑎0 ) Δ 𝑟−𝑚 (𝑎) (−1)𝑚 + ⋅ ⋅ ⋅ −1
+ 𝑏𝑟−𝑚 (𝑎𝑚 𝑎0 ) Δ 1 (𝑎) (−1)𝑟−1 −1
= 0,
+ 𝑏𝑟−𝑚+1 (𝑎𝑚−1 𝑎0 ) Δ 0 (𝑎) (−1)𝑟 ) −1
(−1)𝑟 𝑎𝑟+1−𝑚 𝑎−1 ((𝑎𝑚−1 𝑎0 ) Δ 0 (𝑎)
−1
+ 𝑎−1 (−𝑎0 (𝑏𝑟 𝑏0 ) Δ 𝑟−𝑚+1 (𝑏) (−1)𝑚−1
−1
− (𝑏𝑚−1 𝑏0 ) Δ 0 (𝑏)) = 0,
−1
− 𝑎1 (𝑏𝑟−1 𝑏0 ) Δ 𝑟−𝑚 (𝑏) (−1)𝑚 − ⋅ ⋅ ⋅
−1
(−1)𝑚−1 𝑏0 𝑏−1 ((𝑎𝑟 𝑎0 ) Δ 𝑟+1−𝑚 (𝑎)
−1
− 𝑎𝑟−𝑚 (𝑏𝑚 𝑏0 ) Δ 1 (𝑏) (−1)𝑟−1
−1
− (𝑏𝑟 𝑏0 ) Δ 𝑟+1−𝑚 (𝑏)) = 0,
−1
− 𝑎𝑟−𝑚+1 (𝑏𝑚−1 𝑏0 ) Δ 0 (𝑏) (−1)𝑟 ) + (−1)𝑚+𝑟
−1
(−1)𝑚 𝑏1 𝑏−1 ((𝑎𝑟−1 𝑎0 ) Δ 𝑟−𝑚 (𝑎)
−1
⋅ (𝑎0 𝑏)
−1
− (𝑏𝑟−1 𝑏0 ) Δ 𝑟−𝑚 (𝑏)) = 0,
𝑟+1−𝑚
−1
∑ 𝑎𝑡 𝑏𝑟+1−𝑚−𝑡 − (−1)𝑚+𝑟 (𝑎𝑏0 )
𝑡=0
𝑟+1−𝑚
.. .
(58)
−1
−1
(−1)𝑟−1 𝑏𝑟−𝑚 𝑏−1 ((𝑎𝑚 𝑎0 ) Δ 1 (𝑎) − (𝑏𝑚 𝑏0 ) Δ 1 (𝑏))
−1
(−1)𝑟 𝑏𝑟+1−𝑚 𝑏−1 ((𝑎𝑚−1 𝑎0 ) Δ 0 (𝑎) −1
− (𝑏𝑚−1 𝑏0 ) Δ 0 (𝑏)) = 0. Decompose the determinants Δ 𝑟+1 (𝑎) and Δ 𝑟+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 −1
(𝑎𝑚+𝑟 𝑎0 ) Δ 𝑟+1 (𝑎) = (𝑎𝑚+𝑟 𝑎0 ) ((−1)𝑚+𝑟 𝑎𝑟+1 𝑎𝑚+𝑟 + (−1)𝑟 𝑎𝑟+1−𝑚 𝑎𝑟 Δ 0 (𝑎) + (−1)𝑟−1 𝑎𝑟−𝑚 𝑎𝑟−1 Δ 1 (𝑎) + ⋅ ⋅ ⋅ + (−1)𝑚+1 𝑎2 𝑎𝑚+1 Δ 𝑟−𝑚−1 (𝑎) + (−1)𝑚 𝑎1 𝑎𝑚 Δ 𝑟−𝑚 (𝑎) + (−1)𝑚−1 𝑎0 𝑎𝑚−1 Δ 𝑟−𝑚+1 (𝑎)) ,
⋅ ∑ 𝑎𝑡 𝑏𝑟+1−𝑚−𝑡 = 0. 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+𝑟 𝑎0 )−1 Δ 𝑟+1 (𝑎), (𝑏𝑚+𝑟 𝑏0 )−1 Δ 𝑟+1 (𝑏) and the identity (61), we obtain (𝑎𝑚+𝑟 𝑎0 )−1 Δ 𝑟+1 (𝑎) − (𝑏𝑚+𝑟 𝑏0 )−1 Δ 𝑟+1 (𝑏) = 0. This means that the equality (50) for V = 𝑟+1 is true. The necessity of the conditions of Lemma is proved.
= 0,
−1
(61)
(59)
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 1st, 2-nd, and (𝑟 + 3)-th rows is equal to zero. This means that indicated rows are linearly dependent. The above argument
10
International Journal of Analysis
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 𝑥 𝑥 𝑥 T 10 20 30 be the nonzero solution of the equation (49), where 𝑥10 = 0 or 𝑥20 = 0. Hence we have the equality 𝑏 𝑎𝑏 𝑥20 𝑏 𝑎𝑏 + 𝑎 𝑏 𝑥 = 0 0 0 0 30
(62)
𝑎 𝑎𝑏 𝑥10 𝑎 𝑎𝑏 + 𝑎 𝑏 𝑥 = 0. 0 0 0 30
(63)
for every pair of the roots 𝛼𝑘 , 𝛼𝑙 ∈ M1 such that 2𝑚𝑘 < 𝑛𝑘 , 2𝑚𝑙 < 𝑛𝑙 ; (3) 0 ⋅ ⋅ ⋅ 0 𝑎 𝑎 ⋅ ⋅ ⋅ 𝑎 𝑞0 𝑞1 𝑞,𝑠𝑞 𝑎 d d d d d ... 𝑞 d d d d d 𝑎 𝑞1 1 d d d d 𝑎 𝑞0 𝑚𝑞 +𝑠𝑞 +1 𝑎𝑞 d d d 0 .. d d . 0 𝑎𝑞 0 ⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
or
Since 𝑎, 𝑏, 𝑎0 , 𝑏0 ≠ 0, the determinants of the 2 × 2 T matrices of these equalities are nonzero. Hence, 𝑥20 𝑥30 = T T 0 or 𝑥10 𝑥30 = 0; that is, 𝑥10 𝑥20 𝑥30 = 0 contrary to the our assumption. Therefore in the nonzero solution 𝑥 𝑥 𝑥 T 10 20 30 of (49) necessarily 𝑥10 , 𝑥20 ≠ 0. Lemma is proved.
𝑚𝑞 +𝑠𝑞
0 ⋅ ⋅ ⋅ 0 𝑏 𝑏 ⋅ ⋅ ⋅ 𝑏 𝑞0 𝑞1 𝑞,𝑠𝑞 𝑏 d d d d d ... 𝑞 d d d d d 𝑏 𝑞1 1 d d d d 𝑏𝑞0 = 𝑚 +𝑠 +1 𝑐𝑏𝑞 𝑞 𝑞 d d d 0 .. d d . 0 0 𝑏 𝑞 ⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
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 set M 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:
𝑚𝑞 +𝑠𝑞
for every root 𝛼𝑞 ∈ M such that 𝛼𝑞 ∉ M1 and 𝑚𝑞 < 𝑛𝑞 ; (4) 𝑎𝑟 − 𝑎ℎ 1 𝑏𝑟 − 𝑏ℎ = 𝑎𝑟 𝑎ℎ 𝑐 𝑏𝑟 𝑏ℎ
(1) 𝑎𝑝0 = 𝑐𝑏𝑝0 and 𝑎 𝑎 𝑝1 𝑝2 𝑎𝑝0 𝑎𝑝1 d 0
. . . 𝑎𝑝,𝑠𝑝 −1 . . . 𝑎𝑝,𝑠𝑝 −2 d
.. .
d
𝑎𝑝1 𝑎𝑝0
𝑏 𝑏 𝑝1 𝑝2 𝑏𝑝0 𝑏𝑝1 𝑠𝑝 = 𝑐 d 0
. . . 𝑏𝑝,𝑠𝑝 −2 d
.. .
d
𝑏𝑝1 𝑏𝑝0
(5) 𝑎𝑝,𝑚𝑝 2 𝑎𝑝0
𝑏𝑝,𝑠𝑝 𝑏𝑝,𝑠𝑝 −1 .. . , 𝑏𝑖2 𝑏𝑝1
+
1 1 𝑏𝑝,𝑚𝑝 1 = ( 2 + ) 𝑎𝑞 𝑐 𝑏𝑞 𝑏𝑝0
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 𝑠−𝑚𝑝 −1
𝑚𝑝 +𝑡
𝑎 (𝑥) = 𝑎𝑝 + ∑ 𝑎𝑝𝑡 (𝑥 − 𝛼𝑝 ) 𝑡=0
(2) 2 𝑎𝑘0
−
𝑎𝑙,𝑚𝑙 𝑎𝑙02
𝑏𝑙,𝑚 1 𝑏𝑘,𝑚 = ( 2 𝑘 − 2𝑙) 𝑐 𝑏𝑘0 𝑏𝑙0
(68)
for every pair of the roots 𝛼𝑝 , 𝛼𝑞 ∈ M such that 𝛼𝑝 ∈ M1 , 2𝑚𝑝 < 𝑛𝑝 and 𝛼𝑞 ∉ M1 .
(64)
𝑠𝑝 = 1, . . . , 𝑚𝑝 − 1, 𝑚𝑝 + 1, . . . , 𝑛𝑝 − 𝑚𝑝 − 1, for every root 𝛼𝑝 ∈ M1 such that 𝑚𝑝 < 𝑛𝑝 ; 𝑎𝑘,𝑚𝑘
(67)
for every pair of the roots 𝛼𝑟 , 𝛼ℎ ∈ M such that 𝛼𝑟 , 𝛼ℎ ∉ M1 ;
𝑎𝑝,𝑠𝑝 𝑎𝑝,𝑠𝑝 −1 .. . 𝑎𝑖2 𝑎𝑝1
. . . 𝑏𝑝,𝑠𝑝 −1
(66)
𝑠−𝑚𝑝 −1
(65)
𝑏 (𝑥) = 𝑏𝑝 + ∑ 𝑏𝑝𝑡 (𝑥 − 𝛼𝑝 ) 𝑡=0
𝑚𝑝 +𝑡
, 𝑎𝑝0 ≠ 0,
(69)
, 𝑏𝑝0 ≠ 0,
(70)
International Journal of Analysis
11
we compare the coefficients of equal degrees of binomial 𝑥 − 𝛼𝑝 on both sides of the congruence (15). Then we obtain
into degrees of binomial 𝑥 − 𝛼𝑞 for root 𝛼𝑞 ∈ M such that 𝛼𝑞 ∉ M1 and 𝑚𝑞 < 𝑛𝑞 can be written the system 𝑠22 𝑎𝑞 − 𝑠11 𝑏𝑞 − 𝑠12 𝑎𝑞 𝑏𝑞 = 0,
𝑠22 𝑎𝑝𝑡 − 𝑠11 𝑏𝑝𝑡 = 0, 𝑡 = 0, 1, . . . , 𝑙𝑝 − 𝑚𝑝 − 1, 𝑙𝑝 = min (2𝑚𝑝 , 𝑛𝑝 ) .
𝑠22 𝑎𝑞0 − 𝑠11 𝑏𝑞0 − 𝑠12 (𝑎𝑞0 𝑏𝑞 + 𝑏𝑞0 𝑎𝑞 ) = 0,
(71)
𝑠22 𝑎𝑞1 − 𝑠11 𝑏𝑞1 − 𝑠12 (𝑎𝑞1 𝑏𝑞 + 𝑏𝑞1 𝑎𝑞 ) = 0, .. .
From this it follows that 𝑎𝑝0 = 𝑐𝑏𝑝0 and equality (64), where 𝑐 = 𝑠11 /𝑠22 , for all roots 𝛼𝑝 ∈ M1 such that 2𝑚𝑝 ≥ 𝑛𝑝 . Taking into account the roots 𝛼𝑝 ∈ 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 𝑎 𝑏𝑝0 𝑝0 𝑎𝑝1 𝑏𝑝1 .. .. . . 𝑎𝑝,𝑚 −1 𝑏𝑝,𝑚𝑝 −1 𝑝 𝑏𝑝,𝑚𝑝 𝑎𝑝,𝑚𝑝 𝑎𝑝,𝑚 +1 𝑏𝑝,𝑚𝑝 +1 𝑝 .. .. . . 𝑎 𝑝,𝑛𝑝 −𝑚𝑝 −1 𝑏𝑝,𝑛𝑝 −𝑚𝑝 −1
0 .. . 𝑠22 0 −𝑠 𝑎𝑝0 𝑏𝑝0 11 −𝑠12 (72) 𝑎𝑝0 𝑏𝑝1 + 𝑎𝑝1 𝑏𝑝0 .. . 𝑛𝑝 −2𝑚𝑝 −1 ∑ 𝑎𝑝𝑡 𝑏𝑝,𝑛𝑝 −2𝑚𝑝 −1−𝑡 𝑡=0 0
𝑠22 𝑎𝑞,𝑙𝑞 −𝑚𝑞 −1 − 𝑠11 𝑏𝑞,𝑙𝑞 −𝑚𝑞 −1 − 𝑠12 (𝑎𝑞,𝑙𝑞 −𝑚𝑞 −1 𝑏𝑞 + 𝑏𝑞,𝑙𝑞 −𝑚𝑞 −1 𝑎𝑞 ) = 0, where 𝑙𝑞 = min(2𝑚𝑞 , 𝑛𝑞 ). Excluding from this system 𝑠12 , we have 𝑠22
𝑠22 𝑎𝑘,𝑚𝑘 − 𝑠11 𝑏𝑘,𝑚𝑘 − 𝑠12 𝑎𝑘0 𝑏𝑘0 = 0, 𝑠22 𝑎𝑙,𝑚𝑙 − 𝑠11 𝑏𝑙,𝑚𝑙 − 𝑠12 𝑎𝑙0 𝑏𝑙0 = 0.
(73)
By exclusion from these equalities 𝑠12 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)
𝑎𝑞 𝑏𝑞
𝑏𝑞𝑡 − 𝑠11
𝑏𝑞 𝑎𝑞
𝑎𝑞𝑡 = 0,
(75)
or 𝑎𝑞𝑡 𝑎𝑞2
=
1 𝑏𝑞𝑡 , 𝑐 𝑏𝑞2
𝑡 = 0, 1, . . . , 𝑙𝑞 − 𝑚𝑞 − 1,
(76)
where 𝑐 = 𝑠11 /𝑠22 . This implies the equality (66) for the roots 𝛼𝑞 ∈ M such that 𝛼𝑞 ∉ M1 , 𝑚𝑞 < 𝑛𝑞 , and 2𝑚𝑞 ≥ 𝑛𝑞 . If 2𝑚𝑞 < 𝑛𝑞 , then from congruence (15) the system of equalities can be obtained, which can be written in the matrix form
= 0.
Since 𝑠11 , 𝑠22 ≠ 0, by Lemma 5 from the obtained equality we have that the condition (1) of Theorem is completely satisfied. Let 𝛼𝑘 , 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
(74)
𝑠 22 𝐹 𝐺 𝐻 −𝑠 11 = 0, −𝑠 12 where 𝑎 𝑞 𝑎𝑞0 𝑎𝑞1 . . . 𝑎 𝐹 = 𝑞,𝑚𝑞 −1 , 𝑎𝑞,𝑚 𝑞 𝑎𝑞,𝑚𝑞 +1 .. . 𝑎 𝑞,𝑛𝑞 −𝑚𝑞 −1
(77)
12
International Journal of Analysis 𝑏 𝑞 𝑏 𝑞0 𝑏 𝑞1 . .. 𝐺 = 𝑏𝑞,𝑚𝑞 −1 , 𝑏𝑞,𝑚 𝑞 𝑏𝑞,𝑚𝑞 +1 .. . 𝑏𝑞,𝑛 −𝑚 −1 𝑞 𝑞
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 roots 𝛼𝑟 ∈ M𝑟 , 𝛼ℎ ∈ Mℎ such that 𝑟, ℎ ≠ 1, 𝑟 ≠ ℎ, the congruence (15) implies 𝑠22 𝑎𝑟 − 𝑠11 𝑏𝑟 − 𝑠12 𝑎𝑟 𝑏𝑟 = 0, 𝑠22 𝑎ℎ − 𝑠11 𝑏ℎ − 𝑠12 𝑎ℎ 𝑏ℎ = 0. (78)
𝐻 𝑎𝑞 𝑏𝑞 𝑎 𝑏 + 𝑎 𝑏 𝑞0 𝑞 𝑞 𝑞0 𝑎 𝑏 + 𝑎 𝑏 𝑞1 𝑞 𝑞 𝑞1 .. . (79) 𝑏 + 𝑎 𝑏 𝑎 𝑞,𝑚𝑞 −1 𝑞 𝑞 𝑞,𝑚𝑞 −1 . = 𝑎 𝑏 + 𝑎 𝑏 + 𝑎 𝑏 𝑞,𝑚𝑞 𝑞 𝑞0 𝑞0 𝑞 𝑞,𝑚𝑞 𝑎𝑞,𝑚𝑞 +1 𝑏𝑞 + 𝑎𝑞0 𝑏𝑞1 + 𝑎𝑞1 𝑏𝑞0 + 𝑎𝑞 𝑏𝑞,𝑚𝑞 +1 . .. 𝑛𝑞 −2𝑚𝑞 −1 𝑎𝑞,𝑛 −𝑚 −1 𝑏𝑞 + ∑ 𝑎𝑞𝑡 𝑏𝑞,𝑛 −2𝑚 −1−𝑡 + 𝑎𝑞 𝑏𝑞,𝑛 −𝑚 −1 𝑞 𝑞 𝑞 𝑞 𝑞 𝑞 𝑡+0
Since 𝑠11 , 𝑠22 ≠ 0, by Lemma 6 from the obtained equality, we have 0 ⋅ ⋅ ⋅ 0 𝑎 𝑎 ⋅ ⋅ ⋅ 𝑎 𝑞0 𝑞1 𝑞,𝑠𝑞 𝑎 d d d d d ... 𝑞 d d d d d 𝑎 𝑞1 1 d d d d 𝑎𝑞0 𝑚𝑞 +𝑠𝑞 −1 𝑎𝑞 𝑎𝑞0 d d d 0 .. d d . ⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟ 0 𝑎 0 𝑝 𝑚𝑞 +𝑠𝑞
0 ⋅ ⋅ ⋅ 0 𝑏 𝑏 ⋅ ⋅ ⋅ 𝑏 𝑞0 𝑞1 𝑞,𝑠𝑞 𝑏 d d d d d ... 𝑞 d d d d d 𝑏 𝑞1 1 d d d d 𝑏𝑞0 , = 𝑚 +𝑠 −1 𝑏𝑞 𝑞 𝑞 𝑏𝑞0 d d d 0 .. d d . 0 𝑏𝑝 0 ⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟ 𝑚𝑞 +𝑠𝑞
(80)
(81)
Excluding 𝑠12 it follows the relation (67). The condition (4) of Theorem is proved. Let 𝛼𝑝 , 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 𝑠22 𝑎𝑝,𝑚𝑝 − 𝑠11 𝑏𝑝,𝑚𝑝 − 𝑠12 𝑎𝑝0 𝑏𝑝0 = 0, 𝑠22 𝑎𝑞 − 𝑠11 𝑏𝑞 − 𝑠12 𝑎𝑞 𝑏𝑞 = 0
(82)
can 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 𝑎 𝑏𝑝0 𝑝0 𝑎 𝑏𝑝1 𝑝1 . .. .. . 𝑎 𝑝,𝑚𝑝 −1 𝑏𝑝,𝑚𝑝 −1
0 0 𝑥1 .. 𝑥2 = 0, . 𝑥3 0
𝑎 𝑏𝑝,𝑚𝑝 𝑝,𝑚𝑝 𝑎 𝑠𝑝,𝑚𝑝 +1 𝑝,𝑚𝑝 +1 .. .. . . 𝑎 𝑝,𝑛𝑝 −𝑚𝑝 −1 𝑏𝑝,𝑛𝑝 −𝑚𝑝 −1
(83)
𝑎𝑝0 𝑏𝑝1 + 𝑎𝑝1 𝑏𝑝0 𝑥1 .. 𝑥2 . (84) 𝑥3 𝑛𝑝 −2𝑚𝑝 −1 ∑ 𝑎𝑝𝑡 𝑏𝑝,𝑛𝑝 −2𝑚𝑝 −1−𝑡 𝑡=0 𝑎𝑝0 𝑏𝑝0
=0
T for the indeterminate vector-column 𝑥1 𝑥3 𝑥3 . 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 −𝑐 𝑠 T 12 . 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
International Journal of Analysis
13
matrices of these equations are linearly independent and any other of their rows are linear combination of these rows. Since 1 𝑎 0 𝑝0 𝑏𝑝0 −𝑐 𝑎𝑝,𝑚 𝑏𝑝,𝑚 𝑎𝑝0 𝑏𝑝0 −1 𝑝 𝑝 (85) (𝑐𝑏𝑝,𝑚𝑝 − 𝑎𝑝,𝑚𝑝 ) (𝑎𝑝0 𝑏𝑝0 ) = 0,
= (𝑐𝑏𝑙,𝑚𝑙 − 𝑎𝑙,𝑚𝑙 ) (𝑎𝑙0 𝑏𝑙0 )
(86)
holds 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
−1
−1
(87)
𝑛𝑝
(mod (𝑥 − 𝛼𝑝 ) ) ,
(88)
where −1
𝑐1 = (𝑎𝑝,𝑚𝑝 − 𝑐𝑏𝑝,𝑚𝑝 ) (𝑎𝑝0 𝑏𝑝0 )
(89)
for 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 𝑥 1 𝐹 𝐺 𝐻 𝑥 2 = 0, 𝑥 3
(90)
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 1 𝑎 𝑏 𝑎𝑞 𝑏𝑞 𝑞 𝑞 −𝑐 = 0, 𝑎 𝑏 𝑎 𝑏 + 𝑎 𝑏 𝑞0 𝑞0 𝑞0 𝑞 𝑞 𝑞0 −1 (𝑐𝑏𝑞 − 𝑎𝑞 ) (𝑎𝑞 𝑏𝑞 )
(91)
then the column (92)
(94)
where 𝑐2 = (𝑎𝑞 − 𝑐𝑏𝑞 ) (𝑎𝑞 𝑏𝑞 ) ,
𝑎 (𝑥) − 𝑐𝑏 (𝑥) − 𝑐1 𝑎 (𝑥) 𝑏 (𝑥) ≡ 0
is the solution of (90).
(93)
𝑛𝑞
for 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
−1 T 𝑥 𝑥 𝑥 T 1 3 3 = 1 −𝑐 (𝑐𝑏𝑞 − 𝑎𝑞 ) (𝑎𝑞 𝑏𝑞 )
−1
= (𝑐𝑏ℎ − 𝑎ℎ ) (𝑎ℎ 𝑏ℎ )
(mod (𝑥 − 𝛼𝑞 ) ) ,
is the common nonzero solution of (83) and (84). From the condition (2) we deduce the relation (𝑐𝑏𝑘,𝑚𝑘 − 𝑎𝑘,𝑚𝑘 ) (𝑎𝑘0 𝑏𝑘0 )
−1
(𝑐𝑏𝑟 − 𝑎𝑟 ) (𝑎𝑟 𝑏𝑟 )
𝑎 (𝑥) − 𝑐𝑏 (𝑥) − 𝑐2 𝑎 (𝑥) 𝑏 (𝑥) ≡ 0
this means that 𝑥 𝑥 𝑥 T 1 3 3 −1 T = 1 −𝑐 (𝑐𝑏𝑝,𝑚𝑝 − 𝑎𝑝,𝑚𝑝 ) (𝑎𝑝0 𝑏𝑝0 )
−1
From the condition (4) it follows that the relation
(95)
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 (𝑎𝑝,𝑚𝑝 − 𝑐𝑏𝑝,𝑚𝑝 ) (𝑎𝑝0 𝑏𝑝0 )
−1
−1
= (𝑎𝑞 − 𝑐𝑏𝑞 ) (𝑎𝑞 𝑏𝑞 )
(96)
for 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 Theorem 3. The Theorem is proved.
Conflicts of Interest The author declares that there are no conflicts of interest regarding the publication of this paper.
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14 [7] K. D. Ikramov, “Toeplitz-plus-Hankel circulants are reducible to block diagonal form via unitary congruences,” Linear and Multilinear Algebra, vol. 63, no. 4, pp. 862–867, 2015. [8] D. Bini, T. Ehrhardt, A. Y. Karlovich, and I. Spitkovsky, “Large trunkated toeplitz matrices, toeplitz operators, and related topics,” in Operator Theory: Advances and Applications, vol. 259, Birkhauser, Basel, Switzerland, 2017. [9] L. Baratchart, “Un Theoreme de Factorisation et son Application a la Representation des Systemes Cuclique Causaux,” C. R. Acad. Sci. Paris, Ser. 1, Math, vol. 295, no. 3, pp. 223–226, 1982. [10] J. A. Dias da Silva and T. J. Laffey, “On simultaneous similarity of matrices and related questions,” Linear Algebra and its Applications, vol. 291, no. 1-3, pp. 167–184, 1999. [11] B. Z. Shavarovskii, “A complete system of invariants of a second-order matrix with respect to semiscalar equivalence transformations,” Matematychni Methody i Fizyko-Mekhanichni Polya, vol. 13, pp. 3–12, 1981. [12] B. Z. Shavarovskii, “On invariants and canonical form of matrices of second order with respect to semiscalar equivalence,” Buletinul Academiei de Stiinte a Republicii Moldova. Mathematica, vol. 82, no. 3, pp. 12–23, 2016.
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