Determinants of tridiagonal matrices over some commutative finite chain rings

Somphong Jitman; Yosita Sricharoen

doi:10.1515/spma-2023-0114

Article Open Access

Determinants of tridiagonal matrices over some commutative finite chain rings

Somphong Jitman and Yosita Sricharoen

Published/Copyright: February 24, 2024

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From the journal Special Matrices Volume 12 Issue 1

Abstract

Diagonal matrices and their generalization in terms of tridiagonal matrices have been of interest due to their nice algebraic properties and wide applications. In this article, the determinants of tridiagonal matrices over a finite field F q and a commutative finite chain ring R are studied. The main focus is the enumeration of tridiagonal matrices with prescribed determinant. The number of tridiagonal matrices with prescribed determinant over F q and the number of non-singular tridiagonal matrices with prescribed determinant over R are completely determined. For singular tridiagonal matrices with prescribed determinant over R , bounds on the number of such matrices with prescribed determinant are given. Subsequently, the number of some special tridiagonal matrices with prescribed determinant over F q and R is presented.

Keywords: tridiagonal matrices; determinants; finite fields; finite commutative chain rings; enumeration

MSC 2010: 11C20; 15B33

1 Introduction

Due to their fascinating theoretical properties and practical applications, matrices and determinants have drawn attention and been studied (see [3,12] and references therein). The enumeration of n × n singular matrices and nonsingular matrices over a finite field F q has been established in [14]. The number of n × n matrices over the ring Z m of integers modulo m with prescribed determinant has been first studied in [1] as a generalization of matrices over a finite field Z p . In [12], an alternative and simpler method was used to solve the counting problem in [1]. A commutative finite chain ring (CFCR) and a principal ideal ring are generalizations of the rings Z p and Z m that are useful in applications such as coding theory and cryptography. In [4], the techniques in [12] have been extended to matrices over CFCRs and principal ideal rings. Precisely, the number of n × n matrices over CFCRs of a fixed determinant has been completely determined. Such results have been applied in the enumeration of n × n matrices of a fixed determinant over principal ideal rings. Later, these studies are restricted to some subfamilies of matrices. In [11], the enumeration of diagonal matrices over CFCRs with prescribed determinant established and applied in the enumeration of some circulant matrices of a fixed determinant over CFCRs.

A tridiagonal matrix is a square matrix that has nonzero elements only on the main diagonal, the lower diagonal, and the upper diagonal. Precisely, a tridiagonal matrix over a commutative ring R is of the form

A = a 11 a 12 a 21 a 22 ⋱ ⋱ ⋱ a ( n − 1 ) ( n − 1 ) a ( n − 1 ) n a n ( n − 1 ) a n n ,

where a i j ∈ R . Clearly, a tridiagonal matrix is a generalization of a diagonal matrix. Tridiagonal matrices have been of interest due to their nice algebraic properties and wide applications (see [2,5,6,7] and references therein). Analysis on determinants and inverses of some families of tridiagonal matrices is given in [7,13]. Properties of symmetric tridiagonal matrices are presented in [2]. The eigenvalues of tridiagonal matrices are investigated in [5]. In [6], some applications of tridiagonal matrices are discussed.

The key goal of this article is to establish the enumeration of tridiagonal matrices with prescribed determinant over finite fields and over CFCRs. The number of tridiagonal matrices with prescribed determinant over F q and the number of non-singular tridiagonal matrices with prescribed determinant over CFCRs are completely determined. For singular tridiagonal matrices with prescribed determinant over CFCRs, bounds on the number of such matrices with prescribed determinant are given. Subsequently, the number of some special tridiagonal matrices with prescribed determinant is presented.

The article is organized as follows. Some preliminary results and concepts are recalled in Section 2. The enumeration of tridiagonal matrices with prescribed determinant over F q in Section 3. In Section 4, the number and bounds on the number of tridiagonal matrices with prescribed determinant over CFCRs are presented. Summary and remarks are given in Section 5.

2 Preliminaries and notations

In this section, some preliminary results of rings and tridiagonal matrices are recalled.

2.1 CFCRs

A ring R with identity 1 ≠ 0 is called a CFCR if it is finite, commutative, and its ideals are linearly ordered by inclusion. Let U ( R ) and Z ( R ) denote the set of units and the set of zero-divisors of R , respectively.

To be self-contained, brief algebraic properties of a CFCR are discussed as follows. The reader may refer to [8,9,10] for more details. A CFCR is a principal ideal ring with unique maximal ideal. Let R be a CFCR and let γ be a generator of its maximal ideal. The ideals in R are of the forms

R ⊋ γ R ⊋ γ 2 R ⊋ ⋯ ⊋ γ e − 1 R ⊋ γ e R = { 0 } ,

for some positive integer e . The smallest positive integer e such that γ e = 0 is called the nilpotency index of R . The quotient ring R ∕ γ R forms a finite field and it is called the residue field of R . From [9,10], the following properties are useful in this article.

Lemma 2.1

[9,10] Let R be a CFCR of nilpotency index e and let γ be a generator of the maximal ideal of R. Let V ⊆ R be a set of representatives for the equivalence classes of R under congruence modulo γ . Assume that the residue field R ∕ ⟨ γ ⟩ ≅ F q for some prime power q. Then the following statements hold.

For each r ∈ R , there exist unique a 0 , a 1 , … a e − 1 ∈ V such that
r = a 0 + a 1 γ + ⋯ + a e − 1 γ e − 1 .
∣ V ∣ = q .
∣ γ j R ∣ = q e − j for all 0 ≤ j ≤ e .
U ( R ) = { a + γ b ∣ a ∈ V \ { 0 } a n d b ∈ R } .
∣ U ( R ) ∣ = ( q − 1 ) q e − 1 .
For each 0 ≤ i ≤ e , R ∕ γ i R is a CFCR of nilpotency index i and residue field F q .

2.2 Tridiagonal matrices

Let R be a commutative ring with identity and let n be a positive integer. Denote by M n ( R ) and G L n ( R ) the set of n × n matrices over R and the set of n × n non-singular matrices over R , respectively. Let T n ( R ) denote the set of n × n tridiagonal matrices over R . Let N S T n ( R ) = { A ∈ T n ( R ) ∣ det ( A ) ∈ U ( R ) } . For each a ∈ R , let T n ( R , a ) = { A ∈ T n ( R ) ∣ det ( A ) = a } . It is easily seen that T n ( R , a ) s form a partition of T n ( R ) .

For each r , s ∈ R , let

(1) T n , r , s ( R ) = a 1 s r a 2 ⋱ ⋱ ⋱ a n − 1 s r a n ∣ a i ∈ R for all i ∈ { 1 , 2 , … , n }

be the set of n × n special triagonal matrices over R determined by r and s . Let N S T n , r , s ( R ) = { A ∈ T n , r , s ( R ) ∣ det ( A ) ∈ U ( R ) } . For each a ∈ R , let T n , r , s ( R , a ) = { A ∈ T n , r , s ( R ) ∣ det ( A ) = a } .

The following propositions can be derived directly from the definition.

Proposition 2.2

Let n be a positive integer and R be a commutative ring with identity. Let r , s ∈ R . Then the following statements hold.

T n ( R ) is a group under the addition.
T n , r , s ( R ) is a group under the addition if and only if r = 0 = s .

Proposition 2.3

Let n be a positive integer. If R is a finite commutative ring with identity, then

∣ T n ( R ) ∣ = ∣ R ∣ 3 n − 2

and

∣ T n , r , s ( R ) ∣ = ∣ R ∣ n .

3 Tridiagonal matrices with prescribed determinant over finite fields

In this section, the enumeration of tridiagonal matrices with prescribed determinant over finite fields is established. Precisely, the formula for the number of such matrices is first given in a recursive form and followed by its explicit formula. The results for arbitrary tridiagonal matrices are given in Section 3.1 and the enumeration of tridiagonal matrices of special form is presented in Section 3.2.

3.1 Enumeration of tridiagonal matrices with prescribed determinant over finite fields

In this subsection, we focus on the number of n × n tridiagonal matrices with a fixed determinant over F q .

First, the number ∣ N S T n ( F q ) ∣ of n × n non-singular tridiagonal matrices over F q is presented.

Theorem 3.1

Let n be a positive integer and let F q denote a finite field of order q. Then

(2) ∣ N S T n ( F q ) ∣ = ( q − 1 ) q 2 ∣ N S T n − 1 ( F q ) ∣ + ( q − 1 ) 2 q 3 ∣ N S T n − 2 ( F q ) ∣ ,

where ∣ N S T 1 ( F q ) ∣ = ∣ GL 1 ( F q ) ∣ = q − 1 and ∣ N S T 2 ( F q ) ∣ = ∣ GL 2 ( F q ) ∣ = ( q − 1 ) 2 q ( q + 1 ) .

Proof

We note that, for n ∈ { 1 , 2 } , every n × n matrix is tridiagonal. From [14], we have ∣ N S T 1 ( F q ) ∣ = ∣ GL 1 ( F q ) ∣ = q − 1 and ∣ N S T 2 ( F q ) ∣ = ∣ GL 2 ( F q ) ∣ = ( q − 1 ) 2 q ( q + 1 ) .

Assume that n ≥ 3 . Let

A = a 11 a 12 a 21 a 22 ⋱ ⋱ ⋱ a ( n − 1 ) ( n − 1 ) a ( n − 1 ) n a n ( n − 1 ) a n n ∈ T n ( F q ) .

For each i ∈ { 1 , 2 , … , n } , let R i and C i denote the i th row and the i th column of A , respectively. We consider the following two cases.

Case 1: a n n ≠ 0 . Applying the elementary row and column operations R n − 1 − a ( n − 1 ) n a n n − 1 R n → R n − 1 and C n − 1 − a n ( n − 1 ) a n n − 1 C n → C n − 1 on A , it follows that A is equivalent to B and

det ( A ) = det ( B ) = a n n det ( C ) ,

where B = C 0 0 a n n and C = a 11 a 12 a 21 a 22 ⋱ ⋱ ⋱ a ( n − 2 ) ( n − 2 ) a ( n − 2 ) ( n − 1 ) a ( n − 1 ) ( n − 2 ) a ( n − 1 ) ( n − 1 ) − a ( n ) ( n − 1 ) a ( n − 1 ) ( n ) a n n − 1 . Since a n n ≠ 0 , it follows that det ( A ) ≠ 0 if and only if det ( C ) ≠ 0 .

Let S = { D ∈ T n − 1 ( F q ) ∣ det ( D − diag ( 0 , … , 0 , a n ( n − 1 ) a ( n − 1 ) n a n n − 1 ) ) ≠ 0 } . It is not difficult to see that ∣ S ∣ = ∣ N S T n − 1 ( F q ) ∣ and C ∈ N S T n − 1 ( F q ) if and only if C + diag ( 0 , … , 0 , a n ( n − 1 ) a ( n − 1 ) n a n n − 1 ) ∈ S . Precisely, A ∈ N S T n ( F q ) if and only if a ( n − 1 ) ( n ) , a ( n ) ( n − 1 ) ∈ F q and

a 11 a 12 a 21 a 22 ⋱ ⋱ ⋱ a ( n − 2 ) ( n − 2 ) a ( n − 2 ) ( n − 1 ) a ( n − 1 ) ( n − 2 ) a ( n − 1 ) ( n − 1 ) = C + diag ( 0 , … , 0 , a n ( n − 1 ) a ( n − 1 ) n a n n − 1 ) ∈ S .

In this case, the number of non-singular n × n tridiagonal matrices over F q is

( q − 1 ) q 2 ∣ N S T n − 1 ( F q ) ∣ .

Case 2: a n n = 0 . Then

A = a 11 a 12 a 21 a 22 ⋱ ⋱ ⋱ a ( n − 1 ) ( n − 1 ) a ( n − 1 ) n a n ( n − 1 ) 0

and det ( A ) = − a ( n − 1 ) ( n ) a ( n ) ( n − 1 ) det ( A ′ ) , where

A ′ = a 11 a 12 a 21 a 22 ⋱ ⋱ ⋱ a ( n − 3 ) ( n − 3 ) a ( n − 3 ) ( n − 2 ) a ( n − 2 ) ( n − 3 ) a ( n − 2 ) ( n − 2 ) .

It follows that det ( A ) ≠ 0 if and only if a ( n − 1 ) n ≠ 0 , a n ( n − 1 ) ≠ 0 and det ( A ′ ) ≠ 0 . Since a ( n − 1 ) ( n − 2 ) , a ( n − 2 ) ( n − 1 ) , a ( n − 1 ) ( n − 1 ) ∈ F q , a ( n − 1 ) n , a n ( n − 1 ) ∈ F q \ { 0 } , and A ′ ∈ N S T n − 2 ( F q ) , the number of n × n tridiagonal matrices A over F q with a n n = 0 and det ( A ) ≠ 0 is

( q − 1 ) 2 q 3 ∣ N S T n − 2 ( F q ) ∣ .

From the two cases, we have

∣ N S T n ( F q ) ∣ = ( q − 1 ) q 2 ∣ N S T n − 1 ( F q ) ∣ + ( q − 1 ) 2 q 3 ∣ N S T n − 2 ( F q ) ∣

as desired.□

Theorem 3.2

Let n be a positive integer and let F q denote a finite field of order q. Then

∣ N S T n ( F q ) ∣ = x 1 n − 1 ( q − 1 ) ( ( q 2 − 1 ) q − x 2 ) x 1 − x 2 + x 2 n − 1 ( q − 1 ) ( ( q 2 − 1 ) q − x 1 ) x 2 − x 1 ,

where x 1 = ( q − 1 ) 2 ( q 2 + q 4 + 4 q 3 ) and x 2 = ( q − 1 ) 2 ( q 2 − q 4 + 4 q 3 ) .

Proof

Let a n = ∣ N S T n ( F q ) ∣ . From Theorem 3.1, we have a 1 = q − 1 , a 2 = ( q 2 − 1 ) q ( q − 1 ) , and

(3) a n = q 2 ( q − 1 ) ⋅ a n − 1 + ( q − 1 ) 2 ⋅ q 3 ⋅ a n − 2

for all n ≥ 3 . Then the sequence a n is a homogeneous recurrence relation of degree 2 with characteristic polynomial

(4) x 2 − q 2 ( q − 1 ) x − ( q − 1 ) 2 q 3 = 0 .

Using the quadratic formula, the solutions of (4) are

x = q 2 ( q − 1 ) ± ( q 2 ( q − 1 ) ) 2 + 4 ( q − 1 ) 2 q 3 2 = q 2 ( q − 1 ) ± q 3 ( q − 1 ) 2 ( q + 4 ) 2 = q 2 ( q − 1 ) ± q ( q − 1 ) q ( q + 4 ) 2 = ( q − 1 ) 2 ( q 2 ± q 4 + 4 q 3 ) .

Precisely, (4) has two distinct roots x 1 = ( q − 1 ) 2 ( q 2 + q 4 + 4 q 3 ) and x 2 = ( q − 1 ) 2 ( q 2 − q 4 + 4 q 3 ) . It follows that (5) is of the form

(5) a n = r 1 x 1 n + r 2 x 2 n

for some real numbers r 1 and r 2 .

By substituting a 1 = q − 1 and a 2 = ( q 2 − 1 ) q ( q − 1 ) in (5), we have

r 1 x 1 + r 2 x 2 = q − 1 r 1 x 1 2 + r 2 x 2 2 = ( q 2 − 1 ) q ( q − 1 ) .

Solving the above linear system, it can be deduced that

r 1 = ( q 2 − 1 ) q ( q − 1 ) − ( q − 1 ) x 2 x 1 2 − x 1 x 2

and

r 2 = ( q − 1 ) x 1 − ( q 2 − 1 ) q ( q − 1 ) x 1 x 2 − x 2 2 .

Therefore,

a n = x 1 n − 1 ( q − 1 ) ( ( q 2 − 1 ) q − x 2 ) x 1 − x 2 + x 2 n − 1 ( q − 1 ) ( ( q 2 − 1 ) q − x 1 ) x 2 − x 1 ,

where x 1 = ( q − 1 ) 2 ( q 2 + q 4 + 4 q 3 ) and x 2 = ( q − 1 ) 2 ( q 2 − q 4 + 4 q 3 ) .□

Example 1

For q = 2 , the number of non-singular n × n tridiagonal matrices over F 2 is

∣ N S T n ( F 2 ) ∣ = 3 + 2 3 6 ( 2 + 2 3 ) n − 1 + 3 − 2 3 6 ( 2 − 2 3 ) n − 1

by Theorem 3.2. For n ∈ { 1 , 2 , 3 , 4 } , the number of non-singular n × n tridiagonal matrices over F 2 is as follows: ∣ N S T 1 ( F 2 ) ∣ = 1 , ∣ N S T 2 ( F 2 ) ∣ = 6 , ∣ N S T 3 ( F 2 ) ∣ = 32 , and ∣ N S T 4 ( F 2 ) ∣ = 176 .

Since ∣ T n ( F q ) ∣ = q 3 n − 2 , it follows that

(6) ∣ T n ( F q , 0 ) ∣ = q 3 n − 2 − ∣ N S T n ( F q ) ∣ ,

where ∣ N S T n ( F q ) ∣ is determined explicitly in Theorem 3.2.

For a unit a in F q , the following relation between ∣ T n ( F q , 1 ) ∣ and ∣ T n ( F q , a ) ∣ is useful in the enumeration of n × n tridiagonal matrices over F q of determinant a .

Lemma 3.3

Let n be a positive integer and let F q denote a finite field of order q. Then

∣ T n ( F q , a ) ∣ = ∣ T n ( F q , 1 ) ∣

for all a ∈ F q \ { 0 }

Proof

Let a ∈ F q \ { 0 } and let α : T n ( F q , 1 ) → T n ( F q , a ) be defined by

α ( A ) = diag ( a , 1 , … , 1 ) A .

Let

A = a 11 a 12 a 21 a 22 a 23 a 32 ⋱ ⋱ ⋱ ⋱ a ( n − 1 ) n a ( n + 1 ) n a n n ∈ T n ( F q , 1 ) .

Then

α ( A ) = a ⋅ a 11 a ⋅ a 12 a 21 a 22 a 23 a 32 ⋱ ⋱ ⋱ ⋱ a ( n − 1 ) n a ( n + 1 ) n a n n ∈ T n ( F q )

and

det ( α ( A ) ) = det ( diag ( a , 1 , … , 1 ) ) ⋅ det ( A ) = a det ( A ) = a ⋅ 1 = a ,

which implies that α ( A ) ∈ T n ( F q , a ) . Hence, α is a well-defined function from T n ( F q , 1 ) to T n ( F q , a ) .

Since a is a unit, diag ( a , 1 , … , 1 ) is invertible, which implies that α is injective. For each B ∈ T n ( F q , a ) , it is easily seen that diag ( a − 1 , 1 , … , 1 ) B ∈ T n ( F q , 1 ) and α ( diag ( a − 1 , 1 , … , 1 ) B ) = B . It follows that α is surjective.

Hence, α is a bijection from T n ( F q , 1 ) onto T n ( F q , a ) . Consequently, we have ∣ T n ( F q , 1 ) ∣ = ∣ T n ( F q , a ) ∣ as desired.□

Theorem 3.4

Let n be a positive integer and let F q denote a finite field of order q. Let a ∈ F q \ { 0 } . Then

∣ T n ( F q , a ) ∣ = x 1 n − 1 ( ( q 2 − 1 ) q − x 2 ) x 1 − x 2 + x 2 n − 1 ( ( q 2 − 1 ) q − x 1 ) x 2 − x 1 ,

where x 1 = ( q − 1 ) 2 ( q 2 + q 4 + 4 q 3 ) and x 2 = ( q − 1 ) 2 ( q 2 − q 4 + 4 q 3 ) .

Proof

We note that

N S T n ( F q ) = ⋃ b ∈ F q \ { 0 } T n ( F q , b )

is a disjoint union. By Lemma 3.3, it follows that

∣ N S T n ( F q ) ∣ = ∣ ⋃ b ∈ F q \ { 0 } T n ( F q , b ) ∣ = ∑ b ∈ F q \ { 0 } ∣ T n ( F q , b ) ∣ = ∑ b ∈ F q \ { 0 } ∣ T n ( F q , 1 ) ∣ = ( q − 1 ) ∣ T n ( F q , 1 ) ∣ .

Hence, ∣ T n ( F q , a ) ∣ = ∣ T n ( F q , 1 ) ∣ = ∣ N S T n ( F q ) ∣ q − 1 . By Theorem 3.2, it can be deduced that

∣ T n ( F q , a ) ∣ = ∣ N S T n ( F q ) ∣ q − 1 = x 1 n − 1 ( q − 1 ) ( ( q 2 − 1 ) q − x 2 ) x 1 − x 2 + x 2 n − 1 ( q − 1 ) ( ( q 2 − 1 ) q − x 1 ) x 2 − x 1 q − 1 = x 1 n − 1 ( ( q 2 − 1 ) q − x 2 ) x 1 − x 2 + x 2 n − 1 ( ( q 2 − 1 ) q − x 1 ) x 2 − x 1

as desired.□

3.2 Enumeration of special tridiagonal matrices with prescribed determinant over finite fields

In this subsection, we focus on the number of n × n tridiagonal matrices of special forms defined in (1) with prescribed determinant over F q .

The following lemma can be deduced directly from Proposition 2.3.

Lemma 3.5

Let n be a positive integer and let F q denote a finite field of order q. Then

∣ T n , r , s ( F q ) ∣ = q n

for all r , s ∈ F q .

In the case where r = 0 or s = 0 , the number of tridiagonal matrices in T n , r , s ( F q ) with prescribed determinant coincides with the number of n × n matrices with prescribed determinant over F q given in [11]. The result is summarized as follows.

Lemma 3.6

Let n be a positive integer and let F q be a finite field of order q. Let r , s ∈ F q . If r = 0 or s = 0 , then

∣ N S T n , r , s ( F q ) ∣ = ( q − 1 ) n , ∣ T n , r , s ( F q , 0 ) ∣ = q n − ( q − 1 ) n , a n d ∣ T n , r , s ( F q , a ) ∣ = ∣ T n , r , s ( F q , 1 ) ∣ = ( q − 1 ) n − 1

for all a ∈ F q \ { 0 } .

Next, we focus on the case where r ≠ 0 and s ≠ 0 . A recursive formula for ∣ N S T n , r , s ( F q ) ∣ is given in the next theorem.

Theorem 3.7

Let n be a positive integer and let F q be a finite field of order q. Let r , s ∈ F q \ { 0 } . Then ∣ N S T 1 , r , s ( F q ) ∣ = q − 1 , ∣ N S T 2 , r , s ( F q ) ∣ = q 2 − q + 1 , and

∣ N S T n , r , s ( F q ) ∣ = ( q − 1 ) ∣ N S T n − 1 , r , s ( F q ) ∣ + q ∣ N S T n − 2 , r , s ( F q ) ∣

for all n ≥ 3 .

Proof

For n = 1 and n = 2 , it is clear that

∣ N S T 1 , r , s ( F q ) ∣ = ∣ N S T 1 ( F q ) ∣ = ∣ { [ x 1 ] ∣ det ( [ x 1 ] ) ≠ 0 } ∣ = q − 1

and

∣ N S T 2 , r , s ( F q ) ∣ = x 1 s r x 2 ∣ det x 1 s r x 2 ≠ 0 = ∣ { ( x 1 , x 2 ) ∈ F q 2 ∣ x 1 x 2 ≠ r s } ∣ = ∣ F q 2 \ { ( x 1 , x 2 ) ∈ F q 2 ∣ x 1 x 2 = r s } ∣ = q 2 − ( q − 1 ) = q 2 − q + 1 .

Assume that n ≥ 3 . Let

X = x 1 s r x 2 s r x 3 ⋱ ⋱ ⋱ s r x ( n − 1 ) s r x n ∈ T n , r , s ( F q ) .

The proof is separated into two cases where x n ≠ 0 and where x n = 0 .

Case 1: x n ≠ 0 . By applying the elementary row and column operations R n − 1 − s x n − 1 R n → R n − 1 and C n − 1 − r x n − 1 C n → C n − 1 on X , it follows that

X ∼ Y 0 0 x n ,

where

Y = x 1 s r x 2 s r x 3 ⋱ ⋱ ⋱ s r x ( n − 1 ) − r ⋅ s ⋅ x n − 1 .

Then det ( X ) = x n det ( Y ) , which implies that det ( X ) ≠ 0 if and only if det ( Y ) ≠ 0 .

Let V = { W ∈ T n − 1 , r , s ( F q ) ∣ det ( W − diag ( 0 , … , 0 , r ⋅ s ⋅ x n − 1 ) ) ≠ 0 } . It is not difficult to see that ∣ V ∣ = ∣ N S T n − 1 , r , s ( F q ) ∣ and Y ∈ N S T n − 1 , r , s ( F q ) if and only if

x 1 s r x 2 s r x 3 ⋱ ⋱ ⋱ s r x ( n − 1 ) = Y + diag ( 0 , … , 0 , r ⋅ s ⋅ x n − 1 ) ∈ V .

Since the choice for x n is q − 1 , the number of matrices X ∈ T n , r , s ( F q ) such that x n ≠ 0 and det ( X ) ≠ 0 is

( q − 1 ) ∣ V ∣ = ( q − 1 ) ∣ N S T n − 1 , r , s ( F q ) ∣ .

Case 2: x n = 0 . Then

X = x 1 s r x 2 s r x 3 ⋱ ⋱ ⋱ s r x n − 1 s r 0

and det ( X ) = s ⋅ r ⋅ det ( X ′ ) , where

X ′ = x 1 s r x 2 ⋱ ⋱ ⋱ x n − 3 s r x n − 2 .

It follows that det ( X ) ≠ 0 if and only if det ( X ′ ) ≠ 0 . Precisely, x n − 1 ∈ F q and X ′ ∈ N S T n − 2 , r , s ( F q ) . Hence, the number of matrices X ∈ T n , r , s ( F q ) such that x n = 0 and det ( X ) ≠ 0 is

q ∣ N S T n − 2 , r , s ( F q ) ∣ .

In summary, the number of matrices in N S T n , r , s ( F q ) is

∣ N S T n , r , s ( F q ) ∣ = ( q − 1 ) ∣ N S T n − 1 , r , s ( F q ) ∣ + q ∣ N S T n − 2 , r , s ( F q ) ∣

as desired.□

From the recurrent formula of ∣ N S T n , r , s ( F q ) ∣ in Theorem 3.7, an explicit formula is derived in the next theorem.

Theorem 3.8

Let n be a positive integer and let F q denote a finite field of order q. Let r , s ∈ F q \ { 0 } . Then

∣ N S T n , r , s ( F q ) ∣ = q n + 1 + ( − 1 ) n 1 + q .

Proof

Let c n = ∣ N S T ( n , r , s ) ( F q ) ∣ . From Theorem 3.7, we have c 1 = q − 1 , c 2 = q 2 − q + 1 , and

c n = ( q − 1 ) ⋅ c n − 1 + q ⋅ c n − 2

for all n ≥ 3 . Then c n is a linear recurrent relation of order 2 with the characteristic polynomial

(7) x 2 − ( q − 1 ) x − q = 0 .

It follows that (7) has two distinct real roots x 1 = q and x 2 = − 1 , and hence,

(8) c n = r 1 q n + r 2 ( − 1 ) n

for some real numbers r 1 and r 2 . Substituting c 1 = p − 1 and c 2 = p 2 − p + 1 in (8), we have

p − 1 = r 1 q − r 2 , p 2 − p + 1 = r 1 q 2 + r 2 .

By solving the linear system, we have r 1 = q q + 1 and r 2 = 1 q + 1 . Hence,

c n = q n + 1 + ( − 1 ) n 1 + q

as desired.□

From Theorem 3.8, it is easily seen that the formula for ∣ N S T n , r , s ( F q ) ∣ is independent of r and s .

From Lemma 3.5 and Theorem 3.8, the number

∣ T n , r , s ( F q , 0 ) ∣ = ∣ T n , r , s ( F q ) ∣ − ∣ N S T n , r , s ( F q ) ∣ = q n − q n + 1 + ( − 1 ) n 1 + q = q n − ( − 1 ) n 1 + q

is summarized in the next corollary.

Corollary 3.9

Let n be a positive integer and let q be a prime power. Let r , s ∈ F q \ { 0 } . Then

∣ T n , r , s ( F q , 0 ) ∣ = q n − ( − 1 ) n 1 + q .

Unlike the previous subsection, for an element a ∈ F q \ { 0 } , a relation between T n , r , s ( F q , a ) and T n , r , s ( F q , 1 ) is given only the case where n is odd.

Theorem 3.10

Let n be a positive integer and let F q denote a finite field of order q. Let r , s ∈ F q \ { 0 } . If n is odd, then

∣ T n , r , s ( F q , a ) ∣ = ∣ T n , r , s ( F q , 1 ) ∣

for all a ∈ F q \ { 0 }

Proof

Assume that n is odd. Let a ∈ F q \ { 0 } and let

ϕ : T n , r , s ( F q , 1 ) → T n , r , s ( F q , a )

be defined by

ϕ ( A ) = diag ( a , 1 , a , … , 1 , a ) A diag ( 1 , a − 1 , 1 , … , a − 1 , 1 ) .

Let

A = a 1 s r a 2 s r a 3 ⋱ ⋱ ⋱ s r a ( n − 1 ) s r a n ∈ T n , r , s ( F q , 1 ) .

Then

ϕ ( A ) = a ⋅ a 1 s r a − 1 ⋅ a 2 s r a ⋅ a 3 ⋱ ⋱ ⋱ s r a − 1 ⋅ a n − 1 s r a ⋅ a n ∈ T n , r , s ( F q ) .

Since n is odd, we have

det ( ϕ ( A ) ) = a n + 1 2 ⋅ 1 ⋅ ( a − 1 ) n − 1 2 = a ,

which implies that

ϕ ( A ) ∈ T n , r , s ( F q , a ) .

Hence, ϕ is a well-defined map from T n , r , s ( F q , 1 ) to T n , r , s ( F q , a ) .

Since diag ( a , 1 , a , … , 1 , a ) and diag ( 1 , a − 1 , 1 , … , a − 1 , 1 ) are invertible, the map ϕ is injective. For each Y ∈ T n , r , s ( F q , a ) , it is not difficult to see that X = diag ( a − 1 , 1 , a − 1 , … , 1 , a − 1 ) Y diag ( 1 , a , 1 , … , a , 1 ) ∈ T n , r , s ( F q , 1 ) and ϕ ( X ) = Y . Hence, ϕ surjective.

Therefore, ϕ is a bijection from T n , r , s ( F q , 1 ) onto T n , r , s ( F q , a ) . Consequently, we have ∣ T n , r , s ( F q , a ) ∣ = ∣ T n , r , s ( F q , 1 ) ∣ .□

Using Theorems 3.7 and 3.10, ∣ T n , r , s ( F q , a ) ∣ is determined in the next theorem for all a ∈ F q \ { 0 } and for all odd positive integers n .

Theorem 3.11

Let n be a positive integer and let F q denote a finite field of order q. Let r , s , a ∈ F q \ { 0 } . If n is odd, then

∣ T n , r , s ( F q , a ) ∣ = q n + 1 − 1 q 2 − 1 .

Proof

Assume that n is odd. Since

N S T n , r , s ( F q ) = ⋃ b ∈ F q \ { 0 } T n , r , s ( F q , b )

is a disjoint union, it follows that

∣ N S T n , r , s ( F q ) ∣ = ∣ ⋃ b ∈ F q \ { 0 } T n , r , s ( F q , b ) ∣ = ∑ b ∈ F q \ { 0 } ∣ T n , r , s ( F q , b ) ∣ = ( q − 1 ) ∣ T n , r , s ( F q , 1 ) ∣ .

Applying Theorems 3.7 and 3.10, we have

∣ T n , r , s ( F q , a ) ∣ = ∣ T n , r , s ( F q , 1 ) ∣ = ∣ N S T n , r , s ( F q ) ∣ ( q − 1 ) = q n + 1 + ( − 1 ) n ( q + 1 ) ( q − 1 ) = q n + 1 − 1 q 2 − 1

as desired.□

Remark 1

A direct calculation shows that

∣ T 4 , 1 , 1 ( F 3 , 1 ) ∣ = 35 ≠ 26 = ∣ T 4 , 1 , 1 ( F 3 , 2 ) ∣

and

∣ T 6 , 1 , 1 ( F 3 , 1 ) ∣ = 295 ≠ 313 = ∣ T 6 , 1 , 1 ( F 3 , 2 ) ∣ .

Hence, ∣ T n , r , s ( F q , a ) ∣ and ∣ T n , r , s ( F q , 1 ) ∣ do not need to be equal for an even integer n . The determination of ∣ T n , r , s ( F q , a ) ∣ remains open for an even positive integer n .

4 Determinants of tridiagonal matrices over finite chain rings

In this section, we focus on the enumeration of tridiagonal matrices over a CFCR R with prescribed determinant. The results for arbitrary tridiagonal matrices over R are given in Section 4.1 and the enumeration of special tridiagonal matrices over R is presented in Section 4.2.

Let R be a CFCR of nilpotency index e and residue field F q . It is well known that R is a disjoint union of { 0 } , U ( R ) , and Z ( R ) . Let γ be a generator of the maximal ideal of R . By Lemma 2.1, an element a ∈ R can be written as a = γ s b for some 0 ≤ s ≤ e and b ∈ U ( R ) . Precisely, a = 0 if s = e , a = γ s b ∈ Z ( R ) if 1 ≤ s ≤ e − 1 , and a ∈ ( U ( R ) ) if s = 0 .

Theorem 4.1

Let R be a CFCR of nilpotency index e and residue field F q . Let γ be a generator of the maximal ideal of R. If s is an integer such that 0 ≤ s ≤ e , then

∣ T n ( R , γ s ) ∣ = ∣ T n ( R , b γ s ) ∣

for all units b in U ( R ) .

Proof

Let b be a unit in U ( R ) and let 0 ≤ s ≤ e be an integer. If s = e , then γ s = 0 = γ s b , and hence, we have ∣ T n ( R , γ s ) ∣ = ∣ T n ( R , 0 ) ∣ = ∣ T n ( R , b γ s ) ∣ .

For each 0 ≤ s < e , it is not difficult to see that, for each A ∈ T n ( R , γ s ) , det ( A ) = γ s if and only if det ( diag ( b , 1 , … , 1 ) A ) = b γ s . Hence, the map α : T n ( R , γ s ) → T n ( R , b γ s ) defined by

α ( A ) = diag ( b , 1 , … , 1 ) A

is well-defined. Since b is a unit, the matrix diag ( b , 1 , … , 1 ) is invertible. Using arguments similar to those in Lemma 3.3, it follows that α is a bijection. As desired, we have ∣ T n ( R , γ s ) ∣ = ∣ T n ( R , b γ s ) ∣ .□

For units in R , we have the following relation. The result can be obtained by setting s = 0 in Theorem 4.1.

Corollary 4.2

Let R be a CFCR and let n be a positive integer. Then

∣ T n ( R , a ) ∣ = ∣ T n ( R , 1 ) ∣

for all units a ∈ U ( R ) .

4.1 Enumeration of tridiagonal matrices over finite chain rings

For non-singular tridiagonal matrices, the number of n × n tridiagonal matrices with prescribed determinant a over R is given for positive integers n and for all units a in R . For singular tridiagonal matrices, relations and bounds on the number of n × n tridiagonal matrices with prescribed determinant over R are presented.

4.1.1 Determinants of non-singular tridiagonal matrices over finite chain rings

In this subsection, we focus on ∣ T n ( R , a ) ∣ in the case where a is a unit in U ( R ) . By Corollary 4.2, it is suffice to determine only ∣ T n ( R , 1 ) ∣ .

The number of n × n non-singular tridiagonal matrices over R is derived in the following theorem.

Theorem 4.3

Let R be a CFCR of nilpotency index e and residue field F q and let n be a positive integer. Then

∣ N S T n ( R ) ∣ = q ( e − 1 ) ( 3 n − 2 ) x 1 n − 1 ( q − 1 ) ( ( q 2 − 1 ) q − x 2 ) x 1 − x 2 + x 2 n − 1 ( q − 1 ) ( ( q 2 − 1 ) q − x 1 ) x 2 − x 1 ,

where x 1 = ( q − 1 ) 2 ( q 2 + q 4 + 4 q 3 ) and x 2 = ( q − 1 ) 2 ( q 2 − q 4 + 4 q 3 ) .

Proof

Let γ be a generator of the maximal ideal of R and let ψ : R → F q be the group homomorphism defined by a ↦ a + γ R . Since T n ( R ) and T n ( F q ) are groups under the usual addition of matrices, let Ψ : T n ( R ) → T n ( F q ) be the map defined by

A = [ a i j ] ↦ [ ψ ( a i j ) ] .

It is not difficult to see that Ψ is a surjective group homomorphism. Using the First Isomorphism Theorem, it follows that T n ( F q ) ≅ T n ( R ) ∕ ker ( Ψ ) . Hence,

∣ ker ( Ψ ) ∣ = ∣ T n ( R ) ∣ ∣ T n ( F q ) ∣ = q e ( 3 n − 2 ) q 3 n − 2 = q ( e − 1 ) ( 3 n − 2 ) .

For A ∈ T n ( R ) , it is easily seen that det ( Ψ ( A ) ) = ψ ( det ( A ) ) . Equivalently, det ( A ) ∈ U ( R ) if and only if det ( Ψ ( A ) ) ≠ 0 in F q , which implies that A is non-singular over R if and only if Ψ ( A ) is non-singular over F q . Then the restriction map Ψ ∣ N S T n ( R ) : N S T n ( R ) → N S T n ( F q ) is surjective. From Theorem 3.1, we have

∣ N S T n ( R ) ∣ = ∣ ker ( Ψ ) ∣ ∣ N S T n ( F q ) ∣ = q ( e − 1 ) ( 3 n − 2 ) x 1 n − 1 ( q − 1 ) ( ( q 2 − 1 ) q − x 2 ) x 1 − x 2 + x 2 n − 1 ( q − 1 ) ( ( q 2 − 1 ) q − x 1 ) x 2 − x 1

as desired.□

Corollary 4.4

Let R be a CFCR with residue field F q and nilpotency index e and let n be a positive integer. Then

∣ T n ( R , a ) ∣ = q 3 ( e − 1 ) ( n − 1 ) x 1 n − 1 ( ( q 2 − 1 ) q − x 2 ) x 1 − x 2 + x 2 n − 1 ( q 2 − 1 ) q − x 1 x 2 − x 1 ,

where x 1 = ( q − 1 ) 2 ( q 2 + q 4 + 4 q 3 ) and x 2 = ( q − 1 ) 2 ( q 2 − q 4 + 4 q 3 ) for all a ∈ U ( R ) .

Proof

The set N S T n ( R ) is a disjoint union of T n ( R , a ) for all a ∈ U ( R ) , i.e.,

N S T n ( R ) = ⋃ a ∈ U ( R ) T n ( R , a ) .

By Corollary 4.2, we have ∣ T n ( R , a ) ∣ = ∣ T n ( R , 1 ) ∣ . Consequently,

∣ N S T n ( R ) ∣ = ∣ ⋃ a ∈ U ( R ) T n ( R , a ) ∣ = ∑ a ∈ U ( R ) ∣ T n ( R , a ) ∣ = ∑ a ∈ U ( R ) ∣ T n ( R , 1 ) ∣ = ∣ U ( R ) ∣ ∣ T n ( R , 1 ) ∣ .

From Lemma 2.1, we have ∣ U ( R ) ∣ = ( q − 1 ) q e − 1 . By Theorem 4.3, it can be deduced that

∣ T n ( R , a ) ∣ = ∣ T n ( R , 1 ) ∣ = ∣ N S T n ( R ) ∣ ∣ U ( R ) ∣ = q ( e − 1 ) ( 3 n − 2 ) x 1 n − 1 ( q − 1 ) ( ( q 2 − 1 ) q − x 2 ) x 1 − x 2 + x 2 n − 1 ( q − 1 ) ( ( q 2 − 1 ) q − x 1 ) x 2 − x 1 ( q − 1 ) q e − 1 = q 3 ( e − 1 ) ( n − 1 ) x 1 n − 1 ( ( q 2 − 1 ) q − x 2 ) x 1 − x 2 + x 2 n − 1 ( q 2 − 1 ) q − x 1 x 2 − x 1 .

The proof is completed.□

4.1.2 Determinants of singular tridiagonal matrices over finite chain rings

First, we focus on n × n tridiagonal matrices over R with zero determinant. In this case, a lower bound for ∣ T n ( R , 0 ) ∣ is presented.

Theorem 4.5

Let R be a CFCR with residue field F q and nilpotency index e and let n be a positive integer. Then

∣ T n ( R , 0 ) ∣ ≥ ( q − 1 ) q 2 ( e − 1 ) ( q e + 1 + 1 ) ∣ T n − 1 ( R , 0 ) ∣ + q 3 n − 4 ∣ T n ( R ∕ γ e − 1 R , 0 + γ e − 1 R ) ∣ .

Proof

Let

A = a 11 a 12 a 21 a 22 a 23 a 32 ⋱ ⋱ ⋱ ⋱ a ( n − 1 ) n a ( n + 1 ) n a n n ∈ T n ( R , 0 ) .

We consider the three cases. (1) a n n ∈ U ( R ) , (2) a n n , a ( n − 1 ) n ∉ U ( R ) , and (3) a n n ∉ U ( R ) and a ( n − 1 ) n ∈ U ( R ) .

Case 1: a n n ∈ U ( R ) . Applying the elementary row and column operations R n − 1 − a ( n − 1 ) n a n n − 1 R n → R n − 1 and C n − 1 − a n ( n − 1 ) a n n − 1 C n → C n − 1 on A , it follows that A is equivalent to B and

det ( A ) = det ( B ) = a n n det ( C ) ,

where B = C 0 0 a n n and C = a 11 a 12 a 21 a 22 ⋱ ⋱ ⋱ a ( n − 2 ) ( n − 2 ) a ( n − 2 ) ( n − 1 ) a ( n − 1 ) ( n − 2 ) a ( n − 1 ) ( n − 1 ) − a ( n ) ( n − 1 ) a ( n − 1 ) ( n ) a n n − 1 . Since a n n ∈ U ( R ) , det ( A ) = 0 if and only if det ( C ) = 0 .

Let S = { D ∈ T n − 1 ( R ) ∣ det ( D − diag ( 0 , … , 0 , a n ( n − 1 ) a ( n − 1 ) n a n n − 1 ) ) = 0 } . It is not difficult to see that ∣ S ∣ = ∣ T n − 1 ( R , 0 ) ∣ and C ∈ T n − 1 ( R , 0 ) if and only if C + diag ( 0 , … , 0 , a n ( n − 1 ) a ( n − 1 ) n a n n − 1 ) ∈ S . Precisely, A ∈ T n ( R , 0 ) if and only if a ( n − 1 ) ( n ) , a ( n ) ( n − 1 ) ∈ R and

In this case, the number of n × n tridiagonal matrices A over R with det ( A ) = 0 is

q 3 e − 1 ( q − 1 ) ⋅ ∣ T n − 1 ( R , 0 ) ∣ .

Case 2: a n n , a ( n − 1 ) n ∉ U ( R ) . Then the elements in the last column of A are in the maximal ideal γ R . Let V be defined as in Lemma 2.1 and let B = [ b i j ] ∈ T n ( R ) be defined by

b i j = w i j if ( i , j ) ∈ { ( n − 1 , n ) , ( n , n ) } a i j otherwise,

where a n − 1 , n = γ w n − 1 , n and a n n = γ w n n for some w n − 1 , n , w n n ∈ ∑ j = 0 e − 2 γ j V . Let C = [ c i j ] be the matrix in T n ( R ∕ γ e − 1 R ) defined by c i j = b i j + γ e − 1 R . Then det ( A ) = γ det ( B ) ∈ R and det ( A ) = 0 in R if and only if det ( B ) ∈ γ e − 1 R . Equivalently, det ( C ) = 0 + γ e − 1 R in R ∕ γ e − 1 R . For each C ∈ T n ( R ∕ γ e − 1 R , 0 + R ∕ γ e − 1 R ) , there are q 3 n − 4 corresponding matrices B ∈ T n ( R , 0 ) . The number of matrices C is ∣ T n ( R ∕ γ e − 1 R , 0 + R ∕ γ e − 1 R ) ∣ and A is uniquely determined by B , the number of choices for A is

q 3 n − 4 ∣ T n ( R ∕ γ e − 1 R , 0 + R ∕ γ e − 1 R ) ∣ .

Case 3: a n n ∉ U ( R ) and a ( n − 1 ) n ∈ U ( R ) . Based on the co-factor expansion through the last column, we have

det ( A ) = a n n det a 11 a 12 a 21 a 22 ⋱ ⋱ ⋱ a ( n − 2 ) ( n − 2 ) a ( n − 2 ) ( n − 1 ) a ( n − 1 ) ( n − 2 ) a ( n − 1 ) ( n − 1 ) − a ( n − 1 ) n a n ( n − 1 ) det a 11 a 12 a 21 a 22 ⋱ ⋱ ⋱ a ( n − 3 ) ( n − 3 ) a ( n − 3 ) ( n − 2 ) a ( n − 2 ) ( n − 3 ) a ( n − 2 ) ( n − 2 )

It is easily seen that if det a 11 a 12 a 21 a 22 ⋱ ⋱ ⋱ a ( n − 2 ) ( n − 2 ) a ( n − 2 ) ( n − 1 ) a ( n − 1 ) ( n − 2 ) a ( n − 1 ) ( n − 1 ) = 0 and a n ( n − 1 ) = 0 , then det ( A ) = 0 . Since a n n ∉ U ( R ) , a ( n − 1 ) n ∈ U ( R ) , the number of n × n tridiagonal matrices whose determinant is greater than or equal to

( q − 1 ) q 2 ( e − 1 ) ∣ T n − 1 ( R , 0 ) ∣ .

From the three cases, we have

∣ T n ( R , 0 ) ∣ ≥ ( q − 1 ) q 2 ( e − 1 ) ( q e + 1 + 1 ) ∣ T n − 1 ( R , 0 ) ∣ + q 3 n − 4 ∣ T n ( R ∕ γ e − 1 R , 0 + γ e − 1 R ) ∣

as desired.□

For a CFCR of nilpotency index 2, we have the following bound.

Corollary 4.6

Let R be a CFCR of nilpotency index 2 and residue field F q . If γ is a generator of the maximal ideal of R, then ∣ T 1 ( R , 0 ) ∣ = 1 and

∣ T n ( R , 0 ) ∣ ≥ ( q − 1 ) q 2 ( q 3 + 1 ) ∣ T n − 1 ( R , 0 ) ∣ + q 3 n − 4 ∣ T n ( F q , 0 ) ∣

for all integers n ≥ 2 .

Proof

Clearly, ∣ T 1 ( R , 0 ) ∣ = 1 . Let n ≥ 2 be an integer. We note that R ∕ γ e − 1 R ≅ F q . From Theorem 4.5, we have

∣ T n ( R , 0 ) ∣ ≥ ( q − 1 ) q 2 ( q 3 + 1 ) ∣ T n − 1 ( R , 0 ) ∣ + q 3 n − 4 ∣ T n ( F q , 0 ) ∣

as desired.□

We note that ∣ T n ( F q , 0 ) ∣ is computed explicitly in (6).

Lemma 4.7

Let R be a CFCR of nilpotency index e ≥ 3 and residue field F q and let n be a positive integer. If γ is a generator of the maximal ideal of R, then

∣ T n ( R , γ s ) ∣ = q 3 ( n − 1 ) ∣ T n ( R ∕ γ e − 1 R , γ s + γ e − 1 R ) ∣

for all 1 ≤ s < e − 1 .

Proof

Let s be an integer such that 1 ≤ s < e − 1 and let β : T n ( R ) → T n ( R ∕ γ e − 1 R ) be defined by

β ( A ) = A ¯ ,

where [ a i j ] ¯ ≔ [ a i j + γ e − 1 R ] for all [ a i j ] ∈ T n ( R ) . Then β is an additive group homomorphism. For each A ∈ T n ( R ) , we have that det ( β ( A ) ) = γ s + γ e − 1 R if and only if det ( A ) = γ s + γ e − 1 b for some b ∈ V , where V is given in Lemma 2.1. Since 1 ≤ e − s − 1 < e − 1 , it follows that 1 + γ e − s − 1 b ∈ U ( R ) . Hence,

∣ { A ∈ T n ( R ) ∣ det ( A ) = γ s + γ e − 1 b for some b ∈ V } ∣ = ∣ { A ∈ T n ( R ) ∣ det ( A ) = γ s ( 1 + γ e − s − 1 b ) for some b ∈ V } ∣ = ∣ { A ∈ T n ( R ) ∣ det ( A ) = γ s } ∣ = ∣ T n ( R , γ s ) ∣ .

Equivalently,

(9) ∣ { A ∈ T n ( R ) ∣ det ( β ( A ) ) = γ s + γ e − 1 R } ∣ = ∣ V ∣ ∣ T n ( R , γ s ) ∣ = q ∣ T n ( R , γ s ) ∣ .

Since ∣ ker ( β ) ∣ = q 3 n − 2 , we have

(10) ∣ { A ∈ T n ( R ) ∣ det ( β ( A ) ) = γ s + γ e − 1 R } ∣ = ∣ ker ( β ) ∣ ∣ { B ∈ T n ( R ∕ γ e − 1 R ) ∣ det ( B ) = γ s + γ e − 1 R } ∣ = q 3 n − 2 ∣ T n ( R ∕ γ e − 1 R , γ s + γ e − 1 R ) ∣ .

Combining (9) and (10), it can be concluded that

q ∣ T n ( R , γ s ) ∣ = q 3 n − 2 ∣ T n ( R ∕ γ e − 1 R , γ s + γ e − 1 R ) ∣ .

Therefore,

∣ T n ( R , γ s ) ∣ = q 3 ( n − 1 ) ∣ T n ( R ∕ γ e − 1 R , γ s + γ e − 1 R ) ∣

as desired.□

Applying Lemma 4.7 recursively, the next corollary follows.

Corollary 4.8

Let R be a CFCR of nilpotency index e + f and residue field F q , where 2 ≤ e and 1 ≤ f are integers. If the maximal ideal of R is generated by γ , then

∣ T n ( R , γ s ) ∣ = q 3 f ( n − 1 ) ∣ T n ( R ∕ γ e R , γ s + γ e R ) ∣

for all 1 ≤ s < e .

The number of n × n tridiagonal matrices over a R with determinant γ s is presented in the next theorem.

Theorem 4.9

Let R be a CFCR of nilpotency index e and residue field F q and let n be a positive integer. If the maximal ideal of R is generated by γ , then

∣ T n ( R , γ s ) ∣ = q 3 ( e − s − 1 ) ( n − 1 ) q − 1 ( q 3 n − 2 ∣ T n ( R ∕ γ s R , 0 + γ s R ) ∣ − ∣ T n ( R ∕ γ s + 1 R , 0 + γ s + 1 R ) ∣ ) .

for all integers 1 ≤ s < e .

Proof

Let 1 ≤ s < e be an integer and let μ : T n ( R ∕ γ s + 1 R ) → T n ( R ∕ γ s R ) be a ring homomorphism defined by

μ ( A ) = A ¯

where [ a i j + γ s + 1 R ] ¯ ≔ [ a i j + γ s R ] for all [ a i j + γ s + 1 R ] ∈ T n ( R ∕ γ s + 1 R ) . Then, for each A ∈ T n ( R ∕ γ s + 1 R ) , det ( μ ( A ) ) = 0 + γ s R if and only if det ( A ) = γ s b + γ s + 1 R for some b ∈ V , where V is defined in Lemma 2.1. Since ∣ ker ( μ ) ∣ = q 3 n − 2 , we have

q 3 n − 2 ∣ T n ( R ∕ γ s R , 0 + γ s R ) ∣ = ∣ ker ( μ ) ∣ ∣ T n ( R ∕ γ s R , 0 + γ s R ) ∣ = ∣ T n ( R ∕ γ s + 1 R , 0 + γ s + 1 R ) ∣ + ∑ b ∈ V \ { 0 } ∣ T n ( R ∕ γ s + 1 R , γ s b + γ s + 1 R ) ∣ = ∣ T n ( R ∕ γ s + 1 R , 0 + γ s + 1 R ) ∣ + ( q − 1 ) ∣ T n ( R ∕ γ s + 1 R , γ s + γ s + 1 R ) ∣

by Theorem 4.1. Hence, we have

(11) ∣ T n ( R ∕ γ s + 1 R , γ s + γ s + 1 R ) ∣ = 1 q − 1 ( q 3 n − 2 ∣ T n ( R ∕ γ s R , 0 + γ s R ) ∣ − ∣ T n ( R ∕ γ s + 1 R , 0 + γ s + 1 R ) ∣ ) .

By Corollary 4.8, we have

(12) ∣ T n ( R , γ s ) ∣ = ∣ T n ( R ∕ γ e + 1 + ( s − e − 1 ) R , γ s + γ e + 1 + ( s − e − 1 ) R ) ∣ = q 3 ( e − s − 1 ) ( n − 1 ) ∣ T n ( R ∕ γ s + 1 R , γ s + γ s + 1 R ) ∣ .

Combining (11) and (12), it follows that

∣ T n ( R , γ s ) ∣ = q 3 ( e − s − 1 ) ( n − 1 ) q − 1 ( q 3 n − 2 ∣ T n ( R ∕ γ s R , 0 + γ s R ) ∣ − ∣ T n ( R ∕ γ s + 1 R , 0 + γ s + 1 R ) ∣ ) .

The proof is completed.□

Corollary 4.10

Let R be a CFCR of nilpotency index 2 and residue field F q and let n be a positive integer. If the maximal ideal of R is generated by γ , then

∣ T n ( R , a ) ∣ = ∣ T n ( R , γ ) ∣ ≤ 1 q − 1 ( ( q 2 − 1 ) q 3 n − 4 ∣ T n ( F q , 0 ) ∣ − ( q − 1 ) q 2 ( q 3 + 1 ) ∣ T n − 1 ( R , 0 ) ∣ )

for all a ∈ R \ F q .

Proof

Clearly, ∣ T 1 ( R , a ) ∣ = 1 . Let n ≥ 2 be an integer. By setting s = 1 in (11), we have

∣ T n ( R , a ) ∣ = ∣ T n ( R , γ ) ∣ = 1 q − 1 ( q 3 n − 2 ∣ T n ( R ∕ γ R , 0 + γ R ) ∣ − ∣ T n ( R , 0 ) ∣ ) = 1 q − 1 ( q 3 n − 2 ∣ T n ( F q , 0 ) ∣ − ∣ T n ( R , 0 ) ∣ ) .

Form the proof of Corollary 4.6, we have

∣ T n ( R , 0 ) ∣ ≥ ( q − 1 ) q 2 ( q 3 + 1 ) ∣ T n − 1 ( R , 0 ) ∣ + q 3 n − 4 ∣ T n ( F q , 0 ) ∣ ,

which implies that

∣ T n ( R , a ) ∣ ≤ 1 q − 1 ( q 3 n − 2 ∣ T n ( F q , 0 ) ∣ − ( ( q − 1 ) q 2 ( q 3 + 1 ) ∣ T n − 1 ( R , 0 ) ∣ + q 3 n − 4 ∣ T n ( F q , 0 ) ∣ ) ) = 1 q − 1 ( ( q 3 n − 2 − q 3 n − 4 ) ∣ T n ( F q , 0 ) ∣ − ( q − 1 ) q 2 ( q 3 + 1 ) ∣ T n − 1 ( R , 0 ) ∣ ) = 1 q − 1 ( ( q 2 − 1 ) q 3 n − 4 ∣ T n ( F q , 0 ) ∣ − ( q − 1 ) q 2 ( q 3 + 1 ) ∣ T n − 1 ( R , 0 ) ∣ ) = ( q + 1 ) q 3 n − 4 ∣ T n ( F q , 0 ) ∣ − q 2 ( q 3 + 1 ) ∣ T n − 1 ( R , 0 ) ∣ .

The proof is completed.□

We note that, (1) for a CFCR of nilpotency index e = 2 , a bound on ∣ T n − 1 ( R , 0 ) ∣ is determined recursively in Corollary 4.6, and (2) ∣ T n ( F q , 0 ) ∣ is given in (6).

4.2 Enumeration of special tridiagonal matrices over CFCRs

In this section, we focus on the enumeration of some special non-singular tridiagonal matrices over a CFCR R with prescribed determinant.

From Proposition 2.3, the next lemma follows immediately.

Lemma 4.11

Let R be a CFCR with residue field F q and nilpotency index e and let n be a positive integer. Let r , s ∈ R . Then

∣ T n , r , s ( R ) ∣ = q e n .

First, we determine the number of non-singular matrices in T n , r , s ( R ) .

Lemma 4.12

Let R be a CFCR with residue field F q and nilpotency index e and let n be a positive integer. Let r , s ∈ R . Then

∣ N S T n , r , s ( R ) ∣ = q ( e − 1 ) n ∣ N S T n , ψ ( r ) , ψ ( s ) ( F q ) ∣ ,

where ψ : R → F q is a map defined by a ↦ a + γ R .

Proof

Let ψ : R → F q be the ring homomorphism defined by

(13) a ↦ a + γ R .

Let Ψ : T n , r , s ( R ) → T n , ψ ( r ) , ψ ( s ) ( F q ) be the map defined by

[ a i j ] ↦ [ ψ ( a i j ) ] .

Then Ψ is a surjective map.

Let A ∈ T n , r , s ( R ) . Then for each diagonal entry b of Ψ ( A ) , there are q e − 1 choices of a such that ψ ( a ) = b . Hence, Ψ is a q ( e − 1 ) n to one function which implies that

∣ T n , r , s ( R ) ∣ ∣ T n , ψ ( r ) , ψ ( s ) ( F q ) ∣ = q ( e − 1 ) n .

Since det ( Ψ ( A ) ) = ψ ( det ( A ) ) , it follows that A ∈ N S T n , r , s ( R ) if and only if φ ( A ) ∈ N S T n , ψ ( r ) , ψ ( s ) ( F q ) . Consequently,

∣ N S T n , r , s ( R ) ∣ ∣ N S T n , ψ ( r ) , ψ ( s ) ( F q ) ∣ = q ( e − 1 ) n .

Hence, we have

∣ N S T n , r , s ( R ) ∣ = q ( e − 1 ) n ∣ N S T n , ψ ( r ) , ψ ( s ) ( F q ) ∣

as desired.□

Theorem 4.13

Let R be a CFCR with residue field F q and nilpotency index e and let n be a positive integer. Let r , s ∈ R . Then

∣ N S T n , r , s ( R ) ∣ = q ( e − 1 ) n ( q n + 1 + ( − 1 ) n ) 1 + q i f r , s ∈ U ( R ) q ( e − 1 ) n ( q − 1 ) n − 1 i f r ∉ U ( R ) o r s ∉ U ( R ) .

Proof

Let ψ be defined in (13). We consider the following two cases.

Case 1: r , s ∈ U ( R ) . Then ψ ( r ) , ψ ( s ) ∈ F q \ { 0 } . By Theorem 3.8, we have

∣ N S T n , ψ ( r ) , ψ ( s ) ( F q ) ∣ = q n + 1 + ( − 1 ) n 1 + q .

By Lemma 4.12, it follows that

∣ N S T n , r , s ( R ) ∣ = q ( e − 1 ) n ( q n + 1 + ( − 1 ) n ) 1 + q .

Case 2: r ∉ U ( R ) or s ∉ U ( R ) . Then ψ ( r ) = 0 or ψ ( s ) = { 0 } . By Lemma 3.6, we have

∣ N S T n , ψ ( r ) , ψ ( s ) ( F q ) ∣ = ( q − 1 ) n − 1 .

By Lemma 4.12, it follows that

∣ N S T n , r , s ( R ) ∣ = q ( e − 1 ) n ( q − 1 ) n − 1 .

The proof is completed.□

For an odd positive integer n , we have the following precise results.

Lemma 4.14

Let R be a CFCR of nilpotency index e and residue field F q . Let γ be a generator of the maximal ideal of R. If n is odd and 0 ≤ s ≤ e , then

∣ T n ( R , γ s ) ∣ = ∣ T n ( R , b γ s ) ∣

for all units b in U ( R ) .

Proof

Let b ∈ U ( R ) and let s be an integer such that 0 ≤ s ≤ e . In the case where s = e , we have γ s = 0 = γ s b and ∣ T n , r , s ( R , γ s ) ∣ = ∣ T n , r , s ( R , 0 ) ∣ = ∣ T n ( R , b γ s ) ∣ .

For each integer 0 ≤ s < e , let α : T n , r , s ( R , γ s ) → T n , r , s ( R , b γ s ) be a map defined by

ϕ ( A ) = diag ( b , 1 , b , … , 1 , b ) A diag ( 1 , b − 1 , 1 , … , b − 1 , 1 ) .

Using argument similar to those in the proof of Theorem 3.10, the results follows.□

For an odd positive integer n and a unit a ∈ U ( R ) , we have the relation between ∣ T n , r , s ( R , a ) ∣ and ∣ T n , r , s ( R , 1 ) ∣ in Lemma 4.14.

Corollary 4.15

Let R be a CFCR with residue field F q and nilpotency index e and let n be a positive integer. Let r , s ∈ R . If n is odd, then

∣ T n , r , s ( R , a ) ∣ = ∣ T n , r , s ( R , 1 ) ∣

for all a ∈ U ( R )

Theorem 4.16

Let R be a CFCR with residue field F q and nilpotency index e and let n be a positive integer. Let r , s ∈ R . If n is odd, then

∣ T n , r , s ( R , a ) ∣ = q ( e − 1 ) ( n − 1 ) ( q n + 1 + ( − 1 ) n ) q 2 − 1 i f r , s ∈ U ( R ) q ( e − 1 ) ( n − 1 ) ( q − 1 ) n − 2 i f r ∉ U ( R ) o r s ∉ U ( R )

for all a ∈ U ( R ) .

Proof

Let a ∈ U ( R ) . Similar to the proof of Corollary 4.4, we have

∣ N S T n , r , s ( R ) ∣ = ∣ ⋃ b ∈ U ( R ) T n , r , s ( R , b ) ∣ = ∑ b ∈ U ( R ) ∣ T n , r , s ( R , b ) ∣ = ∑ b ∈ U ( R ) ∣ T n , r , s ( R , 1 ) ∣ = ∣ U ( R ) ∣ ∣ T n , r , s ( R , 1 ) ∣ .

By Theorem 4.13 and Lemma 2.1, it follows that

∣ T n , r , s ( R , a ) ∣ = ∣ T n , r , s ( R , 1 ) ∣ = ∣ N S T n , r , s ( R ) ∣ ∣ U ( R ) ∣ = q ( e − 1 ) n ( q n + 1 + ( − 1 ) n ) 1 + q ( q − 1 ) q e − 1 if r , s ∈ U ( R ) q ( e − 1 ) n ( q − 1 ) n − 1 ( q − 1 ) q e − 1 if r ∉ U ( R ) or s ∉ U ( R ) = q ( e − 1 ) ( n − 1 ) ( q n + 1 + ( − 1 ) n ) q 2 − 1 if r , s ∈ U ( R ) q ( e − 1 ) ( n − 1 ) ( q − 1 ) n − 2 if r ∉ U ( R ) or s ∉ U ( R ) .

The proof is completed.□

5 Conclusion

The enumeration of tridiagonal matrices with prescribed determinant over a finite field F q and a CFCR R has been studied. The number of n × n tridiagonal matrices with prescribed determinant over F q and the number of n × n non-singular tridiagonal matrices with prescribed determinant over R have been completely determined. For n × n singular tridiagonal matrices with prescribed determinant over R , bounds on the number of such matrices with prescribed determinant have been given. Subsequently, the number of special n × n tridiagonal matrices with prescribed determinant over F q and the number of special n × n non-singular tridiagonal matrices with prescribed determinant over R have been presented for all odd positive integers n .

In general, the enumeration of n × n singular tridiagonal matrices with prescribed determinant over R is interesting and still open as well as the enumeration of special n × n singular tridiagonal matrices. For tridiagonal matrices of special forms, the enumeration of n × n tridiagonal matrices with prescribed determinant over F q and over R are open for an even positive integer n .

jitman_s@silpakorn.edu

Acknowledgements

The authors would like to thank the anonymous referees for their helpful comments.

Funding information: This research was funded by National Research Council of Thailand and Silpakorn University under Research Grant N42A650381.
Conflict of interest: The authors declare no conflicts of interest.
Data availability statement: Data sharing is not applicable to this article as no datasets were generated or analysed during the current study.

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Received: 2023-10-06

Revised: 2024-01-06

Accepted: 2024-01-22

Published Online: 2024-02-24

This work is licensed under the Creative Commons Attribution 4.0 International License.

Articles in the same Issue

https://doi.org/10.1515/spma-2023-0114

Keywords for this article

tridiagonal matrices; determinants; finite fields; finite commutative chain rings; enumeration

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