A Matrix-Theoretic Perspective on Some Identities Involving Well-Known Sequences

Abstract

We first consider the companion matrix associated with the characteristic polynomial of a linear recurrence relation, and we investigate its powers. Next, we introduce a new matrix associated with a given linear recurrence sequence, and we get a factorization of this matrix. Finally, we give several applications of our results. Actually, we obtain some identities concerning Fibonacci, Lucas, Pell, and Jacobsthal numbers using matrix theory.

1 Introduction and Notation

Recurrence relations occur in a variety of mathematical contexts, and they play an important role in the field of complexity. They are actually one of the basic mathematical tools of computation. In particular, they can be used to represent mathematical sequences that cannot be easily represented non-recursively. Usually, a sequence generated by a recurrence relation is called a recurrence sequence. Some well-known recurrence sequences are the Fibonacci sequence, Lucas sequence, Pell sequence and Jacobsthal sequence.

In general, a sequence $\lambda =(\lambda _0, \lambda _1, \ldots )$ is said to satisfy a linear recurrence relation of order $k\geqslant 1$ with initial conditions $\lambda _0, \lambda _1, \ldots , \lambda _{k-1}$, if there exist constants $c_0, c_1, \ldots , c_{k-1}$, with $c_0\ne 0$, such that

$$\begin{aligned} \lambda _i=c_0\lambda _{i-k}+c_1\lambda _{i-k+1}+\cdots +c_{k-1}\lambda _{i-1}, \ \ \ \ (i\geqslant k). \end{aligned}$$

(1)

A linear recurrence sequence is a sequence satisfying some linear recurrence as above. Occasionally, the indexing of the terms will start with something other than 0. In particular, we note that the sequence $\lambda$ can be easily extended to negative indices using the rearranged recurrence relation

$$\begin{aligned} \lambda _{i-k}=c^{-1}_{0}(\lambda _i-c_1\lambda _{i-k+1}-\cdots -c_{k-1}\lambda _{i-1}), \ \ \ \ (i < k). \end{aligned}$$

(2)

This lets one define $\lambda _{-1}, \lambda _{-2}, \ldots$, to obtain a doubly infinite sequence

$$\begin{aligned} \ldots , \lambda _{-2}, \lambda _{-1}, \lambda _0, \lambda _1, \lambda _2, \ldots \end{aligned}$$

that now satisfies the same linear recurrence for all integer indices i, positive or negative.

Some well-known linear recurrence sequences are special cases of (1), for instance, we have

The characteristic polynomial of the linear recurrence relation (1) is defined to be the polynomial

$$\begin{aligned} p_\lambda (x)=x^k-c_{k-1}x^{k-1}-\cdots -c_2x^2-c_1x-c_0. \end{aligned}$$

(3)

It is well-known by Cayley-Hamilton theorem that the square matrix

$$\begin{aligned} A=\left[ \begin{array}{c@{\quad }c@{\quad }c@{\quad }c@{\quad }c@{\quad }c} 0 &{} 0 &{} 0&{} 0 &{} \ldots &{} c_{0} \\ 1&{} 0 &{} 0&{} 0 &{} \ldots &{} c_{1} \\ 0&{} 1 &{} 0&{} 0 &{} \ldots &{} c_{2} \\ \vdots &{} \vdots &{} \ddots &{} \vdots &{} \cdots &{} \vdots \\ 0&{} 0 &{} 0&{} 1 &{} \ldots &{} c_{k-2} \\ 0 &{} 0 &{} 0 &{}\ldots &{} 1 &{} c_{k-1}\\ \end{array}\right] \end{aligned}$$

satisfies the equation $p_\lambda (A)=0$; that is

$$\begin{aligned} A^k-c_{k-1}A^{k-1}-\cdots -c_2A^2-c_1A-c_0I=0, \end{aligned}$$

where I is the $k\times k$ identity matrix (see [4]). The matrix A is called the companion matrix of the monic polynomial $p_\lambda (x)$, which will remain fixed throughout the paper.

In what follows, a new approach will be presented to find again this matrix and its powers. To illustrate this, we introduce k new linear recurrence sequences of order k as follows: for any integer $s\in \{0, 1, 2, \ldots , k-1\}$, we define the linear recurrence sequence

$$\begin{aligned} \omega ^{(s)}=\left( \omega ^{(s)}_0, \omega ^{(s)}_1, \omega ^{(s)}_2, \ldots \right) \end{aligned}$$

of order k, by

$$\begin{aligned} \omega ^{(s)}_i=\left\{ \begin{array}{ll} \delta _{i, s} &{}\quad i< k, \\ \sum \limits _{l=0}^{k-1}c_l\omega ^{(s)}_{i-k+l} &{}\quad i\geqslant k, \\ \end{array} \right. \end{aligned}$$

(4)

where $\delta _{i, s}$ is the Kronecker delta.

These sequences and a few first terms of them are shown below:

Again, the sequences $\omega ^{(s)}$ can be extended to negative indices using the rearranged recurrence relation

$$\begin{aligned} \omega ^{(s)}_{i-k}=c^{-1}_{0}\left( \omega ^{(s)}_{i}-\sum _{l=1}^{k-1}c_l\omega ^{(s)}_{i-k+l}\right) , \ \ \ \ (i < k). \end{aligned}$$

(5)

Note that each sequence $\omega ^{(s)}$, $0\leqslant s \leqslant k-1$, is a special case of the sequence $\lambda$ for which the initial conditions $(\lambda _0, \lambda _1, \ldots , \lambda _{k-1})$ are the row vector with all 0s except for a 1 in the sth position.

An Example. Suppose that $k=2$, $\lambda _0=2$ and $\lambda _1=c_0=c_1=1$. Then, we have

In general, for any integer m, we define

$$\begin{aligned} A_m=\left[ \omega _{m+j}^{(i)}\right] _{0\leqslant i,\, j\leqslant k-1}=\left[ \begin{array}{ccccc} \omega _m^{(0)} &{} \quad \omega _{m+1}^{(0)} &{}\quad \omega _{m+2}^{(0)} &{} \quad \cdots &{} \quad \omega _{m+k-1}^{(0)} \\ \omega _{m}^{(1)} &{} \quad \omega _{m+1}^{(1)} &{} \quad \omega _{m+2}^{(1)} &{}\quad \cdots &{} \quad \omega _{m+k-1}^{(1)} \\ \omega _{m}^{(2)} &{} \quad \omega _{m+1}^{(2)} &{} \quad \omega _{m+2}^{(2)} &{} \quad \cdots &{} \omega _{m+k-1}^{(2)} \\ \vdots &{}\quad \vdots &{}\quad \vdots &{} \quad \cdots &{} \quad \vdots \\ \omega _{m}^{(k-1)} &{} \quad \omega _{m+1}^{(k-1)} &{} \quad \omega _{m+2}^{(k-1)} &{} \quad \cdots &{} \quad \omega _{m+k-1}^{(k-1)} \\ \end{array}\right] . \end{aligned}$$

We notice that the companion matrix A can be seen here again (shown in bold type above). Actually, if we take $m=1$, then we have

$$\begin{aligned} A=A_1= & {} \left[ \begin{array}{ccccc} \omega _1^{(0)} &{}\quad \omega _2^{(0)} &{} \quad \omega _3^{(0)} &{}\quad \ldots &{} \quad \omega _k^{(0)} \\ \omega _1^{(1)}&{}\quad \omega _2^{(1)} &{} \quad \omega _3^{(1)} &{}\quad \ldots &{}\quad \omega _k^{(1)} \\ \omega _1^{(2)} &{} \quad \omega _2^{(2)} &{}\quad \omega _3^{(2)} &{} \quad \ldots &{}\quad \omega _k^{(2)} \\ \vdots &{}\quad \vdots &{} \quad \vdots &{}\quad \ddots &{}\quad \vdots \\ \omega _{1}^{(k-2)} &{}\quad \omega _{2}^{(k-2)} &{} \quad \omega _{3}^{(k-2)} &{}\quad \ldots &{}\quad \omega _{k}^{(k-2)} \\ \omega _1^{(k-1)} &{}\quad \omega _2^{(k-1)} &{} \quad \omega _3^{(k-1)} &{} \quad \ldots &{}\quad \omega _k^{(k-1)} \\ \end{array}\right] \\= & {} \left[ \begin{array}{cccccc} 0 &{}\quad 0 &{}\quad 0&{} \quad 0 &{}\quad \ldots &{} \quad c_{0} \\ 1&{}\quad 0 &{}\quad 0&{} \quad 0 &{}\quad \ldots &{}\quad c_{1} \\ 0&{} \quad 1 &{} \quad 0 &{} \quad 0 &{} \quad \ldots &{} \quad c_{2} \\ \vdots &{}\quad \vdots &{} \quad \ddots &{} \quad \vdots &{}\quad \cdots &{}\quad \vdots \\ 0&{} \quad 0 &{}\quad 0 &{} \quad 1 &{} \quad \ldots &{} \quad c_{k-2} \\ 0 &{}\quad 0 &{}\quad 0 &{} \quad \ldots &{} \quad 1 &{} \quad c_{k-1}\\ \end{array}\right] . \end{aligned}$$

For any integer m, it is an easy observation that $A^m=A_m$ (see Theorem 1). Now, using the fact that $\det A^m=(\det A)^m$, we immediately get

$$\begin{aligned} \det A_m=\det (A^m) =(\det A)^m=(-1)^{m(k-1)}c_0^m, \end{aligned}$$

(see also [5, Corollary 2.1]).

In the sequel, we shall introduce some more definitions and notation. Given integers x, $y_i$, and $z_i$, $1\leqslant i\leqslant k-1$, we define the $k\times k$ matrices $R_k(x, y_1, \ldots , y_{k-1}, z_1, \ldots , z_{k-1})$ and $S_k(y_1, \ldots , y_{k-1})$ as follows:

$$\begin{aligned}&R_k(x, y_1, \ldots , y_{k-1}, z_1, \ldots , z_{k-1}):= [\lambda _{x+i{y_i}+j{z_j}}]_{0\leqslant i, j\leqslant k-1},\\&S_k(y_1, \ldots , y_{k-1}):= \left[ \omega _{j{y_j}}^{(i)}\right] _{0\leqslant i, j\leqslant k-1}. \end{aligned}$$

The associated matrices $R_4(x, y_1, y_2, y_3, z_1, z_2, z_3)$ and $S_4(y_1, y_2, y_3)$, for example, are given by

$$\begin{aligned} R_4(x, y_1, y_2, y_3, z_1, z_2, z_3) =\left[ \begin{array}{llll} \lambda _{x} &{} \quad \lambda _{x+z_1}&{} \quad \lambda _{x+2{z_2}}&{} \quad \lambda _{x+3{z_3}} \\ \lambda _{x+{y_1}} &{} \quad \lambda _{x+y_1+z_1}&{} \quad \lambda _{x+y_1+2{z_2}} &{} \quad \lambda _{x+y_1+3{z_3}} \\ \lambda _{x+2{y_2}} &{} \quad \lambda _{x+2{y_2}+z_1}&{} \quad \lambda _{x+2{y_2}+2{z_2}}&{} \quad \lambda _{x+2{y_2}+3{z_3}} \\ \lambda _{x+3{y_3}} &{} \quad \lambda _{x+3{y_3}+z_1}&{} \quad \lambda _{x+3{y_3}+2{z_2}}&{} \quad \lambda _{x+3{y_3}+3{z_3}} \\ \end{array}\right] , \end{aligned}$$

and

$$\begin{aligned} S_4(y_1, y_2, y_3)=\left[ \begin{array}{llll} \omega _{0}^{(0)} &{} \quad \omega _{y_1}^{(0)} &{} \quad \omega _{2{y_2}}^{(0)}&{} \quad \omega _{3{y_3}}^{(0)} \\ \omega _{0}^{(1)} &{} \quad \omega _{y_1}^{(1)} &{} \quad \omega _{2{y_2}}^{(1)}&{} \quad \omega _{3{y_3}}^{(1)} \\ \omega _{0}^{(2)} &{} \quad \omega _{y_1}^{(2)} &{} \quad \omega _{2{y_2}}^{(2)}&{} \quad \omega _{3{y_3}}^{(2)} \\ \omega _{0}^{(3)} &{} \quad \omega _{y_1}^{(3)} &{} \quad \omega _{2{y_2}}^{(3)}&{} \quad \omega _{3{y_3}}^{(3)} \\ \end{array}\right] . \end{aligned}$$

In the case when $x=0$ and $y_i=z_i=1$, $1\leqslant i\leqslant k-1$, we denote the matrix $R_k(0, 1, \ldots , 1, 1, \ldots , 1)$ by $R_k$. In this paper, we will present a matrix factorization for $R_k(x, y_1, \ldots , y_{k-1}, z_1, \ldots , z_{k-1})$ (see Theorem 2). Finally, some applications of the main results are given.

The outline of the paper is as follows. Basic definitions and notation are summarized in Sect. 1. In Sect. 2, we derive some preparatory results. Section 3 contains three theorems and some applications of our findings to obtain many identities involving Fibonacci, Lucas, Pell, and Jacobsthal numbers.

2 Preliminaries

In the following lemma we give a relationship between the terms of sequences $\lambda$ and $\omega ^{(s)}$.

Lemma 1

For every integer m and n, we have

$$\begin{aligned} \sum \limits _{l=0}^{k-1} \lambda _{m+l}\omega _{n}^{(l)}=\lambda _{m+n}. \end{aligned}$$

In particular, if in addition $0\leqslant s\leqslant k-1$ is an integer, we get

$$\begin{aligned} \sum \limits _{l=0}^{k-1} \omega _{m+l}^{(s)}\omega _{n}^{(l)}=\omega _{m+n}^{(s)}. \end{aligned}$$

Proof

First, we assume that $m+n\geqslant 0$. We will proceed by induction with respect to $m+n$. The case when $0\leqslant m+n\leqslant k-1$ is trivial, so we may assume that $m+n\geqslant k$. Therefore, we have

$$\begin{aligned} \lambda _{m+n}= & {} \sum \limits _{l=0}^{k-1}c_l\lambda _{m+n-k+l} \ \ \ \ \ \text{(by } \text{ Eq. } \text{(1)) } \\= & {} \sum \limits _{l=0}^{k-1}c_l\sum \limits _{s=0}^{k-1}\lambda _{m+s}\omega _{n-k+l}^{(s)} \ \ \ \ \ \text{(using } \text{ the } \text{ induction } \text{ hypothesis) }\\= & {} \sum \limits _{s=0}^{k-1}\lambda _{m+s}\sum \limits _{l=0}^{k-1}c_l\omega _{n-k+l}^{(s)} \\= & {} \sum \limits _{s=0}^{k-1}\lambda _{m+s}\omega _{n}^{(s)} \ \ \ \ \ \text{(by } \text{ Eq. } \text{(4)) } \end{aligned}$$

as required. Now, let $m+n < 0$. Again, by induction on $m+n$, we obtain

$$\begin{aligned} \lambda _{m+n}= & {} c^{-1}_{0}\left( \lambda _{m+n+k}-\sum \limits _{l=1}^{k-1}c_l\lambda _{m+n+l}\right) \ \ \ \ \ \text{(by } \text{ Eq. } \text{(2)) } \\= & {} c^{-1}_{0}\left( \sum \limits _{s=0}^{k-1}\lambda _{m+s}\omega ^{(s)}_{n+k}-\sum \limits _{l=1}^{k-1} c_l\sum \limits _{s=0}^{k-1}\lambda _{m+s}\omega ^{(s)}_{n+l}\right) \ \ \ \ \ \text{(using } \text{ the } \text{ induction } \text{ hypothesis) }\\= & {} \sum \limits _{s=0}^{k-1}\lambda _{m+s}c^{-1}_{0}\left( \omega ^{(s)}_{n+k}-\sum \limits _{l=1}^{k-1}c_l\omega ^{(s)}_{n+l}\right) \\= & {} \sum \limits _{s=0}^{k-1}\lambda _{m+s}\omega _{n}^{(s)} \ \ \ \ \ \text{(by } \text{ Eq. } \text{(5)) } \end{aligned}$$

which completes the proof of the first result. The second result is a special case of the first one. $\square$

The following lemma gives some relationships between sequences $\omega ^{(s)}$, $0\leqslant s\leqslant k-1$.

Lemma 2

The following statements hold for any integer i.

• (i) $\omega _{i+1}^{(0)}=c_0\omega _{i}^{(k-1)}$.

• (ii) $\omega _{i+1}^{( s+1)}-\omega _{i}^{(s)}=c_{s+1}\omega _{i}^{(k-1)}$    $(0\leqslant s\leqslant k-2)$.

• (iii) $c_0\omega _{i-1}^{(k-1)}+c_{1}\omega _{i}^{(k-1)}= \omega _{i+1}^{( 1)}$.

Proof

• (i) We apply induction on i. For $0\leqslant i\leqslant k-1$, the proof is obvious, because, if $0\leqslant i\leqslant k-2$, $\omega _{i+1}^{(0)}=c_0\omega _{i}^{(k-1)}=0$, and if $i=k-1$, we have

  $$\begin{aligned} \omega _{k}^{(0)}=\sum _{l=0}^{k-1}c_l\omega _l^{(0)}=c_0=c_0\omega _{k-1}^{(k-1)}. \end{aligned}$$

  Let us for the moment assume that $i\geqslant k$. Using Eq. (4) and the induction hypothesis, we have

  $$\begin{aligned} \omega ^{(0)}_{i+1}=\sum \limits _{l=0}^{k-1}c_l\omega _{i-k+l+1}^{( 0)}=\sum \limits _{l=0}^{k-1}c_l\left( c_0\omega _{i-k+l}^{( k-1)}\right) =c_0\sum \limits _{l=0}^{k-1}c_l\omega _{i-k+l}^{(k-1)}= c_0\omega _{i}^{( k-1)}. \end{aligned}$$

  Now, let $i < 0$. We obtain

  $$\begin{aligned} \omega ^{(0)}_{i+1}= & {} c^{-1}_{0}\left( \omega ^{(0)}_{i+k+1}-\sum \limits _{l=1}^{k-1}c_l\omega ^{(0)}_{i+l+1}\right) \ \ \ \ \ \text{(by } \text{ Eq. } \text{(5)) } \\= & {} c^{-1}_{0}\left( c_{0}\omega ^{(k-1)}_{i+k}-\sum \limits _{l=1}^{k-1}c_{l}c_{0}\omega ^{(k-1)}_{i+l}\right) \ \ \ \ \ \text{(using } \text{ the } \text{ induction } \text{ hypothesis) }\\= & {} \omega ^{(k-1)}_{i+k}-\sum \limits _{l=1}^{k-1}c_{l}\omega ^{(k-1)}_{i+l} \\= & {} c_{0}\omega ^{(k-1)}_i \ \ \ \ \ \text{(by } \text{ Eq. } \text{(5)) } \end{aligned}$$

  which completes the proof of this part.

• (ii) Suppose that $0\leqslant s\leqslant k-2$. Once again, we prove the result by induction on i. Assume first that $0\leqslant i\leqslant k-1$. In this case, we get

  $$\begin{aligned} \begin{array}{lll}\omega _{i}^{(s)}+c_{s+1}\omega _{i}^{( k-1)}&{} = &{}\left\{ \begin{array}{ll} 1 &{} \quad i=s\leqslant k-2, \\ 0 &{} \quad i\ne s\leqslant k-2,\\ c_{s+1} &{} \quad i=k-1, \\ \end{array} \right. \\ &{} = &{} \omega _{i+1}^{( s+1)}.\end{array} \end{aligned}$$

  Assume now that $i\geqslant k$. By a routine calculation, we obtain

  $$\begin{aligned} \omega _{i}^{(s)}+c_{s+1}\omega _{i}^{( k-1)}= & {} \sum \limits _{l=0}^{k-1}c_l\omega _{i-k+l}^{(s)}+c_{s+1}\sum \limits _{l=0}^{k-1}c_l\omega _{i-k+l}^{( k-1)} \ \ \ \ \ \text{(by } \text{ Eq. } \text{(4)) }\\= & {} \sum \limits _{l=0}^{k-1}c_l\left[ \omega _{i-k+l}^{(s)}+c_{s+1}\omega _{i-k+l}^{( k-1)}\right] \\= & {} \sum \limits _{l=0}^{k-1}c_l\omega _{i-k+l+1}^{( s+1)} \ \ \ \ \ \text{(using } \text{ the } \text{ induction } \text{ hypothesis) }\\= & {} \omega _{i+1}^{( s+1)} \ \ \ \ \ \text{(by } \text{ Eq. } \text{(4)) } \end{aligned}$$

  as desired. For the case when $i<0$, we verify the result as follows

  $$\begin{aligned} \omega _{i}^{(s)}+c_{s+1}\omega _{i}^{( k-1)}= & {} c^{-1}_{0}\left( \omega ^{(s)}_{i+k}-\sum \limits _{l=1}^{k-1}c_l\omega ^{(s)}_{i+l}\right) \\&+ c_{s+1}\left\{ c^{-1}_{0}\left( \omega ^{(k-1)}_{i+k}-\sum \limits _{l=1}^{k-1}c_l\omega ^{(k-1)}_{i+l}\right) \right\} \ \ \ \ \ \text{(by } \text{ Eq. } \text{(5)) } \\= & {} c^{-1}_{0}\left\{ \left( \omega ^{(s)}_{i+k}+c_{s+1}\omega ^{(k-1)}_{i+k}\right) -\sum \limits _{l=1}^{k-1}c_l\left( \omega ^{(s)}_{i+l}+c_{s+1}\omega ^{(k-1)}_{i+l}\right) \right\} \\= & {} c^{-1}_{0}\left\{ \omega ^{(s+1)}_{i+k+1}-\sum \limits _{l=1}^{k-1}c_{l}\omega ^{(s+1)}_{i+l+1}\right\} \ \ \ \ \ \text{(using } \text{ the } \text{ induction } \text{ hypothesis) } \\= & {} \omega ^{(s+1)}_{i+1} \ \ \ \ \ \text{(by } \text{ Eq. } \text{(5)) }. \end{aligned}$$

• (iii) We obtain

  $$\begin{aligned} c_0\omega _{i-1}^{( k-1)}+c_{1}\omega _{i}^{( k-1)}= & {} \omega _{i}^{( 0)}+c_{1}\omega _{i}^{( k-1)} \ \ \ \ \ \text{(by } \text{ part }\, (i)) \\= & {} \omega _{i+1}^{( 1)} \ \ \ \ \ \text{(by } \text{ part }\, ({\textit{ii}})\, \text{ for }\, s=0) \end{aligned}$$

  as desired. This completes the proof of the lemma. $\square$

3 Main Results and Some Applications

Before stating our main results we introduce some notation. Let $\lambda =(\lambda _i)_{i \in \mathbb {Z}}$ be a doubly infinite sequence satisfying the linear recurrence relation (1). Given an integer m, we will consider the row vector

$$\begin{aligned} {\mathbf u}_m:=(\lambda _m, \lambda _{m+1}, \ldots , \lambda _{m+k-1}), \end{aligned}$$

which is called the mth state vector of $\lambda$. The state vector ${\mathbf u}_0=(\lambda _0, \lambda _{1}, \ldots , \lambda _{k-1})$ is also referred to as the initial state vector.

We start with the following known result of Er [1], which is obtained by using the approach of Kalman [4]. Here, we provide a proof for the sake of completeness.

Theorem 1

([1]) For any integer m, we have $A^m=A_m$. Furthermore, we have ${\mathbf u}_m={\mathbf u}_0A^m$ and

$$\begin{aligned} A^m[c_0 \ c_1 \ c_2 \ \ldots \ c_{k-1}]^T=\left[ \omega _{k+m}^{(0)} \ \omega _{k+m}^{(1)} \ \omega _{k+m}^{(2)} \ \ldots \ \omega _{k+m}^{(k-1)}\right] ^T, \end{aligned}$$

where T denotes transpose.

Proof

First we assume that $m\geqslant 0$. In this case, we conclude the result by induction on m. For $m=0$ and $m=1$, the result is straightforward. Assume that $m\geqslant 2$. Using the definition of A and induction hypothesis, we have

$$\begin{aligned} A^m= A^{m-1}\cdot A=\left[ \omega _{j+m-1}^{(i)}\right] \cdot \left[ \omega _{j+1}^{(i)}\right] . \end{aligned}$$

Therefore, for $0\leqslant i, j\leqslant k-1$, we have

$$\begin{aligned} (A^m)_{i,j}= & {} \sum \limits _{l=0}^{k-1} \omega _{l+m-1}^{( i)}\omega _{j+1}^{( l)}\\= & {} \omega _{j+m}^{(i)} \ \ \ \ \ \text{(by } \text{ Lemma }\, 1) \end{aligned}$$

as desired. Now, suppose that $m<0$. Since $-m>0$, from the previous part we have

$$\begin{aligned} A^{-m}=\left[ \omega _{j-m}^{(i)}\right] _{0\leqslant i, j\leqslant k-1}. \end{aligned}$$

Thus, using Lemma 1, for $0\leqslant i, j\leqslant k-1$, we get

$$\begin{aligned} \left( \left[ \omega _{j+m}^{(i)}\right] \cdot A^{-m}\right) _{i,j}=\sum \limits _{l=0}^{k-1} \omega _{l+m}^{(i)}\omega _{j-m}^{(l)}=\omega ^{(i)}_j =\left\{ \begin{array}{ll} 1 &{}\quad \ i=j, \\ 0 &{} \quad \ i\ne j.\\ \end{array} \right. \end{aligned}$$

Therefore, we deduce

$$\begin{aligned} A^m=(A^{-m})^{-1}=\left[ \omega _{j+m}^{(i)}\right] _{0\leqslant i, j\leqslant k-1} \end{aligned}$$

which completes the proof of the first part. The rest of conclusion follows immediately from previous part and Lemma 1. $\square$

Theorem 2

For integers x, $y_i$, and $z_i$, $1\leqslant i\leqslant k-1$, we have the following matrix decomposition:

$$\begin{aligned} R_k(x, y_1, \ldots , y_{k-1}, z_1, \ldots , z_{k-1}) =S_k(y_1, \ldots , y_{k-1})^{T}R_kA^xS_k(z_1, \ldots , z_{k-1}), \end{aligned}$$

where $X^T$ is the transpose of X.

Proof

Set $D=S_k(y_1, \ldots , y_{k-1})^{T}R_kA^xS_k(z_1, \ldots , z_{k-1})$. For the proof of the claim we compute the (i, j)-entry of D. Let $0\leqslant i, j\leqslant k-1$. Then, by direct computation, we obtain

$$\begin{aligned} D_{i,j}= & {} \sum \limits _{r=0}^{k-1}\left( S_k(y_1, \ldots , y_{k-1})^T\right) _{i, r}(R_kA^xS_k(z_1, \ldots , z_{k-1}))_{r,j}\\= & {} \sum \limits _{r=0}^{k-1}\omega _{i{y_i}}^{(r)} \sum \limits _{s=0}^{k-1}(R_k)_{r, s}(A^xS_k(z_1, \ldots , z_{k-1}))_{s,j}\\= & {} \sum \limits _{r=0}^{k-1}\omega _{i{y_i}}^{(r)} \sum \limits _{s=0}^{k-1}\lambda _{r+s} \sum \limits _{t=0}^{k-1}(A^x)_{s,t}(S_k(z_1, \ldots , z_{k-1}))_{t,j}\\= & {} \sum \limits _{r=0}^{k-1}\omega _{i{y_i}}^{(r)} \sum \limits _{s=0}^{k-1}\lambda _{r+s}\sum \limits _{t=0}^{k-1} \omega _{t+x}^{(s)}\omega _{j{z_j}}^{(t)} \ \ \ \ \ \text{(by } \text{ Theorem } \text{1) }\\= & {} \sum \limits _{r=0}^{k-1}\omega _{i{y_i}}^{(r)} \sum \limits _{t=0}^{k-1}\Big (\sum \limits _{s=0}^{k-1}\lambda _{r+s} \omega _{t+x}^{(s)}\Big )\omega _{j{z_j}}^{(t)}\\= & {} \sum \limits _{r=0}^{k-1}\omega _{i{y_i}}^{(r)} \sum \limits _{t=0}^{k-1}\lambda _{r+t+x}\omega _{j{z_j}}^{(t)} \ \ \ \ \ \text{(by } \text{ Lemma } \text{1) }\\= & {} \sum \limits _{r=0}^{k-1}\omega _{i{y_i}}^{(r)} \lambda _{r+x+j{z_j}} \ \ \ \ \ \text{(by } \text{ Lemma } \text{1) }\\= & {} \lambda _{x+i{y_i}+j{z_j}} \ \ \ \ \ \text{(by } \text{ Lemma } \text{1) } \\= & {} (R_k(x, y_1, \ldots , y_{k-1}, z_1, \ldots , z_{k-1}))_{i,j}. \end{aligned}$$

This completes the proof of the theorem. $\square$

It is worthwhile to point out that Theorem 2 enables us to find many identities between the terms of a linear recurrence sequence of any order. In the next result, we will illustrate this situation for a linear recurrence sequence of order 2. A similar situation occurs with higher order linear recurrence sequences.

Theorem 3

Let x, y, z, and t be arbitrary integers. Then the following identities hold between the terms of linear recurrence sequences $\lambda$, $\omega ^{(0)}$ and $\omega ^{(1)}$ of order 2:

• (i) $\lambda _{x}\lambda _{x+y+z}-\lambda _{x+y}\lambda _{x+z}= (-c_{0})^{x}(\lambda _{0}\lambda _{2}-\lambda ^{2}_{1}) \omega ^{(1)}_y\omega ^{(1)}_z$.

• (ii) $\lambda _{x+t}\lambda _{x+y+z+t}- \lambda _{x+y+t}\lambda _{x+z+t}= (-c_{0})^{t}(\lambda _{x}\lambda _{x+y+z}- \lambda _{x+y}\lambda _{x+z})$.

Proof

• (i) The identity is obtained by the following calculations:

  $$\begin{aligned}&\lambda _{x}\lambda _{x+y+z}-\lambda _{x+y}\lambda _{x+z}\\&\quad = \det [R_2(x, y, z)]\\&\quad = \det [S_2(y)^{T}R_{2}A^{x}S_2(z)] \ \ \ \ \ \text{(by } \text{ Theorem } \text{2) }\\&\quad = \det [S_2(y)^{T}] \det [R_2] \det [A^x] \det [S_2(z)]\\&\quad = \left( \omega ^{(0)}_0\omega ^{(1)}_y-\omega ^{(1)}_0\omega ^{(0)}_y\right) (\lambda _{0}\lambda _{2}-\lambda ^{2}_{1})(-c_0)^{x} \left( \omega ^{(0)}_0 \omega ^{(1)}_z-\omega ^{(1)}_0\omega ^{(0)}_z\right) \\&\quad = (-c_{0})^{x}(\lambda _{0}\lambda _{2}-\lambda ^{2}_{1}) \omega ^{(1)}_y\omega ^{(1)}_z \ \ \ \ \ \text{(Since }\, \omega ^{(0)}_{0} =1\, \hbox {and}\, \omega ^{(1)}_{0}=0). \end{aligned}$$

• (ii) Again, by a routine calculation, we obtain

  $$\begin{aligned}&\lambda _{x+t}\lambda _{x+y+z+t}- \lambda _{x+y+t}\lambda _{x+z+t}\\&\quad = (-c_{0})^{x+t}(\lambda _{0}\lambda _{2}-\lambda ^{2}_{1}) \omega ^{(1)}_y\omega ^{(1)}_z \ \ \ \ \ \text{(by } \text{ part }\, (i))\\&\quad = (-c_{0})^{t}\left\{ (-c_{0})^{x}(\lambda _{0}\lambda _{2}-\lambda ^{2}_{1}) \omega ^{(1)}_y\omega ^{(1)}_z\right\} \\&\quad = (-c_{0})^{t}(\lambda _{x}\lambda _{x+y+z}- \lambda _{x+y}\lambda _{x+z}) \ \ \ \ \ \text{(by } \text{ part }\, (i)). \end{aligned}$$

  This completes the proof of the theorem. $\square$

In what follows, we have focused on some practical applications of the techniques provided in this paper. First of all, we investigate some relations between Fibonacci, Lucas, Pell, and Jacobsthal numbers.

Lemma 3

Let m and n be two arbitrary integers. Then there hold the following identities:

• (i) ${F}_{m-1}{F}_{m+1}-{F}_{m}^2=(-1)^m$.

• (ii) ${F}_{m-1}{F}_{n}+{F}_{m}{F}_{m+1}={F}_{m+n}$.

• (iii) ${F}_{m-1}+{F}_{m+1}=F_{2m}/F_m$.

Proof

• (i) If $k=2$ and $c_0=c_1=1$, then we have

  $$\begin{aligned} \omega _m^{(0)}={F}_{m-1} \ \ \ \ \text{ and } \ \ \ \ \omega _m^{(1)}={F}_{m}. \end{aligned}$$

  On the other hand, it follows from Theorem 1 that

  $$\begin{aligned} \left[ \begin{array}{ll} 0 &{}\quad 1\\ 1 &{} \quad 1 \end{array}\right] ^m=\left[ \begin{array}{cc} {F}_{m-1} &{}\quad {F}_{m}\\ {F}_{m} &{} \quad {F}_{m+1}\end{array}\right] . \end{aligned}$$

  (6)

  Taking the determinant of both sides yields

  $$\begin{aligned} {F}_{m-1}{F}_{m+1}-{F}_{m}^2=(-1)^m. \end{aligned}$$

• (ii) From the identity $A^{m+n}=A^m\cdot A^n$ and using Eq. (6), we get

  $$\begin{aligned} \left[ \begin{array}{cc} {F}_{m+n-1} &{}\quad {F}_{m+n}\\ {F}_{m+n} &{}\quad {F}_{m+n+1}\end{array}\right]= & {} \left[ \begin{array}{cc} {F}_{m-1} &{} {F}_{m}\\ {F}_{m} &{} \quad {F}_{m+1}\end{array}\right] \left[ \begin{array}{cc} {F}_{n-1} &{} \quad {F}_{n}\\ {F}_{n} &{}\quad {F}_{n+1}\end{array}\right] \\= & {} \left[ \begin{array}{cc} {F}_{m-1} {F}_{n-1}+ {F}_{m} {F}_{n} &{} \quad {F}_{m-1} {F}_{n}+ {F}_{m} {F}_{n+1}\\ {F}_{m} {F}_{n-1}+ {F}_{m+1} {F}_{n} &{} \quad {F}_{m} {F}_{n}+ {F}_{m+1} {F}_{n+1}\end{array}\right] . \end{aligned}$$

  Comparing (0, 1)-entries on both sides of this equation, we obtain

  $$\begin{aligned} {F}_{m+n}={F}_{m-1} {F}_{n}+ {F}_{m} {F}_{n+1}. \end{aligned}$$

• (iii) Setting $n=m$ in the previous part, we conclude that

  $$\begin{aligned} {F}_{m-1}+{F}_{m+1}=F_{2m}/F_m, \end{aligned}$$

  which completes the proof. $\square$

Lemma 4

For any two integers m and n, there hold:

• (i) ${P}_{m-1}{P}_{m+1}-{P}_{m}^2=(-1)^m$.

• (ii) ${P}_{m-1}{P}_{n}+{P}_{m}{P}_{m+1}={P}_{m+n}$.

• (iii) ${P}_{m-1}+{P}_{m+1}=P_{2m}/P_m$.

Proof

By taking $k=2$, $c_0=1$ and $c_1=2$, we conclude that

$$\begin{aligned} \omega _m^{(0)}={P}_{m-1} \ \ \ \ \text{ and } \ \ \ \ \omega _m^{(1)}={P}_{m}. \end{aligned}$$

Moreover, by Theorem 1, we have

$$\begin{aligned} \left[ \begin{array}{ll} 0 &{}\quad 1\\ 1 &{} \quad 2 \end{array}\right] ^m=\left[ \begin{array}{cc} {P}_{m-1} &{}\quad {P}_{m}\\ {P}_{m} &{} \quad {P}_{m+1}\end{array}\right] . \end{aligned}$$

Now, the result follows immediately with arguments analogous to those of Lemma 3. $\square$

Lemma 5

Let m and n be two arbitrary integers. Then the following identities hold true:

• (i) ${J}_{m}+{J}_{m+1}=2^m$.

• (ii) $[{J}_{m}+(-1)^m]{J}_{n}+{J}_{m}{J}_{n+1}={J}_{m+n}$.

• (iii) $2^n{J}_{m}+(-1)^m{J}_{n}={J}_{m+n}$.

• (iv) $J_{2m}/J_m=2^m+(-1)^m$.

• (v) $J_{m}=\left( 2^m+(-1)^{m+1}\right) /3$.

Proof

• (i) If $k=c_0=2$ and $c_1=1$, then we have

  $$\begin{aligned} \omega _m^{(0)}={J}_{m}+(-1)^m \ \ \ \ \text{ and } \ \ \ \ \omega _m^{(1)}={J}_{m}. \end{aligned}$$

  Again by Theorem 1, we obtain

  $$\begin{aligned} \left[ \begin{array}{ll} 0 &{}\quad 2\\ 1 &{}\quad 1 \end{array}\right] ^m=\left[ \begin{array}{ll} {J}_{m}+(-1)^m &{}\quad {J}_{m+1}+(-1)^{m+1} \\ {J}_{m} &{}\quad {J}_{m+1}\end{array}\right] . \end{aligned}$$

  (7)

  Taking the determinant of both sides yields

  $$\begin{aligned} {J}_{m}+{J}_{m+1}=2^m. \end{aligned}$$

  (8)

• (ii) Comparing (0, 1)-entries on both sides of the identity $A^{m+n}=A^m\cdot A^n$, where

  $$\begin{aligned} A=\left[ \begin{array}{ll} 0 &{} \quad 2\\ 1 &{}\quad 1 \end{array}\right] , \end{aligned}$$

  and by Eq. (7), we obtain

  $$\begin{aligned}{}[{J}_{m}+(-1)^m]{J}_n+{J}_{m}{J}_{n+1}={J}_{m+n}. \end{aligned}$$

  (9)

• (iii) Applying Eq. (8), we obtain

  $$\begin{aligned} {J}_{m+n}= & {} [{J}_{m}+(-1)^m]{J}_{m}{J}_{n+1}\\= & {} {J}_{m}[{J}_{n}+{J}_{n+1}]+(-1)^m{J}_{n}\\= & {} 2^n{J}_{m}+(-1)^m{J}_{n}. \end{aligned}$$

• (iv) Setting $n=m$ in part (iii), we get $J_{2m}/J_m=2^m+(-1)^m$.

• (v) We proceed as follows:

  $$\begin{aligned} 2^{m+n}= & {} {J}_{m+n}+{J}_{m+n+1} \ \ \ \ \ \text{(by } \text{ Eq. } \text{(8)) } \\= & {} [{J}_{m}+(-1)^m]{J}_n+{J}_{m}{J}_{n+1}\\&+\, [{J}_{m}+(-1)^m]{J}_{n+1}+{J}_{m}{J}_{n+2} \ \ \ \ \ \text{(by } \text{ Eq. } \text{(9)) } \\= & {} [{J}_{m}+(-1)^m][{J}_n+{J}_{n+1}]+{J}_{m} [{J}_{n+1}+{J}_{n+2}]\\= & {} [{J}_{m}+(-1)^m]2^n+{J}_{m}2^{n+1} \ \ \ \ \ \text{(by } \text{ Eq. } \text{(8)) }, \end{aligned}$$

  and so $2^m={J}_{m}+(-1)^m+2{J}_{m}$, or equivalently

  $$\begin{aligned} {J}_{m}=\left( 2^m+(-1)^{m+1}\right) /3. \end{aligned}$$

  This completes the proof of lemma. $\square$

A consequence of Theorem 3(i) is the following result stated again on Fibonacci, Lucas, Pell, and Jacobsthal numbers.

Corollary 1

For any three integers x, y, and z, the following identities hold true:

• (i) ${F}_{x+y}{F}_{x+z} -{F}_{x}{F}_{x+y+z}= (-1)^{x} {F}_y{F}_z$ (See [2]).

• (ii) ${L}_{x}{L}_{x+y+z}- {L}_{x+y}{L}_{x+z}= (-1)^{x}5 {F}_y{F}_z$.

• (iii) ${P}_{x+y}{P}_{x+z} -{P}_{x}{P}_{x+y+z}= (-1)^{x} {P}_y{P}_z$.

• (iv) ${J}_{x+y}{J}_{x+z} -{J}_{x}{J}_{x+y+z}= (-2)^{x} {J}_y{J}_z$.

• (v) ${L}_{x}{L}_{x+y+z}- {L}_{x+y}{L}_{x+z}= 5({F}_{x+y}{F}_{x+z} -{F}_{x}{F}_{x+y+z})$.

Proof

All parts $(i)-(iv)$ of this corollary can be derived from Theorem 3(i) by choosing suitable values for k, $\lambda _0$, $\lambda _1$, $c_0$, and $c_1$:

• (i) $k=2$, $\lambda _0=0$, $\lambda _1=c_0=c_1=1$: In this case, we have $\lambda _i={F}_i$, $\omega ^{(0)}_i={F}_{i-1}$ and $\omega ^{(1)}_i={F}_{i}$.

• (ii) $k=\lambda _0=2$, $\lambda _1=c_0=c_1=1$: In this case, we have $\lambda _i={L}_i$, $\omega ^{(0)}_i={F}_{i-1}$ and $\omega ^{(1)}_i={F}_{i}$.

• (iii) $k=c_1=2$, $\lambda _0=0$, $\lambda _1=c_0=1$: In this case, we have $\lambda _i={P}_i$, $\omega ^{(0)}_i={P}_{i-1}$ and $\omega ^{(1)}_i={P}_{i}$.

• (iv) $k=c_0=2$, $\lambda _0=0$, $\lambda _1=c_1=1$: In this case, we have $\lambda _i={J}_i$, $\omega ^{(0)}_i={J}_{i}+(-1)^{i}$ and $\omega ^{(1)}_i={J}_{i}$.

Finally, the part (v) is obtained by using parts (i) and (ii). $\square$

In the next corollary, we point out some well-known identities which are special cases of the previous results in Corollary 1, and consequently, Theorem 3(i).

Corollary 2

For any integers m, n, and p, there hold

• (i) ${F}_{m+n}={F}_{m+1}{F}_{n+1} -{F}_{m-1}{F}_{n-1}$ (See [6]).

• (ii) ${F}_{2m+1}{F}_{2n+1} ={F}^{2}_{m+n+1}+{F}^{2}_{m-n}$ (See [8, 9]).

• (iii) ${F}^{2}_{m+2n}-{F}^{2}_{m}= {F}_{2n}{F}_{2m+2n}$ (See [8]).

• (iv) ${F}_{n-1}{F}_{n+1}-{F}^{2}_{n} =(-1)^n$ (Cassini’s identity).

• (v) ${F}_{m+n}{F}_{m-n}-{F}^{2}_{m} =(-1)^{m+n+1}{F}^{2}_{n}$ (Catalan’s identity).

• (vi) ${F}_{m}{F}_{n+p} -{F}_{p}{F}_{n+m}= (-1)^{p}{F}_{m-p}{F}_{n}$ (See [10]).

• (vii) $2{L}_{m+n}={L}_{m}{L}_{n} +5{F}_{m}{F}_{n}$ (See [3]).

• (viii) ${L}_{2m}{L}_{2n} ={L}^{2}_{m+n}+5{F}^{2}_{m-n}$.

In the following table we show how to obtain the identities in Corollaries 1 and 2 directly from Theorem 3(i).

$\lambda _{0}$	$\lambda _{1}$	$c_{0}$	$c_{1}$	x	y	z	Identity	Refs.
0	1	1	1	x	y	z	${F}_{x+y}{F}_{x+z} -{F}_{x}{F}_{x+y+z}= (-1)^{x} {F}_y{F}_z$	[2]
0	1	1	1	2	$m-1$	$n-1$	${F}_{m+n}={F}_{m+1}{F}_{n+1} -{F}_{m-1}{F}_{n-1}$	[6]
0	1	1	1	$2n+1$	$m-n$	$m-n$	${F}_{2m+1}{F}_{2n+1} ={F}^{2}_{m+n+1}+{F}^{2}_{m-n}$	[8, 9]
0	1	1	1	2n	m	m	${F}^{2}_{m+2n}-{F}^{2}_{m}= {F}_{2n}{F}_{2m+2n}$	[8]
0	1	1	1	$n-1$	1	1	${F}_{n-1}{F}_{n+1}-{F}^{2}_{n} =(-1)^n$	Cassini’s identity
0	1	1	1	$m-n$	n	n	${F}_{m+n}{F}_{m-n}-{F}^{2}_{m} =(-1)^{m+n+1}{F}^{2}_{n}$	Catalan’s identity
0	1	1	1	p	n	$m-p$	${F}_{m}{F}_{n+p} -{F}_{p}{F}_{n+m}= (-1)^{p}{F}_{m-p}{F}_{n}$	[10]
2	1	1	1	x	y	z	${L}_{x}{L}_{x+y+z}- {L}_{x+y}{L}_{x+z}= (-1)^{x}5 {F}_y{F}_z$
2	1	1	1	0	m	n	$2{L}_{m+n}={L}_{m}{L}_{n} +5{F}_{m}{F}_{n}$	[3]
2	1	1	1	2n	$m-n$	$m-n$	${L}_{2m}{L}_{2n} ={L}^{2}_{m+n}+5{F}^{2}_{m-n}$
0	1	1	2	x	y	z	${P}_{x+y}{P}_{x+z} -{P}_{x}{P}_{x+y+z}= (-1)^{x} {P}_y{P}_z$
0	1	2	1	x	y	z	${J}_{x+y}{J}_{x+z} -{J}_{x}{J}_{x+y+z}= (-2)^{x} {J}_y{J}_z$