New Properties and Identities for Fibonacci Finite Operator Quaternions

Abstract

In this paper, with the help of the finite operators and Fibonacci numbers, we define a new family of quaternions whose components are the Fibonacci finite operator numbers. We also provide some properties of these types of quaternions. Moreover, we derive many identities related to Fibonacci finite operator quaternions by using the matrix representations.

1. Introduction

The quaternions can be viewed as a four-dimensional vector space defined over real numbers. A quaternion consists of four components, i.e., one real part and three imaginary parts. A quaternion is often represented in the following form:

$$q=q_0+q_1{\mathbf{e}}_1+q_2{\mathbf{e}}_2+q_3{\mathbf{e}}_3$$

where $q_0,$ $q_1,$ $q_2,$ $q_3$ are all real numbers, and the elements $\left\{\mathbf{1},{\mathbf{e}}_1,{\mathbf{e}}_2,{\mathbf{e}}_3\right\}$ form the basis of the quaternion vector space. The $\left\{\mathbf{1},{\mathbf{e}}_1,{\mathbf{e}}_2,{\mathbf{e}}_3\right\}$ obeys, following multiplication rules:

$${\mathbf{e}}_1^2={\mathbf{e}}_2^2={\mathbf{e}}_3^2={\mathbf{e}}_1{\mathbf{e}}_2{\mathbf{e}}_3=-1,$$

and

$${\mathbf{e}}_1{\mathbf{e}}_2=-{\mathbf{e}}_2{\mathbf{e}}_1={\mathbf{e}}_3,{\mathbf{e}}_2{\mathbf{e}}_3=-{\mathbf{e}}_3{\mathbf{e}}_2={\mathbf{e}}_1,{\mathbf{e}}_3{\mathbf{e}}_1=-{\mathbf{e}}_1{\mathbf{e}}_3={\mathbf{e}}_2.$$

The conjugate of a quaternion is defined as:

$$q^{*}=q_0-q_1{\mathbf{e}}_1-q_2{\mathbf{e}}_2-q_3{\mathbf{e}}_3.$$

Quaternions have become increasingly useful for practitioners in research, both in theory and applications. For example, a considerable number—maybe even the majority—of research articles on quaternions frequently appear in journals of mathematical physics, and quantum mechanics based on quaternion analysis is considered mainstream physics. In engineering, quaternions are often used in control systems, and in computer science, they play a role in computer graphics. For implementers in these areas, the following books can serve as valuable reference tool [1,2,3].

One application of quaternions that has been taken up by mathematicians is to define quaternions whose coefficients are special integer sequences or special polynomials and then examine the algebraic features of these quaternion types. Horadam [4] defined the Fibonacci quaternions as

$$\mathbb{Q}{F}_n=F_n+F_{n+1}{\mathbf{e}}_1+F_{n+2}{\mathbf{e}}_2+F_{n+3}{\mathbf{e}}_3,$$

(1)

where $F_n$ is the n-th Fibonacci number defined by

$$F_{n+2}=F_{n+1}+F_n,$$

(2)

for $n\ge 2,$ with the initial values $F_0=0$, $F_1=1.$ Let us just say here that in recent years, some studies related to Fibonacci numbers have been carried out by researchers in connection with other fields of science as well as mathematics [5,6,7]. The Binet formula for the Fibonacci sequence is

$$F_n=\frac{{\alpha }^n-{\beta }^n}{\sqrt{5}},$$

where $\alpha =\frac{1+\sqrt{5}}{2}$ and $\beta =\frac{1-\sqrt{5}}{2}.$ Additionally, the Binet-like formula of the $\mathbb{Q}{F}_n$ is as follows:

$$\mathbb{Q}{F}_n=\frac{\underline{\alpha }{\alpha }^n-\underline{\beta }{\beta }^n}{\sqrt{5}},$$

(3)

where

$$\begin{array}{ccc}\hfill \underline{\alpha }& =& 1+\alpha {\mathbf{e}}_1+{\alpha }^2{\mathbf{e}}_2+{\alpha }^3{\mathbf{e}}_3,\hfill \\ \hfill \underline{\beta }& =& 1+\beta {\mathbf{e}}_1+{\beta }^2{\mathbf{e}}_2+{\beta }^3{\mathbf{e}}_3.\hfill \end{array}$$

This construction has previously been studied by many mathematicians; see, for example, refs. [8,9,10,11,12,13,14,15,16,17,18,19,20,21].

On the other hand, special numbers, special polynomials, and finite operators have been often employed in recent years by a large number of researchers in a variety of fields of science. In particular, special polynomials, combinatorial sums, and generating functions for special numbers and polynomials are the most essential tools for developing mathematical models and methods, computational algorithms, and the other practices. In [22] Simsek defined a nice operator, as follows:

$${\mathbb{Y}}_{\lambda ,\delta }\left[f;a,b\right]\left(x\right)=\lambda E^a\left[f\right]\left(x\right)+\delta E^b\left[f\right]\left(x\right),$$

(4)

where $a,$ b are real parameters, $\lambda ,$ $\delta$ are complex parameters, and $E^a\left[f\right]\left(x\right)=f(x+a).$ For any polynomial sequence $f_n\left(x\right)$ and $i\ge 1,$ i-th finite operator ${\mathbb{Y}}_{\lambda ,\delta }^{\left(i\right)}\left[f_n;a,b\right]\left(x\right)$ (or shortly $f_n^{\left(i\right)}\left(x\right)$) is defined by the following relation:

$${\mathbb{Y}}_{\lambda ,\delta }^{\left(i\right)}\left[f_n;a,b\right]\left(x\right)=f_n^{\left(i\right)}\left(x\right)={\mathbb{Y}}_{\lambda ,\delta }\left[f_n;a,b\right]\left(x\right)\left({\mathbb{Y}}_{\lambda ,\delta }^{(i-1)}\left[f_n;a,b\right]\left(x\right)\right)$$

(5)

where ${\mathbb{Y}}_{\lambda ,\delta }^{\left(1\right)}\left[f_n;a,b\right]\left(x\right)=f_n^{\left(1\right)}\left(x\right)=\lambda f_n(x+a)+\delta f_n(x+b).$ Setting special values for $a,$ $b,$ $\lambda ,$ $\delta$ in Equation (4), Simsek derived the very essential operators used in the theory of finite difference methods for the numerical solution of differential equations, as in

Table 1. Special cases of the new finite operator.

$\mathit{\lambda }$	$\mathit{\delta }$	a	b	Operator
1	0	0	0	$I\left(f\left(x\right)\right)=f\left(x\right)$
1	$-1$	1	0	$\mathrm{\Delta }\left(f\left(x\right)\right)=f(x+1)-f\left(x\right)$
1	$-1$	0	$-1$	$\nabla \left(f\left(x\right)\right)=f\left(x\right)-f(x-1)$
$1/2$	$-1/2$	1	0	$M\left(f\left(x\right)\right)=\frac{1}{2}\left(f(x+1)-f\left(x\right)\right)$
1	$-1$	$a\to a+b$	$b\to a$	$G_{ab}\left(f\left(x\right)\right)=f(x+a+b)-f(x+a)$

.

These operators also have plenty of applications in mathematics, physics, and engineering. Utilizing this operator, Simsek defined two new classes of special polynomials and numbers. Moreover, he extensively examined several identities related to these new special polynomials and numbers, including other special polynomials. For more information related to special polynomials and special numbers, please see [22,23,24,25,26,27,28,29]. In [30] Kızılateş applied the finite operator to Horadam sequences, called Horadam finite operator sequences. Let a and b integer, $\lambda$ and $\delta$ real parameters. The Horadam finite operator numbers are defined by

$${\mathrm{\Delta }}_{\lambda ,\delta ;a,b}^{\left(i\right)}\left(W_n\right)={\mathcal{W}}_n^{\left(i\right)}=\lambda {\mathrm{\Delta }}_{\lambda ,\delta ;a,b}^{(i-1)}\left(W_{n+a}\right)+\delta {\mathrm{\Delta }}_{\lambda ,\delta ;a,b}^{(i-1)}\left(W_{n+b}\right),$$

or

$${\mathrm{\Delta }}_{\lambda ,\delta ;a,b}^{\left(i\right)}\left(W_n\right)={\mathcal{W}}_n^{\left(i\right)}=\sum _{j=0}^i\left(\genfrac{}{}{0pt}{}{i}{j}\right){\lambda }^{i-j}{\delta }^j{W}_{n+jb+(i-j)a},$$

(6)

where

$$W_n=p{W}_{n-1}+q{W}_{n-2},n\ge 2$$

(7)

with the initial values $W_0=r$, $W_1=s$, and p, q are arbitrary integers. The author also examined some algebraic properties and matrix representations of these numbers. If we take $p=q=s=1$ and r$=0$ in Equation (6), we get the Fibonacci finite operator numbers as follows:

$${\mathrm{\Delta }}_{\lambda ,\delta ;a,b}^{\left(i\right)}\left(F_n\right)={\mathcal{F}}_n^{\left(i\right)}=\lambda {\mathrm{\Delta }}_{\lambda ,\delta ;a,b}^{(i-1)}\left(F_{n+a}\right)+\delta {\mathrm{\Delta }}_{\lambda ,\delta ;a,b}^{(i-1)}\left(F_{n+b}\right).$$

(8)

For $n,i\ge 1,$ the Fibonacci finite operator sequence satisfies the recurrence relation as

$${\mathcal{F}}_{n+1}^{\left(i\right)}={\mathcal{F}}_n^{\left(i\right)}+{\mathcal{F}}_{n-1}^{\left(i\right)}.$$

(9)

Additionally, the Binet-like formula for the Fibonacci finite operator sequence is given by

$${\mathcal{F}}_n^{\left(i\right)}={\mathcal{F}}_1^{\left(i\right)}{F}_n+{\mathcal{F}}_0^{\left(i\right)}{F}_{n-1}.$$

(10)

The purpose of this research is to define a quaternion family using Fibonacci finite operators as its components. We also discuss some of the algebraic features of these quaternions. Next, we derive certain results by using the matrix representations of these quaternions. Numerous identities involving Fibonacci finite operator quaternions are established using these matrix representations.

2. Fibonacci Finite Operator Quaternions

In this part of the paper, we define the Fibonacci finite operator quaternions. We also studied some properties of these new types of quaternions.

Definition 1.

The Fibonacci finite operator quaternions, $\mathbb{Q}{\mathcal{F}}_n^{\left(i\right)}$, are defined by

$$\mathbb{Q}{\mathcal{F}}_n^{\left(i\right)}={\mathcal{F}}_n^{\left(i\right)}+{\mathcal{F}}_{n+1}^{\left(i\right)}{\mathbf{e}}_1+{\mathcal{F}}_{n+2}^{\left(i\right)}{\mathbf{e}}_2+{\mathcal{F}}_{n+3}^{\left(i\right)}{\mathbf{e}}_3.$$

(11)

where ${\mathcal{F}}_n^{\left(i\right)}$ is the i-th finite operator numbers.

Note that Definition 1 is much more general to the Fibonacci quaternions defined in Equation (1). Only for $i=1$, some special values of

$$\mathbb{Q}{\mathcal{F}}_n^{\left(1\right)}=\sum _{s=0}^3\left(\lambda F_{n+a+s}+\delta F_{n+b+s}\right){\mathbf{e}}_s,$$

(12)

as follows:

• If we take $\lambda =1$ and $\delta =a=b=0$ in Equation (12), we get the identity operator for Fibonacci quaternion sequence $I\left(\mathbb{Q}{\mathcal{F}}_n^{\left(1\right)}\right)=\mathbb{Q}{F}_n$;
• If we take $\lambda =1$, $\delta =-1$, $a=1$ and $b=0$ in Equation (12), we obtain the forward difference operator for Fibonacci quaternion sequence $\mathrm{\Delta }\left(\mathbb{Q}{\mathcal{F}}_n^{\left(1\right)}\right)=\mathbb{Q}{F}_{n+1}-\mathbb{Q}{F}_n;$
• If we take $\lambda =1$, $\delta =-1$, $a=0$ and $b=-1$ in Equation (12), we obtain the backward difference operator for Fibonacci quaternion sequence $\nabla \left(\mathbb{Q}{\mathcal{F}}_n^{\left(1\right)}\right)=\mathbb{Q}{F}_n-\mathbb{Q}{F}_{n-1};$
• If we take $\lambda =\frac{1}{2},$ $\delta =\frac{-1}{2}$, $a=1$ and $b=0$ in Equation (12), we obtain the means operator for Fibonacci quaternion sequence $M\left(\mathbb{Q}{\mathcal{F}}_n^{\left(1\right)}\right)=\frac{1}{2}\left(\mathbb{Q}{F}_{n+1}-\mathbb{Q}{F}_n\right);$
• If we take $\lambda =1,$ $\delta =-1,$ and substitute $a\to a+b$, $b\to a$ and $ab\ne 0$ in Equation (12), we obtain the Gould operator for Fibonacci quaternion sequence $G_{ab}\left(\mathbb{Q}{\mathcal{F}}_n^{\left(1\right)}\right)=\mathbb{Q}{F}_{n+a+b}-\mathbb{Q}{F}_{n+a}.$

From Equation (11), the Fibonacci finite operator quaternions can be written as

$$\mathbb{Q}{\mathcal{F}}_n^{\left(i\right)}={\mathcal{F}}_n^{\left(i\right)}+u,$$

where $u={\mathcal{F}}_{n+1}^{\left(i\right)}{\mathbf{e}}_1+{\mathcal{F}}_{n+2}^{\left(i\right)}{\mathbf{e}}_2+{\mathcal{F}}_{n+3}^{\left(i\right)}{\mathbf{e}}_3.$

The conjugate of the Fibonacci finite operator quaternions, $\mathbb{Q}{\mathcal{F}}_n^{\left(i\right)}$ is denoted by ${\left(\mathbb{Q}{\mathcal{F}}_n^{\left(i\right)}\right)}^{*}$ as

$${\left(\mathbb{Q}{\mathcal{F}}_n^{\left(i\right)}\right)}^{*}={\mathcal{F}}_n^{\left(i\right)}-u.$$

(13)

For the Fibonacci finite operator quaternions, we can easily obtain that

$$\mathbb{Q}{\mathcal{F}}_n^{\left(i\right)}+{\left(\mathbb{Q}{\mathcal{F}}_n^{\left(i\right)}\right)}^{*}=2{\mathcal{F}}_n^{\left(i\right)}.$$

Theorem 1.

The recurrence relation for the Fibonacci finite operator quaternions; $\mathbb{Q}{\mathcal{F}}_n^{\left(i\right)}$ is

$$\mathbb{Q}{\mathcal{F}}_n^{\left(i\right)}=\mathbb{Q}{\mathcal{F}}_{n-1}^{\left(i\right)}+\mathbb{Q}{\mathcal{F}}_{n-2}^{\left(i\right)}.$$

(14)

Proof. 

Using Equation (9), we find that

$$\begin{array}{cc}\hfill \mathbb{Q}{\mathcal{F}}_n^{\left(i\right)}& ={\mathcal{F}}_n^{\left(i\right)}+{\mathcal{F}}_{n+1}^{\left(i\right)}{\mathbf{e}}_1+{\mathcal{F}}_{n+2}^{\left(i\right)}{\mathbf{e}}_2+{\mathcal{F}}_{n+3}^{\left(i\right)}{\mathbf{e}}_3\hfill \\ \hfill & ={\mathcal{F}}_{n-1}^{\left(i\right)}+{\mathcal{F}}_{n-2}^{\left(i\right)}+\left({\mathcal{F}}_n^{\left(i\right)}+{\mathcal{F}}_{n-1}^{\left(i\right)}\right){\mathbf{e}}_1+\left({\mathcal{F}}_{n+1}^{\left(i\right)}+{\mathcal{F}}_n^{\left(i\right)}\right){\mathbf{e}}_2+\left({\mathcal{F}}_{n+2}^{\left(i\right)}+{\mathcal{F}}_{n+1}^{\left(i\right)}\right){\mathbf{e}}_3\hfill \\ \hfill & =\left({\mathcal{F}}_{n-1}^{\left(i\right)}+{\mathcal{F}}_n^{\left(i\right)}{\mathbf{e}}_1+{\mathcal{F}}_{n+1}^{\left(i\right)}{\mathbf{e}}_2+{\mathcal{F}}_{n+2}^{\left(i\right)}{\mathbf{e}}_3\right)+\left({\mathcal{F}}_{n-2}^{\left(i\right)}+{\mathcal{F}}_{n-1}^{\left(i\right)}{\mathbf{e}}_1+{\mathcal{F}}_n^{\left(i\right)}{\mathbf{e}}_2+{\mathcal{F}}_{n+1}^{\left(i\right)}{\mathbf{e}}_3\right)\hfill \\ \hfill & =\mathbb{Q}{\mathcal{F}}_{n-1}^{\left(i\right)}+\mathbb{Q}{\mathcal{F}}_{n-2}^{\left(i\right)}.\hfill \end{array}$$

 □

Theorem 2.

The Binet-like formula of the Fibonacci finite operator quaternions; $\mathbb{Q}{\mathcal{F}}_n^{\left(i\right)}$ is as follows:

$$\mathbb{Q}{\mathcal{F}}_n^{\left(i\right)}=\frac{\underline{\alpha }{\alpha }^{n-1}\left(\alpha {\mathcal{F}}_1^{\left(i\right)}+{\mathcal{F}}_0^{\left(i\right)}\right)-\underline{\beta }{\beta }^{n-1}\left(\beta {\mathcal{F}}_1^{\left(i\right)}+{\mathcal{F}}_0^{\left(i\right)}\right)}{\sqrt{5}}.$$

(15)

Proof. 

Thanks to Equations (3) and (10), we have

$$\begin{array}{cc}\hfill \mathbb{Q}{\mathcal{F}}_n^{\left(i\right)}& ={\mathcal{F}}_n^{\left(i\right)}+{\mathcal{F}}_{n+1}^{\left(i\right)}{\mathbf{e}}_1+{\mathcal{F}}_{n+2}^{\left(i\right)}{\mathbf{e}}_2+{\mathcal{F}}_{n+3}^{\left(i\right)}{\mathbf{e}}_3\hfill \\ \hfill & ={\mathcal{F}}_1^{\left(i\right)}{F}_n+{\mathcal{F}}_0^{\left(i\right)}{F}_{n-1}+\left({\mathcal{F}}_1^{\left(i\right)}{F}_{n+1}+{\mathcal{F}}_0^{\left(i\right)}{F}_n\right){\mathbf{e}}_1\hfill \\ \hfill & +\left({\mathcal{F}}_1^{\left(i\right)}{F}_{n+2}+{\mathcal{F}}_0^{\left(i\right)}{F}_{n+1}\right){\mathbf{e}}_2+\left({\mathcal{F}}_1^{\left(i\right)}{F}_{n+3}+{\mathcal{F}}_0^{\left(i\right)}{F}_{n+2}\right){\mathbf{e}}_3\hfill \\ \hfill & ={\mathcal{F}}_1^{\left(i\right)}\left(F_n+F_{n+1}{\mathbf{e}}_1+F_{n+2}{\mathbf{e}}_2+F_{n+3}{\mathbf{e}}_3\right)+{\mathcal{F}}_0^{\left(i\right)}\left(F_{n-1}+F_n{\mathbf{e}}_1+F_{n+1}{\mathbf{e}}_2+F_{n+2}{\mathbf{e}}_3\right)\hfill \\ \hfill & ={\mathcal{F}}_1^{\left(i\right)}\mathbb{Q}{F}_n+{\mathcal{F}}_0^{\left(i\right)}\mathbb{Q}{F}_{n-1}\hfill \\ \hfill & ={\mathcal{F}}_1^{\left(i\right)}\frac{\underline{\alpha }{\alpha }^n-\underline{\beta }{\beta }^n}{\sqrt{5}}+{\mathcal{F}}_0^{\left(i\right)}\frac{\underline{\alpha }{\alpha }^{n-1}-\underline{\beta }{\beta }^{n-1}}{\sqrt{5}}\hfill \\ \hfill & =\frac{\underline{\alpha }{\alpha }^{n-1}\left(\alpha {\mathcal{F}}_1^{\left(i\right)}+{\mathcal{F}}_0^{\left(i\right)}\right)-\underline{\beta }{\beta }^{n-1}\left(\beta {\mathcal{F}}_1^{\left(i\right)}+{\mathcal{F}}_0^{\left(i\right)}\right)}{\sqrt{5}}.\hfill \end{array}$$

 □

Theorem 3.

The generating function for the $\mathbb{Q}{\mathcal{F}}_n^{\left(i\right)}$ is as follows:

$$\mathbb{Q}{\mathcal{F}}_n^{\left(i\right)}\left(x\right)=\frac{\mathbb{Q}{\mathcal{F}}_0^{\left(i\right)}+\left(\mathbb{Q}{\mathcal{F}}_1^{\left(i\right)}-\mathbb{Q}{\mathcal{F}}_0^{\left(i\right)}\right)x}{1-x-x^2}.$$

Proof. 

Let $\mathbb{Q}{\mathcal{F}}_n^{\left(i\right)}\left(x\right)$ be the generating function of the Fibonacci finite operator quaternions. Namely,

$$\mathbb{Q}{\mathcal{F}}_n^{\left(i\right)}\left(x\right)=\sum _{n=0}^{\infty }\mathbb{Q}{\mathcal{F}}_n^{\left(i\right)}{x}^n.$$

Then, we have,

$$\begin{array}{cc}\hfill \mathbb{Q}{\mathcal{F}}_n^{\left(i\right)}\left(x\right)& =\mathbb{Q}{\mathcal{F}}_0^{\left(i\right)}+\mathbb{Q}{\mathcal{F}}_1^{\left(i\right)}x+\mathbb{Q}{\mathcal{F}}_2^{\left(i\right)}{x}^2+\cdots +\mathbb{Q}{\mathcal{F}}_n^{\left(i\right)}{x}^n+\cdots \hfill \\ \hfill -x\mathbb{Q}{\mathcal{F}}_n^{\left(i\right)}\left(x\right)& =-\mathbb{Q}{\mathcal{F}}_0^{\left(i\right)}x-\mathbb{Q}{\mathcal{F}}_1^{\left(i\right)}{x}^2-\mathbb{Q}{\mathcal{F}}_2^{\left(i\right)}{x}^3-\cdots -\mathbb{Q}{\mathcal{F}}_n^{\left(i\right)}{x}^{n+1}-\cdots \hfill \\ \hfill -x^2\mathbb{Q}{\mathcal{F}}_n^{\left(i\right)}\left(x\right)& =-\mathbb{Q}{\mathcal{F}}_0^{\left(i\right)}{x}^2-\mathbb{Q}{\mathcal{F}}_1^{\left(i\right)}{x}^3-\mathbb{Q}{\mathcal{F}}_2^{\left(i\right)}{x}^4-\cdots -\mathbb{Q}{\mathcal{F}}_n^{\left(i\right)}{x}^{n+2}-\cdots \hfill \end{array}$$

Using the above identities, we get

$$\left(1-x-x^2\right)\mathbb{Q}{\mathcal{F}}_n^{\left(i\right)}\left(x\right)=\mathbb{Q}{\mathcal{F}}_0^{\left(i\right)}+\left(\mathbb{Q}{\mathcal{F}}_1^{\left(i\right)}-\mathbb{Q}{\mathcal{F}}_0^{\left(i\right)}\right)x+\sum _{n=2}^{\infty }\left(\mathbb{Q}{\mathcal{F}}_n^{\left(i\right)}-\mathbb{Q}{\mathcal{F}}_{n-1}^{\left(i\right)}-\mathbb{Q}{\mathcal{F}}_{n-2}^{\left(i\right)}\right)x^n$$

Following Equation (14), we have

$$\mathbb{Q}{\mathcal{F}}_n^{\left(i\right)}\left(x\right)=\frac{\mathbb{Q}{\mathcal{F}}_0^{\left(i\right)}+\left(\mathbb{Q}{\mathcal{F}}_1^{\left(i\right)}-\mathbb{Q}{\mathcal{F}}_0^{\left(i\right)}\right)x}{1-x-x^2}.$$

 □

Theorem 4.

The exponential generating function for the sequence $\mathbb{Q}{\mathcal{F}}_n^{\left(i\right)}$ is as follows:

$$\sum _{n=0}^{\infty }\mathbb{Q}{\mathcal{F}}_n^{\left(i\right)}\frac{x^n}{n!}=\frac{{\mathcal{F}}_1^{\left(i\right)}\left(\underline{\alpha }{e}^{\alpha x}-\underline{\beta }{e}^{\beta x}\right)-{\mathcal{F}}_0^{\left(i\right)}\left(\underline{\alpha }\beta e^{\alpha x}-\underline{\beta }\alpha e^{\beta x}\right)}{\sqrt{5}}.$$

Proof. 

Using Equation (15), we get

$$\begin{array}{cc}\hfill \sum _{n=0}^{\infty }\mathbb{Q}{\mathcal{F}}_n^{\left(i\right)}\frac{x^n}{n!}& =\sum _{n=0}^{\infty }\left(\frac{\underline{\alpha }{\alpha }^{n-1}\left(\alpha {\mathcal{F}}_1^{\left(i\right)}+{\mathcal{F}}_0^{\left(i\right)}\right)-\underline{\beta }{\beta }^{n-1}\left(\beta {\mathcal{F}}_1^{\left(i\right)}+{\mathcal{F}}_0^{\left(i\right)}\right)}{\sqrt{5}}\right)\frac{x^n}{n!}\hfill \\ \hfill & =\frac{1}{\sqrt{5}}\left(\frac{\underline{\alpha }\left(\alpha {\mathcal{F}}_1^{\left(i\right)}+{\mathcal{F}}_0^{\left(i\right)}\right)}{\alpha }{e}^{\alpha x}-\frac{\underline{\beta }\left(\beta {\mathcal{F}}_1^{\left(i\right)}+{\mathcal{F}}_0^{\left(i\right)}\right)}{\beta }{e}^{\beta x}\right)\hfill \\ \hfill & =\frac{\underline{\alpha }\beta e^{\alpha x}\left(\alpha {\mathcal{F}}_1^{\left(i\right)}+{\mathcal{F}}_0^{\left(i\right)}\right)-\underline{\beta }\alpha e^{\beta x}\left(\beta {\mathcal{F}}_1^{\left(i\right)}+{\mathcal{F}}_0^{\left(i\right)}\right)}{\sqrt{5}\alpha \beta }\hfill \\ \hfill & =\frac{{\mathcal{F}}_1^{\left(i\right)}\left(\underline{\alpha }{e}^{\alpha x}-\underline{\beta }{e}^{\beta x}\right)-{\mathcal{F}}_0^{\left(i\right)}\left(\underline{\alpha }\beta e^{\alpha x}-\underline{\beta }\alpha e^{\beta x}\right)}{\sqrt{5}}.\hfill \end{array}$$

 □

Theorem 5.

For non-negative integer n, we have

$$\sum _{t=0}^n{\left(-1\right)}^t\left(\genfrac{}{}{0pt}{}{n}{t}\right)\mathbb{Q}{\mathcal{F}}_{2t+k}^{\left(i\right)}={\left(-1\right)}^n\mathbb{Q}{\mathcal{F}}_{n+k}^{\left(i\right)}.$$

Proof. 

By virtue of Equation (15), we find that

$$\begin{array}{cc}\hfill \sum _{t=0}^n{\left(-1\right)}^t\left(\genfrac{}{}{0pt}{}{n}{t}\right)\mathbb{Q}{\mathcal{F}}_{2t+k}^{\left(i\right)}& =\sum _{t=0}^n\left(\genfrac{}{}{0pt}{}{n}{t}\right){\left(-1\right)}^t\left(\frac{\underline{\alpha }{\alpha }^{2t+k-1}\left(\alpha {\mathcal{F}}_1^{\left(i\right)}+{\mathcal{F}}_0^{\left(i\right)}\right)-\underline{\beta }{\beta }^{2t+k-1}\left(\beta {\mathcal{F}}_1^{\left(i\right)}+{\mathcal{F}}_0^{\left(i\right)}\right)}{\sqrt{5}}\right)\hfill \\ \hfill & =\frac{\underline{\alpha }{\alpha }^{k-1}\left(\alpha {\mathcal{F}}_1^{\left(i\right)}+{\mathcal{F}}_0^{\left(i\right)}\right)}{\sqrt{5}}{\left(1-{\alpha }^2\right)}^n-\frac{\underline{\beta }{\beta }^{k-1}\left(\beta {\mathcal{F}}_1^{\left(i\right)}+{\mathcal{F}}_0^{\left(i\right)}\right)}{\sqrt{5}}{\left(1-{\beta }^2\right)}^n\hfill \\ \hfill & =\frac{\underline{\alpha }{\alpha }^{k-1}\left(\alpha {\mathcal{F}}_1^{\left(i\right)}+{\mathcal{F}}_0^{\left(i\right)}\right)}{\sqrt{5}}{\left(-\alpha \right)}^n-\frac{\underline{\beta }{\beta }^{k-1}\left(\beta {\mathcal{F}}_1^{\left(i\right)}+{\mathcal{F}}_0^{\left(i\right)}\right)}{\sqrt{5}}{\left(-\beta \right)}^n\hfill \\ \hfill & ={\left(-1\right)}^n\mathbb{Q}{\mathcal{F}}_{n+k}^{\left(i\right)}.\hfill \end{array}$$

 □

Theorem 6.

For non-negative integer n, we have

$$\sum _{t=0}^n\left(\genfrac{}{}{0pt}{}{n}{t}\right)\mathbb{Q}{\mathcal{F}}_t^{\left(i\right)}=\mathbb{Q}{\mathcal{F}}_{2n}^{\left(i\right)}.$$

Proof. 

From Equation (15), we have

$$\begin{array}{cc}\hfill \sum _{t=0}^n\left(\genfrac{}{}{0pt}{}{n}{t}\right)\mathbb{Q}{\mathcal{F}}_t^{\left(i\right)}& =\sum _{t=0}^n\left(\genfrac{}{}{0pt}{}{n}{t}\right)\left(\frac{\underline{\alpha }{\alpha }^{t-1}\left(\alpha {\mathcal{F}}_1^{\left(i\right)}+{\mathcal{F}}_0^{\left(i\right)}\right)-\underline{\beta }{\beta }^{t-1}\left(\beta {\mathcal{F}}_1^{\left(i\right)}+{\mathcal{F}}_0^{\left(i\right)}\right)}{\sqrt{5}}\right)\hfill \\ \hfill & =\frac{\underline{\alpha }{\alpha }^{-1}\left(\alpha {\mathcal{F}}_1^{\left(i\right)}+{\mathcal{F}}_0^{\left(i\right)}\right)}{\sqrt{5}}{\left(1+\alpha \right)}^n-\frac{\underline{\beta }{\beta }^{-1}\left(\beta {\mathcal{F}}_1^{\left(i\right)}+{\mathcal{F}}_0^{\left(i\right)}\right)}{\sqrt{5}}{\left(1+\beta \right)}^n\hfill \\ \hfill & =\frac{\underline{\alpha }{\alpha }^{2n-1}\left(\alpha {\mathcal{F}}_1^{\left(i\right)}+{\mathcal{F}}_0^{\left(i\right)}\right)}{\sqrt{5}}-\frac{\underline{\beta }{\beta }^{2n-1}\left(\beta {\mathcal{F}}_1^{\left(i\right)}+{\mathcal{F}}_0^{\left(i\right)}\right)}{\sqrt{5}}\hfill \\ \hfill & =\mathbb{Q}{\mathcal{F}}_{2n}^{\left(i\right)}.\hfill \end{array}$$

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Theorem 7.

For non-negative integer n, we have

$$\sum _{t=0}^n\left(\genfrac{}{}{0pt}{}{n}{t}\right){\left(\mathbb{Q}{\mathcal{F}}_t^{\left(i\right)}\right)}^2=5^{\frac{n-2}{2}}\left(\left(2\underline{\alpha }-12\alpha -9\right){\alpha }^{n-2}\zeta +{(-1)}^n\left(2\underline{\beta }-12\beta -9\right){\beta }^{n-2}\xi \right)$$

where $\zeta ={\left(\alpha {\mathcal{F}}_1^{\left(i\right)}+{\mathcal{F}}_0^{\left(i\right)}\right)}^2$ and $\xi ={\left(\beta {\mathcal{F}}_1^{\left(i\right)}+{\mathcal{F}}_0^{\left(i\right)}\right)}^2.$

Proof. 

The authors [16] proved that

$${\left(\underline{\alpha }\right)}^2=\left(2\underline{\alpha }-\left(L_5+1\right)\alpha -L_4-L_0\right)=\left(2\underline{\alpha }-12\alpha -9\right)$$

$${\left(\underline{\beta }\right)}^2=\left(2\underline{\beta }-\left(L_5+1\right)\beta -L_4-L_0\right)=\left(2\underline{\beta }-12\beta -9\right),$$

where $L_n$ is the n-th Lucas numbers defined by

$$L_n=L_{n-1}+L_{n-2},$$

for $n\ge 2,$ with the initial values $L_0=2$ and $L_1=1.$ Using (15), after some calculations, we have

$$\begin{array}{cc}\hfill \sum _{t=0}^n\left(\genfrac{}{}{0pt}{}{n}{t}\right){\left(\mathbb{Q}{\mathcal{F}}_t^{\left(i\right)}\right)}^2& =\sum _{t=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{t}}\right){\left(\frac{\underline{\alpha }{\alpha }^{t-1}\left(\alpha {\mathcal{F}}_1^{\left(i\right)}+{\mathcal{F}}_0^{\left(i\right)}\right)-\underline{\beta }{\beta }^{t-1}\left(\beta {\mathcal{F}}_1^{\left(i\right)}+{\mathcal{F}}_0^{\left(i\right)}\right)}{\sqrt{5}}\right)}^2\hfill \\ \hfill & =\frac{1}{5}\sum _{t=0}^n\left(\genfrac{}{}{0pt}{}{n}{t}\right){\left(\underline{\alpha }{\alpha }^{t-1}\left(\alpha {\mathcal{F}}_1^{\left(i\right)}+{\mathcal{F}}_0^{\left(i\right)}\right)-\underline{\beta }{\beta }^{t-1}\left(\beta {\mathcal{F}}_1^{\left(i\right)}+{\mathcal{F}}_0^{\left(i\right)}\right)\right)}^2\hfill \\ \hfill & =\frac{1}{5}\left({\left(\underline{\alpha }\right)}^2{\left(\alpha {\mathcal{F}}_1^{\left(i\right)}+{\mathcal{F}}_0^{\left(i\right)}\right)}^2\sum _{t=0}^n\left(\genfrac{}{}{0pt}{}{n}{t}\right){\alpha }^{2t-2}+{\left(\underline{\beta }\right)}^2{\left(\beta {\mathcal{F}}_1^{\left(i\right)}+{\mathcal{F}}_0^{\left(i\right)}\right)}^2\sum _{t=0}^n\left(\genfrac{}{}{0pt}{}{n}{t}\right){\beta }^{2t-2}\right)\hfill \\ \hfill & =\frac{{\left(\underline{\alpha }\right)}^2{\alpha }^{-2}{\left(\alpha {\mathcal{F}}_1^{\left(i\right)}+{\mathcal{F}}_0^{\left(i\right)}\right)}^2}{5}{\left(1+{\alpha }^2\right)}^n+\frac{{\left(\underline{\beta }\right)}^2{\beta }^{-2}{\left(\beta {\mathcal{F}}_1^{\left(i\right)}+{\mathcal{F}}_0^{\left(i\right)}\right)}^2}{5}{\left(1+{\beta }^2\right)}^n\hfill \\ \hfill & =\frac{{\left(\underline{\alpha }\right)}^2{\alpha }^{-2}{\left(\alpha {\mathcal{F}}_1^{\left(i\right)}+{\mathcal{F}}_0^{\left(i\right)}\right)}^2}{5}{\left(\alpha \sqrt{5}\right)}^n+\frac{{\left(\underline{\beta }\right)}^2{\beta }^{-2}{\left(\beta {\mathcal{F}}_1^{\left(i\right)}+{\mathcal{F}}_0^{\left(i\right)}\right)}^2}{5}{\left(-\beta \sqrt{5}\right)}^n\hfill \\ \hfill & =5^{\frac{n-2}{2}}\left({\left(\underline{\alpha }\right)}^2{\alpha }^{n-2}{\left(\alpha {\mathcal{F}}_1^{\left(i\right)}+{\mathcal{F}}_0^{\left(i\right)}\right)}^2+{(-1)}^n{\left(\underline{\beta }\right)}^2{\beta }^{n-2}{\left(\beta {\mathcal{F}}_1^{\left(i\right)}+{\mathcal{F}}_0^{\left(i\right)}\right)}^2\right)\hfill \\ \hfill & =5^{\frac{n-2}{2}}\left(\left(2\underline{\alpha }-12\alpha -9\right){\alpha }^{n-2}\zeta +{(-1)}^n\left(2\underline{\beta }-12\beta -9\right){\beta }^{n-2}\xi \right).\hfill \end{array}$$

 □

3. Matrix Representations of Fibonacci Finite Operator Quaternions and Their New Properties

From past to present, matrix representations have been studied by many researchers for special integer sequences and various generalizations of these sequences. Halici [8] gave the following matrix representation to obtain the Cassini identity for Fibonacci quaternions defined by Horadam

$$\mathbf{M}=\left[\begin{array}{cc}\mathbb{Q}{F}_2& \mathbb{Q}{F}_1\\ \mathbb{Q}{F}_1& \mathbb{Q}{F}_0\end{array}\right].$$

Similar to Fibonacci quaternion matrices, various matrix representations are given in different quaternion sequences and generalizations of these sequences. The best references here are [31,32,33,34,35,36,37]. Patel and Ray [33] defined the Fibonacci quaternion matrix as follows:

$$\mathbf{M}=\left[\begin{array}{cc}\mathbb{Q}{F}_2& \mathbb{Q}{F}_1\\ \mathbb{Q}{F}_1& \mathbb{Q}{F}_0\end{array}\right]⟹{\mathbf{MU}}^{n-1}=\left[\begin{array}{cc}\mathbb{Q}{F}_{n+1}& \mathbb{Q}{F}_n\\ \mathbb{Q}{F}_n& \mathbb{Q}{F}_{n-1}\end{array}\right],$$

(16)

where the matrix $\mathbf{U}$ satisfies the following matrix relation:

$$\mathbf{U}=\left[\begin{array}{cc}1& 1\\ 1& 0\end{array}\right]⟹{\mathbf{U}}^n=\left[\begin{array}{cc}{F}_{n+1}& F_n\\ F_n& F_{n-1}\end{array}\right].$$

(17)

In the previous section, we have obtained some properties of Fibonacci finite operator quaternions by using the Binet-like formula. In this section of the our paper, based on Tan and Leung’s paper [31] with a similar approach, we give matrix representations for these type of quaternions. Using these representation, we also derive several properties of Fibonacci finite operator quaternions. From the recurrence relation of the Fibonacci finite operator numbers, we can easily see the matrix relation:

$$\mathbf{N}=\left[\begin{array}{cc}{\mathcal{F}}_2^{\left(i\right)}& {\mathcal{F}}_1^{\left(i\right)}\\ {\mathcal{F}}_1^{\left(i\right)}& {\mathcal{F}}_0^{\left(i\right)}\end{array}\right]⟹{\mathbf{NU}}^{n-1}=\left[\begin{array}{cc}{\mathcal{F}}_{n+1}^{\left(i\right)}& {\mathcal{F}}_n^{\left(i\right)}\\ {\mathcal{F}}_n^{\left(i\right)}& {\mathcal{F}}_{n-1}^{\left(i\right)}\end{array}\right].$$

(18)

Considering the matrix equalities in Equations (16) and (18), we have a matrix representation of the Fibonacci finite operator quaternions as follows:

$$\left({\mathbf{NU}}^{n-1}\right)\mathbf{M}=\mathbf{M}\left({\mathbf{NU}}^{n-1}\right)=\left[\begin{array}{cc}\mathbb{Q}{\mathcal{F}}_{n+2}^{\left(i\right)}& \mathbb{Q}{\mathcal{F}}_{n+1}^{\left(i\right)}\\ \mathbb{Q}{\mathcal{F}}_{n+1}^{\left(i\right)}& \mathbb{Q}{\mathcal{F}}_n^{\left(i\right)}\end{array}\right].$$

(19)

Let us point out here that although the matrix multiplication is not commutative, Equality Equation (19) is held. Namely,

$$\begin{array}{cc}\hfill \left({\mathbf{NU}}^{n-1}\right)\mathbf{M}& =\left[\begin{array}{cc}{\mathcal{F}}_{n+1}^{\left(i\right)}& {\mathcal{F}}_n^{\left(i\right)}\\ {\mathcal{F}}_n^{\left(i\right)}& {\mathcal{F}}_{n-1}^{\left(i\right)}\end{array}\right]\left[\begin{array}{cc}\mathbb{Q}{F}_2& \mathbb{Q}{F}_1\\ \mathbb{Q}{F}_1& \mathbb{Q}{F}_0\end{array}\right]\hfill \\ \hfill & =\left[\begin{array}{cc}{\mathcal{F}}_{n+1}^{\left(i\right)}\mathbb{Q}{F}_2+{\mathcal{F}}_n^{\left(i\right)}\mathbb{Q}{F}_1& {\mathcal{F}}_{n+1}^{\left(i\right)}\mathbb{Q}{F}_1+{\mathcal{F}}_n^{\left(i\right)}\mathbb{Q}{F}_0\\ {\mathcal{F}}_n^{\left(i\right)}\mathbb{Q}{F}_2+{\mathcal{F}}_{n-1}^{\left(i\right)}\mathbb{Q}{F}_1& {\mathcal{F}}_n^{\left(i\right)}\mathbb{Q}{F}_1+{\mathcal{F}}_{n-1}^{\left(i\right)}\mathbb{Q}{F}_0\end{array}\right]\hfill \\ \hfill & =\left[\begin{array}{cc}\mathbb{Q}{F}_2{\mathcal{F}}_{n+1}^{\left(i\right)}+\mathbb{Q}{F}_1{\mathcal{F}}_n^{\left(i\right)}& \mathbb{Q}{F}_1{\mathcal{F}}_{n+1}^{\left(i\right)}+\mathbb{Q}{F}_0{\mathcal{F}}_n^{\left(i\right)}\\ \mathbb{Q}{F}_2{\mathcal{F}}_n^{\left(i\right)}+\mathbb{Q}{F}_1{\mathcal{F}}_{n-1}^{\left(i\right)}& \mathbb{Q}{F}_1{\mathcal{F}}_n^{\left(i\right)}+\mathbb{Q}{F}_0{\mathcal{F}}_{n-1}^{\left(i\right)}\end{array}\right]\hfill \\ \hfill & =\left[\begin{array}{cc}\mathbb{Q}{F}_2& \mathbb{Q}{F}_1\\ \mathbb{Q}{F}_1& \mathbb{Q}{F}_0\end{array}\right]\left[\begin{array}{cc}{\mathcal{F}}_{n+1}^{\left(i\right)}& {\mathcal{F}}_n^{\left(i\right)}\\ {\mathcal{F}}_n^{\left(i\right)}& {\mathcal{F}}_{n-1}^{\left(i\right)}\end{array}\right]\hfill \\ \hfill & =\mathbf{M}\left({\mathbf{NU}}^{n-1}\right).\hfill \end{array}$$

We also have another matrix representation for the Fibonacci finite operator quaternions, as follows:

$$\mathbf{O}:=\left[\begin{array}{cc}\mathbb{Q}{\mathcal{F}}_2^{\left(i\right)}& \mathbb{Q}{\mathcal{F}}_1^{\left(i\right)}\\ \mathbb{Q}{\mathcal{F}}_1^{\left(i\right)}& \mathbb{Q}{\mathcal{F}}_0^{\left(i\right)}\end{array}\right]⟹{\mathbf{U}}^n\mathbf{O}={\mathbf{OU}}^n=\left[\begin{array}{cc}\mathbb{Q}{\mathcal{F}}_{n+2}^{\left(i\right)}& \mathbb{Q}{\mathcal{F}}_{n+1}^{\left(i\right)}\\ \mathbb{Q}{\mathcal{F}}_{n+1}^{\left(i\right)}& \mathbb{Q}{\mathcal{F}}_n^{\left(i\right)}\end{array}\right].$$

(20)

Since the quaternion multiplication is non-commutative, the following theorem gives four Cassini’s identities for Fibonacci finite operator quaternions.

Theorem 8.

For non-negative integer $n,$ we find that

$$\mathbb{Q}{\mathcal{F}}_{n+1}^{\left(i\right)}\mathbb{Q}{\mathcal{F}}_{n-1}^{\left(i\right)}-{\left(\mathbb{Q}{\mathcal{F}}_n^{\left(i\right)}\right)}^2={\left(-1\right)}^{n-1}\left(\mathbb{Q}{\mathcal{F}}_2^{\left(i\right)}\mathbb{Q}{\mathcal{F}}_0^{\left(i\right)}-{\left(\mathbb{Q}{\mathcal{F}}_1^{\left(i\right)}\right)}^2\right),$$

(21)

$$\mathbb{Q}{\mathcal{F}}_{n-1}^{\left(i\right)}\mathbb{Q}{\mathcal{F}}_{n+1}^{\left(i\right)}-{\left(\mathbb{Q}{\mathcal{F}}_n^{\left(i\right)}\right)}^2={\left(-1\right)}^{n-1}\left(\mathbb{Q}{\mathcal{F}}_0^{\left(i\right)}\mathbb{Q}{\mathcal{F}}_2^{\left(i\right)}-{\left(\mathbb{Q}{\mathcal{F}}_1^{\left(i\right)}\right)}^2\right),$$

(22)

$$\mathbb{Q}{\mathcal{F}}_{n+1}^{\left(i\right)}\mathbb{Q}{\mathcal{F}}_{n-1}^{\left(i\right)}-{\left(\mathbb{Q}{\mathcal{F}}_n^{\left(i\right)}\right)}^2={\left(-1\right)}^{n-1}\left(\mathbb{Q}{F}_2\mathbb{Q}{F}_0-{\left(\mathbb{Q}{F}_1\right)}^2\right)\left({\left({\mathcal{F}}_1^{\left(i\right)}\right)}^2-{\mathcal{F}}_1^{\left(i\right)}{\mathcal{F}}_0^{\left(i\right)}-{\left({\mathcal{F}}_0^{\left(i\right)}\right)}^2\right),$$

(23)

$$\mathbb{Q}{\mathcal{F}}_{n-1}^{\left(i\right)}\mathbb{Q}{\mathcal{F}}_{n+1}^{\left(i\right)}-{\left(\mathbb{Q}{\mathcal{F}}_n^{\left(i\right)}\right)}^2={\left(-1\right)}^{n-1}\left(\mathbb{Q}{F}_0\mathbb{Q}{F}_2-{\left(\mathbb{Q}{F}_1\right)}^2\right)\left({\left({\mathcal{F}}_1^{\left(i\right)}\right)}^2-{\mathcal{F}}_1^{\left(i\right)}{\mathcal{F}}_0^{\left(i\right)}-{\left({\mathcal{F}}_0^{\left(i\right)}\right)}^2\right).$$

(24)

Proof. 

For Equations (21) and (22), by using Equation (20), we get

$$\left|\begin{array}{cc}\mathbb{Q}{\mathcal{F}}_{n+1}^{\left(i\right)}& \mathbb{Q}{\mathcal{F}}_n^{\left(i\right)}\\ \mathbb{Q}{\mathcal{F}}_n^{\left(i\right)}& \mathbb{Q}{\mathcal{F}}_{n-1}^{\left(i\right)}\end{array}\right|=\left|{\mathbf{U}}^{n-1}\mathbf{O}\right|={\left|\mathbf{U}\right|}^{n-1}\left|\mathbf{O}\right|={\left|\begin{array}{cc}1& 1\\ 1& 0\end{array}\right|}^{n-1}\left|\begin{array}{cc}\mathbb{Q}{\mathcal{F}}_2^{\left(i\right)}& \mathbb{Q}{\mathcal{F}}_1^{\left(i\right)}\\ \mathbb{Q}{\mathcal{F}}_1^{\left(i\right)}& \mathbb{Q}{\mathcal{F}}_0^{\left(i\right)}\end{array}\right|$$

$$\mathbb{Q}{\mathcal{F}}_{n+1}^{\left(i\right)}\mathbb{Q}{\mathcal{F}}_{n-1}^{\left(i\right)}-{\left(\mathbb{Q}{\mathcal{F}}_n^{\left(i\right)}\right)}^2={\left(-1\right)}^{n-1}\left(\mathbb{Q}{\mathcal{F}}_2^{\left(i\right)}\mathbb{Q}{\mathcal{F}}_0^{\left(i\right)}-{\left(\mathbb{Q}{\mathcal{F}}_1^{\left(i\right)}\right)}^2\right),$$

and

$$\mathbb{Q}{\mathcal{F}}_{n-1}^{\left(i\right)}\mathbb{Q}{\mathcal{F}}_{n+1}^{\left(i\right)}-{\left(\mathbb{Q}{\mathcal{F}}_n^{\left(i\right)}\right)}^2={\left(-1\right)}^{n-1}\left(\mathbb{Q}{\mathcal{F}}_0^{\left(i\right)}\mathbb{Q}{\mathcal{F}}_2^{\left(i\right)}-{\left(\mathbb{Q}{\mathcal{F}}_1^{\left(i\right)}\right)}^2\right).$$

Likewise, if we take the determinant on both sides of the matrix Equation (19), we obtain Equations (23) and (24), respectively. □

Theorem 9.

Forinteger $m,n\ge 1$, the following equalities hold:

$${\mathcal{F}}_n^{\left(i\right)}\mathbb{Q}{F}_{m+1}+{\mathcal{F}}_{n-1}^{\left(i\right)}\mathbb{Q}{F}_m=\mathbb{Q}{\mathcal{F}}_{m+n}^{\left(i\right)},$$

(25)

$$F_n\mathbb{Q}{\mathcal{F}}_{m+1}^{\left(i\right)}+F_{n-1}\mathbb{Q}{\mathcal{F}}_m^{\left(i\right)}=\mathbb{Q}{\mathcal{F}}_{m+n}^{\left(i\right)},$$

(26)

$$\mathbb{Q}{\mathcal{F}}_{m+1}^{\left(i\right)}\mathbb{Q}{F}_{n+1}+\mathbb{Q}{\mathcal{F}}_m^{\left(i\right)}\mathbb{Q}{F}_n=\mathbb{Q}{F}_2\mathbb{Q}{\mathcal{F}}_{m+n}^{\left(i\right)}+\mathbb{Q}{F}_1\mathbb{Q}{\mathcal{F}}_{m+n-1}^{\left(i\right)},$$

(27)

$$\mathbb{Q}{\mathcal{F}}_{m+1}^{\left(i\right)}\mathbb{Q}{\mathcal{F}}_{n+1}^{\left(i\right)}+\mathbb{Q}{\mathcal{F}}_m^{\left(i\right)}\mathbb{Q}{\mathcal{F}}_n^{\left(i\right)}=\mathbb{Q}{\mathcal{F}}_2^{\left(i\right)}\mathbb{Q}{\mathcal{F}}_{m+n}^{\left(i\right)}+\mathbb{Q}{\mathcal{F}}_1^{\left(i\right)}\mathbb{Q}{\mathcal{F}}_{m+n-1}^{\left(i\right)}.$$

(28)

Proof. 

Substituting $n\to m+n-1$ into (19) and (20), we have

$$\begin{array}{ccc}\hfill \left[\begin{array}{cc}\mathbb{Q}{\mathcal{F}}_{m+n+1}^{\left(i\right)}& \mathbb{Q}{\mathcal{F}}_{m+n}^{\left(i\right)}\\ \mathbb{Q}{\mathcal{F}}_{m+n}^{\left(i\right)}& \mathbb{Q}{\mathcal{F}}_{m+n-1}^{\left(i\right)}\end{array}\right]& =& \left({\mathbf{NU}}^{m+n-2}\right)\mathbf{M}=\left({\mathbf{NU}}^{n-1}\right)\left({\mathbf{U}}^{m-1}\mathbf{M}\right)\hfill \\ \hfill \left[\begin{array}{cc}\mathbb{Q}{\mathcal{F}}_{m+n+1}^{\left(i\right)}& \mathbb{Q}{\mathcal{F}}_{m+n}^{\left(i\right)}\\ \mathbb{Q}{\mathcal{F}}_{m+n}^{\left(i\right)}& \mathbb{Q}{\mathcal{F}}_{m+n-1}^{\left(i\right)}\end{array}\right]& =& \left[\begin{array}{cc}{\mathcal{F}}_{n+1}^{\left(i\right)}& {\mathcal{F}}_n^{\left(i\right)}\\ {\mathcal{F}}_n^{\left(i\right)}& {\mathcal{F}}_{n-1}^{\left(i\right)}\end{array}\right]\left[\begin{array}{cc}\mathbb{Q}{F}_{m+1}& \mathbb{Q}{F}_m\\ \mathbb{Q}{F}_m& \mathbb{Q}{F}_{m-1}\end{array}\right].\hfill \end{array}$$

If we compare the corresponding entries of both matrix equations, we obtain the desired results of Equation (25).

From Equation (20) we see that

$$\begin{array}{ccc}\hfill \left[\begin{array}{cc}\mathbb{Q}{\mathcal{F}}_{m+n+1}^{\left(i\right)}& \mathbb{Q}{\mathcal{F}}_{m+n}^{\left(i\right)}\\ \mathbb{Q}{\mathcal{F}}_{m+n}^{\left(i\right)}& \mathbb{Q}{\mathcal{F}}_{m+n-1}^{\left(i\right)}\end{array}\right]& =& {\mathbf{U}}^{m+n-1}\mathbf{O}={\mathbf{U}}^{n-1}\left({\mathbf{U}}^m\mathbf{O}\right)\hfill \\ \hfill \left[\begin{array}{cc}\mathbb{Q}{\mathcal{F}}_{m+n+1}^{\left(i\right)}& \mathbb{Q}{\mathcal{F}}_{m+n}^{\left(i\right)}\\ \mathbb{Q}{\mathcal{F}}_{m+n}^{\left(i\right)}& \mathbb{Q}{\mathcal{F}}_{m+n-1}^{\left(i\right)}\end{array}\right]& =& \left[\begin{array}{cc}{F}_n& F_{n-1}\\ F_{n-1}& F_{n-2}\end{array}\right]\left[\begin{array}{cc}\mathbb{Q}{\mathcal{F}}_{m+2}^{\left(i\right)}& \mathbb{Q}{\mathcal{F}}_{m+1}^{\left(i\right)}\\ \mathbb{Q}{\mathcal{F}}_{m+1}^{\left(i\right)}& \mathbb{Q}{\mathcal{F}}_m^{\left(i\right)}\end{array}\right].\hfill \end{array}$$

If we compare the corresponding entries of both matrix equations, we obtain the desired result in Equation (26).

Likewise, substituting $n\to m+n-2$ into Equations (19) and (20), we have

$$\begin{array}{ccc}\hfill \mathbf{M}\left({\mathbf{NU}}^{m+n-3}\right)\mathbf{M}& =& \left({\mathbf{MNU}}^{m-2}\right)\left({\mathbf{MU}}^{n-1}\right),\hfill \\ \hfill \mathbf{O}\left({\mathbf{U}}^{m+n-2}\right)\mathbf{O}& =& \left({\mathbf{OU}}^{m-1}\right)\left({\mathbf{OU}}^{n-1}\right).\hfill \end{array}$$

If we equate the corresponding entries on both sides of the matrix equations, we obtain Equations (27) and (28), respectively. □

Corollary 1.

For $0<n\in \mathbb{Z},$ the following equality holds:

$${\left(\mathbb{Q}{\mathcal{F}}_{n+1}^{\left(i\right)}\right)}^2+{\left(\mathbb{Q}{\mathcal{F}}_n^{\left(i\right)}\right)}^2=\mathbb{Q}{\mathcal{F}}_1^{\left(i\right)}\mathbb{Q}{\mathcal{F}}_{2n+1}^{\left(i\right)}+\mathbb{Q}{\mathcal{F}}_0^{\left(i\right)}\mathbb{Q}{\mathcal{F}}_{2n}^{\left(i\right)}.$$

Proof. 

Substituting $m\to n$ into Equation (28) and using Equation (14), we find that

$$\begin{array}{cc}\hfill {\left(\mathbb{Q}{\mathcal{F}}_{n+1}^{\left(i\right)}\right)}^2+{\left(\mathbb{Q}{\mathcal{F}}_n^{\left(i\right)}\right)}^2& =\mathbb{Q}{\mathcal{F}}_2^{\left(i\right)}\mathbb{Q}{\mathcal{F}}_{2n}^{\left(i\right)}+\mathbb{Q}{\mathcal{F}}_1^{\left(i\right)}\mathbb{Q}{\mathcal{F}}_{2n-1}^{\left(i\right)}\hfill \\ \hfill & =\left(\mathbb{Q}{\mathcal{F}}_1^{\left(i\right)}+\mathbb{Q}{\mathcal{F}}_0^{\left(i\right)}\right)\mathbb{Q}{\mathcal{F}}_{2n}^{\left(i\right)}+\mathbb{Q}{\mathcal{F}}_1^{\left(i\right)}\mathbb{Q}{\mathcal{F}}_{2n-1}^{\left(i\right)}\hfill \\ \hfill & =\mathbb{Q}{\mathcal{F}}_1^{\left(i\right)}\left(\mathbb{Q}{\mathcal{F}}_{2n}^{\left(i\right)}+{\mathcal{F}}_{2n-1}^{\left(i\right)}\right)+\mathbb{Q}{\mathcal{F}}_0^{\left(i\right)}\mathbb{Q}{\mathcal{F}}_{2n}^{\left(i\right)}\hfill \\ \hfill & =\mathbb{Q}{\mathcal{F}}_1^{\left(i\right)}\mathbb{Q}{\mathcal{F}}_{2n+1}^{\left(i\right)}+\mathbb{Q}{\mathcal{F}}_0^{\left(i\right)}\mathbb{Q}{\mathcal{F}}_{2n}^{\left(i\right)}.\hfill \end{array}$$

 □

Corollary 2.

For $0<n\in \mathbb{Z},$ the following equality holds:

$${\left(\mathbb{Q}{\mathcal{F}}_{n+1}^{\left(i\right)}\right)}^2-{\left(\mathbb{Q}{\mathcal{F}}_{n-1}^{\left(i\right)}\right)}^2=\mathbb{Q}{\mathcal{F}}_1^{\left(i\right)}\mathbb{Q}{\mathcal{F}}_{2n}^{\left(i\right)}+\mathbb{Q}{\mathcal{F}}_0^{\left(i\right)}\mathbb{Q}{\mathcal{F}}_{2n-1}^{\left(i\right)}.$$

Proof. 

Firstly, we get

$${\left(\mathbb{Q}{\mathcal{F}}_{n+1}^{\left(i\right)}\right)}^2-{\left(\mathbb{Q}{\mathcal{F}}_{n-1}^{\left(i\right)}\right)}^2=\left({\left(\mathbb{Q}{\mathcal{F}}_{n+1}^{\left(i\right)}\right)}^2+{\left(\mathbb{Q}{\mathcal{F}}_n^{\left(i\right)}\right)}^2\right)-\left({\left(\mathbb{Q}{\mathcal{F}}_n^{\left(i\right)}\right)}^2+{\left(\mathbb{Q}{\mathcal{F}}_{n-1}^{\left(i\right)}\right)}^2\right).$$

(29)

After that, we perform the following computations

$$\begin{array}{cc}\hfill {\left(\mathbb{Q}{\mathcal{F}}_{n+1}^{\left(i\right)}\right)}^2+{\left(\mathbb{Q}{\mathcal{F}}_n^{\left(i\right)}\right)}^2& =\left[\begin{array}{cc}\mathbb{Q}{\mathcal{F}}_{n+1}^{\left(i\right)}& \mathbb{Q}{\mathcal{F}}_n^{\left(i\right)}\end{array}\right]\left[\begin{array}{c}\mathbb{Q}{\mathcal{F}}_{n+1}^{\left(i\right)}\\ \mathbb{Q}{\mathcal{F}}_n^{\left(i\right)}\end{array}\right]\hfill \\ \hfill & =\left[\begin{array}{cc}\mathbb{Q}{\mathcal{F}}_1^{\left(i\right)}& \mathbb{Q}{\mathcal{F}}_0^{\left(i\right)}\end{array}\right]{\mathbf{U}}^n{\mathbf{U}}^n\left[\begin{array}{c}\mathbb{Q}{\mathcal{F}}_1^{\left(i\right)}\\ \mathbb{Q}{\mathcal{F}}_0^{\left(i\right)}\end{array}\right]\hfill \\ \hfill & =\left[\begin{array}{cc}\mathbb{Q}{\mathcal{F}}_1^{\left(i\right)}& \mathbb{Q}{\mathcal{F}}_0^{\left(i\right)}\end{array}\right]{\mathbf{U}}^{2n}\left[\begin{array}{c}\mathbb{Q}{\mathcal{F}}_1^{\left(i\right)}\\ \mathbb{Q}{\mathcal{F}}_0^{\left(i\right)}\end{array}\right].\hfill \end{array}$$

(30)

Similarly, we also have

$$\begin{array}{cc}\hfill {\left(\mathbb{Q}{\mathcal{F}}_n^{\left(i\right)}\right)}^2+{\left(\mathbb{Q}{\mathcal{F}}_{n-1}^{\left(i\right)}\right)}^2& =\left[\begin{array}{cc}\mathbb{Q}{\mathcal{F}}_n^{\left(i\right)}& \mathbb{Q}{\mathcal{F}}_{n-1}^{\left(i\right)}\end{array}\right]\left[\begin{array}{c}\mathbb{Q}{\mathcal{F}}_n^{\left(i\right)}\\ \mathbb{Q}{\mathcal{F}}_{n-1}^{\left(i\right)}\end{array}\right]\hfill \\ \hfill & =\left[\begin{array}{cc}\mathbb{Q}{\mathcal{F}}_1^{\left(i\right)}& \mathbb{Q}{\mathcal{F}}_0^{\left(i\right)}\end{array}\right]{\mathbf{U}}^{n-1}{\mathbf{U}}^{n-1}\left[\begin{array}{c}\mathbb{Q}{\mathcal{F}}_1^{\left(i\right)}\\ \mathbb{Q}{\mathcal{F}}_0^{\left(i\right)}\end{array}\right]\hfill \\ \hfill & =\left[\begin{array}{cc}\mathbb{Q}{\mathcal{F}}_1^{\left(i\right)}& \mathbb{Q}{\mathcal{F}}_0^{\left(i\right)}\end{array}\right]{\mathbf{U}}^{2n-2}\left[\begin{array}{c}\mathbb{Q}{\mathcal{F}}_1^{\left(i\right)}\\ \mathbb{Q}{\mathcal{F}}_0^{\left(i\right)}\end{array}\right].\hfill \end{array}$$

(31)

By using Equations (29)–(31), we have

$$\begin{array}{cc}\hfill {\left(\mathbb{Q}{\mathcal{F}}_{n+1}^{\left(i\right)}\right)}^2-{\left(\mathbb{Q}{\mathcal{F}}_{n-1}^{\left(i\right)}\right)}^2& =\left({\left(\mathbb{Q}{\mathcal{F}}_{n+1}^{\left(i\right)}\right)}^2+{\left(\mathbb{Q}{\mathcal{F}}_n^{\left(i\right)}\right)}^2\right)-\left({\left(\mathbb{Q}{\mathcal{F}}_n^{\left(i\right)}\right)}^2+{\left(\mathbb{Q}{\mathcal{F}}_{n-1}^{\left(i\right)}\right)}^2\right)\hfill \\ \hfill & =\left[\begin{array}{cc}\mathbb{Q}{\mathcal{F}}_1^{\left(i\right)}& \mathbb{Q}{\mathcal{F}}_0^{\left(i\right)}\end{array}\right]{\mathbf{U}}^{2n}\left[\begin{array}{c}\mathbb{Q}{\mathcal{F}}_1^{\left(i\right)}\\ \mathbb{Q}{\mathcal{F}}_0^{\left(i\right)}\end{array}\right]-\left[\begin{array}{cc}\mathbb{Q}{\mathcal{F}}_1^{\left(i\right)}& \mathbb{Q}{\mathcal{F}}_0^{\left(i\right)}\end{array}\right]{\mathbf{U}}^{2n-2}\left[\begin{array}{c}\mathbb{Q}{\mathcal{F}}_1^{\left(i\right)}\\ \mathbb{Q}{\mathcal{F}}_0^{\left(i\right)}\end{array}\right]\hfill \\ \hfill & =\left[\begin{array}{cc}\mathbb{Q}{\mathcal{F}}_1^{\left(i\right)}& \mathbb{Q}{\mathcal{F}}_0^{\left(i\right)}\end{array}\right]\left({\mathbf{U}}^{2n}-{\mathbf{U}}^{2n-2}\right)\left[\begin{array}{c}\mathbb{Q}{\mathcal{F}}_1^{\left(i\right)}\\ \mathbb{Q}{\mathcal{F}}_0^{\left(i\right)}\end{array}\right]\hfill \\ \hfill & =\left[\begin{array}{cc}\mathbb{Q}{\mathcal{F}}_1^{\left(i\right)}& \mathbb{Q}{\mathcal{F}}_0^{\left(i\right)}\end{array}\right]{\mathbf{U}}^{2n-2}\left({\mathbf{U}}^2-I\right)\left[\begin{array}{c}\mathbb{Q}{\mathcal{F}}_1^{\left(i\right)}\\ \mathbb{Q}{\mathcal{F}}_0^{\left(i\right)}\end{array}\right].\hfill \end{array}$$

Following the Cayley–Hamilton theorem, we have

$${\mathbf{U}}^2-\mathbf{U}-I={\left[0\right]}_{2\times 2}.$$

We also get

$$\begin{array}{cc}\hfill {\left(\mathbb{Q}{\mathcal{F}}_{n+1}^{\left(i\right)}\right)}^2-{\left(\mathbb{Q}{\mathcal{F}}_{n-1}^{\left(i\right)}\right)}^2& =\left[\begin{array}{cc}\mathbb{Q}{\mathcal{F}}_1^{\left(i\right)}& \mathbb{Q}{\mathcal{F}}_0^{\left(i\right)}\end{array}\right]{\mathbf{U}}^{2n-2}\mathbf{U}\left[\begin{array}{c}\mathbb{Q}{\mathcal{F}}_1^{\left(i\right)}\\ \mathbb{Q}{\mathcal{F}}_0^{\left(i\right)}\end{array}\right]\hfill \\ \hfill & =\left[\begin{array}{cc}\mathbb{Q}{\mathcal{F}}_1^{\left(i\right)}& \mathbb{Q}{\mathcal{F}}_0^{\left(i\right)}\end{array}\right]{\mathbf{U}}^{2n-1}\left[\begin{array}{c}\mathbb{Q}{\mathcal{F}}_1^{\left(i\right)}\\ \mathbb{Q}{\mathcal{F}}_0^{\left(i\right)}\end{array}\right]\hfill \\ \hfill & =\left[\begin{array}{cc}\mathbb{Q}{\mathcal{F}}_1^{\left(i\right)}& \mathbb{Q}{\mathcal{F}}_0^{\left(i\right)}\end{array}\right]\left[\begin{array}{cc}{F}_{2n}& F_{2n-1}\\ F_{2n-1}& F_{2n}\end{array}\right]\left[\begin{array}{c}\mathbb{Q}{\mathcal{F}}_1^{\left(i\right)}\\ \mathbb{Q}{\mathcal{F}}_0^{\left(i\right)}\end{array}\right]\hfill \\ \hfill & =\left[\begin{array}{cc}\mathbb{Q}{\mathcal{F}}_1^{\left(i\right)}& \mathbb{Q}{\mathcal{F}}_0^{\left(i\right)}\end{array}\right]\left[\begin{array}{c}\mathbb{Q}{\mathcal{F}}_{2n}^{\left(i\right)}\\ \mathbb{Q}{\mathcal{F}}_{2n-1}^{\left(i\right)}\end{array}\right]\hfill \\ \hfill & =\mathbb{Q}{\mathcal{F}}_1^{\left(i\right)}\mathbb{Q}{\mathcal{F}}_{2n}^{\left(i\right)}+\mathbb{Q}{\mathcal{F}}_0^{\left(i\right)}\mathbb{Q}{\mathcal{F}}_{2n-1}^{\left(i\right)}.\hfill \end{array}$$

So the proof is completed. □

Theorem 10.

For the Fibonacci finite operator quaternions, we have

$$\mathbb{Q}{\mathcal{F}}_{n+r}^{\left(i\right)}\mathbb{Q}{\mathcal{F}}_{n+s}^{\left(i\right)}-\mathbb{Q}{\mathcal{F}}_n^{\left(i\right)}\mathbb{Q}{\mathcal{F}}_{n+r+s}^{\left(i\right)}={\left(-1\right)}^n{F}_r\left({\left({\mathcal{F}}_1^{\left(i\right)}\right)}^2-{\mathcal{F}}_0^{\left(i\right)}{\mathcal{F}}_2^{\left(i\right)}\right)\left(\mathbb{Q}{F}_1\mathbb{Q}{F}_s-\mathbb{Q}{F}_0\mathbb{Q}{F}_{s+1}\right).$$

Proof. 

Using Equations (25) and (26), we have the following computation:

$$\begin{array}{ccc}\hfill \left[\begin{array}{cc}\mathbb{Q}{\mathcal{F}}_{n+r}^{\left(i\right)}& \mathbb{Q}{\mathcal{F}}_n^{\left(i\right)}\end{array}\right]& =& \left[\begin{array}{cc}\mathbb{Q}{\mathcal{F}}_{n+1}^{\left(i\right)}& \mathbb{Q}{\mathcal{F}}_n^{\left(i\right)}\end{array}\right]\left[\begin{array}{cc}{F}_r& 0\\ F_{r-1}& 1\end{array}\right]\hfill \\ & =& \left[\begin{array}{cc}\mathbb{Q}{F}_1& \mathbb{Q}{F}_0\end{array}\right]\left[\begin{array}{cc}{\mathcal{F}}_{n+1}^{\left(i\right)}& {\mathcal{F}}_n^{\left(i\right)}\\ {\mathcal{F}}_n^{\left(i\right)}& {\mathcal{F}}_{n-1}^{\left(i\right)}\end{array}\right]\left[\begin{array}{cc}{F}_r& 0\\ F_{r-1}& 1\end{array}\right],\hfill \end{array}$$

and

$$\begin{array}{ccc}\hfill \left[\begin{array}{c}\mathbb{Q}{\mathcal{F}}_{n+s}^{\left(i\right)}\\ -\mathbb{Q}{\mathcal{F}}_{n+r+s}^{\left(i\right)}\end{array}\right]& =& \left[\begin{array}{cc}1& 0\\ -F_{r-1}& F_r\end{array}\right]\left[\begin{array}{c}\mathbb{Q}{\mathcal{F}}_{n+s}^{\left(i\right)}\\ -\mathbb{Q}{\mathcal{F}}_{n+s+1}^{\left(i\right)}\end{array}\right]\hfill \\ & =& \left[\begin{array}{cc}1& 0\\ -F_{r-1}& F_r\end{array}\right]\left[\begin{array}{cc}{\mathcal{F}}_{n-1}^{\left(i\right)}& -{\mathcal{F}}_n^{\left(i\right)}\\ -{\mathcal{F}}_n^{\left(i\right)}& {\mathcal{F}}_{n+1}^{\left(i\right)}\end{array}\right]\left[\begin{array}{c}\mathbb{Q}{F}_s\\ -\mathbb{Q}{F}_{s+1}\end{array}\right].\hfill \end{array}$$

We also have the following computation:

$$\begin{array}{ccc}& & \left[\begin{array}{cc}{\mathcal{F}}_{n+1}^{\left(i\right)}& {\mathcal{F}}_n^{\left(i\right)}\\ {\mathcal{F}}_n^{\left(i\right)}& {\mathcal{F}}_{n-1}^{\left(i\right)}\end{array}\right]\left[\begin{array}{cc}{F}_r& 0\\ F_{r-1}& 1\end{array}\right]\left[\begin{array}{cc}1& 0\\ -F_{r-1}& F_r\end{array}\right]\left[\begin{array}{cc}{\mathcal{F}}_{n-1}^{\left(i\right)}& -{\mathcal{F}}_n^{\left(i\right)}\\ -{\mathcal{F}}_n^{\left(i\right)}& {\mathcal{F}}_{n+1}^{\left(i\right)}\end{array}\right]\hfill \\ & =& \left[\begin{array}{cc}{\mathcal{F}}_{n+1}^{\left(i\right)}& {\mathcal{F}}_n^{\left(i\right)}\\ {\mathcal{F}}_n^{\left(i\right)}& {\mathcal{F}}_{n-1}^{\left(i\right)}\end{array}\right]\left(F_{r}I\right)\left[\begin{array}{cc}{\mathcal{F}}_{n-1}^{\left(i\right)}& -{\mathcal{F}}_n^{\left(i\right)}\\ -{\mathcal{F}}_n^{\left(i\right)}& {\mathcal{F}}_{n+1}^{\left(i\right)}\end{array}\right]\hfill \\ & =& F_r\left[\begin{array}{cc}{\mathcal{F}}_{n+1}^{\left(i\right)}& {\mathcal{F}}_n^{\left(i\right)}\\ {\mathcal{F}}_n^{\left(i\right)}& {\mathcal{F}}_{n-1}^{\left(i\right)}\end{array}\right]\left[\begin{array}{cc}{\mathcal{F}}_{n-1}^{\left(i\right)}& -{\mathcal{F}}_n^{\left(i\right)}\\ -{\mathcal{F}}_n^{\left(i\right)}& {\mathcal{F}}_{n+1}^{\left(i\right)}\end{array}\right]\hfill \\ & =& F_r\left[\begin{array}{cc}{\mathcal{F}}_{n+1}^{\left(i\right)}{\mathcal{F}}_{n-1}^{\left(i\right)}-{\left({\mathcal{F}}_n^{\left(i\right)}\right)}^2& 0\\ 0& -{\left({\mathcal{F}}_n^{\left(i\right)}\right)}^2+{\mathcal{F}}_{n-1}^{\left(i\right)}{\mathcal{F}}_{n+1}^{\left(i\right)}\end{array}\right]\hfill \\ & =& -F_r\left({\left({\mathcal{F}}_n^{\left(i\right)}\right)}^2-{\mathcal{F}}_{n-1}^{\left(i\right)}{\mathcal{F}}_{n+1}^{\left(i\right)}\right)I\hfill \\ & =& {\left(-1\right)}^n{F}_r\left({\left({\mathcal{F}}_1^{\left(i\right)}\right)}^2-{\mathcal{F}}_0^{\left(i\right)}{\mathcal{F}}_2^{\left(i\right)}\right)I.\hfill \end{array}$$

Then, we get

$$\begin{array}{ccc}\hfill \mathbb{Q}{\mathcal{F}}_{n+r}^{\left(i\right)}\mathbb{Q}{\mathcal{F}}_{n+s}^{\left(i\right)}-\mathbb{Q}{\mathcal{F}}_n^{\left(i\right)}\mathbb{Q}{\mathcal{F}}_{n+r+s}^{\left(i\right)}& =& \left[\begin{array}{cc}\mathbb{Q}{\mathcal{F}}_{n+r}^{\left(i\right)}& \mathbb{Q}{\mathcal{F}}_n^{\left(i\right)}\end{array}\right]\left[\begin{array}{c}\mathbb{Q}{\mathcal{F}}_{n+s}^{\left(i\right)}\\ -\mathbb{Q}{\mathcal{F}}_{n+r+s}^{\left(i\right)}\end{array}\right]\hfill \\ & =& {\left(-1\right)}^n{F}_r\left({\left({\mathcal{F}}_1^{\left(i\right)}\right)}^2-{\mathcal{F}}_0^{\left(i\right)}{\mathcal{F}}_2^{\left(i\right)}\right)\left[\begin{array}{cc}\mathbb{Q}{F}_1& \mathbb{Q}{F}_0\end{array}\right]\left[\begin{array}{c}\mathbb{Q}{F}_s\\ -\mathbb{Q}{F}_{s+1}\end{array}\right]\hfill \\ & =& {\left(-1\right)}^n{F}_r\left({\left({\mathcal{F}}_1^{\left(i\right)}\right)}^2-{\mathcal{F}}_0^{\left(i\right)}{\mathcal{F}}_2^{\left(i\right)}\right)\left(\mathbb{Q}{F}_1\mathbb{Q}{F}_s-\mathbb{Q}{F}_0\mathbb{Q}{F}_{s+1}\right).\hfill \end{array}$$

 □

Corollary 3.

In the above theorem, substituting $n\to n-1$ and $r=s=1$, we obtain Equations (23) and (24).

4. Conclusions

In this paper, we introduced the Fibonacci finite operator quaternions by means of the finite operators and Fibonacci numbers. We also gave some properties of this new type of quaternions, such as recurrence relation, Binet-like formula, generating function, exponential generating function, and some sum formulas, including the Fibonacci finite operator quaternions. We obtained many identities related to Fibonacci finite operator quaternions by using the matrix representations. Indeed, for the interested readers of this work, the results presented here have the potential to motivate further research on the subject of the Fibonacci finite operator hyper complex numbers (Kızılateş and Kone [20]) or Fibonacci finite operator hybrid numbers (Szynal and Wloch [38]).