The Hadamard Products for bi-periodic Fibonacci and bi-periodic Lucas Generating matrices

TL;DR

This paper introduces matrix representations for bi-periodic Fibonacci and Lucas polynomials, explores their properties, and examines the Hadamard products of their generating matrices, revealing new structural insights.

Contribution

It defines new matrix forms for bi-periodic Fibonacci polynomials and investigates the properties of Hadamard products of their generating matrices, a novel approach in this area.

Findings

Derived nth power and determinant formulas for the matrices.

Established properties of Hadamard products of bi-periodic Fibonacci and Lucas matrices.

Provided new insights into the structure of bi-periodic Fibonacci and Lucas sequences.

Abstract

In this paper, firstly, we define the Qq-generating matrix for bi-periodic Fibonacci polynomial. And we give nth power, determinant and some properties of the bi-periodic Fibonacci polynomial by considering this matrix representation. Also, we introduce the Hadamard products for bi-periodic Fibonacci Qnq generating matrix and bi-periodic Lucas Qnl generating matrix of which entries is bi-periodic Fibonacci and Lucas numbers. Then, we investigate some properties of these products.

Equations

$$F_{n+1}=F_n+F_{n-1},L_{n+1}=L_n+L_{n-1}$$

$$q_{n}(x)=\left\{\begin{array}[]{c}axq_{n-1}(x)+q_{n-2}(x),\ \ \text{if }n\text{ is odd}\\ bxq_{n-1}(x)+q_{n-2}(x),\ \ \text{if }n\text{ is even}\end{array}\right.$$

$$q_{n}=\left\{\begin{array}[]{c}aq_{n-1}+q_{n-2},\ \ \text{if }n\text{ is even}\\ bq_{n-1}+q_{n-2},\ \ \text{if }n\text{ is odd}\end{array}\right.\ ,$$

$$l_{n}=\left\{\begin{array}[]{c}bl_{n-1}+l_{n-2},\ \ \text{if }n\text{ is even}\\ al_{n-1}+l_{n-2},\ \ \text{if }n\text{ is odd}\end{array}\right.\ ,$$

$$(ab+4)q_n$$

$$l_n$$

$$a^{1-\epsilon (n)}{b}^{\epsilon (n)}{q}_{n+1}{q}_{n-1}-a^{\epsilon (n)}{b}^{1-\epsilon (n)}{q}_n^2$$

$${\left(\frac{b}{a}\right)}^{1-\epsilon (n)}{l}_{n+1}{l}_{n-1}-{\left(\frac{b}{a}\right)}^{\epsilon (n)}{l}_n^2$$

$$Q=\left[\begin{array}[]{cc}1&1\\ 1&0\end{array}\right].$$

$$Q_{l}=\left[\begin{array}[]{cc}a^{2}+2\frac{a}{b}&\frac{a^{2}}{b}\\ a&2\frac{a}{b}\end{array}\right].$$

$$Q_{l}^{n}=\left\{\begin{array}[]{c}\left(\frac{a}{b}\right)^{n}(ab+4)^{\frac{n}{2}}\left[\begin{array}[]{cc}q_{n+1}&q_{n}\\ \frac{b}{a}q_{n}&q_{n-1}\end{array}\right],n\text{ }even\\ \left(\frac{a}{b}\right)^{n}(ab+4)^{\frac{n-1}{2}}\left[\begin{array}[]{cc}l_{n+1}&l_{n}\\ \frac{b}{a}l_{n}&l_{n-1}\end{array}\right],n\text{ }odd\end{array}\right.,\text{ }\det\left(\frac{a^{2}}{b^{2}}\left(ab+4\right)\right)^{n}.$$

$$Q_{q}=\left[\begin{array}[]{cc}bx&\frac{b}{a}\\ 1&0\end{array}\right].$$

$$Q_{q}^{n}=\left(\frac{b}{a}\right)^{\left\lfloor\frac{n}{2}\right\rfloor}\left[\begin{array}[]{cc}\left(\frac{b}{a}\right)^{\varepsilon(n)}q_{n+1}(x)&\frac{b}{a}q_{n}(x)\\ q_{n}(x)&\left(\frac{b}{a}\right)^{\varepsilon(n)}q_{n-1}(x)\end{array}\right]$$

$$Q_{q}=\left[\begin{array}[]{cc}bx&\frac{b}{a}\\ 1&0\end{array}\right]=\left(\frac{b}{a}\right)^{\left\lfloor\frac{1}{2}\right\rfloor}\left[\begin{array}[]{cc}\left(\frac{b}{a}\right)^{\varepsilon(1)}q_{2}(x)&\frac{b}{a}q_{1}(x)\\ q_{1}(x)&\left(\frac{b}{a}\right)^{\varepsilon(1)}q_{0}(x)\end{array}\right].$$

$$Q_{q}^{2}=\left(\begin{array}[]{cc}b^{2}x^{2}+\frac{b}{a}&\frac{b^{2}}{a}x\\ bx&\frac{b}{a}\end{array}\right)=\left(\frac{b}{a}\right)^{\left\lfloor\frac{2}{2}\right\rfloor}\left[\begin{array}[]{cc}\left(\frac{b}{a}\right)^{\varepsilon(2)}q_{3}(x)&\frac{b}{a}q_{2}(x)\\ q_{2}(x)&\left(\frac{b}{a}\right)^{\varepsilon(2)}q_{1}(x)\end{array}\right].$$

$$Q_{q}^{k}=\left(\frac{b}{a}\right)^{\left\lfloor\frac{k}{2}\right\rfloor}\left[\begin{array}[]{cc}\left(\frac{b}{a}\right)^{\varepsilon(k)}q_{k+1}(x)&\frac{b}{a}q_{k}(x)\\ q_{k}(x)&\left(\frac{b}{a}\right)^{\varepsilon(k)}q_{k-1}(x)\end{array}\right].$$

$$Q_{q}^{k+2}=Q_{q}^{2}Q_{q}^{k}=\left(\frac{b}{a}\right)^{\frac{k+2}{2}}\left[\begin{array}[]{cc}q_{k+3}(x)&\frac{b}{a}q_{k+2}(x)\\ q_{k+2}(x)&q_{k+1}(x)\end{array}\right].$$

$$Q_{q}^{k+2}=Q_{q}^{2}Q_{q}^{k}=\left(\frac{b}{a}\right)^{\frac{k+1}{2}}\left[\begin{array}[]{cc}\frac{b}{a}q_{k+3}(x)&\frac{b}{a}q_{k+2}(x)\\ q_{k+2}(x)&\frac{b}{a}q_{k+1}(x)\end{array}\right].$$

$$Q_{q}^{k+2}=\left(\frac{b}{a}\right)^{\left\lfloor\frac{k+2}{2}\right\rfloor}\left[\begin{array}[]{cc}\left(\frac{b}{a}\right)^{\varepsilon(k+2)}q_{k+3}(x)&\frac{b}{a}q_{k+2}(x)\\ q_{k+2}(x)&\left(\frac{b}{a}\right)^{\varepsilon(k+2)}q_{k+1}(x)\end{array}\right].$$

$$\det \left(Q_q^n\right)={\left(-\frac{b}{a}\right)}^n.$$

$$q_n(x)=\frac{a^{1-\epsilon (n)}}{{\left(ab\right)}^{\lfloor \frac{n}{2}\rfloor }{x}^{n-1}}\frac{{\alpha }^n(x)-{\beta }^n(x)}{\alpha (x)-\beta (x)}$$

$${\lambda }_1=\frac{\alpha (x)}{ax},{\lambda }_2=\frac{\beta (x)}{ax},u_1=\left[\frac{b}{a},-\frac{\beta (x)}{ax}\right],u_2=\left[\frac{b}{a},-\frac{\alpha (x)}{ax}\right].$$

$$V=U^{-1}{Q}_{q}U$$

$$U=\left[\begin{array}[]{cc}u_{1}^{T}&u_{2}^{T}\end{array}\right]=\left[\begin{array}[]{cc}\frac{b}{a}&\frac{b}{a}\\ -\frac{\beta(x)}{ax}&-\frac{\alpha(x)}{ax}\end{array}\right],\text{ }V=diag\left[\lambda_{1},\lambda_{2}\right]=\left[\begin{array}[]{cc}\frac{\alpha}{ax}&0\\ 0&\frac{\beta}{ax}\end{array}\right].$$

$$Q_q^n=U{V}^n{U}^{-1}.$$

$$Q_q^n$$

$$\frac{1}{\alpha (x)-\beta (x)}\frac{{\alpha }^n(x)-{\beta }^n(x)}{a^{n-1}{x}^{n-1}}={\left(\frac{b}{a}\right)}^{\lfloor \frac{n}{2}\rfloor }{q}_n(x).$$

$$Q_{q}^{m+n}=\left(\frac{b}{a}\right)^{\left\lfloor\frac{m+n}{2}\right\rfloor}\left[\begin{array}[]{cc}\left(\frac{b}{a}\right)^{\varepsilon(m+n)}q_{m+n+1}(x)&\frac{b}{a}q_{m+n}(x)\\ q_{m+n}(x)&\left(\frac{b}{a}\right)^{\varepsilon(n)}q_{m+n-1}(x)\end{array}\right].$$

$$Q_{q}^{m}Q_{q}^{n}=\left(\frac{b}{a}\right)^{\frac{m+n}{2}-1}\left[\begin{array}[]{cc}\left(\frac{b}{a}\right)^{2}q_{m+1}(x)q_{n+1}(x)+\frac{b}{a}q_{m}(x)q_{n}(x)&\left(\frac{b}{a}\right)^{2}\left\{q_{m+1}(x)q_{n}(x)+q_{m}(x)q_{n-1}(x)\right\}\\ \frac{b}{a}\left\{q_{m}(x)q_{n+1}(x)+q_{m-1}(x)q_{n}(x)\right\}&\frac{b}{a}q_{m}(x)q_{n}(x)+\left(\frac{b}{a}\right)^{2}q_{m-1}(x)q_{n-1}(x)\end{array}\right].$$

$$q_{m+n}(x)=q_{m+1}(x)q_n(x)+q_m(x)q_{n-1}(x).$$

$$q_{m+n}(x)=q_m(x)q_{n+1}(x)+q_{m-1}(x)q_n(x).$$

Full text

Keywords and Phrases: bi-periodic Fibonacci numbers, bi-periodic Fibonacci polynomial, bi-periodic Lucas numbers, Fibonacci $Q$ matrix, generating matrix, matrix method.

2010 Mathematics Subject Classification: 11B25; 11B37; 11B39; 15A24.

1 Introduction and Preliminaries

The special sequences and their properties have been investigated in many articles and books (see, for example [1, 3, 5, 6, 8, 9], [14]-[17] and the references cited therein). The Fibonacci and Lucas numbers have attracted the attention of mathematicians because of their intrinsic theory and applications. Fibonacci $\left\{F_{n}\right\}_{n\in\mathbb{N}}$ and Lucas $\left\{L_{n}\right\}_{n\in\mathbb{N}}$ sequences was defined

[TABLE]

with initial conditions $F_{1}=\ F_{2}=1,$ $L_{1}=2,$ $L_{2}=1.$

Many authors have generalized Fibonacci sequence in different ways. In the one of those generalizations, in [17], we define the bi-periodic Fibonacci $\left\{q_{n}(x)\right\}_{n\in\mathbb{N}}$ polynomial as in the form

[TABLE]

where $q_{0}(x)=0,\ q_{1}(x)=1$ and $a,b$ are nonzero real numbers.

Also, the Binet formula and Cassini identity of this polynomial was given.

In [5], for $a$ and $b$ are nonzero real numbers, the authors defined the bi-periodic Fibonacci $\left\{q_{n}\right\}_{n\in\mathbb{N}}$ sequence

[TABLE]

where $q_{0}=0,\ q_{1}=1.$ In [1], for $a$ and $b$ are nonzero real numbers, it is defined the bi-periodic Lucas $\left\{l_{n}\right\}_{n\in\mathbb{N}}$ sequence as

[TABLE]

where $l_{0}=2,\ l_{1}=a.$ The author also gave in the following relations

[TABLE]

[TABLE]

where $\varepsilon(n)=n-2\left\lfloor\frac{n}{2}\right\rfloor.$

On the other hand, it have been studied the matrix representation of special sequences ([2, 4, 7, 11, 12]). In [12], Sylvester gave Fibonacci $Q$-matrix as

[TABLE]

Then, he says that some properties of Fibonacci numbers can be founded by using this matrix. Considering this matrix, in [8], the author obtained some equalities for $Q^{n}.$

And, using this properties in [10], the authors defined the Hadamard product $Q^{n}\circ Q^{-n}$. Similarly, in [13], the author gave some properties for the Hadamard products of its Adjoint matrix with a square matrix.

In [4], we defined the $Q_{l}$ matrix as

[TABLE]

And, we gave

[TABLE]

This study consists of three sections. In Section 2, we define the $Q_{q}$-generating matrix as the first time in the literature. This matrix is generalization form of well-known $Q$-Fibonacci matrix. By using this generalized matrix, we find some equalities for bi-periodic Fibonacci polynomial. In the third part of our work, we define the Hadamard products $Q_{q}^{n}\circ Q_{q}^{-n}$**  and $Q_{l}^{n}\circ Q_{l}^{-n}$** . And, we get some properties of these generalized Hadamard products.

2 The properties of $Q_{q}$-generating matrix

Definition 2.1

Bi-periodic Fibonacci $Q_{q}$-generating matrix is defined by

[TABLE]

Theorem 2.2

Let $Q_{q}$-generating matrix be as in equation (2.1).Then, we have

[TABLE]

where $\varepsilon(n)=n-\left\lfloor\frac{n}{2}\right\rfloor$ and $q_{n}(x)$ is $n$th bi-periodic Fibonacci polynomial.

Proof. We use mathematical induction on $n$, we can write

[TABLE]

[TABLE]

which show that the equation (2.2) is true for $n=1$ and $n=2$. Now, we suppose that equation (2.2) is true for $n=k$, that is

[TABLE]

If we supposed that k is even, by using properties of the bi-periodic Fibonacci polynomial, we get

[TABLE]

Similarly, for k is odd, we can write

[TABLE]

By combining this equalities, we obtain

[TABLE]

$\square$

Corollary 2.3

Let $Q_{q}^{n}$ be as in equation (2.1). Then the following equality is true for all positive integers

[TABLE]

Proof. By using the Cassini identity for bi-periodic Fibonacci polnomial, the desired result is obtained. $\square$

The Binet formula for bi-periodic Fibonacci polynomial, given in [17], can also be obtained by using $Q_{q}$ matrix.

Theorem 2.4

Let $n$ be any integer. The Binet formula of bi-periodic Fibonacci polynomial is

[TABLE]

where $\alpha(x)$ and $\beta(x)$ are roots of $r^{2}-abx^{2}r-abx^{2}=0.$

Proof. Let the $Q_{q}$ matrix be as in equation (2.1). Characteristic equation of $Q_{q}$-generating matrix is $\lambda^{2}-bx\lambda-\frac{b}{a}=0$. Then, eigenvalues and eigenvectors of the $Q_{q}$ are

[TABLE]

The generating matrix can be diagonalized by using

[TABLE]

which

[TABLE]

From properties of similar matrices, for $n$ is any integer, we obtain

[TABLE]

Thus we get

[TABLE]

Taking into account the Theorem (2.2), we have

[TABLE]

Making the necessary arrangements, the desired result is obtained. $\square$

In the following, we give some properties of bi-periodic Fibonacci polynomial.

Theorem 2.5

For every $m,n\in\mathbb{N},$ the following statements are valid

1. $i)$

q_{m+n}(x)=\left\{\begin{array}[]{c}\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ }q_{m+1}(x)q_{n}(x)+q_{m}(x)q_{n-1}(x),\text{ \ \ \ \ \ \ \ \ \ }m+n\text{ even}\\ \left(\frac{b}{a}\right)^{\varepsilon(m)}q_{m+1}(x)q_{n}(x)+\left(\frac{b}{a}\right)^{\varepsilon(n)}q_{m}(x)q_{n-1}(x),\text{ \ \ }m+n\text{ odd}\end{array}\right., 2. $ii)$

q_{m+n}(x)=\left\{\begin{array}[]{c}\text{ \ \ \ \ \ \ \ \ \ \ \ \ }q_{m}(x)q_{n+1}(x)+q_{m-1}(x)q_{n}(x),\text{ \ \ \ \ \ \ \ \ \ \ }m+n\text{ even}\\ \left(\frac{b}{a}\right)^{\varepsilon(n)}q_{m}(x)q_{n+1}(x)+\left(\frac{b}{a}\right)^{\varepsilon(m)}q_{m-1}(x)q_{n}(x),\text{ \ \ }m+n\text{ odd}\end{array}\right., 3. $iii)$

q_{m-n}(x)=\left\{\begin{array}[]{c}\text{ \ \ \ \ \ }\left(-1\right)^{n+1}\left\{q_{m+1}(x)q_{n}(x)-q_{m}(x)q_{n+1}(x)\right\},\text{ \ \ \ \ \ \ \ }m+n\text{ even}\\ \left(-\frac{b}{a}\right)^{\varepsilon(m)}q_{m+1}(x)q_{n}(x)+\left(-\frac{b}{a}\right)^{\varepsilon(n)}q_{m}(x)q_{n+1}(x),\text{ \ \ }m+n\text{ odd}\end{array}\right., 4. $iv)$

q_{m-n}(x)=\left\{\begin{array}[]{c}\text{ \ \ \ \ \ }\left(-1\right)^{n+1}\left\{q_{m-1}(x)q_{n}(x)-q_{m}(x)q_{n-1}(x)\right\},\text{ \ \ \ \ \ \ \ }m+n\text{ even}\\ \left(-\frac{b}{a}\right)^{\varepsilon(n)}q_{m}(x)q_{n-1}(x)+\left(-\frac{b}{a}\right)^{\varepsilon(m)}q_{m-1}(x)q_{n}(x),\text{ \ \ }m+n\text{ odd}\end{array}\right..

Proof. By using equation (2.2), $Q_{q}^{m+n}$ can be written as

[TABLE]

For case of odd $m$ and $n$, we can write

[TABLE]

If we compare the $1st$ row and $2nd$ column entries of the matrices (2.6) and (2.7), we get

[TABLE]

On the other hand, comparing the $2nd$ row and $1st$ column, we obtain

[TABLE]

Similarly, for the case of even $m$ and $n$, we have

[TABLE]

And, for the case of odd $m$ and even $n$ (or case of even $m$ and odd $n$), we have

[TABLE]

Thus, the proof of $i)$ and $ii)$ is obtained.

Now, we give the proof of $iii)$ and $iv).$ By calculating inverse of the matrix $Q_{q}^{n}$ in (2.2), we conclude

[TABLE]

Benefitting from the equality $Q_{q}^{m-n}=Q_{q}^{m}Q_{q}^{-n}$ and by comparing the entries, the desired result can be obtained. That is, for the case of odd $m$ and $n$, we get

[TABLE]

Similarly, for the case of even $m$ and $n$, we obtain

[TABLE]

And, for the case of odd $m$ and even $n$ (or case of even $m$ and odd $n$), we have

[TABLE]

Thus, we have the desired expressions. $\square$

3 On the **Hadamard Products $Q_{q}^{n}\circ Q_{q}^{-n}$  and $Q_{l}^{n}\circ Q_{l}^{-n}$ for Bi-periodic

Fibonacci and Bi-periodic Lucas Generating Matrices**

We will accept $x=1$ in $Q_{q}^{n}$ matrix throughout this section. Then, we can give the following Theorem for the generating matrices of bi-periodic Fibonacci and Lucas numbers.

Theorem 3.1

For any integer $n,$ we have

• $i)$

$Q_{q}^{n}=bQ_{q}^{n-1}+\frac{b}{a}Q_{q}^{n-2},$

• $ii)$

$Q_{l}^{n}=\frac{b}{a(ab+4)}Q_{l}^{n+1}+\frac{a}{b}Q_{l}^{n-1}.$

Proof. We will omit the proof of $(ii)$ because it is similar to the proof of $(i).$

By considering the Equation (2.2), for even $n$, we can write

[TABLE]

Similarly, for odd $n$, we have

[TABLE]

Thus, the desired equality is obtained. $\square$

Now, we define the Hadamard products for bi-periodic Fibonacci and Lucas generating matrices by considering the determinants. Thus, we can write

[TABLE]

and

[TABLE]

In the following theorem, we give the determinants of the Hadamard products.

Theorem 3.2

For any integer $n,$ we have

[TABLE]

Proof. We are doing the proof for $\det(Q_{q}^{n}\circ Q_{q}^{-n})$ because the proof of $\det(Q_{l}^{n}\circ Q_{l}^{-n})$ can be done in similar way. By using the Equation (3.5), for even $n$, we can write

[TABLE]

And, we have

[TABLE]

Similarly, for odd $n$, we get

[TABLE]

[TABLE]

Thus, we get the desired result. $\square$

Using the above definition and theorem, we obtain traces, eigenvalues, eigenvectors and inverses for Hadamard products as in the following.

Corollary 3.3

We have

$i)$

trace(Q_{q}^{n}\circ Q_{q}^{-n})=\left\{\begin{array}[]{c}2\left(1+\frac{b}{a}q_{n}^{2}\right),\text{ }n\text{ }even\\ 2\left(1-q_{n}^{2}\right),n\text{ }odd\end{array}\right.,

trace(Q_{l}^{n}\circ Q_{l}^{-n})=\left\{\begin{array}[]{c}2\left(1+\frac{b}{a}q_{n}^{2}\right),\text{ }n\text{ }even\\ 2\left(1+\frac{b}{a\left(ab+4\right)}l_{n}^{2}\right),n\text{ }odd\end{array}\right..

$ii)$

The eigenvalues of the matrix $Q_{q}^{n}\circ Q_{q}^{-n}$ are given as

[TABLE]

and the eigenvalues of the matrix $Q_{l}^{n}\circ Q_{l}^{-n}$ are given as

[TABLE]

$iii)$

The eigenvectors corresponding to the eigenvalues of the matrices $Q_{q}^{n}\circ Q_{q}^{-n}$ and $Q_{l}^{n}\circ Q_{l}^{-n}$ are

[TABLE]

respectively.

$iv)$

The matrices $Q_{q}^{n}\circ Q_{q}^{-n}$ and $Q_{l}^{n}\circ Q_{l}^{-n}$ are invertible and

[TABLE]

4 Conclusion

In Section 2, we present some properties of bi-periodic Fibonacci polynomial by using the $Q_{q}$ matrix. Then, we express that well-known matrix representations are special cases of this generalized matrices.

If we choose $x=1,$ then we get generating matrix for bi-periodic Fibonacci sequence and properties of this sequence.

Thus, if we choose the different values of $a$ and $b$, then we obtain generating matrices for well-known matrix sequences in the literature:

• •

If we replace $a=b=1$ in $Q_{q}$, we obtain the generating matrices for Fibonacci sequence.

• •

If we replace $a=b=2$ in $Q_{q}$, we obtain the generating matrices for Pell sequence.

• •

If we replace $a=b=k$ in $Q_{q}$, we obtain the generating matrices for $k$-Fibonacci sequence.

And, in Section 3, we define the Hadamard products for the generating matrices of bi-periodic Fibonacci and Lucas sequence and give some properties of these new matrices. By taking into account these generalized matrices, it also can be obtained properties of some special matrices. Namely, for different values of $a$ and $b$, we can rewrite some properties for the well-known matrices in the literature such as Fibonacci $Q$.