# On the bounds for the spectral norms of geometric circulant matrices

## Abstract

In this paper, we define a geometric circulant matrix whose entries are the generalized Fibonacci numbers and hyperharmonic Fibonacci numbers. Then we give upper and lower bounds for the spectral norms of these matrices.

## Introduction

The circulant and r-circulant matrices have important applications in numerical analysis, probability, coding theory, and so on. An $$n\times n$$ matrix $$C_{r}$$ is called an r-circulant matrix if it is defined as follows:

$$C_{r}= \begin{pmatrix} c_{0} & c_{1} & c_{2} & \ldots & c_{n-2} & c_{n-1} \\ rc_{n-1} & c_{0} & c_{1} & \ldots & c_{n-3} & c_{n-2} \\ rc_{n-2} & rc_{n-1} & c_{0} & \ldots & c_{n-4} & c_{n-3} \\ \vdots & \vdots & \vdots & & \vdots & \vdots \\ rc_{1} & rc_{2} & rc_{3} & \ldots & rc_{n-1} & c_{0} \end{pmatrix} .$$

The matrix $$C_{r}$$ is determined by its first row elements and r, we denote $$C_{r}=\operatorname{Circ}_{r}(c_{0},c_{1},c_{2}, \ldots ,c_{n-1})$$. In particular for $$r=1$$

$$C= \begin{pmatrix} c_{0} & c_{1} & c_{2} & \ldots & c_{n-2} & c_{n-1} \\ c_{n-1} & c_{0} & c_{1} & \ldots & c_{n-3} & c_{n-2} \\ c_{n-2} & c_{n-1} & c_{0} & \ldots & c_{n-4} & c_{n-3} \\ \vdots & \vdots & \vdots & & \vdots & \vdots \\ c_{1} & c_{2} & c_{3} & \ldots & c_{n-1} & c_{0} \end{pmatrix}$$

is called a circulant matrix.

Circulant and r-circulant matrices with the special numbers have been studied by many researchers in last decade. For example, in , Solak has studied the spectral norms of circulant matrices with the Fibonacci and Lucas numbers. In , Kocer et al. obtained norms of circulant and semicirculant matrices with Horadam numbers. In , Shen and Cen have given upper and lower bounds for the spectral norms of r-circulant matrices with the Fibonacci and Lucas numbers. In , Bahsi computed the spectral norms of circulant and r-circulant matrices with the hyperharmonic numbers. Moreover, in , Bahsi and Solak studied norms of circulant and r-circulant matrices with the hyper-Fibonacci and hyper-Lucas numbers. In , Jiang and Zhou studied spectral norms of even order r-circulant matrices. In [7, 8], Tuglu and Kızılateş have calculated Euclidean norm by using the finite difference operator and given spectral norms of circulant, r-circulant and some special matrices with the harmonic Fibonacci and hyperharmonic Fibonacci numbers. In , Yazlık and Taskara have presented new upper and lower bounds for the spectral norms of an r-circulant matrix with the generalized k-Horadam numbers. In , He et al. gave the upper bound estimation of the spectral norm for r-circulant matrices with Fibonacci and Lucas numbers.

In view of the above papers, we define a new circulant matrix which is called geometric circulant matrix and give upper and lower bounds for the spectral norms of this matrix with the generalized Fibonacci and hyperharmonic Fibonacci numbers by using the same method given in .

## Preliminaries

The well-known Fibonacci and Lucas sequences are defined by the following recurrence relations: for $$n\geq 0$$,

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

and

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

where $$F_{0}=0$$, $$F_{1}=1$$, $$L_{0}=2$$ and $$L_{1}=1$$, respectively. The generalized Fibonacci and Lucas sequences, $$\{ U_{n} \}$$ and $$\{ V_{n} \}$$, are defined by the following recurrence relations: for $$n\geq 0$$, and any non-zero integer p,

$$U_{n+2}=pU_{n+1}+U_{n}$$

and

$$V_{n+2}=pV_{n+1}+V_{n},$$

where $$U_{0}=0$$, $$U_{1}=1$$, $$V_{0}=2$$ and $$V_{1}=p$$. If we take $$p=1$$, then $$U_{n}=F_{n}$$ and $$V_{n}=L_{n}$$. Let α and β be the roots of the characteristic equation $$x^{2}-px-1=0$$. Then the Binet formulas for the sequences $$\{ U_{n} \}$$ and $$\{ V_{n} \}$$ are given by

$$U_{n}=\frac{\alpha ^{n}-\beta ^{n}}{\alpha -\beta }$$

and

$$V_{n}=\alpha ^{n}+\beta ^{n},$$

where $$\alpha =\frac{p+\sqrt{p^{2}+4}}{2}$$ and $$\beta =\frac{p-\sqrt{p^{2}+4}}{2}$$.

On the other hand, Yazlık and Taskara examined the generalized k-Horadam numbers via the following recurrence relations:

$$H_{k,n+2}=f(k)H_{k,n+1}+g(k)H_{k,n}$$
(1)

with the initial values $$H_{k,0}=a$$, $$H_{k,1}=b$$. Moreover they calculated sum of squares of k-Horadam numbers (see ). If we take $$f(k)=p$$, $$g(k)=1$$, $$a=0$$ and $$b=1$$ in (1), we get

$$\sum_{i=0}^{n-1}U_{i}^{2}= \frac{U_{n}^{2}-U_{n-1}^{2}+(-1)^{n}}{p^{2}}$$
(2)

and if we take $$f(k)=p$$, $$g(k)=1$$, $$a=2$$ and $$b=p$$ in (1), we have

$$\sum_{i=0}^{n-1}V_{i}^{2}= \frac{V_{n}^{2}-V_{n-1}^{2}+p^{2}-4+(1-(-1)^{n})(p^{2}+4)}{p^{2}} .$$
(3)

In , Tuglu et al. defined hyperharmonic Fibonacci numbers for $$n,r\geq 1$$,

$$\mathbb{F}_{n}^{(r)}=\sum_{k=1}^{n} \mathbb{F}_{k}^{(r-1)},$$

where $$\mathbb{F}_{n}^{(0)}=\frac{1}{F_{n}}$$ and $$\mathbb{F}_{0}=0$$. Then they gave for the sum of the squares of hyperharmonic Fibonacci numbers as follows:

$$\frac{1}{\sqrt{n}}\mathbb{F}_{n-1}^{(r+1)}\leq \sqrt{\sum _{k=0}^{n-1} \bigl(\mathbb{F}_{k}^{(r)} \bigr) ^{2}}\leq \mathbb{F}_{n-1}^{(r+1)}.$$
(4)

Now we give some definitions and Lemmas related to our study.

### Definition 1

An $$n\times n$$ matrix $$C_{r^{\ast }}$$ is called a geometric circulant matrix if it is of the form

$$C_{r^{\ast }}= \begin{pmatrix} c_{0} & c_{1} & c_{2} & \ldots & c_{n-2} & c_{n-1} \\ rc_{n-1} & c_{0} & c_{1} & \ldots & c_{n-3} & c_{n-2} \\ r^{2}c_{n-2} & rc_{n-1} & c_{0} & \ldots & c_{n-4} & c_{n-3} \\ \vdots & \vdots & \vdots & & \vdots & \vdots \\ r^{n-1}c_{1} & r^{n-2}c_{2} & r^{n-3}c_{3} & \ldots & rc_{n-1} & c_{0} \end{pmatrix} .$$

We denote it for brevity by $$C_{r^{\ast }}=\operatorname{Circ}_{r^{\ast}}(c_{0},c_{1},c_{2},\ldots ,c_{n-1})$$. Note that, for $$r=1$$, geometric circulant matrix turns into circulant matrix given in [11, 12]. In fact, in [11, 12], the authors calculated the spectral norms of the circulant matrices with the generalized Fibonacci and hyperharmonic Fibonacci numbers.

### Definition 2

Let $$A=(a_{ij})$$ be any $$m\times n$$ matrix. The Euclidean norm of matrix A is

$$\Vert A\Vert _{E}=\sqrt{ \Biggl(\sum _{i=1}^{m}\sum_{j=1}^{n} \vert a_{ij}\vert ^{2} \Biggr) }.$$

### Definition 3

Let $$A=(a_{ij})$$ be any $$m\times n$$ matrix. The spectral norm of matrix A is

$$\Vert A\Vert _{2}=\sqrt{\max_{1\leq i\leq n}\lambda _{i}\bigl(A^{H}A\bigr)},$$

where $$\lambda _{i}(A^{H}A)$$ is eigenvalue of $$A^{H}A$$ and $$A^{H}$$ is conjugate transpose of matrix A.

Then the following inequalities hold between the Euclidean norm and the spectral norm:

\begin{aligned}& \frac{1}{\sqrt{n}} \Vert A\Vert _{E}\leq \Vert A\Vert _{2}\leq \Vert A\Vert _{E}, \end{aligned}
(5)
\begin{aligned}& \Vert A\Vert _{2}\leq \Vert A\Vert _{E}\leq \sqrt{n} \Vert A\Vert _{2}. \end{aligned}
(6)

### Definition 4

Let $$A=(a_{ij})$$ and $$B=(b_{ij})$$ are each $$m\times n$$ matrices, then their Hadamard product is the $$m\times n$$ matrix of elementwise products

$$A\circ B=(a_{ij}b_{ij}).$$

### Lemma 1

 Let A and B be two $$m\times n$$ matrices. We have

$$\Vert A\circ B\Vert _{2}\leq r_{1}(A)c_{1}(B),$$

where

\begin{aligned}& r_{1}(A)=\max_{1\leq i\leq m}\sqrt{\sum _{j=1}^{n}\vert a_{ij}\vert ^{2}},\\& c_{1}(B)=\max_{1\leq j\leq n}\sqrt{\sum _{i=1}^{m}\vert b_{ij}\vert ^{2}}. \end{aligned}

## Main results

### Theorem 1

Let $$U_{r^{\ast }}=\operatorname{Circ}_{r^{\ast }}(U_{0},U_{1},U_{2},\ldots,U_{n-1})$$ be an $$n\times n$$ geometric circulant matrix.

1. (i)

If $$\vert r\vert >1$$, then

$$\sqrt{\frac{U_{n}^{2}-U_{n-1}^{2}+(-1)^{n}}{p^{2}}}\leq \Vert U_{r^{\ast }}\Vert _{2}\leq \sqrt{\frac{(\vert r\vert ^{2}-\vert r\vert ^{2n})(U_{n}^{2}-U_{n-1}^{2}+(-1)^{n})}{ (1-\vert r\vert ^{2} ) p^{2}}}.$$
2. (ii)

If $$\vert r\vert <1$$, then

\begin{aligned} &\frac{\vert r\vert }{\sqrt{p^{2}+4}}\sqrt{\frac{2\vert r\vert ^{2n+2}-\vert r\vert ^{2n}(p^{2}+2)-\vert r\vert ^{2}V_{2n}+V_{2n-2}}{\vert r\vert ^{4}-\vert r\vert ^{2}(p^{2}+2)+1}-2\frac{\vert r\vert ^{2n}-(-1)^{n}}{ \vert r\vert ^{2}+1}} \\ &\quad \leq \Vert U_{r^{\ast }}\Vert _{2}\leq \sqrt{\frac{(n-1) ( U_{n}^{2}-U_{n-1}^{2}+(-1)^{n} ) }{p^{2}}}. \end{aligned}

### Proof

We have the matrix

$$U_{r^{\ast }}= \begin{pmatrix} U_{0} & U_{1} & U_{2} & \ldots & U_{n-2} & U_{n-1} \\ rU_{n-1} & U_{0} & U_{1} & \ldots & U_{n-3} & U_{n-2} \\ r^{2}U_{n-2} & rU_{n-1} & U_{0} & \ldots & U_{n-4} & U_{n-3} \\ \vdots & \vdots & \vdots & & \vdots & \vdots \\ r^{n-1}U_{1} & r^{n-2}U_{2} & r^{n-3}U_{3} & \ldots & rU_{n-1} & U_{0} \end{pmatrix} .$$

(i) From $$\vert r\vert >1$$ and definition of Euclidean norm, we have

\begin{aligned} \Vert U_{r^{\ast }}\Vert _{E}^{2}&=\sum _{k=0}^{n-1}(n-k)U_{k}^{2}+\sum _{k=1}^{n-1}k\bigl\vert r^{n-k}\bigr\vert ^{2}U_{k}^{2} \\ &\geq \sum_{k=0}^{n-1}(n-k)U_{k}^{2}+ \sum_{k=1}^{n-1}kU_{k}^{2} \\ &= n\sum_{k=0}^{n-1}U_{k}^{2} \\ &= n\frac{U_{n}^{2}-U_{n-1}^{2}+(-1)^{n}}{p^{2}}, \end{aligned}

that is,

$$\frac{1}{\sqrt{n}} \Vert U_{r^{\ast }}\Vert _{E}\geq \sqrt {\frac{U_{n}^{2}-U_{n-1}^{2}+(-1)^{n}}{p^{2}}};$$

from (5), we have

$$\sqrt{\frac{U_{n}^{2}-U_{n-1}^{2}+(-1)^{n}}{p^{2}}}\leq \Vert U_{r^{\ast }}\Vert _{2}.$$

On the other hand, let the matrices A and B be defined by

$$A= \begin{pmatrix} U_{0} & 1 & 1 & \ldots & 1 & 1 \\ r & U_{0} & 1 & \ldots & 1 & 1 \\ r^{2} & r & U_{0} & \ldots & 1 & 1 \\ \vdots & \vdots & \vdots & & \vdots & \vdots \\ r^{n-1} & r^{n-2} & r^{n-3} & \ldots & r & U_{0} \end{pmatrix}$$

and

$$B= \begin{pmatrix} U_{0} & U_{1} & U_{2} & \ldots & U_{n-2} & U_{n-1} \\ U_{n-1} & U_{0} & U_{1} & \ldots & U_{n-3} & U_{n-2} \\ U_{n-2} & U_{n-1} & U_{0} & \ldots & U_{n-4} & U_{n-3} \\ \vdots & \vdots & \vdots & & \vdots & \vdots \\ U_{1} & U_{2} & U_{3} & \ldots & U_{n-1} & U_{0} \end{pmatrix} .$$

That is, $$U_{r^{\ast }}=A\circ B$$. Then we obtain

\begin{aligned} r_{1}(A) &=\max_{1\leq i\leq n}\sqrt{\sum _{j=1}^{n}\vert a_{ij}\vert ^{2}} \\ &=\sqrt{\bigl\vert r^{1}\bigr\vert ^{2}+\cdots +\bigl\vert r^{n-1}\bigr\vert ^{2}} \\ &=\sqrt{\frac{\vert r\vert ^{2}-\vert r\vert ^{2n}}{1-\vert r\vert ^{2}}} \end{aligned}

and

\begin{aligned} c_{1}(B) &=\max_{1\leq j\leq n}\sqrt{\sum _{i=1}^{n}\vert b_{ij}\vert ^{2}} \\ &=\sqrt{\sum_{k=0}^{n-1}U_{k}^{2}} \\ &=\sqrt{\frac{U_{n}^{2}-U_{n-1}^{2}+(-1)^{n}}{p^{2}}}. \end{aligned}

From Lemma 1, we have

$$\Vert U_{r^{\ast }}\Vert _{2}\leq \sqrt{ \frac{(\vert r\vert ^{2}-\vert r\vert ^{2n})(U_{n}^{2}-U_{n-1}^{2}+(-1)^{n})}{ ( 1-\vert r\vert ^{2} ) p^{2}}}.$$

Thus, we have

$$\sqrt{\frac{U_{n}^{2}-U_{n-1}^{2}+(-1)^{n}}{p^{2}}}\leq \Vert U_{r^{\ast }}\Vert _{2}\leq \sqrt{\frac{(\vert r\vert ^{2}-\vert r\vert ^{2n})(U_{n}^{2}-U_{n-1}^{2}+(-1)^{n})}{ ( 1-\vert r\vert ^{2} ) p^{2}}}.$$

(ii) From $$\vert r\vert <1$$, we have

\begin{aligned} &{\Vert U_{r^{\ast }}\Vert _{E}^{2} =\sum _{k=0}^{n-1}(n-k)U_{k}^{2}+\sum _{k=1}^{n-1}k\bigl\vert r^{n-k}\bigr\vert ^{2}U_{k}^{2}} \\ &{\hphantom{\Vert U_{r^{\ast }}\Vert _{E}^{2} }\geq \sum_{k=0}^{n-1}(n-k)\bigl\vert r^{n-k}\bigr\vert ^{2}U_{k}^{2}+\sum _{k=1}^{n-1}k\bigl\vert r^{n-k}\bigr\vert ^{2}U_{k}^{2}} \\ &{\hphantom{\Vert U_{r^{\ast }}\Vert _{E}^{2} }= n\vert r\vert ^{2n}\sum_{k=0}^{n-1} \biggl( \frac{U_{k}}{\vert r\vert ^{k}} \biggr) ^{2}} \\ &{\hphantom{\Vert U_{r^{\ast }}\Vert _{E}^{2} }=\frac{n\vert r\vert ^{2n}}{p^{2}+4}\sum_{k=0}^{n-1} \biggl( \frac{\alpha ^{k}-\beta ^{k}}{\vert r\vert ^{k}} \biggr) ^{2}} \\ &{\hphantom{\Vert U_{r^{\ast }}\Vert _{E}^{2} }=\frac{n\vert r\vert ^{2n}}{p^{2}+4} \biggl( \frac{1- ( \frac{\alpha ^{2}}{\vert r\vert ^{2}} ) ^{n}}{1- ( \frac{\alpha ^{2}}{\vert r\vert ^{2}} ) }+\frac{1- ( \frac{\beta ^{2}}{\vert r\vert ^{2}} ) ^{n}}{1- ( \frac{\beta^{2}}{\vert r\vert ^{2}} ) }-2 \frac{1- ( \frac{-1}{\vert r\vert ^{2}} ) ^{n}}{1+\frac{1}{\vert r\vert ^{2}}} \biggr)} \\ &{\hphantom{\Vert U_{r^{\ast }}\Vert _{E}^{2} }=\frac{n\vert r\vert ^{2}}{p^{2}+4}\sqrt{\frac{2\vert r\vert ^{2n+2}-\vert r\vert ^{2n}(p^{2}+2)-\vert r\vert ^{2}V_{2n}+V_{2n-2}}{\vert r\vert ^{4}-\vert r\vert ^{2}(p^{2}+2)+1}-2\frac{\vert r\vert ^{2n}-(-1)^{n}}{\vert r\vert ^{2}+1}},}\\ &{\frac{1}{\sqrt{n}} \Vert U_{r^{\ast }}\Vert _{E} \geq \frac{\vert r\vert }{\sqrt{p^{2}+4}}\sqrt{\frac{2\vert r\vert ^{2n+2}-\vert r\vert ^{2n}(p^{2}+2)-\vert r\vert ^{2}V_{2n}+V_{2n-2}}{\vert r\vert ^{4}-\vert r\vert ^{2}(p^{2}+2)+1}-2\frac{\vert r\vert ^{2n}-(-1)^{n}}{\vert r\vert ^{2}+1}}.} \end{aligned}

From (5)

$$\frac{\vert r\vert }{\sqrt{p^{2}+4}}\sqrt{\frac{2\vert r\vert ^{2n+2}-\vert r\vert ^{2n}(p^{2}+2)-\vert r\vert ^{2}V_{2n}+V_{2n-2}}{\vert r\vert ^{4}-\vert r\vert ^{2}(p^{2}+2)+1}-2\frac{\vert r\vert ^{2n}-(-1)^{n}}{\vert r\vert ^{2}+1}}\leq \Vert U_{r^{\ast }}\Vert _{2}.$$

On the other hand, let the matrices A and B be defined by

$$A= \begin{pmatrix} U_{0} & 1 & 1 & \ldots & 1 & 1 \\ r & U_{0} & 1 & \ldots & 1 & 1 \\ r^{2} & r & U_{0} & \ldots & 1 & 1 \\ \vdots & \vdots & \vdots & & \vdots & \vdots \\ r^{n-1} & r^{n-2} & r^{n-3} & \ldots & r & U_{0} \end{pmatrix}$$

and

$$B= \begin{pmatrix} U_{0} & U_{1} & U_{2} & \ldots & U_{n-2} & U_{n-1} \\ U_{n-1} & U_{0} & U_{1} & \ldots & U_{n-3} & U_{n-2} \\ U_{n-2} & U_{n-1} & U_{0} & \ldots & U_{n-4} & U_{n-3} \\ \vdots & \vdots & \vdots & & \vdots & \vdots \\ U_{1} & U_{2} & U_{3} & \ldots & U_{n-1} & U_{0} \end{pmatrix}.$$

That is, $$U_{r^{\ast }}=A\circ B$$. Then we obtain

\begin{aligned} r_{1}(A) &=\max_{1\leq i\leq n}\sqrt{\sum _{j=1}^{n}\vert a_{ij}\vert ^{2}} \\ &=\sqrt{U_{0}^{2}+n-1} \\ &=\sqrt{n-1} \end{aligned}

and

\begin{aligned} c_{1}(B) &=\max_{1\leq j\leq n}\sqrt{\sum _{i=1}^{n}\vert b_{ij}\vert ^{2}} \\ &=\sqrt{\sum_{k=0}^{n-1}U_{k}^{2}} \\ &=\sqrt{\frac{U_{n}^{2}-U_{n-1}^{2}+(-1)^{n}}{p^{2}}}. \end{aligned}

Hence, from Lemma 1, we have

$$\Vert U_{r^{\ast }}\Vert _{2}\leq \sqrt{ \frac{(n-1) (U_{n}^{2}-U_{n-1}^{2}+(-1)^{n} ) }{p^{2}}}.$$

Thus, we have

\begin{aligned} &\frac{\vert r\vert }{\sqrt{p^{2}+4}}\sqrt{\frac{2\vert r\vert ^{2n+2}-\vert r\vert ^{2n}(p^{2}+2)-\vert r\vert ^{2}V_{2n}+V_{2n-2}}{\vert r\vert ^{4}-\vert r\vert ^{2}(p^{2}+2)+1}-2\frac{\vert r\vert ^{2n}-(-1)^{n}}{\vert r\vert ^{2}+1}} \\ &\quad\leq \Vert U_{r^{\ast }}\Vert _{2} \leq \sqrt{\frac{(n-1) ( U_{n}^{2}-U_{n-1}^{2}+(-1)^{n} ) }{p^{2}}}. \end{aligned}

□

### Theorem 2

Let $$V_{r^{\ast }}=\operatorname{Circ}_{r^{\ast}}(V_{0},V_{1},V_{2},\ldots ,V_{n-1})$$ be an $$n\times n$$ geometric circulant matrix.

1. (i)

If $$\vert r\vert >1$$, then

\begin{aligned} &\sqrt{\frac{V_{n}^{2}-V_{n-1}^{2}+p^{2}-4+(1-(-1)^{n})(p^{2}+4)}{p^{2}}} \\ &\quad \leq \Vert V_{r^{\ast }}\Vert _{2}\leq \sqrt{\frac{1-\vert r\vert ^{2n}}{1-\vert r\vert ^{2}}\frac{V_{n}^{2}-V_{n-1}^{2}+p^{2}-4+(1-(-1)^{n})(p^{2}+4)}{p^{2}}}. \end{aligned}
2. (ii)

If $$\vert r\vert <1$$, then

\begin{aligned} &\vert r\vert \sqrt{\frac{2\vert r\vert ^{2n+2}-\vert r\vert ^{2n}(p^{2}+2)-\vert r\vert ^{2}V_{2n}+V_{2n-2}}{\vert r\vert ^{4}-\vert r\vert ^{2}(p^{2}+2)+1}+2\frac{\vert r\vert ^{2n}-(-1)^{n}}{\vert r\vert ^{2}+1}} \\ & \quad\leq \Vert V_{r^{\ast }}\Vert _{2}\leq \sqrt{\frac{n (V_{n}^{2}-V_{n-1}^{2}+p^{2}-4+(1-(-1)^{n})(p^{2}+4) ) }{p^{2}}}. \end{aligned}

### Proof

We have the matrix

$$V_{r^{\ast }}= \begin{pmatrix} V_{0} & V_{1} & V_{2} & \ldots & V_{n-2} & V_{n-1} \\ rV_{n-1} & V_{0} & V_{1} & \ldots & V_{n-3} & V_{n-2} \\ r^{2}V_{n-2} & rV_{n-1} & V_{0} & \ldots & V_{n-4} & V_{n-3} \\ \vdots & \vdots & \vdots & & \vdots & \vdots \\ r^{n-1}V_{1} & r^{n-2}V_{2} & r^{n-3}V_{3} & \ldots & rV_{n-1} & V_{0} \end{pmatrix}.$$

(i) From $$\vert r\vert >1$$, we have

\begin{aligned} \Vert V_{r^{\ast }}\Vert _{E}^{2}&=\sum _{k=0}^{n-1}(n-k)V_{k}^{2}+\sum _{k=1}^{n-1}k\bigl\vert r^{n-k}\bigr\vert ^{2}V_{k}^{2} \\ &\geq n\sum_{k=0}^{n-1}V_{k}^{2} \\ &=\frac{n ( V_{n}^{2}-V_{n-1}^{2}+p^{2}-4+(1-(-1)^{n})(p^{2}+4) ) }{p^{2}}, \end{aligned}

that is,

$$\frac{1}{\sqrt{n}} \Vert V_{r^{\ast }}\Vert _{E}\geq \sqrt {\frac{V_{n}^{2}-V_{n-1}^{2}+p^{2}-4+(1-(-1)^{n})(p^{2}+4)}{p^{2}}},$$

from (5), we have

$$\sqrt{\frac{V_{n}^{2}-V_{n-1}^{2}+p^{2}-4+(1-(-1)^{n})(p^{2}+4)}{p^{2}}}\leq \Vert V_{r^{\ast }}\Vert _{2}.$$

On the other hand, let the matrices A and B be defined by

$$A= \begin{pmatrix} 1 & 1 & 1 & \ldots & 1 & 1 \\ r & 1 & 1 & \ldots & 1 & 1 \\ r^{2} & r & 1 & \ldots & 1 & 1 \\ \vdots & \vdots & \vdots & & \vdots & \vdots \\ r^{n-1} & r^{n-2} & r^{n-3} & \ldots & r & 1 \end{pmatrix}$$

and

$$B= \begin{pmatrix} V_{0} & V_{1} & V_{2} & \ldots & V_{n-2} & V_{n-1} \\ V_{n-1} & V_{0} & V_{1} & \ldots & V_{n-3} & V_{n-2} \\ V_{n-2} & V_{n-1} & V_{0} & \ldots & V_{n-4} & V_{n-3} \\ \vdots & \vdots & \vdots & & \vdots & \vdots \\ V_{1} & V_{2} & V_{3} & \ldots & V_{n-1} & V_{0} \end{pmatrix}.$$

That is, $$V_{r^{\ast }}=A\circ B$$. Then we obtain

\begin{aligned} r_{1}(A) &=\max_{1\leq i\leq n}\sqrt{\sum _{j=1}^{n}\vert a_{ij}\vert ^{2}} \\ &=\sqrt{1+\bigl\vert r^{1}\bigr\vert ^{2}+\cdots +\bigl\vert r^{n-1}\bigr\vert ^{2}} \\ &=\sqrt{\frac{1-\vert r\vert ^{2n}}{1-\vert r\vert ^{2}}} \end{aligned}

and

\begin{aligned} c_{1}(B) &=\max_{1\leq j\leq n}\sqrt{\sum _{i=1}^{n}\vert b_{ij}\vert ^{2}} \\ &=\sqrt{\sum_{k=0}^{n-1}V_{k}^{2}} \\ &=\sqrt{\frac{V_{n}^{2}-V_{n-1}^{2}+p^{2}-4+(1-(-1)^{n})(p^{2}+4)}{p^{2}}}. \end{aligned}

From Lemma 1, we have

$$\Vert V_{r^{\ast }}\Vert _{2}\leq \sqrt{ \frac{1-\vert r\vert ^{2n}}{1-\vert r\vert ^{2}}\frac{ V_{n}^{2}-V_{n-1}^{2}+p^{2}-4+(1-(-1)^{n})(p^{2}+4)}{p^{2}}}.$$

Thus, we have

\begin{aligned} &\sqrt{\frac{V_{n}^{2}-V_{n-1}^{2}+p^{2}-4+(1-(-1)^{n})(p^{2}+4)}{p^{2}}} \\ & \quad\leq \Vert V_{r^{\ast }}\Vert _{2}\leq \sqrt{\frac{1-\vert r\vert ^{2n}}{1-\vert r\vert ^{2}}\frac{V_{n}^{2}-V_{n-1}^{2}+p^{2}-4+(1-(-1)^{n})(p^{2}+4)}{p^{2}}}. \end{aligned}

(ii) From $$\vert r\vert <1$$, we have

\begin{aligned} &{\Vert V_{r^{\ast }}\Vert _{E}^{2}=\sum _{k=0}^{n-1}(n-k)V_{k}^{2}+\sum _{k=1}^{n-1}k\bigl\vert r^{n-k}\bigr\vert ^{2}V_{k}^{2}} \\ &{\hphantom{\Vert V_{r^{\ast }}\Vert _{E}^{2}}\geq \sum_{k=0}^{n-1}(n-k)\bigl\vert r^{n-k}\bigr\vert ^{2}V_{k}^{2}+\sum _{k=1}^{n-1}k\bigl\vert r^{n-k}\bigr\vert ^{2}V_{k}^{2}} \\ &{\hphantom{\Vert V_{r^{\ast }}\Vert _{E}^{2}}= n\vert r\vert ^{2n}\sum_{k=0}^{n-1} \biggl( \frac{V_{k}}{\vert r\vert ^{k}} \biggr) ^{2}} \\ &{\hphantom{\Vert V_{r^{\ast }}\Vert _{E}^{2}}= n\vert r\vert ^{2n} \Biggl( \sum_{k=0}^{n-1} \frac{\alpha ^{2k}}{\vert r\vert ^{2k}}+\sum_{k=0}^{n-1} \frac{\beta ^{2k}}{\vert r\vert ^{2k}}+2\sum_{k=0}^{n-1} \frac{(-1)^{k}}{\vert r\vert ^{2k}} \Biggr)} \\ &{\hphantom{\Vert V_{r^{\ast }}\Vert _{E}^{2}}= n\vert r\vert ^{2n} \biggl( \frac{1- ( \frac{\alpha ^{2}}{\vert r\vert ^{2}} ) ^{n}}{1- ( \frac{\alpha ^{2}}{\vert r\vert ^{2}} ) }+ \frac{1- ( \frac{\beta ^{2}}{\vert r\vert ^{2}} ) ^{n}}{1- ( \frac{\beta ^{2}}{\vert r\vert ^{2}} ) }+2\frac{1- ( \frac{-1}{\vert r\vert ^{2}} ) ^{n}}{1+\frac{1}{\vert r\vert ^{2}}} \biggr)} \\ &{\hphantom{\Vert V_{r^{\ast }}\Vert _{E}^{2}}= n\vert r\vert ^{2} \biggl( \frac{2\vert r\vert ^{2n+2}-\vert r\vert ^{2n}(p^{2}+2)-\vert r\vert ^{2}V_{2n}+V_{2n-2}}{\vert r\vert ^{4}-\vert r\vert ^{2}(p^{2}+2)+1}+2 \frac{\vert r\vert ^{2n}-(-1)^{n}}{\vert r\vert ^{2}+1} \biggr) ,}\\ &{\frac{1}{\sqrt{n}} \Vert V_{r^{\ast }}\Vert _{E} \geq \vert r\vert \sqrt{\frac{2\vert r\vert ^{2n+2}-\vert r\vert ^{2n}(p^{2}+2)-\vert r\vert ^{2}V_{2n}+V_{2n-2}}{\vert r\vert ^{4}-\vert r\vert ^{2}(p^{2}+2)+1}+2\frac{\vert r\vert ^{2n}-(-1)^{n}}{\vert r\vert ^{2}+1}}.} \end{aligned}

From (5),

$$\vert r\vert \sqrt{\frac{2\vert r\vert ^{2n+2}-\vert r\vert ^{2n}(p^{2}+2)-\vert r\vert ^{2}V_{2n}+V_{2n-2}}{\vert r\vert ^{4}-\vert r\vert ^{2}(p^{2}+2)+1}+2\frac{\vert r\vert ^{2n}-(-1)^{n}}{\vert r\vert ^{2}+1}}\leq \Vert V_{r^{\ast }}\Vert _{2}.$$

On the other hand, let the matrices A and B be defined by

$$A= \begin{pmatrix} 1 & 1 & 1 & \ldots & 1 & 1 \\ r & 1 & 1 & \ldots & 1 & 1 \\ r^{2} & r & 1 & \ldots & 1 & 1 \\ \vdots & \vdots & \vdots & & \vdots & \vdots \\ r^{n-1} & r^{n-2} & r^{n-3} & \ldots & r & 1 \end{pmatrix}$$

and

$$B= \begin{pmatrix} V_{0} & V_{1} & V_{2} & \ldots & V_{n-2} & V_{n-1} \\ V_{n-1} & V_{0} & V_{1} & \ldots & V_{n-3} & V_{n-2} \\ V_{n-2} & V_{n-1} & V_{0} & \ldots & V_{n-4} & V_{n-3} \\ \vdots & \vdots & \vdots & & \vdots & \vdots \\ V_{1} & V_{2} & V_{3} & \ldots & V_{n-1} & V_{0} \end{pmatrix}.$$

That is, $$V_{r^{\ast }}=A\circ B$$. Then we obtain

$$r_{1}(A)=\max_{1\leq i\leq n}\sqrt{\sum _{j=1}^{n}\vert a_{ij}\vert ^{2}}=\sqrt{n}$$

and

\begin{aligned} c_{1}(B) &=\max_{1\leq j\leq n}\sqrt{\sum _{i=1}^{n}\vert b_{ij}\vert ^{2}} \\ &=\sqrt{\sum_{k=0}^{n-1}V_{k}^{2}} \\ &=\sqrt{\frac{V_{n}^{2}-V_{n-1}^{2}+p^{2}-4+(1-(-1)^{n})(p^{2}+4)}{p^{2}}}. \end{aligned}

From Lemma 1, we have

$$\Vert V_{r^{\ast }}\Vert _{2}\leq \sqrt{ \frac{n (V_{n}^{2}-V_{n-1}^{2}+p^{2}-4+(1-(-1)^{n})(p^{2}+4) ) }{p^{2}}}.$$

Thus we have

\begin{aligned} \vert r\vert & \sqrt{\frac{2\vert r\vert ^{2n+2}-\vert r\vert ^{2n}(p^{2}+2)-\vert r\vert ^{2}V_{2n}+V_{2n-2}}{\vert r\vert ^{4}-\vert r\vert ^{2}(p^{2}+2)+1}+2\frac{\vert r\vert ^{2n}-(-1)^{n}}{\vert r\vert ^{2}+1}} \\ & \quad\leq \Vert V_{r^{\ast }}\Vert _{2}\leq \sqrt{\frac{n (V_{n}^{2}-V_{n-1}^{2}+p^{2}-4+(1-(-1)^{n})(p^{2}+4) ) }{p^{2}}}. \end{aligned}

□

### Theorem 3

Let $$\mathbb{F}_{r^{\ast }}^{(k)}=\operatorname{Circ}_{r^{\ast }}(\mathbb{F}_{0}^{(k)},\mathbb{F}_{1}^{(k)},\mathbb{F}_{2}^{(k)},\ldots ,\mathbb{F}_{n-1}^{(k)})$$ be an $$n\times n$$ geometric circulant matrix.

1. (i)

If $$\vert r\vert >1$$, then

$$\frac{1}{\sqrt{n}}\mathbb{F}_{n-1}^{(k+1)}\leq \bigl\Vert \mathbb{F}_{r^{\ast }}^{(k)}\bigr\Vert _{2}\leq \sqrt{ \frac{\vert r\vert ^{2}-\vert r\vert ^{2n}}{1-\vert r\vert ^{2}}}\mathbb{F}_{n-1}^{(k+1)}.$$
2. (ii)

If $$\vert r\vert <1$$, then

$$\frac{\vert r\vert ^{n}}{\sqrt{n}}\mathbb{F}_{n-1}^{(k+1)}\leq \bigl\Vert \mathbb{F}_{r^{\ast }}^{(k)}\bigr\Vert _{2}\leq \sqrt{n-1} \mathbb{F}_{n-1}^{(k+1)}.$$

### Proof

Since the $$\mathbb{F}_{r^{\ast }}^{(k)}$$ is of the form

$$\mathbb{F}_{r^{\ast }}^{(k)}= \begin{pmatrix} \mathbb{F}_{0}^{(k)} & \mathbb{F}_{1}^{(k)} & \mathbb{F}_{2}^{(k)} & \ldots& \mathbb{F}_{n-2}^{(k)} & \mathbb{F}_{n-1}^{(k)} \\ r\mathbb{F}_{n-1}^{(k)} & \mathbb{F}_{0}^{(k)} & \mathbb{F}_{1}^{(k)} & \ldots & \mathbb{F}_{n-3}^{(k)} & \mathbb{F}_{n-2}^{(k)} \\ r^{2}\mathbb{F}_{n-2}^{(k)} & r\mathbb{F}_{n-1}^{(k)} & \mathbb{F}_{0}^{(k)}& \ldots & \mathbb{F}_{n-4}^{(k)} &\mathbb{F}_{n-3}^{(k)} \\ \vdots & \vdots & \vdots & & \vdots & \vdots \\ r^{n-1}\mathbb{F}_{1}^{(k)} & r^{n-2}\mathbb{F}_{2}^{(k)} & r^{n-3}\mathbb{F}_{3}^{(k)} & \ldots & r\mathbb{F}_{n-1}^{(k)} & \mathbb{F}_{0}^{(k)} \end{pmatrix}$$

and from the definition of Euclidean norm, we have

$$\bigl\Vert \mathbb{F}_{r^{\ast }}^{(k)}\bigr\Vert _{E}^{2}=\sum_{s=0}^{n-1}(n-s) \bigl( \mathbb{F}_{s}^{(k)} \bigr)^{2}+\sum _{s=1}^{n-1}s\bigl\vert r^{n-s}\bigr\vert ^{2} \bigl( \mathbb{F}_{s}^{(k)} \bigr) ^{2}.$$

(i) From $$\vert r\vert >1$$, we have

\begin{aligned} \bigl\Vert \mathbb{F}_{r^{\ast }}^{(k)}\bigr\Vert _{E}^{2} &\geq \sum_{s=0}^{n-1}(n-s) \bigl( \mathbb{F}_{s}^{(k)} \bigr) ^{2}+\sum _{s=1}^{n-1}s \bigl( \mathbb{F}_{s}^{(k)} \bigr) ^{2} \\ &= n\sum_{s=0}^{n-1} \bigl( \mathbb{F}_{s}^{(k)} \bigr) ^{2}. \end{aligned}

Thus from (5) and (4),

$$\frac{1}{\sqrt{n}}\mathbb{F}_{n-1}^{(k+1)}\leq \bigl\Vert \mathbb{F}_{r^{\ast }}^{(k)}\bigr\Vert _{2}.$$

On the other hand let the matrices $$A^{(k)}$$ and $$B^{(k)}$$ be defined by

$$A^{(k)}= \begin{pmatrix} \mathbb{F}_{0}^{(k)} & 1 & 1 & \ldots & 1 & 1 \\ r & \mathbb{F}_{0}^{(k)} & 1 & \ldots & 1 & 1 \\ r^{2} & r & \mathbb{F}_{0}^{(k)} & \ldots & 1 & 1 \\ \vdots & \vdots & \vdots & & \vdots & \vdots \\ r^{n-1} & r^{n-2} & r^{n-3} & \ldots & r & \mathbb{F}_{0}^{(k)} \end{pmatrix}$$

and

$$B^{(k)}= \begin{pmatrix} \mathbb{F}_{0}^{(k)} & \mathbb{F}_{1}^{(k)} & \mathbb{F}_{2}^{(k)} & \ldots& \mathbb{F}_{n-2}^{(k)} & \mathbb{F}_{n-1}^{(k)} \\ \mathbb{F}_{n-1}^{(k)} & \mathbb{F}_{0}^{(k)} & \mathbb{F}_{1}^{(k)} & \ldots & \mathbb{F}_{n-3}^{(k)} & \mathbb{F}_{n-2}^{(k)} \\ \mathbb{F}_{n-2}^{(k)} & \mathbb{F}_{n-1}^{(k)} & \mathbb{F}_{0}^{(k)} & \ldots & \mathbb{F}_{n-4}^{(k)} & \mathbb{F}_{n-3}^{(k)} \\ \vdots & \vdots & \vdots & & \vdots & \vdots \\ \mathbb{F}_{1}^{(k)} & \mathbb{F}_{2}^{(k)} & \mathbb{F}_{3}^{(k)} & \ldots & \mathbb{F}_{n-1}^{(k)} & \mathbb{F}_{0}^{(k)} \end{pmatrix}.$$

That is, $$\mathbb{F}_{r^{\ast }}^{(k)}=A^{(k)}\circ B^{(k)}$$. Then we obtain

$$r_{1}\bigl(A^{(k)}\bigr)=\max_{1\leq i\leq n}\sqrt {\sum_{j=1}^{n}\bigl\vert a_{ij}^{(k)}\bigr\vert ^{2}}=\sqrt{ \frac{\vert r\vert ^{2}-\vert r\vert ^{2n}}{1-\vert r\vert ^{2}}}$$

and

$$c_{1}\bigl(B^{(k)}\bigr)=\max_{1\leq j\leq n}\sqrt {\sum_{i=1}^{n}\bigl\vert b_{ij}^{(k)}\bigr\vert ^{2}}=\sqrt{\sum _{s=0}^{n-1} \bigl( \mathbb{F}_{s}^{(k)} \bigr) ^{2}}.$$

From Lemma 1 and (4), we have

$$\bigl\Vert \mathbb{F}_{r^{\ast }}^{(k)}\bigr\Vert _{2} \leq \sqrt{\frac{\vert r\vert ^{2}-\vert r\vert ^{2n}}{1-\vert r\vert ^{2}}}\mathbb{F}_{n-1}^{(k+1)},$$

which is desired result.

(ii) From $$\vert r\vert <1$$, we have

\begin{aligned} \bigl\Vert \mathbb{F}_{r^{\ast }}^{(k)}\bigr\Vert _{E}^{2} &\geq \sum_{s=0}^{n-1}(n-s) \bigl\vert r^{n-s}\bigr\vert ^{2} \bigl( \mathbb{F}_{s}^{(k)} \bigr) ^{2}+\sum _{s=1}^{n-1}s\bigl\vert r^{n-s}\bigr\vert ^{2} \bigl( \mathbb{F}_{s}^{(k)} \bigr) ^{2} \\ &\geq n\vert r\vert ^{2n}\sum_{s=0}^{n-1} \bigl( \mathbb{F}_{s}^{(k)} \bigr) ^{2}. \end{aligned}

From (5) and (4),

$$\frac{\vert r\vert ^{n}}{\sqrt{n}}\mathbb{F}_{n-1}^{(k+1)}\leq \bigl\Vert \mathbb{F}_{r^{\ast }}^{(k)}\bigr\Vert _{2}.$$

On the other hand, let the matrices $$A^{(k)}$$ and $$B^{(k)}$$ be defined by

$$A^{(k)}= \begin{pmatrix} \mathbb{F}_{0}^{(k)} & 1 & 1 & \ldots & 1 & 1 \\ r & \mathbb{F}_{0}^{(k)} & 1 & \ldots & 1 & 1 \\ r^{2} & r & \mathbb{F}_{0}^{(k)} & \ldots & 1 & 1 \\ \vdots & \vdots & \vdots & & \vdots & \vdots \\ r^{n-1} & r^{n-2} & r^{n-3} & \ldots & r & \mathbb{F}_{0}^{(k)} \end{pmatrix}$$

and

$$B^{(k)}= \begin{pmatrix} \mathbb{F}_{0}^{(k)} & \mathbb{F}_{1}^{(k)} & \mathbb{F}_{2}^{(k)} & \ldots & \mathbb{F}_{n-2}^{(k)} & \mathbb{F}_{n-1}^{(k)} \\ \mathbb{F}_{n-1}^{(k)} & \mathbb{F}_{0}^{(k)} & \mathbb{F}_{1}^{(k)} & \ldots & \mathbb{F}_{n-3}^{(k)} & \mathbb{F}_{n-2}^{(k)} \\ \mathbb{F}_{n-2}^{(k)} & \mathbb{F}_{n-1}^{(k)} & \mathbb{F}_{0}^{(k)} & \ldots & \mathbb{F}_{n-4}^{(k)} & \mathbb{F}_{n-3}^{(k)} \\ \vdots & \vdots & \vdots & & \vdots & \vdots \\ \mathbb{F}_{1}^{(k)} & \mathbb{F}_{2}^{(k)} & \mathbb{F}_{3}^{(k)} & \ldots & \mathbb{F}_{n-1}^{(k)} & \mathbb{F}_{0}^{(k)} \end{pmatrix}.$$

That is, $$\mathbb{F}_{r^{\ast }}^{(k)}=A^{(k)}\circ B^{(k)}$$. Then we obtain

$$r_{1}\bigl(A^{(k)}\bigr)=\max_{1\leq i\leq n}\sqrt {\sum_{j=1}^{n}\bigl\vert a_{ij}^{(k)}\bigr\vert ^{2}}=\sqrt{n-1}$$

and

$$c_{1}\bigl(B^{(k)}\bigr)=\max_{1\leq j\leq n}\sqrt {\sum_{i=1}^{n}\bigl\vert b_{ij}^{(k)}\bigr\vert ^{2}}=\sqrt{\sum _{s=0}^{n-1} \bigl( \mathbb{F}_{s}^{(k)} \bigr) ^{2}}.$$

From Lemma 1 and (4), we have

$$\bigl\Vert \mathbb{F}_{r^{\ast }}^{(k)}\bigr\Vert _{2} \leq \sqrt{n-1}\mathbb{F}_{n-1}^{(k+1)}.$$

Thus we have

$$\frac{\vert r\vert ^{n}}{\sqrt{n}}\mathbb{F}_{n-1}^{(k+1)}\leq \bigl\Vert \mathbb{F}_{r^{\ast }}^{(k)}\bigr\Vert _{2}\leq \sqrt{n-1} \mathbb{F}_{n-1}^{(k+1)}.$$

□

## Conclusion

In this paper we approximated lower and upper bounds of the spectral norms of geometric circulant matrices with the generalized Fibonacci and hyperharmonic Fibonacci numbers. If we take $$p=1$$ and $$p=2$$ in Theorem 1, we obtain lower and upper bounds of the spectral norms of geometric circulant matrices with the Fibonacci and Pell numbers, respectively. Similarly if we take $$p=1$$ and $$p=2$$ in Theorem 2, we obtain lower and upper bounds of the spectral norms of geometric circulant matrices with the Lucas and Pell-Lucas numbers, respectively.

In the future it may be possible that one can generalize our results to the Horadam, tribonacci and tribonacci-like sequences.

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## Acknowledgements

The authors are grateful to two anonymous referees and the associate editor for their careful reading, helpful comments, and constructive suggestions, which improved the presentation of results.

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Correspondence to Can Kızılateş.

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The authors declare that they have no competing interests.

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Each of the authors contributed to each part of this work equally and read and approved the final version of the manuscript.

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