The upper bound estimation on the spectral norm of r-circulant matrices with the Fibonacci and Lucas numbers

Abstract

Let us define $A=\operatorname{Circ}_{r}(a_{0},a_{1},\ldots,a_{n-1})$ to be a $n\times n$ r-circulant matrix. The entries in the first row of $A=\operatorname{Circ}_{r}(a_{0},a_{1},\ldots,a_{n-1})$ are $a_{i}=F_{i}$, or $a_{i}=L_{i}$, or $a_{i}=F_{i}L_{i}$, or $a_{i}=F_{i}^{2}$, or $a_{i}=L_{i}^{2}$ ($i=0,1,\ldots,n-1$), where $F_{i}$ and $L_{i}$ are the ith Fibonacci and Lucas numbers, respectively. This paper gives an upper bound estimation of the spectral norm for r-circulant matrices with Fibonacci and Lucas numbers. The result is more accurate than the corresponding results of S Solak and S Shen, and of J Cen, and the numerical examples have provided further proof.

1 Introduction

For $n>0$, the Fibonacci sequence $\{F_{n}\}$ is defined by $F_{n+1}=F_{n}+F_{n-1}$, where $F_{0}=0$ and $F_{1}=1$. If we start by zero, then the sequence is given by

$$\begin{aligned} \begin{array}{@{}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{}} n & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & \cdots \\ F_{n} & 0 & 1 & 1 & 2 & 3 & 5 & 8 & 13 & 21 & \cdots \end{array} \end{aligned}$$

(1)

If we deduce from $F_{n+1}$ that $L_{n+1}=L_{n}+L_{n-1}$, and let $L_{0}=2$, $L_{1}=1$, then we obtain the Lucas sequence $\{L_{n}\}$,

$$\begin{aligned} \begin{array}{@{}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{}} n & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & \cdots\\ L_{n} & 2 & 1 & 3 & 4 & 7 & 11 & 18 & 29 & 47 & \cdots \end{array} \end{aligned}$$

(2)

Furthermore, the sequences $\{F_{n}\}$ and $\{L_{n}\}$ satisfy the following recursion:

$$\begin{aligned} F_{n}+L_{n}=2F_{n+1}. \end{aligned}$$

(3)

Definition 1.1

A matrix A is an r-circulant matrix if it is of the form

$$A= \begin{pmatrix} a_{0} & a_{1} & \cdots& a_{n-2} & a_{n-1}\\ ra_{n-1} & a_{0} & \cdots& a_{n-3} & a_{n-2}\\ \cdots& \cdots& \cdots& \cdots& \cdots\\ ra_{2} & ra_{3} & \cdots& a_{0} & a_{1}\\ ra_{1} & ra_{2} & \cdots& ra_{n-1} & a_{0} \end{pmatrix}. $$

Obviously, the elements of this r-circulant matrix are determined by its first row elements $a_{0},a_{1},\ldots,a_{n-1}$ and the parameter r, thus we denote $A=\operatorname{Circ}_{r}(a_{0}, a_{1}, \ldots, a_{n-1})$. Especially when $r=1$, we obtain $A=\operatorname{Circ}(a_{0}, a_{1}, \ldots, a_{n-1})$.

Definition 1.2

A matrix A is called a symmetric r-circulant matrix if it is of the form

$$A= \begin{pmatrix} a_{0} & a_{1} & \cdots& a_{n-2} & a_{n-1}\\ a_{1} & a_{2} & \cdots& a_{n-1} & ra_{0}\\ \cdots& \cdots& \cdots& \cdots& \cdots\\ a_{n-2} & a_{n-1} & \cdots& ra_{n-4} & ra_{n-3}\\ a_{n-1} & ra_{0} & \cdots& ra_{n-3} & ra_{n-2} \end{pmatrix}. $$

Obviously, the elements of this r-circulant matrix are determined by its first row elements $a_{0},a_{1},\ldots,a_{n-1}$ and the parameter r; thus we denote $A=\operatorname{SCirc}_{r}(a_{0}, a_{1}, \ldots, a_{n-1})$. Especially when $r=1$, we obtain $A=\operatorname{SCirc}(a_{0}, a_{1}, \ldots, a_{n-1})$.

For any $A=(a_{ij})_{m\times n}$, the well-known spectral norm of the matrix A is

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

in which $\lambda_{i} ( A^{H} A )$ is the eigenvalue of $A^{H}A$ and $A^{H}$ is the conjugate transpose of matrix A.

Define the maximum column length norm $c_{1}(\cdot)$ and the maximum row length norm $r_{1}(\cdot)$ of any matrix A by

$$c_{1}(A)=\max_{j}\sqrt{\sum _{i}|a_{ij}|^{2}} $$

and

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

respectively.

Let A, B, and C be $m \times n$ matrices. If $A=B\circ C$, then in accordance with [1] we have

$$\begin{aligned} \|A\|_{2}\leq r_{1}(B)c_{1}(C) \end{aligned}$$

(4)

and

$$\begin{aligned} \|A\|_{2}\leq\|B\|_{2}\|C\|_{2}. \end{aligned}$$

(5)

Here, we define $B=(b_{ij})_{m\times n}$, $C=(c_{ij})_{m\times n}$, and we let $B\circ C$ be the Hadamard product of B and C.

In recent years, many authors (see [2–6]) were concerned with r-circulant matrices associated with a number sequence. References [2–4] calculate and estimate the Frobenius norm and the spectral norm of a circulant matrix where the elements of the r-circulant matrix are Fibonacci numbers and Lucas numbers; the authors found more accurate results of the upper bound estimated, and the numerical examples also have provided further proof.

Theorem 1.3

(see [2])

Let $A=\operatorname{Circ}(F_{0}, F_{1}, \ldots, F_{n-1})$ be a circulant matrix, then we have

$$\|A\|_{2}\leq F_{n}F_{n-1}, $$

where $\|\cdot\|_{2}$ is the spectral norm and $F_{n}$ denotes the nth Fibonacci number.

Theorem 1.4

(see [3])

Let $A=\operatorname{Circ}(L_{0}, L_{1}, \ldots, L_{n-1})$ be a circulant matrix, then we have

$${{\|A\|_{2}\leq \begin{cases} \sqrt{[F_{n}F_{n-1}+4F_{n-1}^{2}+4F_{n-1}F_{n-2}+4]\times [F_{n}F_{n-1}+4F_{n-1}^{2}+4F_{n-1}F_{n-2}+4]}, & n\textit{ odd},\\ \sqrt{[F_{n}F_{n-1}+4F_{n-1}^{2}+4F_{n-1}F_{n-2}]\times [F_{n}F_{n-1}+4F_{n-1}^{2}+4F_{n-1}F_{n-2}-3]}, & n\textit{ even}, \end{cases}}} $$

where $\|\cdot\|_{2}$ is the spectral norm, and $L_{n}$ and $F_{n}$ denote the nth Lucas and Fibonacci numbers, respectively.

Theorem 1.5

(see [4])

Let $A=\operatorname{Circ}_{r}(F_{0}, F_{1}, \ldots, F_{n-1})$ be a r-circulant matrix, in which $|r|\geq1$, and then

$$\|A\|_{2}\leq|r|F_{n}F_{n-1}, $$

where $r\in\mathbb{C}$, $\|\cdot\|_{2}$ is the spectral norm and $F_{n}$ denotes the nth Fibonacci number.

Theorem 1.6

(see [4])

Let $A=\operatorname{Circ}_{r}(L_{0}, L_{1}, \ldots, L_{n-1})$ be a r-circulant matrix and $|r|\geq1$, then we obtain

$$\|A\|_{2}\leq \begin{cases} \sqrt{(5|r|^{2}F_{n}F_{n-1}+4)(5F_{n}F_{n-1}+1)}, & n\textit{ odd},\\ \sqrt{[5|r|^{2}F_{n}F_{n-1}+4(1-|r|^{2})](5F_{n}F_{n-1}-3)}, & n\textit{ even}, \end{cases} $$

where $r\in\mathbb{C}$, $\|\cdot\|_{2}$ is the spectral norm, and $L_{n}$ and $F_{n}$ denote the nth Lucas and Fibonacci numbers, respectively.

2 Main results

Theorem 2.1

Let $A=\operatorname{Circ}(F_{0}, F_{1}, \ldots, F_{n-1})$ be a circulant matrix, then we have

$$\|A\|_{2}\leq\sqrt{(n-1)F_{n}F_{n-1}}, $$

where $\|\cdot\|_{2}$ is the spectral norm and $F_{n}$ denotes the nth Fibonacci number.

Proof

Since $A=\operatorname{Circ}(F_{0}, F_{1}, \ldots, F_{n-1})$ is a circulant matrix, let the matrices B and C be

$$B= \begin{pmatrix} F_{0} & 1 & \cdots& 1\\ 1 & F_{0} & \cdots& 1\\ \cdots& \cdots& \cdots& \cdots\\ 1 & 1 & \cdots& F_{0} \end{pmatrix},\qquad C= \begin{pmatrix} F_{0} & F_{1} & \cdots& F_{n-2} & F_{n-1}\\ F_{n-1} & F_{0} & \cdots& F_{n-3} & F_{n-2}\\ \cdots& \cdots& \cdots& \cdots& \cdots\\ F_{2} & F_{3} & \cdots& F_{0} & F_{1}\\ F_{1} & F_{2} & \cdots& F_{n-1} & F_{0} \end{pmatrix} , $$

we get $A=B\circ C$.

For

$$r_{1}(B)=\max_{i}\sqrt{\sum _{j}|b_{ij}|^{2}}=\sqrt{n-1} $$

and

$$c_{1}(C)=\max_{j}\sqrt{\sum _{i}|c_{ij}|^{2}}=\max _{j}\sqrt{\sum_{i=1}^{n}|c_{in}|^{2}}= \sqrt{\sum_{s=0}^{n-1}F_{s}^{2}}= \sqrt{F_{n}F_{n-1}}. $$

From (4), we have

$$\|A\|_{2}\leq\sqrt{(n-1)F_{n}F_{n-1}}. $$

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Corollary 2.2

Let $A=\operatorname{SCirc}(F_{0}, F_{1}, \ldots, F_{n-1})$ be a symmetric circulant matrix, then we have

$$\|A\|_{2}\leq\sqrt{(n-1)F_{n}F_{n-1}}, $$

where $\|\cdot\|_{2}$ is the spectral norm and $F_{n}$ denotes the nth Fibonacci number.

Corollary 2.3

Let $A=\operatorname{Circ}(F_{0}^{2}, F_{1}^{2}, \ldots, F_{n-1}^{2})$ be a circulant matrix, then we have

$$\|A\|_{2}\leq(n-1)F_{n}F_{n-1}, $$

where $\|\cdot\|_{2}$ is the spectral norm and $F_{n}$ denotes the nth Fibonacci number.

Proof

Since $A=\operatorname{Circ}(F_{0}^{2}, F_{1}^{2}, \ldots, F_{n-1}^{2})$ is a circulant matrix, if the matrices $B=\operatorname{Circ}(F_{0}, F_{1}, \ldots, F_{n-1})$, we get $A=B\circ B$; thus from (5) and Theorem 2.1 we obtain

$$\|A\|_{2}\leq(n-1)F_{n}F_{n-1}. $$

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Theorem 2.4

Let $A=\operatorname{Circ}(L_{0}, L_{1}, \ldots, L_{n-1})$ be a circulant matrix, then we have

$$\|A\|_{2}\leq \begin{cases} \sqrt{5nF_{n}F_{n-1}+4n}, & n\textit{ odd},\\ \sqrt{5nF_{n}F_{n-1}}, & n\textit{ even}, \end{cases} $$

where $\|\cdot\|_{2}$ is the spectral norm and $L_{n}$ denotes the Lucas number.

Proof

Since $A=\operatorname{Circ}(L_{0}, L_{1}, \ldots, L_{n-1})$ is a circulant matrix, let the following matrices be defined:

$$B= \begin{pmatrix} 1 & 1 & \cdots& 1\\ 1 & 1 & \cdots& 1\\ \cdots& \cdots& \cdots& \cdots\\ 1 & 1 & \cdots& 1 \end{pmatrix} ,\qquad C= \begin{pmatrix} L_{0} & L_{1} & \cdots& L_{n-2} & L_{n-1}\\ L_{n-1} & L_{0} & \cdots& L_{n-3} & L_{n-2}\\ \cdots& \cdots& \cdots& \cdots& \cdots\\ L_{2} & L_{3} & \cdots& L_{0} & L_{1}\\ L_{1} & L_{2} & \cdots& L_{n-1} & L_{0} \end{pmatrix}, $$

then $A=B\circ C$.

We have

$$r_{1}(B)=\max_{i}\sqrt{\sum _{j}|b_{ij}|^{2}}=\sqrt{n} $$

and

$$c_{1}(C)=\max_{j}\sqrt{\sum _{i}|c_{ij}|^{2}}=\sqrt {\sum _{i=1}^{n}|c_{in}|^{2}}= \sqrt{\sum_{s=0}^{n-1}L_{s}^{2}}= \sqrt {\sum_{s=0}^{n-1}(F_{s}+2F_{s-1})^{2}}. $$

Here

$$\sum_{s=0}^{n-1}F_{s}^{2}=F_{n}F_{n-1}, \qquad \sum_{s=0}^{n-1}F_{s}F_{s-1}= \begin{cases} F_{n-1}^{2} ,& n\mbox{ odd},\\ F_{n-1}^{2}-1, & n\mbox{ even}, \end{cases}\qquad \sum_{s=0}^{n-1}F_{s-1}^{2}=F_{n-1}F_{n-2}+1, $$

thus

$$c_{1}(C)= \begin{cases} \sqrt{5F_{n}F_{n-1}+4},& n\mbox{ odd},\\ \sqrt{5F_{n}F_{n-1}}, & n\mbox{ even}, \end{cases} $$

and from (4) we obtain

$$\|A\|_{2}\leq \begin{cases} \sqrt{5nF_{n}F_{n-1}+4n}, & n\mbox{ odd},\\ \sqrt{5nF_{n}F_{n-1}}, & n\mbox{ even}. \end{cases} $$

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Corollary 2.5

Let $A=\operatorname{SCirc}(L_{0}, L_{1}, \ldots, L_{n-1})$ be a symmetric circulant matrix, then we have

$$\|A\|_{2}\leq \begin{cases} \sqrt{5nF_{n}F_{n-1}+4n}, & n\textit{ odd},\\ \sqrt{5nF_{n}F_{n-1}}, & n\textit{ even}, \end{cases} $$

where $\|\cdot\|_{2}$ is the spectral norm, and $L_{n}$ and $F_{n}$ denote the nth Lucas and Fibonacci numbers, respectively.

Corollary 2.6

Let $A=\operatorname{Circ}(L_{0}^{2}, L_{1}^{2}, \ldots, L_{n-1}^{2})$ be circulant matrices, then

$$\|A\|_{2}\leq \begin{cases} 5nF_{n}F_{n-1}+4n, & n\textit{ odd},\\ 5nF_{n}F_{n-1}, & n\textit{ even}, \end{cases} $$

where $\|\cdot\|_{2}$ is the spectral norm, and $L_{n}$ and $F_{n}$ denote the nth Lucas and Fibonacci numbers, respectively.

Proof

Since $A=\operatorname{Circ}(L_{0}^{2}, L_{1}^{2}, \ldots, L_{n-1}^{2})$ is a circulant matrix, if the matrices $B=\operatorname{Circ}(L_{0}, L_{1}, \ldots, L_{n-1})$, we get $A=B\circ B$; thus from (5) and Theorem 2.4, we obtain

$$\|A\|_{2}\leq \begin{cases} 5nF_{n}F_{n-1}+4n, & n\mbox{ odd},\\ 5nF_{n}F_{n-1}, & n\mbox{ even}. \end{cases} $$

 □

Corollary 2.7

Let $A=\operatorname{Circ}(F_{0}L_{0}, F_{1}L_{1}, \ldots, F_{n-1}L_{n-1})$ be circulant matrices, then

$$\|A\|_{2}\leq \begin{cases} \sqrt{(n-1)nF_{n}F_{n-1}(5F_{n}F_{n-1}+4)}, & n\textit{ odd},\\ \sqrt{5(n-1)n}F_{n}F_{n-1}, & n\textit{ even}, \end{cases} $$

where $\|\cdot\|_{2}$ is the spectral norm, and $L_{n}$ and $F_{n}$ denote the nth Lucas and Fibonacci numbers, respectively.

Proof

Since $A=\operatorname{Circ}(F_{0}L_{0}, F_{1}L_{1}, \ldots, F_{n-1}L_{n-1})$ is a circulant matrix, if the matrices $B=\operatorname{Circ}(F_{0}, F_{1}, \ldots, F_{n-1})$ and $C=\operatorname{Circ}(L_{0}, L_{1}, \ldots, L_{n-1})$, we get $A=B\circ C$; thus from (5), Theorems 2.1, and 2.4, we obtain

$$\|A\|_{2}\leq \begin{cases} \sqrt{(n-1)nF_{n}F_{n-1}(5F_{n}F_{n-1}+4)}, & n\mbox{ odd},\\ \sqrt{5(n-1)n}F_{n}F_{n-1}, & n\mbox{ even}. \end{cases} $$

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Theorem 2.8

Let $A=\operatorname{Circ}_{r}(F_{0}, F_{1}, \ldots, F_{n-1})$ be a r-circulant matrix, in which $|r|\geq1$, and then

$$\|A\|_{2}\leq\sqrt{(n-1)|r|^{2}F_{n}F_{n-1}}, $$

where $r\in\mathbb{C}$, $\|\cdot\|_{2}$ is the spectral norm and $F_{n}$ denotes the nth Fibonacci number.

Proof

Since $A=\operatorname{Circ}_{r}(F_{0}, F_{1}, \ldots, F_{n-1})$ is a r-circulant matrix, let B and C, respectively, be

$$B= \begin{pmatrix} F_{0} & 1 & 1 & \cdots& 1\\ r & F_{0} & 1 & \cdots& 1\\ r & r & F_{0} & \cdots& 1\\ \cdots& \cdots& \cdots& \cdots& \cdots\\ r & r & r & \cdots& F_{0} \end{pmatrix} ,\qquad C= \begin{pmatrix} F_{0} & F_{1} & F_{2} & \cdots& F_{n-1}\\ F_{n-1} & F_{0} & F_{1} & \cdots& F_{n-2}\\ F_{n-2} & F_{n-1} & F_{0} & \cdots& F_{n-3}\\ \cdots& \cdots& \cdots& \cdots& \cdots\\ F_{1} & F_{2} & F_{3} & \cdots& F_{0} \end{pmatrix}, $$

then $A=B\circ C$.

For

$$r_{1}(B)=\max_{i}\sqrt{\sum _{j}|b_{ij}|^{2}}=\sqrt{(n-1)|r|^{2}} $$

and

$$c_{1}(C)=\max_{j}\sqrt{\sum _{i}|c_{ij}|^{2}}=\sqrt {\sum _{i=1}^{n}|c_{in}|^{2}}= \sqrt{\sum_{s=0}^{n-1}F_{s}^{2}}= \sqrt {F_{n}F_{n-1}}, $$

from (4), we have

$$\|A\|_{2}\leq\sqrt{(n-1)|r|^{2}F_{n}F_{n-1}}. $$

 □

Corollary 2.9

Let $A=\operatorname{SCirc}_{r}(F_{0}, F_{1}, \ldots, F_{n-1})$ be a symmetric r-circulant matrix, in which $|r|\geq1$, and then

$$\|A\|_{2}\leq\sqrt{(n-1)|r|^{2}F_{n}F_{n-1}}, $$

where $r\in\mathbb{C}$, $\|\cdot\|_{2}$ is the spectral norm and $F_{n}$ denotes the nth Fibonacci number.

Corollary 2.10

Let $A=\operatorname{Circ}_{r}(F_{0}^{2}, F_{1}^{2}, \ldots, F_{n-1}^{2})$ be a r-circulant matrix, while $|r|\geq1$, then we obtain

$$\|A\|_{2}\leq(n-1)|r|F_{n}F_{n-1}, $$

where $r\in\mathbb{C}$, $\|\cdot\|_{2}$ is the spectral norm and $F_{n}$ denotes the Fibonacci number.

Proof

Since $A=\operatorname{Circ}_{r}(F_{0}^{2}, F_{1}^{2}, \ldots, F_{n-1}^{2})$ is a r-circulant matrix, if the matrices $B=\operatorname{Circ}_{r}(F_{0}, F_{1}, \ldots, F_{n-1})$ and $C=\operatorname{Circ}(F_{0}, F_{1}, \ldots, F_{n-1})$, we get $A=B\circ C$; thus from (5), Theorems 2.1, and 2.8, we obtain

$$\|A\|_{2}\leq(n-1)|r|F_{n}F_{n-1}. $$

 □

Corollary 2.11

Let $A=\operatorname{Circ}_{r}(F_{0}L_{0}, F_{1}L_{1}, \ldots, F_{n-1}L_{n-1})$ be a r-circulant matrix, while $|r|\geq1$, then we obtain

$$\|A\|_{2} \leq \begin{cases} \sqrt{(n-1)n|r|^{2}F_{n}F_{n-1}(5F_{n}F_{n-1}+4)}, & n\textit{ odd},\\ F_{n}F_{n-1} \sqrt{5|r|^{2}(n-1)n}, & n\textit{ even}, \end{cases} $$

where $r\in\mathbb{C}$, $\|\cdot\|_{2}$ is the spectral norm, and $L_{n}$ and $F_{n}$ denote the nth Lucas and Fibonacci numbers, respectively.

Proof

Since $A=\operatorname{Circ}_{r}(F_{0}L_{0}, F_{1}L_{1}, \ldots, F_{n-1}L_{n-1})$ is a r-circulant matrix, if the matrices $B=\operatorname{Circ}_{r}(F_{0}, F_{1}, \ldots, F_{n-1})$ and $C=\operatorname{Circ}(L_{0}, L_{1}, \ldots, L_{n-1})$, we get $A=B\circ C$; thus from (5), Theorems 2.4, and 2.8, we obtain

$$\|A\|_{2}\leq \begin{cases} \sqrt{(n-1)n|r|^{2}F_{n}F_{n-1}(5F_{n}F_{n-1}+4)}, & n\mbox{ odd},\\ F_{n}F_{n-1} \sqrt{5|r|^{2}(n-1)n},& n\mbox{ even}. \end{cases} $$

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Theorem 2.12

Let $A=\operatorname{Circ}_{r}(L_{0}, L_{1}, \ldots, L_{n-1})$ be a r-circulant matrix and $|r|\geq1$, then we obtain

$$\|A\|_{2}\leq \begin{cases} \sqrt{(n-1)|r|^{2}+1}\times\sqrt{5F_{n}F_{n-1}+4}, & n\textit{ odd},\\ \sqrt{(n-1)|r|^{2}+1}\times\sqrt{5F_{n}F_{n-1}}, & n\textit{ even}, \end{cases} $$

where $r\in\mathbb{C}$, $\|\cdot\|_{2}$ is the spectral norm, and $L_{n}$ and $F_{n}$ denote the nth Lucas and Fibonacci numbers, respectively.

Proof

Since $A=\operatorname{Circ}_{r}(L_{0}, L_{1}, \ldots, L_{n-1})$ is a r-circulant matrix, let B and C, respectively, be

$$B= \begin{pmatrix} 1 & 1 & 1 & \cdots& 1\\ r & 1 & 1 & \cdots& 1\\ r & r & 1 & \cdots& 1\\ \cdots& \cdots& \cdots& \cdots& \cdots\\ r & r & r & \cdots& 1 \end{pmatrix} ,\qquad C= \begin{pmatrix} L_{0} & L_{1} & \cdots& L_{n-2}& L_{n-1}\\ L_{n-1} & L_{0} & \cdots& L_{n-3} & L_{n-2}\\ \cdots& \cdots& \cdots& \cdots& \cdots\\ L_{2} & L_{3} & \cdots& L_{0} & L_{1}\\ L_{1} & L_{2} & \cdots& L_{n-1} & L_{0} \end{pmatrix} , $$

and then $A=B\circ C$.

We have

$$r_{1}(B)=\max_{i}\sqrt{\sum _{j}|b_{ij}|^{2}}=\sqrt {(n-1)|r|^{2}+1} $$

and

$$c_{1}(C)=\max_{j}\sqrt{\sum _{i}|c_{ij}|^{2}}=\sqrt {\sum _{i=1}^{n}|c_{in}|^{2}}= \sqrt{\sum_{s=0}^{n-1}L_{s}^{2}}= \sqrt {\sum_{s=0}^{n-1}(F_{s}+2F_{s-1})^{2}}, $$

in which

$$\sum_{s=0}^{n-1}F_{s}^{2}=F_{n}F_{n-1}, \qquad \sum_{s=0}^{n-1}F_{s-1}F_{s}= \begin{cases} F_{n-1}^{2}, & n\mbox{ odd},\\ F_{n-1}^{2}-1, & n\mbox{ even}, \end{cases}\qquad \sum_{s=0}^{n-1}F_{s-1}^{2}=F_{n-1}F_{n-2}+1, $$

and we get

$$c_{1}(C)= \begin{cases} \sqrt{5F_{n}F_{n-1}+4}, & n\mbox{ odd},\\ \sqrt{5F_{n}F_{n-1}}, & n\mbox{ even}. \end{cases} $$

From (4), we further infer

$$\|A\|_{2}\leq \begin{cases} \sqrt{(n-1)|r|^{2}+1}\times\sqrt{5F_{n}F_{n-1}+4}, & n\mbox{ odd},\\ \sqrt{(n-1)|r|^{2}+1}\times\sqrt{5F_{n}F_{n-1}}, & n\mbox{ even}. \end{cases} $$

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Corollary 2.13

Let $A=\operatorname{SCirc}_{r}(L_{0}, L_{1}, \ldots, L_{n-1})$ be a symmetric r-circulant matrix and $|r|\geq1$, then we obtain

$$\|A\|_{2}\leq \begin{cases} \sqrt{(n-1)|r|^{2}+1}\times\sqrt{5F_{n}F_{n-1}+4}, & n\textit{ odd},\\ \sqrt{(n-1)|r|^{2}+1}\times\sqrt{5F_{n}F_{n-1}}, & n\textit{ even}, \end{cases} $$

where $r\in\mathbb{C}$, $\|\cdot\|_{2}$ is the spectral norm, and $L_{n}$ and $F_{n}$ denote the nth Lucas and Fibonacci numbers, respectively.

Corollary 2.14

Let $A=\operatorname{Circ}_{r}(L_{0}^{2}, L_{1}^{2}, \ldots, L_{n-1}^{2})$ be a r-circulant matrix and $|r|\geq1$, then

$$\|A\|_{2}\leq \begin{cases} (5F_{n}F_{n-1}+4)\sqrt{n[(n-1)|r|^{2}+1]}, & n\textit{ odd},\\ 5F_{n}F_{n-1}\sqrt{n[(n-1)|r|^{2}+1]}, & n\textit{ even}, \end{cases} $$

where $r\in\mathbb{C}$, $\|\cdot\|_{2}$ is the spectral norm, and $L_{n}$ and $F_{n}$ denote the nth Lucas and Fibonacci numbers, respectively.

Proof

Since $A=\operatorname{Circ}_{r}(L_{0}^{2}, L_{1}^{2}, \ldots, L_{n-1}^{2})$ is a r-circulant matrix, if the matrices $B=\operatorname{Circ}(L_{0}, L_{1}, \ldots, L_{n-1})$ and $C=\operatorname{Circ}_{r}(L_{0}, L_{1}, \ldots, L_{n-1})$, we get $A=B\circ C$; thus from (5), Theorems 2.4, and 2.12, we obtain

$$\|A\|_{2}\leq \begin{cases} (5F_{n}F_{n-1}+4)\sqrt{n[(n-1)|r|^{2}+1]}, & n\mbox{ odd},\\ 5F_{n}F_{n-1}\sqrt{n[(n-1)|r|^{2}+1]}, & n\mbox{ even}. \end{cases} $$

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3 Examples

Example 1

Let $A=\operatorname{Circ}(F_{0}, F_{1}, \ldots, F_{n-1})$ be a circulant matrix, in which $F_{i}$ ($i=0,1,\ldots,n-1$) denotes the Fibonacci number.

From Table 1, it is easy to find that the upper bounds for the spectral norm, of Theorem 2.1 are more accurate than Theorem 1.3 when $n\geq4$.

Table 1 Numerical results of $\pmb{a_{i}=F_{i}}$ , $\pmb{r=1}$

Example 2

Let $A=\operatorname{Circ}(L_{0}, L_{1}, \ldots, L_{n-1})$ be a circulant matrix, where $L_{i}$ ($i=0,1,\ldots,n-1$) denotes the Lucas sequence.

Let $n\geq3$, and it is easy to find that the upper bounds for the spectral norm of Theorem 2.4 are more accurate than Theorem 1.4 (see Table 2).

Table 2 Numerical results of $\pmb{a_{i}=L_{i}}$ , $\pmb{r=1}$

Example 3

Let $A=\operatorname{Circ}_{2}(F_{0}, F_{1}, \ldots, F_{n-1})$ be a 2-circulant matrix, in which $F_{i}$ ($i=0,1,\ldots,n-1$) denotes the Fibonacci number.

Let $n\geq4$, and it is easy to find that the upper bounds for the spectral norm of Theorem 2.8 are more precise than Theorem 1.5 (see Table 3).

Table 3 Numerical results of $\pmb{a_{i}=F_{i}}$ , $\pmb{r=2}$

Example 4

Let $A=\operatorname{Circ}_{2}(L_{0}, L_{1}, \ldots, L_{n-1})$ be a 2-circulant matrix where $L_{i}$ ($i=0,1,\ldots, n-1$) denotes the Lucas sequence.

It can be seen from Table 4 that the upper bounds for the spectral norm of Theorem 2.12 are more precise than Theorem 1.6 when $n\geq3$.

Table 4 Numerical results of $\pmb{a_{i}=L_{i}}$ , $\pmb{r=2}$