Open Access

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

  • Chengyuan He1Email author,
  • Jiangming Ma,
  • Kunpeng Zhang and
  • Zhenghua Wang
Journal of Inequalities and Applications20152015:72

https://doi.org/10.1186/s13660-015-0596-5

Received: 25 September 2014

Accepted: 12 February 2015

Published: 28 February 2015

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.

Keywords

r-circulant matrices Fibonacci number Lucas number spectral norm estimation

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 [26]) were concerned with r-circulant matrices associated with a number sequence. References [24] 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
$$\|A\|_{2}\leq\sqrt{(n-1)F_{n}F_{n-1}}. $$
 □

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

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

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

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

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

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

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 \(\pmb{a_{i}=F_{i}}\) , \(\pmb{r=1}\)

n

Theorem  2.1

Theorem  1.3

\(\frac{\mathbf{Third\ column}}{\mathbf{Second\ column}}\)

2

1

1

\(\frac{1}{1}=1\)

3

2

2

\(\frac{2}{2}=1\)

4

\(3\sqrt{2}\)

6

\(\frac{6}{3\sqrt{2}}=\sqrt{2}\)

5

\(\sqrt{60}\)

15

\(\frac{15}{\sqrt{60}}\approx1.936\)

6

\(\sqrt{200}\)

40

\(\frac{40}{\sqrt{200}}=2\sqrt{2}\)

n

\(\sqrt{(n-1)F_{n}F_{n-1}}\)

\(F_{n}F_{n-1}\)

\(\frac {F_{n}F_{n-1}}{\sqrt{(n-1)F_{n}F_{n-1}}}=\sqrt{\frac {F_{n}F_{n-1}}{n-1}}\)

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 \(\pmb{a_{i}=L_{i}}\) , \(\pmb{r=1}\)

n

Theorem  2.4

Theorem  1.4

\(\frac{\mathbf{Third\ column}}{\mathbf{Second\ column}}\)

1

2

2

\(\frac{2}{2}=1\)

2

\(\sqrt{10}\)

\(\sqrt{10}\)

\(\frac{\sqrt{10}}{\sqrt{10}}=1\)

3

\(\sqrt{42}\)

\(\sqrt{154}\)

\(\frac{\sqrt{154}}{\sqrt {42}}\approx 1.915\)

4

\(\sqrt{120}\)

\(\sqrt{810}\)

\(\frac{\sqrt{810}}{\sqrt {120}}\approx 2.598\)

5

\(\sqrt{395}\)

\(\sqrt{6{,}004}\)

\(\frac{\sqrt{6{,}004}}{\sqrt {395}}\approx3.899\)

6

\(\sqrt{1{,}200}\)

\(\sqrt{39{,}400}\)

\(\frac{\sqrt{39{,}400}}{\sqrt {1{,}200}}\approx5.730\)

n

  

\(\sqrt{n^{-1}(5F_{n}F_{n-1}+1)}\)n odd,

\(\sqrt{n^{-1}(5F_{n}F_{n-1}-3)}\)n even

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 \(\pmb{a_{i}=F_{i}}\) , \(\pmb{r=2}\)

n

Theorem  2.8

Theorem  1.5

\(\frac{\mathbf{Third\ column}}{\mathbf{Second\ column}}\)

2

2

2

\(\frac{2}{2}=1\)

3

4

4

\(\frac{4}{4}=1\)

4

\(6\sqrt{2}\)

12

\(\frac{12}{6\sqrt{2}}=\sqrt{2}\)

5

\(4\sqrt{15}\)

\(\sqrt{30}\)

\(\frac{30}{4\sqrt{15}}\approx 1.936\)

6

\(20\sqrt{2}\)

80

\(\frac{80}{20\sqrt{2}}=2\sqrt{2}\)

n

\(\sqrt{(n-1)|r|^{2}F_{n}F_{n-1}}\)

\(|r|F_{n}F_{n-1}\)

\(\sqrt {(n-1)^{-1}F_{n}F_{n-1}}\)

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 \(\pmb{a_{i}=L_{i}}\) , \(\pmb{r=2}\)

n

Theorem  2.12

Theorem  1.6

\(\frac{\mathbf{Third\ column}}{\mathbf{Second\ column}}\)

1

2

2

\(\frac{2}{2}=1\)

2

5

4

\(\frac{4}{5}=\frac{4}{5}\)

3

\(3\sqrt{14}\)

\(2\sqrt{231}\)

\(\frac{2\sqrt{231}}{3\sqrt {14}}\approx2.708\)

4

\(\sqrt{390}\)

54

\(\frac{54}{\sqrt{390}}\approx2.734\)

5

\(\sqrt{1{,}343}\)

152

\(\frac{152}{\sqrt{1{,}343}}\approx4.418\)

6

\(10\sqrt{42}\)

394

\(\frac{394}{10\sqrt{342}}\approx6.080\)

Declarations

Acknowledgements

Project supported by Applied Fundamental Research Plan of Sichuan Province (No. 2013JY0178).

Open Access This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/4.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly credited.

Authors’ Affiliations

(1)
School of Mathematics and Computer Engineering, Xihua University Chengdu

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© He et al.; licensee Springer. 2015