On the spectral norm of r -circulant matrices with the Pell and Pell-Lucas numbers

Ramazan Türkmen; Hasan Gökbaş

doi:10.1186/s13660-016-0997-0

Research
Open Access

On the spectral norm of r-circulant matrices with the Pell and Pell-Lucas numbers

Ramazan Türkmen¹Email author and
Hasan Gökbaş²

Journal of Inequalities and Applications20162016:65

https://doi.org/10.1186/s13660-016-0997-0

Received: 29 September 2015
Accepted: 1 February 2016
Published: 16 February 2016

Abstract

Let us define $A=C_{r}(a_{0},a_{1},\ldots,a_{n-1})$ to be a $n \times n$ r-circulant matrix. The entries in the first row of $A=C_{r}(a_{0},a_{1},\ldots,a_{n-1})$ are $a_{i}=P_{i}$, $a_{i}=Q_{i}$, $a_{i}=P_{i}^{2}$ or $a_{i}=Q_{i}^{2}$ ($i=0, 1, 2, \ldots, n-1$), where $P_{i}$ and $Q_{i}$ are the ith Pell and Pell-Lucas numbers, respectively. We find some bounds estimation of the spectral norm for r-Circulant matrices with Pell and Pell-Lucas numbers.

Keywords

Pell numbers
Pell-Lucas numbers
r-circulant matrix
spectral norm

1 Introduction

Special matrices is a widely studied subject in matrix analysis. Especially special matrices whose entries are well-known number sequences have become a very interesting research subject in recent years and many authors have obtained some good results in this area. For example, Bahşi and Solak have studied the norms of r-circulant matrices with the hyper-Fibonacci and Lucas numbers [1], Bozkurt and Tam have obtained some results belong to determinants and inverses of r-circulant matrices associated with a number sequence [2], Shen and Cen have made a similar study by using r-circulant matrices with the Fibonacci and Lucas numbers [3, 4] and He et al. have established on the spectral norm inequalities on r-circulant matrices with Fibonacci and Lucas numbers [5].

Lots of article have been written so far, which concern estimates for spectral norms of circulant and r-circulant matrices, which have connections with signal and image processing, time series analysis and many other problems.

In this paper, we derive expressions of spectral norms for r-circulant matrices. We explain some preliminaries and well-known results. We thicken the identities of estimations for spectral norms of r-circulant matrices with the Pell and Pell-Lucas numbers.

The Pell and Pell-Lucas sequences $P_{n}$ and $Q_{n}$ are defined by the recurrence relations

$$ P_{0}=0,\qquad P_{1}=1,\qquad P_{n}=2P_{n-1}+P_{n-2} \quad \mbox{for } n \geq2 $$

and

$$ Q_{0}=2,\qquad Q_{1}=2,\qquad Q_{n}=2Q_{n-1}+Q_{n-2} \quad \mbox{for } n \geq2. $$

If we start from $n=0$, then the Pell and Pell-Lucas sequence are given by

$$\textstyle\begin{array}{@{}l@{\qquad}l@{\quad}l@{\quad}l@{\quad}l@{\quad}l@{\quad}l@{\quad}l@{\quad}l@{\quad}l@{}} n:&0&1&2&3&4&5&6&7&\cdots\\ P_{n}:&0&1&2&5&12&29&70&169&\cdots\\ Q_{n}:&2&2&6&14&34&82&198&478&\cdots \end{array} $$

The following sum formulas for the Pell and Pell-Lucas numbers are well known [6, 7]:

$$ \sum_{k=1}^{n}P_{k}^{2}= \frac{P_{n}P_{n+1}}{2} $$

and

$$ \sum_{k=1}^{n}Q_{k}^{2}= \frac{Q_{2n+1}+2(-1)^{n}-4}{2}. $$

A matrix $C= [c_{ij} ] \in M_{n,n} (\mathbb{C} )$ is called a r-circulant matrix if it is of the form

$$ c_{ij}=\left \{ \textstyle\begin{array}{@{}l@{\quad}l} c_{j-i},&j \geq i,\\ rc_{n+j-i},&j< i. \end{array}\displaystyle \right . $$

Obviously, the r-circulant matrix C is determined by the parameter r and its first row elements $c_{0}, c_{1}, \ldots, c_{n-1}$, thus we denote $C=C_{r} (c_{0}, c_{1}, \ldots, c_{n-1} )$. Especially, let $r=1$, the matrix C is called a circulant matrix [3].

The Euclidean norm of the matrix A is defined as

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

The singular values of the matrix A are

$$ \sigma_{i} =\sqrt{\lambda_{i} \bigl(A^{*}A} \bigr), $$

where $\lambda_{i}$ is an eigenvalue of $A^{*}A$ and $A^{*}$ is conjugate transpose of matrix A. The square roots of the maximum eigenvalues of $A^{*}A$ are called the spectral norm of A and are induced by $\Vert A\Vert _{2}$.

The following inequality holds:

$$ \frac{1}{\sqrt{n}} \Vert A\Vert _{E}\leq \Vert A\Vert _{2}\leq \Vert A\Vert _{E}. $$

Define the maximum column length norm $c_{1}$, and the maximum row length norm $r_{1}$ of any matrix A by

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

and

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

respectively. Let A, B, and C be $m\times n$ matrices. If $A=B\circ C$ then

$$ \Vert A\Vert _{2}\leq r_{1} (B )c_{1} (C )\quad \mbox{[8]}. $$

2 Result and discussion

Theorem 1

Let $A=C_{r}(P_{0},P_{1},\ldots,P_{n-1})$ be a r-circulant matrix, where $r \in\mathbb{C}$. We have

$$\begin{aligned}& \hphantom{i}(\mathrm{i})\quad \vert r \vert \geq1,\quad \sqrt{ \frac{P_{n}P_{n-1}}{2}}\leq \Vert A \Vert _{2} \leq \vert r \vert \sqrt{(n-1)\frac {P_{n}P_{n-1}}{2}}, \\& (\mathrm{ii})\quad \vert r \vert < 1, \quad \vert r \vert \sqrt{ \frac {P_{n}P_{n-1}}{2}}\leq \Vert A \Vert _{2} \leq\sqrt{(n-1) \frac {P_{n}P_{n-1}}{2}}. \end{aligned}$$

Proof

The matrix A is of the form

$$A= \begin{bmatrix} P_{0}&P_{1}&\ldots&P_{n-2}&P_{n-1}\\ rP_{n-1}&P_{0}&\ldots&P_{n-3}&P_{n-2}\\ \vdots&\vdots&\ddots&\vdots&\vdots\\ rP_{2}&rP_{3}&\ldots&P_{0}&P_{1}\\ rP_{1}&rP_{2}&\ldots&rP_{n-1}&P_{0} \end{bmatrix}. $$

Then we have

$$\Vert A \Vert _{E}^{2}=\sum _{i=0}^{n-1}(n-i)P_{i}^{2}+\sum _{i=1}^{n-1}i\vert r \vert ^{2}P_{i}^{2}; $$

hence, when $\vert r \vert \geq1 $ we obtain

$$\Vert A \Vert _{E}^{2} \geq\sum _{i=0}^{n-1}(n-i)P_{i}^{2}+\sum _{i=1}^{n-1}iP_{i}^{2}=n \sum_{i=0}^{n-1}P_{i}^{2}=n \frac {P_{n}P_{n-1}}{2}, $$

that is,

$$ \frac{1}{\sqrt{n}} \Vert A \Vert _{E} \geq\sqrt{\frac {P_{n}P_{n-1}}{2}} \quad\Rightarrow\quad \Vert A \Vert _{2} \geq\sqrt{\frac {P_{n}P_{n-1}}{2}}. $$

On the other hand, let the matrices B and C be

$$B= \begin{bmatrix} P_{0}&1&\ldots&1&1\\ r&P_{0}&\ldots&1&1\\ \vdots&\vdots&\ddots&\vdots&\vdots\\ r&r&\ldots&P_{0}&1\\ r&r&\ldots&r&P_{0} \end{bmatrix}\quad \mbox{and} \quad C= \begin{bmatrix} P_{0}&P_{1}&\ldots&P_{n-2}&P_{n-1}\\ P_{n-1}&P_{0}&\ldots&P_{n-3}&P_{n-2}\\ \vdots&\vdots&\ddots&\vdots&\vdots\\ P_{2}&P_{3}&\ldots&P_{0}&P_{1}\\ P_{1}&P_{2}&\ldots&P_{n-1}&P_{0} \end{bmatrix} $$

such that $A=B\circ C$. Then

$$\begin{aligned}& r_{1}(B)=\max_{i}\sqrt{\sum _{j}\vert b_{nj}\vert ^{2}}=\sqrt { \vert r \vert ^{2}(n-1)}=\vert r \vert \sqrt{(n-1)}\quad \mbox{and} \\& c_{1}(C)=\max_{j}\sqrt{\sum _{i}\vert c_{in}\vert ^{2}}=\sqrt {\sum_{i=0}^{n-1}P_{i}^{2}}= \sqrt{\frac{P_{n}P_{n-1}}{2}}. \end{aligned}$$

We have

$$ \Vert A \Vert _{2} \leq \vert r \vert \sqrt{(n-1) \frac {P_{n}P_{n-1}}{2}}. $$

When $\vert r \vert <1 $ we also obtain

$$\Vert A \Vert _{E}^{2} \geq\sum _{i=0}^{n-1}(n-i)\vert r \vert ^{2}P_{i}^{2}+ \sum_{i=1}^{n-1}i\vert r \vert ^{2}P_{i}^{2}=n \vert r \vert ^{2} \frac{P_{n}P_{n-1}}{2}, $$

that is,

$$ \frac{1}{\sqrt{n}} \Vert A \Vert _{E} \geq \vert r \vert \sqrt { \frac{P_{n}P_{n-1}}{2}} \quad\Rightarrow\quad \Vert A \Vert _{2} \geq \vert r \vert \sqrt{\frac{P_{n}P_{n-1}}{2}}. $$

On the other hand, let the matrices B and C be

$$B= \begin{bmatrix} P_{0}&1&\ldots&1&1\\ r&P_{0}&\ldots&1&1\\ \vdots&\vdots&\ddots&\vdots&\vdots\\ r&r&\ldots&P_{0}&1\\ r&r&\ldots&r&P_{0} \end{bmatrix} \quad\mbox{and}\quad C= \begin{bmatrix} P_{0}&P_{1}&\ldots&P_{n-2}&P_{n-1}\\ P_{n-1}&P_{0}&\ldots&P_{n-3}&P_{n-2}\\ \vdots&\vdots&\ddots&\vdots&\vdots\\ P_{2}&P_{3}&\ldots&P_{0}&P_{1}\\ P_{1}&P_{2}&\ldots&P_{n-1}&P_{0} \end{bmatrix} $$

such that $A=B\circ C$. Then

$$\begin{aligned}& r_{1}(B)=\max_{i}\sqrt{\sum _{j}\vert b_{ij}\vert ^{2}}=\sqrt {\sum_{j=0}^{n-1} \vert b_{nj} \vert ^{2}}=\sqrt{n-1}\quad \mbox{and} \\& c_{1}(C)=\max_{j}\sqrt{\sum _{i}\vert c_{ij}\vert ^{2}}=\sqrt {\sum_{i=0}^{n-1} \vert c_{in} \vert ^{2}}=\sqrt{\sum _{i=0}^{n-1}P_{i}^{2}}=\sqrt{ \frac{P_{n}P_{n-1}}{2}}. \end{aligned}$$

We have

$$ \Vert A \Vert _{2} \leq\sqrt{(n-1)\frac{P_{n}P_{n-1}}{2}}. $$

Thus, the proof is completed. □

Corollary 2

Let $A=C_{r}(P_{0}^{2},P_{1}^{2},\ldots,P_{n-1}^{2})$ be a r-circulant matrix, where $r \in\mathbb{C}$, $\vert r \vert \geq1$; we have

$$ \Vert A \Vert _{2} \leq(n-1)\vert r \vert \frac{P_{n}P_{n-1}}{2}, $$

where $\Vert \cdot \Vert _{2}$ is the spectral norm and $P_{n}$ denotes the nth Pell number.

Proof

Since $A=C_{r}(P_{0}^{2},P_{1}^{2},\ldots,P_{n-1}^{2})$ is a r-circulant matrix, if the matrices $B=C_{r}(P_{0},P_{1}, \ldots,P_{n-1})$ and $C=C(P_{0}^{2},P_{1}^{2},\ldots,P_{n-1}^{2})$ we get $A=B\circ C$; thus, we obtain

$$ \Vert A \Vert _{2} \leq(n-1) \vert r \vert \frac{P_{n}P_{n-1}}{2}. $$

□

Theorem 3

Let $A=C_{r}(Q_{0},Q_{1},\ldots,Q_{n-1})$ be a r-circulant matrix, where $r \in\mathbb{C}$.

$$\begin{aligned}& \hphantom{i}(\mathrm{i})\quad \vert r \vert \geq1, \quad \left \{ \textstyle\begin{array}{@{}l@{\quad}l} \sqrt{\frac{Q_{2n-1}+6}{2}}\leq \Vert A \Vert _{2} \leq \vert r \vert \sqrt{n\frac{Q_{2n-1}+6}{2}},& n \textit{ odd},\\ \sqrt{\frac{Q_{2n-1}+2}{2}}\leq \Vert A \Vert _{2} \leq \vert r \vert \sqrt{n\frac{Q_{2n-1}+2}{2}},& n\textit{ even}, \end{array}\displaystyle \right . \\& (\mathrm{ii})\quad \vert r \vert < 1, \quad \left \{ \textstyle\begin{array}{@{}l@{\quad}l} \vert r \vert \sqrt{\frac {Q_{2n-1}+6}{2}}\leq \Vert A \Vert _{2} \leq\sqrt{n\frac {Q_{2n-1}+6}{2}},& n\textit{ odd}, \\ \vert r \vert \sqrt{\frac {Q_{2n-1}+2}{2}}\leq \Vert A \Vert _{2} \leq \sqrt{n\frac {Q_{2n-1}+2}{2}},& n\textit{ even}. \end{array}\displaystyle \right . \end{aligned}$$

Proof

The matrix A is of the form

$$A= \begin{bmatrix} Q_{0}&Q_{1}&\ldots&Q_{n-2}&Q_{n-1}\\ rQ_{n-1}&Q_{0}&\ldots&Q_{n-3}&Q_{n-2}\\ \vdots&\vdots&\ddots&\vdots&\vdots\\ rQ_{2}&rQ_{3}&\ldots&Q_{0}&Q_{1}\\ rQ_{1}&rQ_{2}&\ldots&rQ_{n-1}&Q_{0} \end{bmatrix}. $$

Then we have

$$\Vert A \Vert _{E}^{2}=\sum _{i=0}^{n-1}(n-i)Q_{i}^{2}+\sum _{i=1}^{n-1}i\vert r \vert ^{2}Q_{i}^{2}; $$

hence, when $\vert r \vert \geq1$ we obtain

$$\Vert A \Vert _{E}^{2} \geq\sum _{i=0}^{n-1}(n-i)Q_{i}^{2}+\sum _{i=1}^{n-1}iQ_{i}^{2}=n \sum_{i=0}^{n-1}Q_{i}^{2}= \left \{ \textstyle\begin{array}{@{}l@{\quad}l} \sqrt{n\frac{Q_{2n-1}+6}{2}},& n\mbox{ odd},\\ \sqrt{n\frac{Q_{2n-1}+2}{2}},& n\mbox{ even}, \end{array}\displaystyle \right . $$

that is,

$$ \frac{1}{\sqrt{n}} \Vert A \Vert _{E} \geq \Vert A \Vert _{2} \geq \left \{ \textstyle\begin{array}{@{}l@{\quad}l} \sqrt{\frac{Q_{2n-1}+6}{2}},& n\mbox{ odd},\\ \sqrt{\frac{Q_{2n-1}+2}{2}},& n\mbox{ even}. \end{array}\displaystyle \right . $$

On the other hand, let the matrices B and C be

$$B= \begin{bmatrix} 1&1&\ldots&1&1\\ r&1&\ldots&1&1\\ \vdots&\vdots&\ddots&\vdots&\vdots\\ r&r&\ldots&1&1\\ r&r&\ldots&r&1 \end{bmatrix} \quad\mbox{and}\quad C= \begin{bmatrix} Q_{0}&Q_{1}&\ldots&Q_{n-2}&Q_{n-1}\\ Q_{n-1}&Q_{0}&\ldots&Q_{n-3}&Q_{n-2}\\ \vdots&\vdots&\ddots&\vdots&\vdots\\ Q_{2}&Q_{3}&\ldots&Q_{0}&Q_{1}\\ Q_{1}&Q_{2}&\ldots&Q_{n-1}&Q_{0} \end{bmatrix} $$

such that $A=B\circ C$. Then

$$\begin{aligned}& r_{1}(B)=\max_{i}\sqrt{\sum _{j}\vert b_{ij}\vert ^{2}}=\sqrt {\sum_{j=0}^{n-1} \vert b_{nj} \vert ^{2}}=\sqrt{\vert r \vert ^{2}(n-1)+1}\quad \mbox{and} \\& c_{1}(C)=\max_{j}\sqrt{\sum _{i}\vert c_{ij}\vert ^{2}}=\sqrt {\sum_{i=0}^{n-1} \vert c_{in} \vert ^{2}}=\sqrt{\sum _{i=0}^{n-1}Q_{i}^{2}}= \left \{ \textstyle\begin{array}{@{}l@{\quad}l} \sqrt{\frac{Q_{2n-1}+6}{2}},& n\mbox{ odd},\\ \sqrt{\frac{Q_{2n-1}+2}{2}},& n\mbox{ even}. \end{array}\displaystyle \right . \end{aligned}$$

We have

$$ \Vert A \Vert _{2} \leq \left \{ \textstyle\begin{array}{@{}l@{\quad}l} \sqrt{(\vert r \vert ^{2}(n-1)+1)(\frac {Q_{2n-1}+6}{2})},& n\mbox{ odd},\\ \sqrt{(\vert r \vert ^{2}(n-1)+1)(\frac{Q_{2n-1}+2}{2})},& n\mbox{ even}. \end{array}\displaystyle \right . $$

When $\vert r \vert <1$ we also obtain

$$\Vert A \Vert _{E}^{2} \geq\sum _{i=0}^{n-1}(n-i)\vert r \vert ^{2}Q_{i}^{2}+ \sum_{i=1}^{n-1}i\vert r \vert ^{2}Q_{i}^{2}=\left \{ \textstyle\begin{array}{@{}l@{\quad}l} \vert r \vert \sqrt{n(\frac {Q_{2n-1}+6}{2})},& n\mbox{ odd},\\ \vert r \vert \sqrt {n(\frac{Q_{2n-1}+2}{2})},& n\mbox{ even}, \end{array}\displaystyle \right . $$

that is,

$$ \frac{1}{\sqrt{n}} \Vert A \Vert _{E} \geq \Vert A \Vert _{2} \geq \left \{ \textstyle\begin{array}{@{}l@{\quad}l} \vert r \vert \sqrt{\frac {Q_{2n-1}+6}{2}},& n\mbox{ odd},\\ \vert r \vert \sqrt{\frac {Q_{2n-1}+2}{2}},& n\mbox{ even}. \end{array}\displaystyle \right . $$

On the other hand, let the matrices B and C be

$$B= \begin{bmatrix} 1&1&\ldots&1&1\\ r&1&\ldots&1&1\\ \vdots&\vdots&\ddots&\vdots&\vdots\\ r&r&\ldots&1&1\\ r&r&\ldots&r&1 \end{bmatrix}\quad \mbox{and} \quad C= \begin{bmatrix} Q_{0}&Q_{1}&\ldots&Q_{n-2}&Q_{n-1}\\ Q_{n-1}&Q_{0}&\ldots&Q_{n-3}&Q_{n-2}\\ \vdots&\vdots&\ddots&\vdots&\vdots\\ Q_{2}&Q_{3}&\ldots&Q_{0}&Q_{1}\\ Q_{1}&Q_{2}&\ldots&Q_{n-1}&Q_{0} \end{bmatrix} $$

such that $A=B\circ C$. Then

$$\begin{aligned}& r_{1}(B)=\max_{i}\sqrt{\sum _{j}\vert b_{ij}\vert ^{2}}=\sqrt {\sum_{j=0}^{n-1} \vert b_{nj} \vert ^{2}}=\sqrt{n}\quad \mbox{and} \\& c_{1}(C)=\max_{j}\sqrt{\sum _{i}\vert c_{ij}\vert ^{2}}=\sqrt {\sum_{i=0}^{n-1} \vert c_{in} \vert ^{2}}=\sqrt{\sum _{i=0}^{n-1}Q_{i}^{2}}= \left \{ \textstyle\begin{array}{@{}l@{\quad}l} \sqrt{\frac{Q_{2n-1}+6}{2}},& n\mbox{ odd},\\ \sqrt{\frac{Q_{2n-1}+2}{2}},& n\mbox{ even}. \end{array}\displaystyle \right . \end{aligned}$$

We have

$$ \Vert A \Vert _{2} \leq \left \{ \textstyle\begin{array}{@{}l@{\quad}l} \sqrt{n\frac{Q_{2n-1}+6}{2}},& n\mbox{ odd},\\ \sqrt{n\frac{Q_{2n-1}+2}{2}},& n\mbox{ even}. \end{array}\displaystyle \right . $$

Thus, the proof is completed. □

Corollary 4

Let $A=C_{r}(Q_{0}^{2},Q_{1}^{2},\ldots,Q_{n-1}^{2})$ be a r-circulant matrix, where $r \in\mathbb{C}$, $\vert r \vert \geq1$,

$$ \Vert A \Vert _{2} \leq \left \{ \textstyle\begin{array}{@{}l@{\quad}l} n \vert r \vert \frac{Q_{2n-1}+6}{2},& n\textit{ odd}, \\ n \vert r \vert \frac{Q_{2n-1}+2}{2},& n\textit{ even}, \end{array}\displaystyle \right . $$

where $\Vert \cdot \Vert _{2}$ is the spectral norm and $Q_{n}$ denotes the nth Pell-Lucas number.

Proof

Since $A=C_{r}(Q_{0}^{2},Q_{1}^{2},\ldots,Q_{n-1}^{2})$ is a r-circulant matrix, if the matrices $B=C_{r}(Q_{0},Q_{1}, \ldots,Q_{n-1})$ and $C=C(Q_{0}^{2},Q_{1}^{2},\ldots,Q_{n-1}^{2})$ we get $A=B\circ C$; thus, we obtain

$$ \Vert A \Vert _{2} \leq \left \{ \textstyle\begin{array}{@{}l@{\quad}l} n \vert r \vert \frac{Q_{2n-1}+6}{2},& n\mbox{ odd},\\ n \vert r \vert \frac{Q_{2n-1}+2}{2},& n\mbox{ even}. \end{array}\displaystyle \right . $$

□

Declarations

Acknowledgements

The authors wish to express their heartfelt thanks to the referees for their detailed and helpful suggestions for revising the manuscript.

Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.

Authors’ Affiliations

(1)

Science Faculty, Selcuk University, Konya, 42031, Turkey

(2)

Semsi Tebrizi Anatolian Religious Vocational High School, Konya, Turkey

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