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

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.

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

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

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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

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

 □