On the spectral norms of some circulant matrices with the trigonometric functions

Abstract

In this paper, we use the properties of an r-circulant matrix and a geometric circulant matrix to study the spectral norms of the r-circulant matrix and the geometric circulant matrix involving trigonometric functions by some algebra methods.

Introduction

In 1885, circulant matrix was first proposed by Muir, and he did some basic research. Until 1950–1955, Good et al. began to study the inverse, determinants and characteristic values of circulant matrices; these efforts have opened the door to study circulant matrices. A circulant matrix is a kind of matrix with a special structure, which has been widely used in algebra, geometry, signal processing and coding theory. In recent years, the circulant matrix is still a topic of focus in the research of matrix theory. Especially, some scholars studied the norms of r-circulant matrices and geometric circulant matrices with some famous numbers and polynomials, for example, on the spectral norms of circulant matrices, r-circulant matrices, geometric circulant matrices with Fibonacci number, Lucas number, generalized Fibonacci and Lucas numbers, generalized k-Horadam numbers, the biperiodic Fibonacci and Lucas numbers have been studied [1,2,3,4,5,6,7,8,9,10,11,12,13]. To the best of our knowledge, no one has studied the upper and lower estimate problems for the spectral norms involving trigonometric functions $\cos (\frac{k \pi }{n} )$, $\sin (\frac{k\pi }{n} )$ yet by using exponential sum.

A $n\times n$ r-circulant matrix $C_{r}$ is defined by [8]

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

Kızılateş and Tuglu [9] defined geometric circulant matrices by the form

$$\begin{aligned} C_{r^{*}}= \begin{pmatrix} c_{0}&c_{1}&c_{2}&\cdots &c_{n-2}&c_{n-1} \\ rc_{n-1}&c_{0}&c_{1}&\cdots &c_{n-3}&c_{n-2} \\ r^{2}c_{n-2}&rc_{n-1}&c_{0}&\cdots &c_{n-4}&c_{n-3} \\ \vdots &\vdots &\vdots & &\vdots &\vdots \\ r^{n-1}c_{1}&r^{n-2}c_{2}&r^{n-3}c_{3}&\cdots &rc_{n-1}&c_{0} \end{pmatrix}_{n\times n}. \end{aligned}$$

Obviously, when the parameter satisfies $r=1$, we can get the classical circulant matrix. Inspired by [7], in this paper, we shall use identities of the trigonometric functions and power sums of $\cos (\frac{k\pi }{n} )$, $\sin (\frac{k\pi }{n} )$ to study the norms of the r-circulant matrices

$$\begin{aligned} &A=\mathrm{Circ}_{r} \biggl(\cos \frac{0\cdot \pi }{n},\cos \frac{1\cdot \pi }{n},\cos \frac{2\cdot \pi }{n},\ldots, \cos \frac{(n-1) \cdot \pi }{n} \biggr), \\ &B=\mathrm{Circ}_{r} \biggl(\sin \frac{0\cdot \pi }{n},\sin \frac{1\cdot \pi }{n},\sin \frac{2\cdot \pi }{n},\ldots, \sin \frac{(n-1) \cdot \pi }{n} \biggr), \end{aligned}$$

and then we obtain the norms of geometric circulant matrices

$$\begin{aligned} &P_{r^{*}}=\mathrm{Circ}_{r^{*}} \biggl(\cos \frac{0\cdot \pi }{n},\cos \frac{1 \cdot \pi }{n},\cos \frac{2\cdot \pi }{n}, \ldots, \cos \frac{(n-1) \cdot \pi }{n} \biggr), \\ &R_{r^{*}}=\mathrm{Circ}_{r^{*}} \biggl(\sin \frac{0\cdot \pi }{n},\sin \frac{1 \cdot \pi }{n},\sin \frac{2\cdot \pi }{n}, \ldots, \sin \frac{(n-1) \cdot \pi }{n} \biggr). \end{aligned}$$

Then we get some interesting and concise results which are stated by the following theorems.

Theorem 1

Let $A=C_{r} (\cos \frac{0\cdot \pi }{n}, \cos \frac{1\cdot \pi }{n},\cos \frac{2\cdot \pi }{n},\ldots, \cos \frac{(n-1) \cdot \pi }{n} )$ be an r-circulant matrix, then we have

$$\begin{aligned} &\vert r \vert \geq 1,\quad \frac{\sqrt{2}}{2}\leq \Vert A \Vert _{2}\leq \sqrt{ \frac{n}{2}}\sqrt{(n-1) \vert r \vert ^{2}+1}; \\ &\vert r \vert < 1,\quad \frac{\sqrt{2}}{2} \vert r \vert \leq \Vert A \Vert _{2}\leq \frac{\sqrt{2}}{2}n. \end{aligned}$$

Theorem 2

Let $B=C_{r} (\sin \frac{0\cdot \pi }{n}, \sin \frac{1\cdot \pi }{n},\sin \frac{2\cdot \pi }{n},\ldots, \sin \frac{(n-1) \cdot \pi }{n} )$ be an r-circulant matrix, then we have

$$\begin{aligned} & \vert r \vert \geq 1, \quad\frac{\sqrt{2}}{2}\leq \Vert B \Vert _{2}\leq \vert r \vert \sqrt{ \frac{n(n-1)}{2}}; \\ & \vert r \vert < 1,\quad \frac{\sqrt{2}}{2} \vert r \vert \leq \Vert B \Vert _{2}\leq \sqrt{ \frac{n(n-1)}{2}}. \end{aligned}$$

Theorem 3

Let $P_{r^{*}}=C_{r^{*}} (\cos \frac{0 \cdot \pi }{n},\cos \frac{1\cdot \pi }{n},\cos \frac{2\cdot \pi }{n}, \ldots, \cos \frac{(n-1)\cdot \pi }{n} )$ be a geometric circulant matrix, we have

$$\begin{aligned} &\vert r \vert > 1, \quad\frac{\sqrt{2}}{2}\leq \Vert P_{r^{*}} \Vert _{2}\leq \sqrt{ \frac{n}{2}} \sqrt{ \frac{1- \vert r \vert ^{2n}}{1- \vert r \vert ^{2}}}; \\ &\vert r \vert < 1,\quad \vert r \vert ^{n} \sqrt{N_{1}} \leq \Vert P_{r^{*}} \Vert _{2} \leq \frac{ \sqrt{2}}{2}n, \end{aligned}$$

where $N_{1}=\frac{1-r^{-2}-r^{-2n+2}}{4}+ \frac{1-r^{-2n}}{2(1-r^{-2})}$.

Theorem 4

Let $R_{r^{*}}=C_{r^{*}} (\sin \frac{0 \cdot \pi }{n},\sin \frac{1\cdot \pi }{n},\sin \frac{2\cdot \pi }{n}, \ldots, \sin \frac{(n-1)\cdot \pi }{n} )$ be a geometric circulant matrix, we have

$$\begin{aligned} & \vert r \vert > 1,\quad \frac{\sqrt{2}}{2}\leq \Vert R_{r^{*}} \Vert _{2}\leq \sqrt{ \frac{n}{2}} \sqrt{ \frac{ \vert r \vert ^{2}- \vert r \vert ^{2n}}{1- \vert r \vert ^{2}}}; \\ &\vert r \vert < 1,\quad \vert r \vert ^{n} \sqrt{N_{2}} \leq \Vert R_{r^{*}} \Vert _{2} \leq \sqrt{ \frac{n(n-1)}{2}}, \end{aligned}$$

where $N_{2}=\frac{1-r^{-2n}}{2(1-r^{-2})}- \frac{1-r^{-2}-r^{-2n+2}}{4}$.

Preliminaries

Definition 1

([9])

Let any matrix $A=(a_{ij})\in M_{m \times n}(C)$, the spectral norm and the Euclidean norm of matrix A are defined by

$$\begin{aligned} \Vert A \Vert _{2}=\sqrt{ \max_{1\leq i\leq n}\lambda _{i} \bigl(A^{H}A \bigr)}, \qquad \Vert A \Vert _{E}= \Biggl( \sum^{m}_{i=1} \sum^{n}_{j=1} \vert a_{ij} \vert ^{2} \Biggr)^{\frac{1}{2}}, \quad \text{respectively}, \end{aligned}$$

where the $\lambda _{i}(A^{H}A)$ are the eigenvalues of matrices $A^{H}A$ and $A^{H}$ is the conjugate transpose of A.

The following important inequalities hold between the Euclidean norm and spectral norm:

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

(1)

Definition 2

([9])

Let both $A=(a_{ij})$ and $B=(b_{ij})$ be $m \times n$ matrices, then the Hadamard product of A and B is the $m \times n$ matrix of elementwise products, namely $A\circ B=(a_{ij}b _{ij})$.

Then we have the following inequalities:

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

(2)

Lemma 1

([7])

For any positive integer $n\geq 2$, we have

$$\begin{aligned} \sum^{n-1}_{k=0}\cos ^{2} \biggl(\frac{k\pi }{n} \biggr)= \sum^{n-1}_{k=0} \sin ^{2} \biggl(\frac{k\pi }{n} \biggr)=\frac{n}{2}. \end{aligned}$$

Lemma 2

For any positive integer $n\geq 2$, we can get

$$\begin{aligned} &\sum^{n-1}_{k=0}r^{-2k}\cos ^{2} \biggl(\frac{k\pi }{n} \biggr)=\frac{1-r ^{-2n}}{2(1-r^{-2})}+ \frac{1-r^{-2}-r^{-2n+2}}{4}=N_{1}, \\ &\sum^{n-1}_{k=0}r^{-2k}\sin ^{2} \biggl(\frac{k\pi }{n} \biggr)=\frac{1-r ^{-2n}}{2(1-r^{-2})}- \frac{1-r^{-2}-r^{-2n+2}}{4}=N_{2}. \end{aligned}$$

Proof

By the properties of $\cos 2\theta =2\cos ^{2}\theta -1=1-2 \sin ^{2}\theta$, $e^{i\theta }=\cos \theta +i\sin \theta$, we can easily get $\cos \theta =\frac{e^{i\theta }+e^{-i\theta }}{2}$; let $e(x)=e^{2\pi ix}$, note that $e(1)=e(-1)=1$, using the properties of the trigonometric sums $\sum^{n-1}_{k=0}e (\frac{k}{n} )=0$. Hence,

$$\begin{aligned} \sum^{n-1}_{k=0}r^{-2k}\cos ^{2} \biggl(\frac{k\pi }{n} \biggr) &= \sum ^{n-1}_{k=0}r^{-2k}\frac{1+\cos (\frac{2k \pi }{n} )}{2} \\ &=\frac{1-r^{-2n}}{2(1-r^{-2})}+\frac{1}{4} \sum^{n-1}_{k=0}r^{-2k} \biggl(e \biggl( \frac{k}{n} \biggr)+e \biggl(\frac{-k}{n} \biggr) \biggr). \end{aligned}$$

Taking

$$\begin{aligned} &S_{1} = \sum^{n-1}_{k=0}r^{-2k}e \biggl(\frac{k}{n} \biggr) \\ &\phantom{S_{1}}=r^{-2\cdot 0}1+r^{-2\cdot 1}e \biggl( \frac{1}{n} \biggr)+r^{-2 \cdot 2}e \biggl(\frac{2}{n} \biggr)+\cdots +r^{-2\cdot (n-2)}e \biggl(\frac{n-2}{n} \biggr)+r ^{-2\cdot (n-1)}e \biggl(\frac{n-1}{n} \biggr), \\ &e \biggl(\frac{1}{n} \biggr)S_{1}=r^{-2\cdot 0}e \biggl(\frac{1}{n} \biggr)+r ^{-2\cdot 1}e \biggl( \frac{2}{n} \biggr)+r^{-2\cdot 2}e \biggl(\frac{3}{n} \biggr)+ \cdots \\ &\phantom{e \biggl(\frac{1}{n} \biggr)S_{1}=}{}+r^{-2\cdot (n-2)}e \biggl(\frac{n-1}{n} \biggr)+r^{-2\cdot (n-1)}e(1). \end{aligned}$$

Therefore,

$$\begin{aligned} \biggl(1-e \biggl(\frac{1}{n} \biggr) \biggr)S_{1}=1+r^{-2} \sum^{n-1}_{k=1}e \biggl( \frac{k}{n} \biggr)-r^{-2n+2}=1-r^{-2}-r^{-2n+2}, \end{aligned}$$

that is $S_{1}= \sum^{n-1}_{k=0}r^{-2k}e (\frac{k}{n} )=\frac{1-r^{-2}-r ^{-2n+2}}{1-e (\frac{1}{n} )}$, as the same time, $\sum^{n-1}_{k=0}r^{-2k}e (\frac{-k}{n} )=\frac{1-r^{-2}-r ^{-2n+2}}{1-e (\frac{-1}{n} )}$.

So,

$$\begin{aligned} \sum^{n-1}_{k=0}r^{-2k}\cos ^{2} \biggl(\frac{k\pi }{n} \biggr) &=\frac{1-r ^{-2n}}{2(1-r^{-2})}+ \frac{1-r^{-2}-r^{-2n+2}}{4} \biggl(\frac{1}{1-e (\frac{1}{n} )}+\frac{1}{1-e (\frac{-1}{n} )} \biggr) \\ &=\frac{1-r^{-2n}}{2(1-r^{-2})}+\frac{1-r^{-2}-r^{-2n+2}}{4}=N_{1}. \end{aligned}$$

Using the same methods, note that

$$\begin{aligned} S_{2}&= \sum^{n-1}_{k=0}r^{-2k} \sin ^{2} \biggl(\frac{k\pi }{n} \biggr) =\sum ^{n-1}_{k=0}r^{-2k}\frac{1-\cos (\frac{2k \pi }{n} )}{2} \\ &=\frac{1}{2} \sum^{n-1}_{k=0}r^{-2k}- \sum^{n-1}_{k=0}r^{-2k}\cos \biggl(\frac{2k\pi }{n} \biggr) \\ &=\frac{1-r^{-2n}}{2(1-r^{-2})}-\frac{1-r^{-2}-r^{-2n+2}}{4}=N_{2}. \end{aligned}$$

 □

Proofs of theorems

Proof of Theorem 1

The matrix $A=C_{r} (\cos \frac{0\cdot \pi }{n}, \cos \frac{1\cdot \pi }{n},\cos \frac{2\cdot \pi }{n},\ldots, \cos \frac{(n-1) \cdot \pi }{n} )$ is of the following form:

$$\begin{aligned} A= \begin{pmatrix} \cos \frac{0\cdot \pi }{n}&\cos \frac{1\cdot \pi }{n}&\cos \frac{2 \cdot \pi }{n}&\cdots &\cos \frac{(n-1)\cdot \pi }{n} \\ r\cos \frac{(n-1)\cdot \pi }{n}&\cos \frac{0\cdot \pi }{n}&\cos \frac{1 \cdot \pi }{n}&\cdots &\cos \frac{(n-2)\cdot \pi }{n} \\ r\cos \frac{(n-2)\cdot \pi }{n}&r\cos \frac{(n-1)\cdot \pi }{n}& \cos \frac{0\cdot \pi }{n}&\cdots &\cos \frac{(n-3)\cdot \pi }{n} \\ \vdots &\vdots &\vdots &\ddots &\vdots \\ r\cos \frac{1\cdot \pi }{n}&r\cos \frac{2\cdot \pi }{n}&r\cos \frac{3 \cdot \pi }{n}&\cdots &\cos \frac{0\cdot \pi }{n} \end{pmatrix}_{n\times n}. \end{aligned}$$

(i) From $|r|\geq 1$, using the definition of Euclidean norm and Lemma 1, we have

$$\begin{aligned} \Vert A \Vert ^{2}_{E} &=\sum ^{n-1}_{k=0} (n-k )\cos ^{2} \biggl( \frac{k\cdot \pi }{n} \biggr)+ \sum^{n-1}_{k=1}k \vert r \vert ^{2}\cos ^{2} \biggl( \frac{k\cdot \pi }{n} \biggr) \\ &\geq \sum_{k=0}^{n-1} (n-k )\cos ^{2} \biggl(\frac{k\cdot \pi }{n} \biggr)+\sum ^{n-1}_{k=1}k\cos ^{2} \biggl( \frac{k\cdot \pi }{n} \biggr) \\ &= n\sum^{n-1}_{k=0}\cos ^{2} \biggl(\frac{k\cdot \pi }{n} \biggr)=\frac{n ^{2}}{2}, \end{aligned}$$

by (1), that is to say,

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

Moreover, let the matrices E and F be defined by

$$\begin{aligned} E= \begin{pmatrix} 1&1&1&\cdots &1 \\ r&1&1&\cdots &1 \\ r&r&1&\cdots &1 \\ \vdots &\vdots &\vdots & &\vdots \\ r&r&r&\cdots &1 \end{pmatrix}_{n\times n} \end{aligned}$$

and

$$\begin{aligned} F= \begin{pmatrix} \cos \frac{0\cdot \pi }{n}&\cos \frac{1\cdot \pi }{n}&\cos \frac{2 \cdot \pi }{n}&\cdots &\cos \frac{(n-1)\cdot \pi }{n} \\ \cos \frac{(n-1)\cdot \pi }{n}&\cos \frac{0\cdot \pi }{n}&\cos \frac{1 \cdot \pi }{n}&\cdots &\cos \frac{(n-2)\cdot \pi }{n} \\ \cos \frac{(n-2)\cdot \pi }{n}&\cos \frac{(n-1)\cdot \pi }{n}&\cos \frac{0 \cdot \pi }{n}&\cdots &\cos \frac{(n-3)\cdot \pi }{n} \\ \vdots &\vdots &\vdots &\ddots &\vdots \\ \cos \frac{1\cdot \pi }{n}&\cos \frac{2\cdot \pi }{n}&\cos \frac{3 \cdot \pi }{n}&\cdots &\cos \frac{0\cdot \pi }{n} \end{pmatrix}_{n\times n}, \end{aligned}$$

then $A=E\circ F$. So $\|A\|_{2}=\|E\circ F\|_{2}\leq r_{1}(E)C_{1}(F)$,

$$\begin{aligned} &r_{1}(E)=\max_{1\leq i\leq n}\sqrt{\sum _{j=1}^{n} \vert e_{ij} \vert ^{2}}=\sqrt{(n-1)r ^{2}+1}; \\ &c_{1}(F)=\max_{1\leq j\leq n}\sqrt{\sum _{i=1}^{n} \vert f_{ij} \vert ^{2}}=\sqrt{ \sum^{n-1}_{k=0} \cos ^{2}\frac{k\cdot \pi }{n}}=\sqrt{ \frac{n}{2}}. \end{aligned}$$

Therefore, we have

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

Thus, we can obtain the inequality

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

(ii) From $|r|<1$,

$$\begin{aligned} \Vert A \Vert ^{2}_{E} &= \sum ^{n-1}_{k=0} (n-k )\cos ^{2} \biggl( \frac{k\cdot \pi }{n} \biggr)+ \sum^{n-1}_{k=1}k \vert r \vert ^{2}\cos ^{2} \biggl( \frac{k\cdot \pi }{n} \biggr) \\ &\geq \sum_{k=0}^{n-1} (n-k ) \vert r \vert ^{2}\cos ^{2} \biggl(\frac{k \cdot \pi }{n} \biggr)+\sum^{n-1}_{k=1}k \vert r \vert ^{2}\cos ^{2} \biggl(\frac{k \cdot \pi }{n} \biggr) \\ &= n \vert r \vert ^{2}\sum^{n-1}_{k=0} \cos ^{2} \biggl(\frac{k\cdot \pi }{n} \biggr)=\frac{ \vert r \vert ^{2}n ^{2}}{2}, \end{aligned}$$

we can get

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

Moreover, for the matrices E and F as mentioned above, $A=E\circ F$. So $\|A\|_{2}=\|E\circ F\|_{2}\leq r_{1}(E)C_{1}(F)= \frac{\sqrt{2}}{2}n$.

Therefore, we have $\frac{\sqrt{2}}{2}|r|\leq \|A\|_{2}\leq \frac{ \sqrt{2}}{2}n$.

This proves Theorem 1. □

Now we prove Theorem 2.

Proof

$$\begin{aligned} B= \begin{pmatrix} \sin \frac{0\cdot \pi }{n}&\sin \frac{1\cdot \pi }{n}&\sin \frac{2 \cdot \pi }{n}&\cdots &\sin \frac{(n-1)\cdot \pi }{n} \\ r\sin \frac{(n-1)\cdot \pi }{n}&\sin \frac{0\cdot \pi }{n}&\sin \frac{1 \cdot \pi }{n}&\cdots &\sin \frac{(n-2)\cdot \pi }{n} \\ r\sin \frac{(n-2)\cdot \pi }{n}&r\sin \frac{(n-1)\cdot \pi }{n}& \sin \frac{0\cdot \pi }{n}&\cdots &\sin \frac{(n-3)\cdot \pi }{n} \\ \vdots &\vdots &\vdots &\ddots &\vdots \\ r\sin \frac{1\cdot \pi }{n}&r\sin \frac{2\cdot \pi }{n}&r\sin \frac{3 \cdot \pi }{n}&\cdots &\sin \frac{0\cdot \pi }{n} \end{pmatrix}_{n\times n}. \end{aligned}$$

(i) From $|r|\geq 1$, using the definition of Euclidean norm and Lemma 1, we have

$$\begin{aligned} \Vert B \Vert ^{2}_{E} &= \sum ^{n-1}_{k=0} (n-k )\sin ^{2} \biggl( \frac{k\cdot \pi }{n} \biggr)+ \sum^{n-1}_{k=1}k \vert r \vert ^{2}\sin ^{2} \biggl( \frac{k\cdot \pi }{n} \biggr) \\ &\geq \sum_{k=0}^{n-1} (n-k )\sin ^{2} \biggl(\frac{k\cdot \pi }{n} \biggr)+\sum ^{n-1}_{k=1}k\sin ^{2} \biggl( \frac{k\cdot \pi }{n} \biggr) \\ &= n\sum^{n-1}_{k=0}\sin ^{2} \biggl(\frac{k\cdot \pi }{n} \biggr)=\frac{n ^{2}}{2}, \end{aligned}$$

that is,

$$ \Vert B \Vert _{2}\geq \frac{1}{\sqrt{n}} \Vert B \Vert _{E}\geq \frac{\sqrt{2}}{2}. $$

Moreover, let the matrices C and D be defined by

$$\begin{aligned} C= \begin{pmatrix} \sin \frac{0\cdot \pi }{n}&1&1&\cdots &1 \\ r&\sin \frac{0\cdot \pi }{n}&1&\cdots &1 \\ r&r&\sin \frac{0\cdot \pi }{n}&\cdots &1 \\ \vdots &\vdots &\vdots & &\vdots \\ r&r&r&\cdots &\sin \frac{0\cdot \pi }{n} \end{pmatrix}_{n\times n} \end{aligned}$$

and

$$\begin{aligned} D= \begin{pmatrix} \sin \frac{0\cdot \pi }{n}&\sin \frac{1\cdot \pi }{n}&\sin \frac{2 \cdot \pi }{n}&\cdots &\sin \frac{(n-1)\cdot \pi }{n} \\ \sin \frac{(n-1)\cdot \pi }{n}&\sin \frac{0\cdot \pi }{n}&\sin \frac{1 \cdot \pi }{n}&\cdots &\sin \frac{(n-2)\cdot \pi }{n} \\ \sin \frac{(n-2)\cdot \pi }{n}&\sin \frac{(n-1)\cdot \pi }{n}&\sin \frac{0 \cdot \pi }{n}&\cdots &\sin \frac{(n-3)\cdot \pi }{n} \\ \vdots &\vdots &\vdots &\ddots &\vdots \\ \sin \frac{1\cdot \pi }{n}&\sin \frac{2\cdot \pi }{n}&\sin \frac{3 \cdot \pi }{n}&\cdots &\sin \frac{0\cdot \pi }{n} \end{pmatrix}_{n\times n}, \end{aligned}$$

then $B=C\circ D$. So $\|B\|_{2}=\|C\circ D\|_{2}\leq r_{1}(C)C_{1}(D)$,

$$\begin{aligned} &r_{1}(C)=\max_{1\leq i\leq n}\sqrt{\sum _{j=1}^{n} \vert c_{ij} \vert ^{2}}=\sqrt{(n-1)r ^{2}}; \\ &c_{1}(D)=\max_{1\leq j\leq n}\sqrt{\sum _{i=1}^{n} \vert d_{ij} \vert ^{2}}=\sqrt{ \sum^{n-1}_{k=0} \sin ^{2}\frac{k\cdot \pi }{n}}=\sqrt{ \frac{n}{2}}. \end{aligned}$$

Therefore, we have

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

Thus, we can obtain

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

(ii) From $|r|<1$,

$$\begin{aligned} \Vert B \Vert ^{2}_{E} &= \sum ^{n-1}_{k=0} (n-k )\sin ^{2} \biggl( \frac{k\cdot \pi }{n} \biggr)+ \sum^{n-1}_{k=1}k \vert r \vert ^{2}\sin ^{2} \biggl( \frac{k\cdot \pi }{n} \biggr) \\ &\geq \sum_{k=0}^{n-1} (n-k ) \vert r \vert ^{2}\sin ^{2} \biggl(\frac{k \cdot \pi }{n} \biggr)+\sum^{n-1}_{k=1}k \vert r \vert ^{2}\sin ^{2} \biggl(\frac{k \cdot \pi }{n} \biggr) \\ &= n \vert r \vert ^{2}\sum^{n-1}_{k=0} \sin ^{2} \biggl(\frac{k\cdot \pi }{n} \biggr)=\frac{ \vert r \vert ^{2}n ^{2}}{2}, \end{aligned}$$

we can get

$$ \Vert B \Vert _{2}\geq \frac{1}{\sqrt{n}} \Vert B \Vert _{E}\geq \frac{\sqrt{2}}{2} \vert r \vert . $$

On the other hand, for the matrices C and D as mentioned above, $B=C\circ D$. So $\|B\|_{2}=\|C\circ D\|_{2}\leq r_{1}(C)C_{1}(D)=\sqrt{ \frac{n(n-1)}{2}}$.

Therefore, we have $\frac{\sqrt{2}}{2}|r|\leq \|B\|_{2}\leq \sqrt{ \frac{n(n-1)}{2}}$.

This proves Theorem 2. □

Now we prove Theorem 3 and Theorem 4.

Proof

$$\begin{aligned} P_{r^{*}}= \begin{pmatrix} \cos \frac{0\cdot \pi }{n}&\cos \frac{1\cdot \pi }{n}&\cos \frac{2 \cdot \pi }{n}&\cdots &\cos \frac{(n-1)\cdot \pi }{n} \\ r\cos \frac{(n-1)\cdot \pi }{n}&\cos \frac{0\cdot \pi }{n}&\cos \frac{1 \cdot \pi }{n}&\cdots &\cos \frac{(n-2)\cdot \pi }{n} \\ r^{2}\cos \frac{(n-2)\cdot \pi }{n}&r\cos \frac{(n-1)\cdot \pi }{n}& \cos \frac{0\cdot \pi }{n}&\cdots &\cos \frac{(n-3)\cdot \pi }{n} \\ \vdots &\vdots &\vdots &\ddots &\vdots \\ r^{n-1}\cos \frac{1\cdot \pi }{n}&r^{n-2}\cos \frac{2\cdot \pi }{n}&r ^{n-3}\cos \frac{3\cdot \pi }{n}&\cdots &\cos \frac{0\cdot \pi }{n} \end{pmatrix}_{n\times n}. \end{aligned}$$

(i) On the one hand, $|r|>1$ and by using the definition of Euclidean norm, we can obtain

$$\begin{aligned} \Vert P_{r^{*}} \Vert ^{2}_{E} &= \sum ^{n-1}_{k=0} (n-k )\cos ^{2} \biggl(\frac{k\cdot \pi }{n} \biggr)+ \sum^{n-1}_{k=1}i \bigl\vert r^{n-k} \bigr\vert ^{2}\cos ^{2} \biggl(\frac{k\cdot \pi }{n} \biggr) \\ &\geq \sum_{k=0}^{n-1} (n-k )\cos ^{2} \biggl(\frac{k\cdot \pi }{n} \biggr)+\sum ^{n-1}_{k=1}k\cos ^{2} \biggl( \frac{k\cdot \pi }{n} \biggr) \\ &= n\sum^{n-1}_{k=0}\cos ^{2} \biggl(\frac{k\cdot \pi }{n} \biggr)=\frac{n ^{2}}{2}. \end{aligned}$$

That is,

$$ \Vert P_{r^{*}} \Vert _{2}\geq \frac{1}{\sqrt{n}} \Vert P_{r^{*}} \Vert _{E}\geq \frac{ \sqrt{2}}{2}n. $$

On the other hand, let the matrices S and Q be represented by

$$\begin{aligned} S= \begin{pmatrix} 1&1&1&\cdots &1&1 \\ r&1&1&\cdots &1&1 \\ r^{2}&r&1&\cdots &1&1 \\ \vdots &\vdots &\vdots & &\vdots &\vdots \\ r^{n-1}&r^{n-2}&r^{n-3}&\cdots &r&1 \end{pmatrix}_{n\times n} \end{aligned}$$

and

$$\begin{aligned} Q= \begin{pmatrix} \cos \frac{0\cdot \pi }{n}&\cos \frac{1\cdot \pi }{n}&\cos \frac{2 \cdot \pi }{n}&\cdots &\cos \frac{(n-1)\cdot \pi }{n} \\ \cos \frac{(n-1)\cdot \pi }{n}&\cos \frac{0\cdot \pi }{n}&\cos \frac{1 \cdot \pi }{n}&\cdots &\cos \frac{(n-2)\cdot \pi }{n} \\ \cos \frac{(n-2)\cdot \pi }{n}&\cos \frac{(n-1)\cdot \pi }{n}&\cos \frac{0 \cdot \pi }{n}&\cdots &\cos \frac{(n-3)\cdot \pi }{n} \\ \vdots &\vdots &\vdots &\ddots &\vdots \\ \cos \frac{1\cdot \pi }{n}&\cos \frac{2\cdot \pi }{n}&\cos \frac{3 \cdot \pi }{n}&\cdots &\cos \frac{0\cdot \pi }{n} \end{pmatrix}_{n\times n}, \end{aligned}$$

then $P_{r^{*}}=S\circ Q$. So $\|P_{r^{*}}\|_{2}=\|S\circ Q\|_{2} \leq r_{1}(S)C_{1}(Q)$,

$$\begin{aligned} &r_{1}(S) =\max_{1\leq i\leq n}\sqrt{\sum _{j=1}^{n} \vert s_{ij} \vert ^{2}}=\sqrt{1+ \vert r \vert ^{2}+ \cdots + \bigl\vert r^{n-1} \bigr\vert ^{2}}=\sqrt{ \frac{1- \vert r \vert ^{2n}}{1- \vert r \vert ^{2}}}, \\ &c_{1}(Q)=\max_{1\leq j\leq n}\sqrt{\sum _{i=1}^{n} \vert q_{ij} \vert ^{2}}=\sqrt{ \sum^{n-1}_{k=0} \cos ^{2} \biggl(\frac{k\cdot \pi }{n} \biggr)}=\sqrt{ \frac{n}{2}}. \end{aligned}$$

Therefore,

$$ \Vert P_{r^{*}} \Vert _{2}\leq r_{1}(S)c_{1}(Q)= \sqrt{ \frac{1- \vert r \vert ^{2n}}{1- \vert r \vert ^{2}}}\sqrt{\frac{n}{2}}. $$

(ii) For $|r|<1$,

$$\begin{aligned} \Vert P_{r^{*}} \Vert ^{2}_{E} &=\sum ^{n-1}_{k=0} (n-k )\cos ^{2} \biggl(\frac{k\cdot \pi }{n} \biggr)+ \sum^{n-1}_{k=1}k \bigl\vert r^{n-k} \bigr\vert ^{2}\cos ^{2} \biggl(\frac{k\cdot \pi }{n} \biggr) \\ &\geq \sum_{k=0}^{n-1} (n-k ) \bigl\vert r^{n-k} \bigr\vert ^{2}\cos ^{2} \biggl(\frac{k \cdot \pi }{n} \biggr)+\sum^{n-1}_{k=1}i \bigl\vert r^{n-k} \bigr\vert ^{2}\cos ^{2} \biggl(\frac{k \cdot \pi }{n} \biggr) \\ &= n \vert r \vert ^{2n}\sum^{n-1}_{k=0} \vert r \vert ^{-2k}\cos ^{2} \biggl( \frac{k\cdot \pi }{n} \biggr)=n \vert r \vert ^{2n}N_{1}. \end{aligned}$$

So

$$ \Vert P_{r^{*}} \Vert _{2}\geq \frac{1}{\sqrt{n}} \Vert P_{r^{*}} \Vert _{E}\geq \vert r \vert ^{n}\sqrt{N _{1}}, $$

where $N_{1}=\frac{1-r^{-2n}}{2(1-r^{-2})}+ \frac{1-r^{-2}-r^{-2n+2}}{4}$.

Moreover, for the matrices S and Q as mentioned above, in this case, $P_{r^{*}}=S\circ Q$. So $\|P_{r^{*}}\|_{2}=\|S\circ Q\|_{2}\leq r _{1}(S)C_{1}(Q)$,

$$\begin{aligned} &r_{1}(S) =\max_{1\leq i\leq n}\sqrt{\sum _{j=1}^{n} \vert s_{ij} \vert ^{2}}= \sqrt{n}, \\ &c_{1}(Q)=\max_{1\leq j\leq n}\sqrt{\sum _{i=1}^{n} \vert q_{ij} \vert ^{2}}=\sqrt{ \sum^{n-1}_{i=0} \cos ^{2} \biggl(\frac{k\cdot \pi }{n} \biggr)}=\sqrt{ \frac{n}{2}}, \end{aligned}$$

$\|P_{r^{*}}\|_{2}\leq \frac{\sqrt{2}}{2}n$.

Therefore, we have

$$ \vert r \vert ^{n}\sqrt{N_{1}}\leq \Vert P_{r^{*}} \Vert _{2}\leq \frac{\sqrt{2}}{2}n. $$

By the same methods, using Lemma 2 and Theorem 2, we can get Theorem 4.

This completes all of the theorems. □

Remark

Lemma 2 of this paper gave a new method to compute the power sums of the trigonometric functions.

Conclusion

By the same methods as of this paper, we can also get determinants and norms of some other special circulant matrices involving trigonometric functions $\cos (\frac{k\pi }{n} )$, $\sin (\frac{k \pi }{n} )$.