1 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 [1‐13]. 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]
Kızılateş and Tuglu [9] defined geometric circulant matrices by the form
$$\begin{aligned} C_{r}= \begin{pmatrix} c_{0}&c_{1}&c_{2}&\cdots &c_{n-2}&c_{n-1} \\ rc_{n-1}&c_{0}&c_{1}&\cdots &c_{n-3}&c_{n-2} \\ rc_{n-2}&rc_{n-1}&c_{0}&\cdots &c_{n-4}&c_{n-3} \\ \vdots &\vdots &\vdots & &\vdots &\vdots \\ rc_{1}&rc_{2}&rc_{3}&\cdots &rc_{n-1}&c_{0} \end{pmatrix}_{n\times n}. \end{aligned}$$
$$\begin{aligned} C_{r^{*}}= \begin{pmatrix} c_{0}&c_{1}&c_{2}&\cdots &c_{n-2}&c_{n-1} \\ rc_{n-1}&c_{0}&c_{1}&\cdots &c_{n-3}&c_{n-2} \\ r^{2}c_{n-2}&rc_{n-1}&c_{0}&\cdots &c_{n-4}&c_{n-3} \\ \vdots &\vdots &\vdots & &\vdots &\vdots \\ r^{n-1}c_{1}&r^{n-2}c_{2}&r^{n-3}c_{3}&\cdots &rc_{n-1}&c_{0} \end{pmatrix}_{n\times n}. \end{aligned}$$
Anzeige
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
and then we obtain the norms of geometric circulant matrices
Then we get some interesting and concise results which are stated by the following theorems.
$$\begin{aligned} &A=\mathrm{Circ}_{r} \biggl(\cos \frac{0\cdot \pi }{n},\cos \frac{1\cdot \pi }{n},\cos \frac{2\cdot \pi }{n},\ldots, \cos \frac{(n-1) \cdot \pi }{n} \biggr), \\ &B=\mathrm{Circ}_{r} \biggl(\sin \frac{0\cdot \pi }{n},\sin \frac{1\cdot \pi }{n},\sin \frac{2\cdot \pi }{n},\ldots, \sin \frac{(n-1) \cdot \pi }{n} \biggr), \end{aligned}$$
$$\begin{aligned} &P_{r^{*}}=\mathrm{Circ}_{r^{*}} \biggl(\cos \frac{0\cdot \pi }{n},\cos \frac{1 \cdot \pi }{n},\cos \frac{2\cdot \pi }{n}, \ldots, \cos \frac{(n-1) \cdot \pi }{n} \biggr), \\ &R_{r^{*}}=\mathrm{Circ}_{r^{*}} \biggl(\sin \frac{0\cdot \pi }{n},\sin \frac{1 \cdot \pi }{n},\sin \frac{2\cdot \pi }{n}, \ldots, \sin \frac{(n-1) \cdot \pi }{n} \biggr). \end{aligned}$$
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
where
\(N_{1}=\frac{1-r^{-2}-r^{-2n+2}}{4}+ \frac{1-r^{-2n}}{2(1-r^{-2})}\).
$$\begin{aligned} &\vert r \vert > 1, \quad\frac{\sqrt{2}}{2}\leq \Vert P_{r^{*}} \Vert _{2}\leq \sqrt{ \frac{n}{2}} \sqrt{ \frac{1- \vert r \vert ^{2n}}{1- \vert r \vert ^{2}}}; \\ &\vert r \vert < 1,\quad \vert r \vert ^{n} \sqrt{N_{1}} \leq \Vert P_{r^{*}} \Vert _{2} \leq \frac{ \sqrt{2}}{2}n, \end{aligned}$$
Anzeige
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
where
\(N_{2}=\frac{1-r^{-2n}}{2(1-r^{-2})}- \frac{1-r^{-2}-r^{-2n+2}}{4}\).
$$\begin{aligned} & \vert r \vert > 1,\quad \frac{\sqrt{2}}{2}\leq \Vert R_{r^{*}} \Vert _{2}\leq \sqrt{ \frac{n}{2}} \sqrt{ \frac{ \vert r \vert ^{2}- \vert r \vert ^{2n}}{1- \vert r \vert ^{2}}}; \\ &\vert r \vert < 1,\quad \vert r \vert ^{n} \sqrt{N_{2}} \leq \Vert R_{r^{*}} \Vert _{2} \leq \sqrt{ \frac{n(n-1)}{2}}, \end{aligned}$$
2 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
where the \(\lambda _{i}(A^{H}A)\) are the eigenvalues of matrices \(A^{H}A\) and \(A^{H}\) is the conjugate transpose of A.
$$\begin{aligned} \Vert A \Vert _{2}=\sqrt{ \max_{1\leq i\leq n}\lambda _{i} \bigl(A^{H}A \bigr)}, \qquad \Vert A \Vert _{E}= \Biggl( \sum^{m}_{i=1} \sum^{n}_{j=1} \vert a_{ij} \vert ^{2} \Biggr)^{\frac{1}{2}}, \quad \text{respectively}, \end{aligned}$$
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,
Taking
Therefore,
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, \(\sum^{n-1}_{k=0}r^{-2k}e (\frac{-k}{n} )=\frac{1-r^{-2}-r ^{-2n+2}}{1-e (\frac{-1}{n} )}\).
$$\begin{aligned} \sum^{n-1}_{k=0}r^{-2k}\cos ^{2} \biggl(\frac{k\pi }{n} \biggr) &= \sum ^{n-1}_{k=0}r^{-2k}\frac{1+\cos (\frac{2k \pi }{n} )}{2} \\ &=\frac{1-r^{-2n}}{2(1-r^{-2})}+\frac{1}{4} \sum^{n-1}_{k=0}r^{-2k} \biggl(e \biggl( \frac{k}{n} \biggr)+e \biggl(\frac{-k}{n} \biggr) \biggr). \end{aligned}$$
$$\begin{aligned} &S_{1} = \sum^{n-1}_{k=0}r^{-2k}e \biggl(\frac{k}{n} \biggr) \\ &\phantom{S_{1}}=r^{-2\cdot 0}1+r^{-2\cdot 1}e \biggl( \frac{1}{n} \biggr)+r^{-2 \cdot 2}e \biggl(\frac{2}{n} \biggr)+\cdots +r^{-2\cdot (n-2)}e \biggl(\frac{n-2}{n} \biggr)+r ^{-2\cdot (n-1)}e \biggl(\frac{n-1}{n} \biggr), \\ &e \biggl(\frac{1}{n} \biggr)S_{1}=r^{-2\cdot 0}e \biggl(\frac{1}{n} \biggr)+r ^{-2\cdot 1}e \biggl( \frac{2}{n} \biggr)+r^{-2\cdot 2}e \biggl(\frac{3}{n} \biggr)+ \cdots \\ &\phantom{e \biggl(\frac{1}{n} \biggr)S_{1}=}{}+r^{-2\cdot (n-2)}e \biggl(\frac{n-1}{n} \biggr)+r^{-2\cdot (n-1)}e(1). \end{aligned}$$
$$\begin{aligned} \biggl(1-e \biggl(\frac{1}{n} \biggr) \biggr)S_{1}=1+r^{-2} \sum^{n-1}_{k=1}e \biggl( \frac{k}{n} \biggr)-r^{-2n+2}=1-r^{-2}-r^{-2n+2}, \end{aligned}$$
So,
Using the same methods, note that
□
$$\begin{aligned} \sum^{n-1}_{k=0}r^{-2k}\cos ^{2} \biggl(\frac{k\pi }{n} \biggr) &=\frac{1-r ^{-2n}}{2(1-r^{-2})}+ \frac{1-r^{-2}-r^{-2n+2}}{4} \biggl(\frac{1}{1-e (\frac{1}{n} )}+\frac{1}{1-e (\frac{-1}{n} )} \biggr) \\ &=\frac{1-r^{-2n}}{2(1-r^{-2})}+\frac{1-r^{-2}-r^{-2n+2}}{4}=N_{1}. \end{aligned}$$
$$\begin{aligned} S_{2}&= \sum^{n-1}_{k=0}r^{-2k} \sin ^{2} \biggl(\frac{k\pi }{n} \biggr) =\sum ^{n-1}_{k=0}r^{-2k}\frac{1-\cos (\frac{2k \pi }{n} )}{2} \\ &=\frac{1}{2} \sum^{n-1}_{k=0}r^{-2k}- \sum^{n-1}_{k=0}r^{-2k}\cos \biggl(\frac{2k\pi }{n} \biggr) \\ &=\frac{1-r^{-2n}}{2(1-r^{-2})}-\frac{1-r^{-2}-r^{-2n+2}}{4}=N_{2}. \end{aligned}$$
3 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:
(i) From \(|r|\geq 1\), using the definition of Euclidean norm and Lemma 1, we have
by (1), that is to say,
$$\begin{aligned} A= \begin{pmatrix} \cos \frac{0\cdot \pi }{n}&\cos \frac{1\cdot \pi }{n}&\cos \frac{2 \cdot \pi }{n}&\cdots &\cos \frac{(n-1)\cdot \pi }{n} \\ r\cos \frac{(n-1)\cdot \pi }{n}&\cos \frac{0\cdot \pi }{n}&\cos \frac{1 \cdot \pi }{n}&\cdots &\cos \frac{(n-2)\cdot \pi }{n} \\ r\cos \frac{(n-2)\cdot \pi }{n}&r\cos \frac{(n-1)\cdot \pi }{n}& \cos \frac{0\cdot \pi }{n}&\cdots &\cos \frac{(n-3)\cdot \pi }{n} \\ \vdots &\vdots &\vdots &\ddots &\vdots \\ r\cos \frac{1\cdot \pi }{n}&r\cos \frac{2\cdot \pi }{n}&r\cos \frac{3 \cdot \pi }{n}&\cdots &\cos \frac{0\cdot \pi }{n} \end{pmatrix}_{n\times n}. \end{aligned}$$
$$\begin{aligned} \Vert A \Vert ^{2}_{E} &=\sum ^{n-1}_{k=0} (n-k )\cos ^{2} \biggl( \frac{k\cdot \pi }{n} \biggr)+ \sum^{n-1}_{k=1}k \vert r \vert ^{2}\cos ^{2} \biggl( \frac{k\cdot \pi }{n} \biggr) \\ &\geq \sum_{k=0}^{n-1} (n-k )\cos ^{2} \biggl(\frac{k\cdot \pi }{n} \biggr)+\sum ^{n-1}_{k=1}k\cos ^{2} \biggl( \frac{k\cdot \pi }{n} \biggr) \\ &= n\sum^{n-1}_{k=0}\cos ^{2} \biggl(\frac{k\cdot \pi }{n} \biggr)=\frac{n ^{2}}{2}, \end{aligned}$$
$$ \Vert A \Vert _{2}\geq \frac{1}{\sqrt{n}} \Vert A \Vert _{E}\geq \frac{\sqrt{2}}{2}. $$
Moreover, let the matrices E and F be defined by
and
then \(A=E\circ F\). So \(\|A\|_{2}=\|E\circ F\|_{2}\leq r_{1}(E)C_{1}(F)\),
Therefore, we have
Thus, we can obtain the inequality
(ii) From \(|r|<1\),
we can get
$$\begin{aligned} E= \begin{pmatrix} 1&1&1&\cdots &1 \\ r&1&1&\cdots &1 \\ r&r&1&\cdots &1 \\ \vdots &\vdots &\vdots & &\vdots \\ r&r&r&\cdots &1 \end{pmatrix}_{n\times n} \end{aligned}$$
$$\begin{aligned} F= \begin{pmatrix} \cos \frac{0\cdot \pi }{n}&\cos \frac{1\cdot \pi }{n}&\cos \frac{2 \cdot \pi }{n}&\cdots &\cos \frac{(n-1)\cdot \pi }{n} \\ \cos \frac{(n-1)\cdot \pi }{n}&\cos \frac{0\cdot \pi }{n}&\cos \frac{1 \cdot \pi }{n}&\cdots &\cos \frac{(n-2)\cdot \pi }{n} \\ \cos \frac{(n-2)\cdot \pi }{n}&\cos \frac{(n-1)\cdot \pi }{n}&\cos \frac{0 \cdot \pi }{n}&\cdots &\cos \frac{(n-3)\cdot \pi }{n} \\ \vdots &\vdots &\vdots &\ddots &\vdots \\ \cos \frac{1\cdot \pi }{n}&\cos \frac{2\cdot \pi }{n}&\cos \frac{3 \cdot \pi }{n}&\cdots &\cos \frac{0\cdot \pi }{n} \end{pmatrix}_{n\times n}, \end{aligned}$$
$$\begin{aligned} &r_{1}(E)=\max_{1\leq i\leq n}\sqrt{\sum _{j=1}^{n} \vert e_{ij} \vert ^{2}}=\sqrt{(n-1)r ^{2}+1}; \\ &c_{1}(F)=\max_{1\leq j\leq n}\sqrt{\sum _{i=1}^{n} \vert f_{ij} \vert ^{2}}=\sqrt{ \sum^{n-1}_{k=0} \cos ^{2}\frac{k\cdot \pi }{n}}=\sqrt{ \frac{n}{2}}. \end{aligned}$$
$$ \Vert A \Vert _{2}\leq \sqrt{(n-1)r^{2}+1}\sqrt{ \frac{n}{2}}. $$
$$ \frac{\sqrt{2}}{2}\leq \Vert A \Vert _{2}\leq \sqrt{ \frac{n}{2}} \sqrt{(n-1) \vert r \vert ^{2}+1}. $$
$$\begin{aligned} \Vert A \Vert ^{2}_{E} &= \sum ^{n-1}_{k=0} (n-k )\cos ^{2} \biggl( \frac{k\cdot \pi }{n} \biggr)+ \sum^{n-1}_{k=1}k \vert r \vert ^{2}\cos ^{2} \biggl( \frac{k\cdot \pi }{n} \biggr) \\ &\geq \sum_{k=0}^{n-1} (n-k ) \vert r \vert ^{2}\cos ^{2} \biggl(\frac{k \cdot \pi }{n} \biggr)+\sum^{n-1}_{k=1}k \vert r \vert ^{2}\cos ^{2} \biggl(\frac{k \cdot \pi }{n} \biggr) \\ &= n \vert r \vert ^{2}\sum^{n-1}_{k=0} \cos ^{2} \biggl(\frac{k\cdot \pi }{n} \biggr)=\frac{ \vert r \vert ^{2}n ^{2}}{2}, \end{aligned}$$
$$ \Vert A \Vert _{2}\geq \frac{1}{\sqrt{n}} \Vert A \Vert _{E}\geq \frac{\sqrt{2}}{2} \vert r \vert . $$
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
$$\begin{aligned} B= \begin{pmatrix} \sin \frac{0\cdot \pi }{n}&\sin \frac{1\cdot \pi }{n}&\sin \frac{2 \cdot \pi }{n}&\cdots &\sin \frac{(n-1)\cdot \pi }{n} \\ r\sin \frac{(n-1)\cdot \pi }{n}&\sin \frac{0\cdot \pi }{n}&\sin \frac{1 \cdot \pi }{n}&\cdots &\sin \frac{(n-2)\cdot \pi }{n} \\ r\sin \frac{(n-2)\cdot \pi }{n}&r\sin \frac{(n-1)\cdot \pi }{n}& \sin \frac{0\cdot \pi }{n}&\cdots &\sin \frac{(n-3)\cdot \pi }{n} \\ \vdots &\vdots &\vdots &\ddots &\vdots \\ r\sin \frac{1\cdot \pi }{n}&r\sin \frac{2\cdot \pi }{n}&r\sin \frac{3 \cdot \pi }{n}&\cdots &\sin \frac{0\cdot \pi }{n} \end{pmatrix}_{n\times n}. \end{aligned}$$
$$\begin{aligned} \Vert B \Vert ^{2}_{E} &= \sum ^{n-1}_{k=0} (n-k )\sin ^{2} \biggl( \frac{k\cdot \pi }{n} \biggr)+ \sum^{n-1}_{k=1}k \vert r \vert ^{2}\sin ^{2} \biggl( \frac{k\cdot \pi }{n} \biggr) \\ &\geq \sum_{k=0}^{n-1} (n-k )\sin ^{2} \biggl(\frac{k\cdot \pi }{n} \biggr)+\sum ^{n-1}_{k=1}k\sin ^{2} \biggl( \frac{k\cdot \pi }{n} \biggr) \\ &= n\sum^{n-1}_{k=0}\sin ^{2} \biggl(\frac{k\cdot \pi }{n} \biggr)=\frac{n ^{2}}{2}, \end{aligned}$$
$$ \Vert B \Vert _{2}\geq \frac{1}{\sqrt{n}} \Vert B \Vert _{E}\geq \frac{\sqrt{2}}{2}. $$
Moreover, let the matrices C and D be defined by
and
then \(B=C\circ D\). So \(\|B\|_{2}=\|C\circ D\|_{2}\leq r_{1}(C)C_{1}(D)\),
Therefore, we have
Thus, we can obtain
(ii) From \(|r|<1\),
we can get
$$\begin{aligned} C= \begin{pmatrix} \sin \frac{0\cdot \pi }{n}&1&1&\cdots &1 \\ r&\sin \frac{0\cdot \pi }{n}&1&\cdots &1 \\ r&r&\sin \frac{0\cdot \pi }{n}&\cdots &1 \\ \vdots &\vdots &\vdots & &\vdots \\ r&r&r&\cdots &\sin \frac{0\cdot \pi }{n} \end{pmatrix}_{n\times n} \end{aligned}$$
$$\begin{aligned} D= \begin{pmatrix} \sin \frac{0\cdot \pi }{n}&\sin \frac{1\cdot \pi }{n}&\sin \frac{2 \cdot \pi }{n}&\cdots &\sin \frac{(n-1)\cdot \pi }{n} \\ \sin \frac{(n-1)\cdot \pi }{n}&\sin \frac{0\cdot \pi }{n}&\sin \frac{1 \cdot \pi }{n}&\cdots &\sin \frac{(n-2)\cdot \pi }{n} \\ \sin \frac{(n-2)\cdot \pi }{n}&\sin \frac{(n-1)\cdot \pi }{n}&\sin \frac{0 \cdot \pi }{n}&\cdots &\sin \frac{(n-3)\cdot \pi }{n} \\ \vdots &\vdots &\vdots &\ddots &\vdots \\ \sin \frac{1\cdot \pi }{n}&\sin \frac{2\cdot \pi }{n}&\sin \frac{3 \cdot \pi }{n}&\cdots &\sin \frac{0\cdot \pi }{n} \end{pmatrix}_{n\times n}, \end{aligned}$$
$$\begin{aligned} &r_{1}(C)=\max_{1\leq i\leq n}\sqrt{\sum _{j=1}^{n} \vert c_{ij} \vert ^{2}}=\sqrt{(n-1)r ^{2}}; \\ &c_{1}(D)=\max_{1\leq j\leq n}\sqrt{\sum _{i=1}^{n} \vert d_{ij} \vert ^{2}}=\sqrt{ \sum^{n-1}_{k=0} \sin ^{2}\frac{k\cdot \pi }{n}}=\sqrt{ \frac{n}{2}}. \end{aligned}$$
$$ \Vert B \Vert _{2}\leq \vert r \vert \sqrt{ \frac{n(n-1)}{2}}. $$
$$ \frac{\sqrt{2}}{2}\leq \Vert B \Vert _{2}\leq \vert r \vert \sqrt{\frac{n(n-1)}{2}}. $$
$$\begin{aligned} \Vert B \Vert ^{2}_{E} &= \sum ^{n-1}_{k=0} (n-k )\sin ^{2} \biggl( \frac{k\cdot \pi }{n} \biggr)+ \sum^{n-1}_{k=1}k \vert r \vert ^{2}\sin ^{2} \biggl( \frac{k\cdot \pi }{n} \biggr) \\ &\geq \sum_{k=0}^{n-1} (n-k ) \vert r \vert ^{2}\sin ^{2} \biggl(\frac{k \cdot \pi }{n} \biggr)+\sum^{n-1}_{k=1}k \vert r \vert ^{2}\sin ^{2} \biggl(\frac{k \cdot \pi }{n} \biggr) \\ &= n \vert r \vert ^{2}\sum^{n-1}_{k=0} \sin ^{2} \biggl(\frac{k\cdot \pi }{n} \biggr)=\frac{ \vert r \vert ^{2}n ^{2}}{2}, \end{aligned}$$
$$ \Vert B \Vert _{2}\geq \frac{1}{\sqrt{n}} \Vert B \Vert _{E}\geq \frac{\sqrt{2}}{2} \vert r \vert . $$
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. □
Proof
$$\begin{aligned} P_{r^{*}}= \begin{pmatrix} \cos \frac{0\cdot \pi }{n}&\cos \frac{1\cdot \pi }{n}&\cos \frac{2 \cdot \pi }{n}&\cdots &\cos \frac{(n-1)\cdot \pi }{n} \\ r\cos \frac{(n-1)\cdot \pi }{n}&\cos \frac{0\cdot \pi }{n}&\cos \frac{1 \cdot \pi }{n}&\cdots &\cos \frac{(n-2)\cdot \pi }{n} \\ r^{2}\cos \frac{(n-2)\cdot \pi }{n}&r\cos \frac{(n-1)\cdot \pi }{n}& \cos \frac{0\cdot \pi }{n}&\cdots &\cos \frac{(n-3)\cdot \pi }{n} \\ \vdots &\vdots &\vdots &\ddots &\vdots \\ r^{n-1}\cos \frac{1\cdot \pi }{n}&r^{n-2}\cos \frac{2\cdot \pi }{n}&r ^{n-3}\cos \frac{3\cdot \pi }{n}&\cdots &\cos \frac{0\cdot \pi }{n} \end{pmatrix}_{n\times n}. \end{aligned}$$
$$\begin{aligned} \Vert P_{r^{*}} \Vert ^{2}_{E} &= \sum ^{n-1}_{k=0} (n-k )\cos ^{2} \biggl(\frac{k\cdot \pi }{n} \biggr)+ \sum^{n-1}_{k=1}i \bigl\vert r^{n-k} \bigr\vert ^{2}\cos ^{2} \biggl(\frac{k\cdot \pi }{n} \biggr) \\ &\geq \sum_{k=0}^{n-1} (n-k )\cos ^{2} \biggl(\frac{k\cdot \pi }{n} \biggr)+\sum ^{n-1}_{k=1}k\cos ^{2} \biggl( \frac{k\cdot \pi }{n} \biggr) \\ &= n\sum^{n-1}_{k=0}\cos ^{2} \biggl(\frac{k\cdot \pi }{n} \biggr)=\frac{n ^{2}}{2}. \end{aligned}$$
$$ \Vert P_{r^{*}} \Vert _{2}\geq \frac{1}{\sqrt{n}} \Vert P_{r^{*}} \Vert _{E}\geq \frac{ \sqrt{2}}{2}n. $$
$$\begin{aligned} S= \begin{pmatrix} 1&1&1&\cdots &1&1 \\ r&1&1&\cdots &1&1 \\ r^{2}&r&1&\cdots &1&1 \\ \vdots &\vdots &\vdots & &\vdots &\vdots \\ r^{n-1}&r^{n-2}&r^{n-3}&\cdots &r&1 \end{pmatrix}_{n\times n} \end{aligned}$$
$$\begin{aligned} Q= \begin{pmatrix} \cos \frac{0\cdot \pi }{n}&\cos \frac{1\cdot \pi }{n}&\cos \frac{2 \cdot \pi }{n}&\cdots &\cos \frac{(n-1)\cdot \pi }{n} \\ \cos \frac{(n-1)\cdot \pi }{n}&\cos \frac{0\cdot \pi }{n}&\cos \frac{1 \cdot \pi }{n}&\cdots &\cos \frac{(n-2)\cdot \pi }{n} \\ \cos \frac{(n-2)\cdot \pi }{n}&\cos \frac{(n-1)\cdot \pi }{n}&\cos \frac{0 \cdot \pi }{n}&\cdots &\cos \frac{(n-3)\cdot \pi }{n} \\ \vdots &\vdots &\vdots &\ddots &\vdots \\ \cos \frac{1\cdot \pi }{n}&\cos \frac{2\cdot \pi }{n}&\cos \frac{3 \cdot \pi }{n}&\cdots &\cos \frac{0\cdot \pi }{n} \end{pmatrix}_{n\times n}, \end{aligned}$$
$$\begin{aligned} &r_{1}(S) =\max_{1\leq i\leq n}\sqrt{\sum _{j=1}^{n} \vert s_{ij} \vert ^{2}}=\sqrt{1+ \vert r \vert ^{2}+ \cdots + \bigl\vert r^{n-1} \bigr\vert ^{2}}=\sqrt{ \frac{1- \vert r \vert ^{2n}}{1- \vert r \vert ^{2}}}, \\ &c_{1}(Q)=\max_{1\leq j\leq n}\sqrt{\sum _{i=1}^{n} \vert q_{ij} \vert ^{2}}=\sqrt{ \sum^{n-1}_{k=0} \cos ^{2} \biggl(\frac{k\cdot \pi }{n} \biggr)}=\sqrt{ \frac{n}{2}}. \end{aligned}$$
$$ \Vert P_{r^{*}} \Vert _{2}\leq r_{1}(S)c_{1}(Q)= \sqrt{ \frac{1- \vert r \vert ^{2n}}{1- \vert r \vert ^{2}}}\sqrt{\frac{n}{2}}. $$
(ii) For \(|r|<1\),
So
where \(N_{1}=\frac{1-r^{-2n}}{2(1-r^{-2})}+ \frac{1-r^{-2}-r^{-2n+2}}{4}\).
$$\begin{aligned} \Vert P_{r^{*}} \Vert ^{2}_{E} &=\sum ^{n-1}_{k=0} (n-k )\cos ^{2} \biggl(\frac{k\cdot \pi }{n} \biggr)+ \sum^{n-1}_{k=1}k \bigl\vert r^{n-k} \bigr\vert ^{2}\cos ^{2} \biggl(\frac{k\cdot \pi }{n} \biggr) \\ &\geq \sum_{k=0}^{n-1} (n-k ) \bigl\vert r^{n-k} \bigr\vert ^{2}\cos ^{2} \biggl(\frac{k \cdot \pi }{n} \biggr)+\sum^{n-1}_{k=1}i \bigl\vert r^{n-k} \bigr\vert ^{2}\cos ^{2} \biggl(\frac{k \cdot \pi }{n} \biggr) \\ &= n \vert r \vert ^{2n}\sum^{n-1}_{k=0} \vert r \vert ^{-2k}\cos ^{2} \biggl( \frac{k\cdot \pi }{n} \biggr)=n \vert r \vert ^{2n}N_{1}. \end{aligned}$$
$$ \Vert P_{r^{*}} \Vert _{2}\geq \frac{1}{\sqrt{n}} \Vert P_{r^{*}} \Vert _{E}\geq \vert r \vert ^{n}\sqrt{N _{1}}, $$
Moreover, for the matrices S and Q as mentioned above, in this case, \(P_{r^{*}}=S\circ Q\). So \(\|P_{r^{*}}\|_{2}=\|S\circ Q\|_{2}\leq r _{1}(S)C_{1}(Q)\),
\(\|P_{r^{*}}\|_{2}\leq \frac{\sqrt{2}}{2}n\).
$$\begin{aligned} &r_{1}(S) =\max_{1\leq i\leq n}\sqrt{\sum _{j=1}^{n} \vert s_{ij} \vert ^{2}}= \sqrt{n}, \\ &c_{1}(Q)=\max_{1\leq j\leq n}\sqrt{\sum _{i=1}^{n} \vert q_{ij} \vert ^{2}}=\sqrt{ \sum^{n-1}_{i=0} \cos ^{2} \biggl(\frac{k\cdot \pi }{n} \biggr)}=\sqrt{ \frac{n}{2}}, \end{aligned}$$
Therefore, we have
By the same methods, using Lemma 2 and Theorem 2, we can get Theorem 4.
$$ \vert r \vert ^{n}\sqrt{N_{1}}\leq \Vert P_{r^{*}} \Vert _{2}\leq \frac{\sqrt{2}}{2}n. $$
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.
4 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 \(\cos (\frac{k\pi }{n} )\), \(\sin (\frac{k \pi }{n} )\).
Acknowledgements
The author is grateful to anonymous referees and the associate editor for their careful reading, helpful comments, and constructive suggestions, which improved the presentation of the results.
Competing interests
The author declares to have no competing interests.
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