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.