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