1 Introduction
The circulant and r-circulant matrices have important applications in numerical analysis, probability, coding theory, and so on. An \(n\times n\) matrix \(C_{r}\) is called an r-circulant matrix if it is defined as follows:
The matrix \(C_{r}\) is determined by its first row elements and r, we denote \(C_{r}=\operatorname{Circ}_{r}(c_{0},c_{1},c_{2}, \ldots ,c_{n-1})\). In particular for \(r=1\)
is called a circulant matrix.
$$ C_{r}= \begin{pmatrix} c_{0} & c_{1} & c_{2} & \ldots & c_{n-2} & c_{n-1} \\ rc_{n-1} & c_{0} & c_{1} & \ldots & c_{n-3} & c_{n-2} \\ rc_{n-2} & rc_{n-1} & c_{0} & \ldots & c_{n-4} & c_{n-3} \\ \vdots & \vdots & \vdots & & \vdots & \vdots \\ rc_{1} & rc_{2} & rc_{3} & \ldots & rc_{n-1} & c_{0} \end{pmatrix} . $$
$$ C= \begin{pmatrix} c_{0} & c_{1} & c_{2} & \ldots & c_{n-2} & c_{n-1} \\ c_{n-1} & c_{0} & c_{1} & \ldots & c_{n-3} & c_{n-2} \\ c_{n-2} & c_{n-1} & c_{0} & \ldots & c_{n-4} & c_{n-3} \\ \vdots & \vdots & \vdots & & \vdots & \vdots \\ c_{1} & c_{2} & c_{3} & \ldots & c_{n-1} & c_{0} \end{pmatrix} $$
Circulant and r-circulant matrices with the special numbers have been studied by many researchers in last decade. For example, in [1], Solak has studied the spectral norms of circulant matrices with the Fibonacci and Lucas numbers. In [2], Kocer et al. obtained norms of circulant and semicirculant matrices with Horadam numbers. In [3], Shen and Cen have given upper and lower bounds for the spectral norms of r-circulant matrices with the Fibonacci and Lucas numbers. In [4], Bahsi computed the spectral norms of circulant and r-circulant matrices with the hyperharmonic numbers. Moreover, in [5], Bahsi and Solak studied norms of circulant and r-circulant matrices with the hyper-Fibonacci and hyper-Lucas numbers. In [6], Jiang and Zhou studied spectral norms of even order r-circulant matrices. In [7, 8], Tuglu and Kızılateş have calculated Euclidean norm by using the finite difference operator and given spectral norms of circulant, r-circulant and some special matrices with the harmonic Fibonacci and hyperharmonic Fibonacci numbers. In [9], Yazlık and Taskara have presented new upper and lower bounds for the spectral norms of an r-circulant matrix with the generalized k-Horadam numbers. In [10], He et al. gave the upper bound estimation of the spectral norm for r-circulant matrices with Fibonacci and Lucas numbers.
Anzeige
In view of the above papers, we define a new circulant matrix which is called geometric circulant matrix and give upper and lower bounds for the spectral norms of this matrix with the generalized Fibonacci and hyperharmonic Fibonacci numbers by using the same method given in [3].
2 Preliminaries
The well-known Fibonacci and Lucas sequences are defined by the following recurrence relations: for \(n\geq 0\),
and
where \(F_{0}=0\), \(F_{1}=1\), \(L_{0}=2\) and \(L_{1}=1\), respectively. The generalized Fibonacci and Lucas sequences, \(\{ U_{n} \} \) and \(\{ V_{n} \}\), are defined by the following recurrence relations: for \(n\geq 0\), and any non-zero integer p,
and
where \(U_{0}=0\), \(U_{1}=1\), \(V_{0}=2\) and \(V_{1}=p\). If we take \(p=1\), then \(U_{n}=F_{n}\) and \(V_{n}=L_{n}\). Let α and β be the roots of the characteristic equation \(x^{2}-px-1=0\). Then the Binet formulas for the sequences \(\{ U_{n} \} \) and \(\{ V_{n} \} \) are given by
and
where \(\alpha =\frac{p+\sqrt{p^{2}+4}}{2}\) and \(\beta =\frac{p-\sqrt{p^{2}+4}}{2}\).
$$ F_{n+2}=F_{n+1}+F_{n} $$
$$ L_{n+2}=L_{n+1}+L_{n}, $$
$$ U_{n+2}=pU_{n+1}+U_{n} $$
$$ V_{n+2}=pV_{n+1}+V_{n}, $$
$$ U_{n}=\frac{\alpha ^{n}-\beta ^{n}}{\alpha -\beta } $$
$$ V_{n}=\alpha ^{n}+\beta ^{n}, $$
On the other hand, Yazlık and Taskara examined the generalized k-Horadam numbers via the following recurrence relations:
with the initial values \(H_{k,0}=a\), \(H_{k,1}=b\). Moreover they calculated sum of squares of k-Horadam numbers (see [9]). If we take \(f(k)=p\), \(g(k)=1\), \(a=0\) and \(b=1\) in (1), we get
and if we take \(f(k)=p\), \(g(k)=1\), \(a=2\) and \(b=p\) in (1), we have
In [11], Tuglu et al. defined hyperharmonic Fibonacci numbers for \(n,r\geq 1\),
where \(\mathbb{F}_{n}^{(0)}=\frac{1}{F_{n}}\) and \(\mathbb{F}_{0}=0\). Then they gave for the sum of the squares of hyperharmonic Fibonacci numbers as follows:
Now we give some definitions and Lemmas related to our study.
$$ H_{k,n+2}=f(k)H_{k,n+1}+g(k)H_{k,n} $$
(1)
$$ \sum_{i=0}^{n-1}U_{i}^{2}= \frac{U_{n}^{2}-U_{n-1}^{2}+(-1)^{n}}{p^{2}} $$
(2)
$$ \sum_{i=0}^{n-1}V_{i}^{2}= \frac{V_{n}^{2}-V_{n-1}^{2}+p^{2}-4+(1-(-1)^{n})(p^{2}+4)}{p^{2}} . $$
(3)
$$ \mathbb{F}_{n}^{(r)}=\sum_{k=1}^{n} \mathbb{F}_{k}^{(r-1)}, $$
$$ \frac{1}{\sqrt{n}}\mathbb{F}_{n-1}^{(r+1)}\leq \sqrt{\sum _{k=0}^{n-1} \bigl(\mathbb{F}_{k}^{(r)} \bigr) ^{2}}\leq \mathbb{F}_{n-1}^{(r+1)}. $$
(4)
Definition 1
An \(n\times n\) matrix \(C_{r^{\ast }}\) is called a geometric circulant matrix if it is of the form
$$ C_{r^{\ast }}= \begin{pmatrix} c_{0} & c_{1} & c_{2} & \ldots & c_{n-2} & c_{n-1} \\ rc_{n-1} & c_{0} & c_{1} & \ldots & c_{n-3} & c_{n-2} \\ r^{2}c_{n-2} & rc_{n-1} & c_{0} & \ldots & c_{n-4} & c_{n-3} \\ \vdots & \vdots & \vdots & & \vdots & \vdots \\ r^{n-1}c_{1} & r^{n-2}c_{2} & r^{n-3}c_{3} & \ldots & rc_{n-1} & c_{0} \end{pmatrix} . $$
Anzeige
We denote it for brevity by \(C_{r^{\ast }}=\operatorname{Circ}_{r^{\ast}}(c_{0},c_{1},c_{2},\ldots ,c_{n-1})\). Note that, for \(r=1\), geometric circulant matrix turns into circulant matrix given in [11, 12]. In fact, in [11, 12], the authors calculated the spectral norms of the circulant matrices with the generalized Fibonacci and hyperharmonic Fibonacci numbers.
Definition 2
Let \(A=(a_{ij})\) be any \(m\times n\) matrix. The Euclidean norm of matrix A is
$$ \Vert A\Vert _{E}=\sqrt{ \Biggl(\sum _{i=1}^{m}\sum_{j=1}^{n} \vert a_{ij}\vert ^{2} \Biggr) }. $$
Definition 3
Let \(A=(a_{ij})\) be any \(m\times n\) matrix. The spectral norm of matrix A is
where \(\lambda _{i}(A^{H}A)\) is eigenvalue of \(A^{H}A\) and \(A^{H}\) is conjugate transpose of matrix A.
$$ \Vert A\Vert _{2}=\sqrt{\max_{1\leq i\leq n}\lambda _{i}\bigl(A^{H}A\bigr)}, $$
Then the following inequalities hold between the Euclidean norm and the spectral norm:
$$\begin{aligned}& \frac{1}{\sqrt{n}} \Vert A\Vert _{E}\leq \Vert A\Vert _{2}\leq \Vert A\Vert _{E}, \end{aligned}$$
(5)
$$\begin{aligned}& \Vert A\Vert _{2}\leq \Vert A\Vert _{E}\leq \sqrt{n} \Vert A\Vert _{2}. \end{aligned}$$
(6)
Definition 4
Let \(A=(a_{ij})\) and \(B=(b_{ij})\) are each \(m\times n\) matrices, then their Hadamard product is the \(m\times n\) matrix of elementwise products
$$ A\circ B=(a_{ij}b_{ij}). $$
Lemma 1
[13] Let
A
and
B
be two
\(m\times n\)
matrices. We have
where
$$ \Vert A\circ B\Vert _{2}\leq r_{1}(A)c_{1}(B), $$
$$\begin{aligned}& r_{1}(A)=\max_{1\leq i\leq m}\sqrt{\sum _{j=1}^{n}\vert a_{ij}\vert ^{2}},\\& c_{1}(B)=\max_{1\leq j\leq n}\sqrt{\sum _{i=1}^{m}\vert b_{ij}\vert ^{2}}. \end{aligned}$$
3 Main results
Theorem 1
Let
\(U_{r^{\ast }}=\operatorname{Circ}_{r^{\ast }}(U_{0},U_{1},U_{2},\ldots,U_{n-1})\)
be an
\(n\times n\)
geometric circulant matrix.
(i)
If
\(\vert r\vert >1\), then
$$ \sqrt{\frac{U_{n}^{2}-U_{n-1}^{2}+(-1)^{n}}{p^{2}}}\leq \Vert U_{r^{\ast }}\Vert _{2}\leq \sqrt{\frac{(\vert r\vert ^{2}-\vert r\vert ^{2n})(U_{n}^{2}-U_{n-1}^{2}+(-1)^{n})}{ (1-\vert r\vert ^{2} ) p^{2}}}. $$
(ii)
If
\(\vert r\vert <1\), then
$$\begin{aligned} &\frac{\vert r\vert }{\sqrt{p^{2}+4}}\sqrt{\frac{2\vert r\vert ^{2n+2}-\vert r\vert ^{2n}(p^{2}+2)-\vert r\vert ^{2}V_{2n}+V_{2n-2}}{\vert r\vert ^{4}-\vert r\vert ^{2}(p^{2}+2)+1}-2\frac{\vert r\vert ^{2n}-(-1)^{n}}{ \vert r\vert ^{2}+1}} \\ &\quad \leq \Vert U_{r^{\ast }}\Vert _{2}\leq \sqrt{\frac{(n-1) ( U_{n}^{2}-U_{n-1}^{2}+(-1)^{n} ) }{p^{2}}}. \end{aligned}$$
Proof
We have the matrix
$$ U_{r^{\ast }}= \begin{pmatrix} U_{0} & U_{1} & U_{2} & \ldots & U_{n-2} & U_{n-1} \\ rU_{n-1} & U_{0} & U_{1} & \ldots & U_{n-3} & U_{n-2} \\ r^{2}U_{n-2} & rU_{n-1} & U_{0} & \ldots & U_{n-4} & U_{n-3} \\ \vdots & \vdots & \vdots & & \vdots & \vdots \\ r^{n-1}U_{1} & r^{n-2}U_{2} & r^{n-3}U_{3} & \ldots & rU_{n-1} & U_{0} \end{pmatrix} . $$
(i) From \(\vert r\vert >1\) and definition of Euclidean norm, we have
that is,
from (5), we have
On the other hand, let the matrices A and B be defined by
and
That is, \(U_{r^{\ast }}=A\circ B\). Then we obtain
and
From Lemma 1, we have
Thus, we have
$$\begin{aligned} \Vert U_{r^{\ast }}\Vert _{E}^{2}&=\sum _{k=0}^{n-1}(n-k)U_{k}^{2}+\sum _{k=1}^{n-1}k\bigl\vert r^{n-k}\bigr\vert ^{2}U_{k}^{2} \\ &\geq \sum_{k=0}^{n-1}(n-k)U_{k}^{2}+ \sum_{k=1}^{n-1}kU_{k}^{2} \\ &= n\sum_{k=0}^{n-1}U_{k}^{2} \\ &= n\frac{U_{n}^{2}-U_{n-1}^{2}+(-1)^{n}}{p^{2}}, \end{aligned}$$
$$ \frac{1}{\sqrt{n}} \Vert U_{r^{\ast }}\Vert _{E}\geq \sqrt {\frac{U_{n}^{2}-U_{n-1}^{2}+(-1)^{n}}{p^{2}}}; $$
$$ \sqrt{\frac{U_{n}^{2}-U_{n-1}^{2}+(-1)^{n}}{p^{2}}}\leq \Vert U_{r^{\ast }}\Vert _{2}. $$
$$ A= \begin{pmatrix} U_{0} & 1 & 1 & \ldots & 1 & 1 \\ r & U_{0} & 1 & \ldots & 1 & 1 \\ r^{2} & r & U_{0} & \ldots & 1 & 1 \\ \vdots & \vdots & \vdots & & \vdots & \vdots \\ r^{n-1} & r^{n-2} & r^{n-3} & \ldots & r & U_{0} \end{pmatrix} $$
$$ B= \begin{pmatrix} U_{0} & U_{1} & U_{2} & \ldots & U_{n-2} & U_{n-1} \\ U_{n-1} & U_{0} & U_{1} & \ldots & U_{n-3} & U_{n-2} \\ U_{n-2} & U_{n-1} & U_{0} & \ldots & U_{n-4} & U_{n-3} \\ \vdots & \vdots & \vdots & & \vdots & \vdots \\ U_{1} & U_{2} & U_{3} & \ldots & U_{n-1} & U_{0} \end{pmatrix} . $$
$$\begin{aligned} r_{1}(A) &=\max_{1\leq i\leq n}\sqrt{\sum _{j=1}^{n}\vert a_{ij}\vert ^{2}} \\ &=\sqrt{\bigl\vert r^{1}\bigr\vert ^{2}+\cdots +\bigl\vert r^{n-1}\bigr\vert ^{2}} \\ &=\sqrt{\frac{\vert r\vert ^{2}-\vert r\vert ^{2n}}{1-\vert r\vert ^{2}}} \end{aligned}$$
$$\begin{aligned} c_{1}(B) &=\max_{1\leq j\leq n}\sqrt{\sum _{i=1}^{n}\vert b_{ij}\vert ^{2}} \\ &=\sqrt{\sum_{k=0}^{n-1}U_{k}^{2}} \\ &=\sqrt{\frac{U_{n}^{2}-U_{n-1}^{2}+(-1)^{n}}{p^{2}}}. \end{aligned}$$
$$ \Vert U_{r^{\ast }}\Vert _{2}\leq \sqrt{ \frac{(\vert r\vert ^{2}-\vert r\vert ^{2n})(U_{n}^{2}-U_{n-1}^{2}+(-1)^{n})}{ ( 1-\vert r\vert ^{2} ) p^{2}}}. $$
$$ \sqrt{\frac{U_{n}^{2}-U_{n-1}^{2}+(-1)^{n}}{p^{2}}}\leq \Vert U_{r^{\ast }}\Vert _{2}\leq \sqrt{\frac{(\vert r\vert ^{2}-\vert r\vert ^{2n})(U_{n}^{2}-U_{n-1}^{2}+(-1)^{n})}{ ( 1-\vert r\vert ^{2} ) p^{2}}}. $$
(ii) From \(\vert r\vert <1\), we have
From (5)
On the other hand, let the matrices A and B be defined by
and
That is, \(U_{r^{\ast }}=A\circ B\). Then we obtain
and
Hence, from Lemma 1, we have
$$\begin{aligned} &{\Vert U_{r^{\ast }}\Vert _{E}^{2} =\sum _{k=0}^{n-1}(n-k)U_{k}^{2}+\sum _{k=1}^{n-1}k\bigl\vert r^{n-k}\bigr\vert ^{2}U_{k}^{2}} \\ &{\hphantom{\Vert U_{r^{\ast }}\Vert _{E}^{2} }\geq \sum_{k=0}^{n-1}(n-k)\bigl\vert r^{n-k}\bigr\vert ^{2}U_{k}^{2}+\sum _{k=1}^{n-1}k\bigl\vert r^{n-k}\bigr\vert ^{2}U_{k}^{2}} \\ &{\hphantom{\Vert U_{r^{\ast }}\Vert _{E}^{2} }= n\vert r\vert ^{2n}\sum_{k=0}^{n-1} \biggl( \frac{U_{k}}{\vert r\vert ^{k}} \biggr) ^{2}} \\ &{\hphantom{\Vert U_{r^{\ast }}\Vert _{E}^{2} }=\frac{n\vert r\vert ^{2n}}{p^{2}+4}\sum_{k=0}^{n-1} \biggl( \frac{\alpha ^{k}-\beta ^{k}}{\vert r\vert ^{k}} \biggr) ^{2}} \\ &{\hphantom{\Vert U_{r^{\ast }}\Vert _{E}^{2} }=\frac{n\vert r\vert ^{2n}}{p^{2}+4} \biggl( \frac{1- ( \frac{\alpha ^{2}}{\vert r\vert ^{2}} ) ^{n}}{1- ( \frac{\alpha ^{2}}{\vert r\vert ^{2}} ) }+\frac{1- ( \frac{\beta ^{2}}{\vert r\vert ^{2}} ) ^{n}}{1- ( \frac{\beta^{2}}{\vert r\vert ^{2}} ) }-2 \frac{1- ( \frac{-1}{\vert r\vert ^{2}} ) ^{n}}{1+\frac{1}{\vert r\vert ^{2}}} \biggr)} \\ &{\hphantom{\Vert U_{r^{\ast }}\Vert _{E}^{2} }=\frac{n\vert r\vert ^{2}}{p^{2}+4}\sqrt{\frac{2\vert r\vert ^{2n+2}-\vert r\vert ^{2n}(p^{2}+2)-\vert r\vert ^{2}V_{2n}+V_{2n-2}}{\vert r\vert ^{4}-\vert r\vert ^{2}(p^{2}+2)+1}-2\frac{\vert r\vert ^{2n}-(-1)^{n}}{\vert r\vert ^{2}+1}},}\\ &{\frac{1}{\sqrt{n}} \Vert U_{r^{\ast }}\Vert _{E} \geq \frac{\vert r\vert }{\sqrt{p^{2}+4}}\sqrt{\frac{2\vert r\vert ^{2n+2}-\vert r\vert ^{2n}(p^{2}+2)-\vert r\vert ^{2}V_{2n}+V_{2n-2}}{\vert r\vert ^{4}-\vert r\vert ^{2}(p^{2}+2)+1}-2\frac{\vert r\vert ^{2n}-(-1)^{n}}{\vert r\vert ^{2}+1}}.} \end{aligned}$$
$$ \frac{\vert r\vert }{\sqrt{p^{2}+4}}\sqrt{\frac{2\vert r\vert ^{2n+2}-\vert r\vert ^{2n}(p^{2}+2)-\vert r\vert ^{2}V_{2n}+V_{2n-2}}{\vert r\vert ^{4}-\vert r\vert ^{2}(p^{2}+2)+1}-2\frac{\vert r\vert ^{2n}-(-1)^{n}}{\vert r\vert ^{2}+1}}\leq \Vert U_{r^{\ast }}\Vert _{2}. $$
$$ A= \begin{pmatrix} U_{0} & 1 & 1 & \ldots & 1 & 1 \\ r & U_{0} & 1 & \ldots & 1 & 1 \\ r^{2} & r & U_{0} & \ldots & 1 & 1 \\ \vdots & \vdots & \vdots & & \vdots & \vdots \\ r^{n-1} & r^{n-2} & r^{n-3} & \ldots & r & U_{0} \end{pmatrix}$$
$$ B= \begin{pmatrix} U_{0} & U_{1} & U_{2} & \ldots & U_{n-2} & U_{n-1} \\ U_{n-1} & U_{0} & U_{1} & \ldots & U_{n-3} & U_{n-2} \\ U_{n-2} & U_{n-1} & U_{0} & \ldots & U_{n-4} & U_{n-3} \\ \vdots & \vdots & \vdots & & \vdots & \vdots \\ U_{1} & U_{2} & U_{3} & \ldots & U_{n-1} & U_{0} \end{pmatrix}. $$
$$\begin{aligned} r_{1}(A) &=\max_{1\leq i\leq n}\sqrt{\sum _{j=1}^{n}\vert a_{ij}\vert ^{2}} \\ &=\sqrt{U_{0}^{2}+n-1} \\ &=\sqrt{n-1} \end{aligned}$$
$$\begin{aligned} c_{1}(B) &=\max_{1\leq j\leq n}\sqrt{\sum _{i=1}^{n}\vert b_{ij}\vert ^{2}} \\ &=\sqrt{\sum_{k=0}^{n-1}U_{k}^{2}} \\ &=\sqrt{\frac{U_{n}^{2}-U_{n-1}^{2}+(-1)^{n}}{p^{2}}}. \end{aligned}$$
$$ \Vert U_{r^{\ast }}\Vert _{2}\leq \sqrt{ \frac{(n-1) (U_{n}^{2}-U_{n-1}^{2}+(-1)^{n} ) }{p^{2}}}. $$
Thus, we have
□
$$\begin{aligned} &\frac{\vert r\vert }{\sqrt{p^{2}+4}}\sqrt{\frac{2\vert r\vert ^{2n+2}-\vert r\vert ^{2n}(p^{2}+2)-\vert r\vert ^{2}V_{2n}+V_{2n-2}}{\vert r\vert ^{4}-\vert r\vert ^{2}(p^{2}+2)+1}-2\frac{\vert r\vert ^{2n}-(-1)^{n}}{\vert r\vert ^{2}+1}} \\ &\quad\leq \Vert U_{r^{\ast }}\Vert _{2} \leq \sqrt{\frac{(n-1) ( U_{n}^{2}-U_{n-1}^{2}+(-1)^{n} ) }{p^{2}}}. \end{aligned}$$
Theorem 2
Let
\(V_{r^{\ast }}=\operatorname{Circ}_{r^{\ast}}(V_{0},V_{1},V_{2},\ldots ,V_{n-1})\)
be an
\(n\times n\)
geometric circulant matrix.
(i)
If
\(\vert r\vert >1\), then
$$\begin{aligned} &\sqrt{\frac{V_{n}^{2}-V_{n-1}^{2}+p^{2}-4+(1-(-1)^{n})(p^{2}+4)}{p^{2}}} \\ &\quad \leq \Vert V_{r^{\ast }}\Vert _{2}\leq \sqrt{\frac{1-\vert r\vert ^{2n}}{1-\vert r\vert ^{2}}\frac{V_{n}^{2}-V_{n-1}^{2}+p^{2}-4+(1-(-1)^{n})(p^{2}+4)}{p^{2}}}. \end{aligned}$$
(ii)
If
\(\vert r\vert <1\), then
$$\begin{aligned} &\vert r\vert \sqrt{\frac{2\vert r\vert ^{2n+2}-\vert r\vert ^{2n}(p^{2}+2)-\vert r\vert ^{2}V_{2n}+V_{2n-2}}{\vert r\vert ^{4}-\vert r\vert ^{2}(p^{2}+2)+1}+2\frac{\vert r\vert ^{2n}-(-1)^{n}}{\vert r\vert ^{2}+1}} \\ & \quad\leq \Vert V_{r^{\ast }}\Vert _{2}\leq \sqrt{\frac{n (V_{n}^{2}-V_{n-1}^{2}+p^{2}-4+(1-(-1)^{n})(p^{2}+4) ) }{p^{2}}}. \end{aligned}$$
Proof
We have the matrix
$$ V_{r^{\ast }}= \begin{pmatrix} V_{0} & V_{1} & V_{2} & \ldots & V_{n-2} & V_{n-1} \\ rV_{n-1} & V_{0} & V_{1} & \ldots & V_{n-3} & V_{n-2} \\ r^{2}V_{n-2} & rV_{n-1} & V_{0} & \ldots & V_{n-4} & V_{n-3} \\ \vdots & \vdots & \vdots & & \vdots & \vdots \\ r^{n-1}V_{1} & r^{n-2}V_{2} & r^{n-3}V_{3} & \ldots & rV_{n-1} & V_{0} \end{pmatrix}. $$
(i) From \(\vert r\vert >1\), we have
that is,
from (5), we have
On the other hand, let the matrices A and B be defined by
and
That is, \(V_{r^{\ast }}=A\circ B\). Then we obtain
and
From Lemma 1, we have
Thus, we have
$$\begin{aligned} \Vert V_{r^{\ast }}\Vert _{E}^{2}&=\sum _{k=0}^{n-1}(n-k)V_{k}^{2}+\sum _{k=1}^{n-1}k\bigl\vert r^{n-k}\bigr\vert ^{2}V_{k}^{2} \\ &\geq n\sum_{k=0}^{n-1}V_{k}^{2} \\ &=\frac{n ( V_{n}^{2}-V_{n-1}^{2}+p^{2}-4+(1-(-1)^{n})(p^{2}+4) ) }{p^{2}}, \end{aligned}$$
$$ \frac{1}{\sqrt{n}} \Vert V_{r^{\ast }}\Vert _{E}\geq \sqrt {\frac{V_{n}^{2}-V_{n-1}^{2}+p^{2}-4+(1-(-1)^{n})(p^{2}+4)}{p^{2}}}, $$
$$ \sqrt{\frac{V_{n}^{2}-V_{n-1}^{2}+p^{2}-4+(1-(-1)^{n})(p^{2}+4)}{p^{2}}}\leq \Vert V_{r^{\ast }}\Vert _{2}. $$
$$ A= \begin{pmatrix} 1 & 1 & 1 & \ldots & 1 & 1 \\ r & 1 & 1 & \ldots & 1 & 1 \\ r^{2} & r & 1 & \ldots & 1 & 1 \\ \vdots & \vdots & \vdots & & \vdots & \vdots \\ r^{n-1} & r^{n-2} & r^{n-3} & \ldots & r & 1 \end{pmatrix} $$
$$ B= \begin{pmatrix} V_{0} & V_{1} & V_{2} & \ldots & V_{n-2} & V_{n-1} \\ V_{n-1} & V_{0} & V_{1} & \ldots & V_{n-3} & V_{n-2} \\ V_{n-2} & V_{n-1} & V_{0} & \ldots & V_{n-4} & V_{n-3} \\ \vdots & \vdots & \vdots & & \vdots & \vdots \\ V_{1} & V_{2} & V_{3} & \ldots & V_{n-1} & V_{0} \end{pmatrix}. $$
$$\begin{aligned} r_{1}(A) &=\max_{1\leq i\leq n}\sqrt{\sum _{j=1}^{n}\vert a_{ij}\vert ^{2}} \\ &=\sqrt{1+\bigl\vert r^{1}\bigr\vert ^{2}+\cdots +\bigl\vert r^{n-1}\bigr\vert ^{2}} \\ &=\sqrt{\frac{1-\vert r\vert ^{2n}}{1-\vert r\vert ^{2}}} \end{aligned}$$
$$\begin{aligned} c_{1}(B) &=\max_{1\leq j\leq n}\sqrt{\sum _{i=1}^{n}\vert b_{ij}\vert ^{2}} \\ &=\sqrt{\sum_{k=0}^{n-1}V_{k}^{2}} \\ &=\sqrt{\frac{V_{n}^{2}-V_{n-1}^{2}+p^{2}-4+(1-(-1)^{n})(p^{2}+4)}{p^{2}}}. \end{aligned}$$
$$ \Vert V_{r^{\ast }}\Vert _{2}\leq \sqrt{ \frac{1-\vert r\vert ^{2n}}{1-\vert r\vert ^{2}}\frac{ V_{n}^{2}-V_{n-1}^{2}+p^{2}-4+(1-(-1)^{n})(p^{2}+4)}{p^{2}}}. $$
$$\begin{aligned} &\sqrt{\frac{V_{n}^{2}-V_{n-1}^{2}+p^{2}-4+(1-(-1)^{n})(p^{2}+4)}{p^{2}}} \\ & \quad\leq \Vert V_{r^{\ast }}\Vert _{2}\leq \sqrt{\frac{1-\vert r\vert ^{2n}}{1-\vert r\vert ^{2}}\frac{V_{n}^{2}-V_{n-1}^{2}+p^{2}-4+(1-(-1)^{n})(p^{2}+4)}{p^{2}}}. \end{aligned}$$
(ii) From \(\vert r\vert <1\), we have
From (5),
On the other hand, let the matrices A and B be defined by
and
That is, \(V_{r^{\ast }}=A\circ B\). Then we obtain
and
From Lemma 1, we have
Thus we have
□
$$\begin{aligned} &{\Vert V_{r^{\ast }}\Vert _{E}^{2}=\sum _{k=0}^{n-1}(n-k)V_{k}^{2}+\sum _{k=1}^{n-1}k\bigl\vert r^{n-k}\bigr\vert ^{2}V_{k}^{2}} \\ &{\hphantom{\Vert V_{r^{\ast }}\Vert _{E}^{2}}\geq \sum_{k=0}^{n-1}(n-k)\bigl\vert r^{n-k}\bigr\vert ^{2}V_{k}^{2}+\sum _{k=1}^{n-1}k\bigl\vert r^{n-k}\bigr\vert ^{2}V_{k}^{2}} \\ &{\hphantom{\Vert V_{r^{\ast }}\Vert _{E}^{2}}= n\vert r\vert ^{2n}\sum_{k=0}^{n-1} \biggl( \frac{V_{k}}{\vert r\vert ^{k}} \biggr) ^{2}} \\ &{\hphantom{\Vert V_{r^{\ast }}\Vert _{E}^{2}}= n\vert r\vert ^{2n} \Biggl( \sum_{k=0}^{n-1} \frac{\alpha ^{2k}}{\vert r\vert ^{2k}}+\sum_{k=0}^{n-1} \frac{\beta ^{2k}}{\vert r\vert ^{2k}}+2\sum_{k=0}^{n-1} \frac{(-1)^{k}}{\vert r\vert ^{2k}} \Biggr)} \\ &{\hphantom{\Vert V_{r^{\ast }}\Vert _{E}^{2}}= n\vert r\vert ^{2n} \biggl( \frac{1- ( \frac{\alpha ^{2}}{\vert r\vert ^{2}} ) ^{n}}{1- ( \frac{\alpha ^{2}}{\vert r\vert ^{2}} ) }+ \frac{1- ( \frac{\beta ^{2}}{\vert r\vert ^{2}} ) ^{n}}{1- ( \frac{\beta ^{2}}{\vert r\vert ^{2}} ) }+2\frac{1- ( \frac{-1}{\vert r\vert ^{2}} ) ^{n}}{1+\frac{1}{\vert r\vert ^{2}}} \biggr)} \\ &{\hphantom{\Vert V_{r^{\ast }}\Vert _{E}^{2}}= n\vert r\vert ^{2} \biggl( \frac{2\vert r\vert ^{2n+2}-\vert r\vert ^{2n}(p^{2}+2)-\vert r\vert ^{2}V_{2n}+V_{2n-2}}{\vert r\vert ^{4}-\vert r\vert ^{2}(p^{2}+2)+1}+2 \frac{\vert r\vert ^{2n}-(-1)^{n}}{\vert r\vert ^{2}+1} \biggr) ,}\\ &{\frac{1}{\sqrt{n}} \Vert V_{r^{\ast }}\Vert _{E} \geq \vert r\vert \sqrt{\frac{2\vert r\vert ^{2n+2}-\vert r\vert ^{2n}(p^{2}+2)-\vert r\vert ^{2}V_{2n}+V_{2n-2}}{\vert r\vert ^{4}-\vert r\vert ^{2}(p^{2}+2)+1}+2\frac{\vert r\vert ^{2n}-(-1)^{n}}{\vert r\vert ^{2}+1}}.} \end{aligned}$$
$$ \vert r\vert \sqrt{\frac{2\vert r\vert ^{2n+2}-\vert r\vert ^{2n}(p^{2}+2)-\vert r\vert ^{2}V_{2n}+V_{2n-2}}{\vert r\vert ^{4}-\vert r\vert ^{2}(p^{2}+2)+1}+2\frac{\vert r\vert ^{2n}-(-1)^{n}}{\vert r\vert ^{2}+1}}\leq \Vert V_{r^{\ast }}\Vert _{2}. $$
$$ A= \begin{pmatrix} 1 & 1 & 1 & \ldots & 1 & 1 \\ r & 1 & 1 & \ldots & 1 & 1 \\ r^{2} & r & 1 & \ldots & 1 & 1 \\ \vdots & \vdots & \vdots & & \vdots & \vdots \\ r^{n-1} & r^{n-2} & r^{n-3} & \ldots & r & 1 \end{pmatrix} $$
$$ B= \begin{pmatrix} V_{0} & V_{1} & V_{2} & \ldots & V_{n-2} & V_{n-1} \\ V_{n-1} & V_{0} & V_{1} & \ldots & V_{n-3} & V_{n-2} \\ V_{n-2} & V_{n-1} & V_{0} & \ldots & V_{n-4} & V_{n-3} \\ \vdots & \vdots & \vdots & & \vdots & \vdots \\ V_{1} & V_{2} & V_{3} & \ldots & V_{n-1} & V_{0} \end{pmatrix}. $$
$$ r_{1}(A)=\max_{1\leq i\leq n}\sqrt{\sum _{j=1}^{n}\vert a_{ij}\vert ^{2}}=\sqrt{n} $$
$$\begin{aligned} c_{1}(B) &=\max_{1\leq j\leq n}\sqrt{\sum _{i=1}^{n}\vert b_{ij}\vert ^{2}} \\ &=\sqrt{\sum_{k=0}^{n-1}V_{k}^{2}} \\ &=\sqrt{\frac{V_{n}^{2}-V_{n-1}^{2}+p^{2}-4+(1-(-1)^{n})(p^{2}+4)}{p^{2}}}. \end{aligned}$$
$$ \Vert V_{r^{\ast }}\Vert _{2}\leq \sqrt{ \frac{n (V_{n}^{2}-V_{n-1}^{2}+p^{2}-4+(1-(-1)^{n})(p^{2}+4) ) }{p^{2}}}. $$
$$\begin{aligned} \vert r\vert & \sqrt{\frac{2\vert r\vert ^{2n+2}-\vert r\vert ^{2n}(p^{2}+2)-\vert r\vert ^{2}V_{2n}+V_{2n-2}}{\vert r\vert ^{4}-\vert r\vert ^{2}(p^{2}+2)+1}+2\frac{\vert r\vert ^{2n}-(-1)^{n}}{\vert r\vert ^{2}+1}} \\ & \quad\leq \Vert V_{r^{\ast }}\Vert _{2}\leq \sqrt{\frac{n (V_{n}^{2}-V_{n-1}^{2}+p^{2}-4+(1-(-1)^{n})(p^{2}+4) ) }{p^{2}}}. \end{aligned}$$
Theorem 3
Let
\(\mathbb{F}_{r^{\ast }}^{(k)}=\operatorname{Circ}_{r^{\ast }}(\mathbb{F}_{0}^{(k)},\mathbb{F}_{1}^{(k)},\mathbb{F}_{2}^{(k)},\ldots ,\mathbb{F}_{n-1}^{(k)})\)
be an
\(n\times n\)
geometric circulant matrix.
(i)
If
\(\vert r\vert >1\), then
$$ \frac{1}{\sqrt{n}}\mathbb{F}_{n-1}^{(k+1)}\leq \bigl\Vert \mathbb{F}_{r^{\ast }}^{(k)}\bigr\Vert _{2}\leq \sqrt{ \frac{\vert r\vert ^{2}-\vert r\vert ^{2n}}{1-\vert r\vert ^{2}}}\mathbb{F}_{n-1}^{(k+1)}. $$
(ii)
If
\(\vert r\vert <1\), then
$$ \frac{\vert r\vert ^{n}}{\sqrt{n}}\mathbb{F}_{n-1}^{(k+1)}\leq \bigl\Vert \mathbb{F}_{r^{\ast }}^{(k)}\bigr\Vert _{2}\leq \sqrt{n-1} \mathbb{F}_{n-1}^{(k+1)}. $$
Proof
Since the \(\mathbb{F}_{r^{\ast }}^{(k)}\) is of the form
and from the definition of Euclidean norm, we have
$$ \mathbb{F}_{r^{\ast }}^{(k)}= \begin{pmatrix} \mathbb{F}_{0}^{(k)} & \mathbb{F}_{1}^{(k)} & \mathbb{F}_{2}^{(k)} & \ldots& \mathbb{F}_{n-2}^{(k)} & \mathbb{F}_{n-1}^{(k)} \\ r\mathbb{F}_{n-1}^{(k)} & \mathbb{F}_{0}^{(k)} & \mathbb{F}_{1}^{(k)} & \ldots & \mathbb{F}_{n-3}^{(k)} & \mathbb{F}_{n-2}^{(k)} \\ r^{2}\mathbb{F}_{n-2}^{(k)} & r\mathbb{F}_{n-1}^{(k)} & \mathbb{F}_{0}^{(k)}& \ldots & \mathbb{F}_{n-4}^{(k)} &\mathbb{F}_{n-3}^{(k)} \\ \vdots & \vdots & \vdots & & \vdots & \vdots \\ r^{n-1}\mathbb{F}_{1}^{(k)} & r^{n-2}\mathbb{F}_{2}^{(k)} & r^{n-3}\mathbb{F}_{3}^{(k)} & \ldots & r\mathbb{F}_{n-1}^{(k)} & \mathbb{F}_{0}^{(k)} \end{pmatrix} $$
$$ \bigl\Vert \mathbb{F}_{r^{\ast }}^{(k)}\bigr\Vert _{E}^{2}=\sum_{s=0}^{n-1}(n-s) \bigl( \mathbb{F}_{s}^{(k)} \bigr)^{2}+\sum _{s=1}^{n-1}s\bigl\vert r^{n-s}\bigr\vert ^{2} \bigl( \mathbb{F}_{s}^{(k)} \bigr) ^{2}. $$
(i) From \(\vert r\vert >1\), we have
Thus from (5) and (4),
On the other hand let the matrices \(A^{(k)}\) and \(B^{(k)}\) be defined by
and
That is, \(\mathbb{F}_{r^{\ast }}^{(k)}=A^{(k)}\circ B^{(k)}\). Then we obtain
and
From Lemma 1 and (4), we have
which is desired result.
$$\begin{aligned} \bigl\Vert \mathbb{F}_{r^{\ast }}^{(k)}\bigr\Vert _{E}^{2} &\geq \sum_{s=0}^{n-1}(n-s) \bigl( \mathbb{F}_{s}^{(k)} \bigr) ^{2}+\sum _{s=1}^{n-1}s \bigl( \mathbb{F}_{s}^{(k)} \bigr) ^{2} \\ &= n\sum_{s=0}^{n-1} \bigl( \mathbb{F}_{s}^{(k)} \bigr) ^{2}. \end{aligned}$$
$$ \frac{1}{\sqrt{n}}\mathbb{F}_{n-1}^{(k+1)}\leq \bigl\Vert \mathbb{F}_{r^{\ast }}^{(k)}\bigr\Vert _{2}. $$
$$ A^{(k)}= \begin{pmatrix} \mathbb{F}_{0}^{(k)} & 1 & 1 & \ldots & 1 & 1 \\ r & \mathbb{F}_{0}^{(k)} & 1 & \ldots & 1 & 1 \\ r^{2} & r & \mathbb{F}_{0}^{(k)} & \ldots & 1 & 1 \\ \vdots & \vdots & \vdots & & \vdots & \vdots \\ r^{n-1} & r^{n-2} & r^{n-3} & \ldots & r & \mathbb{F}_{0}^{(k)} \end{pmatrix} $$
$$ B^{(k)}= \begin{pmatrix} \mathbb{F}_{0}^{(k)} & \mathbb{F}_{1}^{(k)} & \mathbb{F}_{2}^{(k)} & \ldots& \mathbb{F}_{n-2}^{(k)} & \mathbb{F}_{n-1}^{(k)} \\ \mathbb{F}_{n-1}^{(k)} & \mathbb{F}_{0}^{(k)} & \mathbb{F}_{1}^{(k)} & \ldots & \mathbb{F}_{n-3}^{(k)} & \mathbb{F}_{n-2}^{(k)} \\ \mathbb{F}_{n-2}^{(k)} & \mathbb{F}_{n-1}^{(k)} & \mathbb{F}_{0}^{(k)} & \ldots & \mathbb{F}_{n-4}^{(k)} & \mathbb{F}_{n-3}^{(k)} \\ \vdots & \vdots & \vdots & & \vdots & \vdots \\ \mathbb{F}_{1}^{(k)} & \mathbb{F}_{2}^{(k)} & \mathbb{F}_{3}^{(k)} & \ldots & \mathbb{F}_{n-1}^{(k)} & \mathbb{F}_{0}^{(k)} \end{pmatrix}. $$
$$ r_{1}\bigl(A^{(k)}\bigr)=\max_{1\leq i\leq n}\sqrt {\sum_{j=1}^{n}\bigl\vert a_{ij}^{(k)}\bigr\vert ^{2}}=\sqrt{ \frac{\vert r\vert ^{2}-\vert r\vert ^{2n}}{1-\vert r\vert ^{2}}} $$
$$ c_{1}\bigl(B^{(k)}\bigr)=\max_{1\leq j\leq n}\sqrt {\sum_{i=1}^{n}\bigl\vert b_{ij}^{(k)}\bigr\vert ^{2}}=\sqrt{\sum _{s=0}^{n-1} \bigl( \mathbb{F}_{s}^{(k)} \bigr) ^{2}}. $$
$$ \bigl\Vert \mathbb{F}_{r^{\ast }}^{(k)}\bigr\Vert _{2} \leq \sqrt{\frac{\vert r\vert ^{2}-\vert r\vert ^{2n}}{1-\vert r\vert ^{2}}}\mathbb{F}_{n-1}^{(k+1)}, $$
(ii) From \(\vert r\vert <1\), we have
From (5) and (4),
On the other hand, let the matrices \(A^{(k)}\) and \(B^{(k)}\) be defined by
and
That is, \(\mathbb{F}_{r^{\ast }}^{(k)}=A^{(k)}\circ B^{(k)}\). Then we obtain
and
From Lemma 1 and (4), we have
Thus we have
□
$$\begin{aligned} \bigl\Vert \mathbb{F}_{r^{\ast }}^{(k)}\bigr\Vert _{E}^{2} &\geq \sum_{s=0}^{n-1}(n-s) \bigl\vert r^{n-s}\bigr\vert ^{2} \bigl( \mathbb{F}_{s}^{(k)} \bigr) ^{2}+\sum _{s=1}^{n-1}s\bigl\vert r^{n-s}\bigr\vert ^{2} \bigl( \mathbb{F}_{s}^{(k)} \bigr) ^{2} \\ &\geq n\vert r\vert ^{2n}\sum_{s=0}^{n-1} \bigl( \mathbb{F}_{s}^{(k)} \bigr) ^{2}. \end{aligned}$$
$$ \frac{\vert r\vert ^{n}}{\sqrt{n}}\mathbb{F}_{n-1}^{(k+1)}\leq \bigl\Vert \mathbb{F}_{r^{\ast }}^{(k)}\bigr\Vert _{2}. $$
$$ A^{(k)}= \begin{pmatrix} \mathbb{F}_{0}^{(k)} & 1 & 1 & \ldots & 1 & 1 \\ r & \mathbb{F}_{0}^{(k)} & 1 & \ldots & 1 & 1 \\ r^{2} & r & \mathbb{F}_{0}^{(k)} & \ldots & 1 & 1 \\ \vdots & \vdots & \vdots & & \vdots & \vdots \\ r^{n-1} & r^{n-2} & r^{n-3} & \ldots & r & \mathbb{F}_{0}^{(k)} \end{pmatrix} $$
$$ B^{(k)}= \begin{pmatrix} \mathbb{F}_{0}^{(k)} & \mathbb{F}_{1}^{(k)} & \mathbb{F}_{2}^{(k)} & \ldots & \mathbb{F}_{n-2}^{(k)} & \mathbb{F}_{n-1}^{(k)} \\ \mathbb{F}_{n-1}^{(k)} & \mathbb{F}_{0}^{(k)} & \mathbb{F}_{1}^{(k)} & \ldots & \mathbb{F}_{n-3}^{(k)} & \mathbb{F}_{n-2}^{(k)} \\ \mathbb{F}_{n-2}^{(k)} & \mathbb{F}_{n-1}^{(k)} & \mathbb{F}_{0}^{(k)} & \ldots & \mathbb{F}_{n-4}^{(k)} & \mathbb{F}_{n-3}^{(k)} \\ \vdots & \vdots & \vdots & & \vdots & \vdots \\ \mathbb{F}_{1}^{(k)} & \mathbb{F}_{2}^{(k)} & \mathbb{F}_{3}^{(k)} & \ldots & \mathbb{F}_{n-1}^{(k)} & \mathbb{F}_{0}^{(k)} \end{pmatrix}. $$
$$ r_{1}\bigl(A^{(k)}\bigr)=\max_{1\leq i\leq n}\sqrt {\sum_{j=1}^{n}\bigl\vert a_{ij}^{(k)}\bigr\vert ^{2}}=\sqrt{n-1} $$
$$ c_{1}\bigl(B^{(k)}\bigr)=\max_{1\leq j\leq n}\sqrt {\sum_{i=1}^{n}\bigl\vert b_{ij}^{(k)}\bigr\vert ^{2}}=\sqrt{\sum _{s=0}^{n-1} \bigl( \mathbb{F}_{s}^{(k)} \bigr) ^{2}}. $$
$$ \bigl\Vert \mathbb{F}_{r^{\ast }}^{(k)}\bigr\Vert _{2} \leq \sqrt{n-1}\mathbb{F}_{n-1}^{(k+1)}. $$
$$ \frac{\vert r\vert ^{n}}{\sqrt{n}}\mathbb{F}_{n-1}^{(k+1)}\leq \bigl\Vert \mathbb{F}_{r^{\ast }}^{(k)}\bigr\Vert _{2}\leq \sqrt{n-1} \mathbb{F}_{n-1}^{(k+1)}. $$
4 Conclusion
In this paper we approximated lower and upper bounds of the spectral norms of geometric circulant matrices with the generalized Fibonacci and hyperharmonic Fibonacci numbers. If we take \(p=1\) and \(p=2\) in Theorem 1, we obtain lower and upper bounds of the spectral norms of geometric circulant matrices with the Fibonacci and Pell numbers, respectively. Similarly if we take \(p=1\) and \(p=2\) in Theorem 2, we obtain lower and upper bounds of the spectral norms of geometric circulant matrices with the Lucas and Pell-Lucas numbers, respectively.
In the future it may be possible that one can generalize our results to the Horadam, tribonacci and tribonacci-like sequences.
Acknowledgements
The authors are grateful to two anonymous referees and the associate editor for their careful reading, helpful comments, and constructive suggestions, which improved the presentation of results.
Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.
Competing interests
The authors declare that they have no competing interests.
Authors’ contributions
Each of the authors contributed to each part of this work equally and read and approved the final version of the manuscript.