1 Introduction

The topic of matrices has become one of the most important and interesting research areas in applied and computational mathematics in recent years. Several researchers have worked in different areas of application of matrices and they continue to work on matrices. In particular, studying determinants, norms and some properties of different types of matrices is a hot and important topic in matrix theory [1]. Many authors studied the norms of circulant matrices, r-circulant matrices, geometric circulant matrices, Hankel matrices, and Toeplitz matrices involving some well known famous number sequences [2,3,4,5,6,7,8,9,10,11,12,13]. Many special types of matrices have been introduced in the literature [14,15,16,17,18,19,20]. Two of these are r-Hankel and r-Toeplitz matrices.

A \(n\times n\) r-Hankel matrix \(H_{r}=\left( h_{ij}\right) _{i,j=0}^{n-1}\) is defined by [14]

$$\begin{aligned} H_{r}=\left( \begin{array}{ccccc} rh_{0} &{} rh_{1} &{} \cdots &{} rh_{n-2} &{} h_{n-1} \\ rh_{1} &{} rh_{2} &{} \cdots &{} h_{n-1} &{} h_{n} \\ \vdots &{} \vdots &{} \cdots &{} \vdots &{} \vdots \\ rh_{n-2} &{} h_{n-1} &{} \cdots &{} h_{2n-4} &{} h_{2n-3} \\ h_{n-1} &{} h_{n} &{} \cdots &{} h_{2n-3} &{} h_{2n-2} \end{array} \right) _{n\times n}. \end{aligned}$$

Namely,

$$\begin{aligned} h_{ij}=\left\{ \begin{array}{ll} rh_{i+j}, &{} \hbox {if }\ i+j< n-1 \\ &{} \\ h_{i+j}, &{} \hbox {if }\ i+j\ge n-1 \end{array} \right. , \end{aligned}$$

and the matrix \(H_{r}\) is symmetric and then normal matrix.

A matrix \(T_{r}=\left( t_{ij}\right) _{i,j=0}^{n-1}\) is called r-Toeplitz matrix [15] defined by

$$\begin{aligned} T_{r}=\left( \begin{array}{ccccc} t_{0} &{} t_{-1} &{} \cdots &{} t_{2-n} &{} t_{1-n} \\ rt_{1} &{} t_{0} &{} \cdots &{} t_{3-n} &{} t_{2-n} \\ \vdots &{} \vdots &{} &{} \vdots &{} \vdots \\ rt_{n-2} &{} rt_{n-3} &{} \cdots &{} t_{0} &{} t_{-1} \\ rt_{n-1} &{} rt_{n-2} &{} \cdots &{} rt_{1} &{} t_{0} \end{array} \right) _{n\times n}, \end{aligned}$$

where the elements of the r-Toeplitz is generated by the rule:

$$\begin{aligned} T_{r}=\left\{ \begin{array}{ll} t_{i-j}, &{} \hbox {if } i\le j\\ &{} \\ rt_{i-j}, &{} \hbox {if }\ i>j \end{array} \right. . \end{aligned}$$

Obviously, when the parameter satisfies \(r=1\), we can get the classical Hankel and Toeplitz matrices.

Many authors have examined r-Hankel and r-Toeplitz matrices. In [14, 15], the authors studied the norms of r-Hankel and r-Toeplitz matrices involving Fibonacci and Lucas numbers, respectively. In 2019, Shi [22] studied the norms of r-Hankel and r-Toeplitz matrices with exponential forms \(e(x)=e^{2\pi ix}\).

On the other hand, the concept of the generalized inverse of a matrix was introduced by Moore, who proposed a unique generalized inverse using projectors of matrices [23]. Prior to the mid-1950s, there was limited research on this topic. However, the increasing use of matrix inverses in solving linear equation systems sparked renewed interest in the subject. In 1955, R. Penrose proposed a generalization of the inverse of a non-singular matrix as the unique solution of a specific set of equations [24]. This work inspired new studies on generalized inverses and attracted the attention of numerous researchers. The inverse is now known as the Moore-Penrose inverse, named after the contributions of E. H. Moore and R. Penrose. Recently, Moore-Penrose inverse of some matrices whose entries are special number sequences have been studied in detail. For example, Radi čić’s papers are an important reference [25,26,27,28]. Please see also Köme and Yazlik [29].

Let us recall that a geometric sequence is a sequence that has the following form:

$$\begin{aligned} g_{0}=g,\text { }g_{1}=gq,\text { }g_{2}=gq^{2},\text { }g_{2}=gq^{3},\ldots \end{aligned}$$
(1.1)

where \(g\in \mathbb {C} \setminus \left\{ 0\right\} \) and \(q\in \mathbb {R} {\setminus } \left\{ 0\right\} \) i.e. \(g_{l}=gq^{l},\) \(l\in \mathbb {N} _{0}= \mathbb {N} \cup \left\{ 0\right\} .\) For \(q=1,\) (1.1) is a constant sequence. For \(q=-1,\) (1.1) is an alternating sequence. Inspired by the above ideas, in what follows, we shall consider the similar inverse and generalized inverse (Moore-Penrose inverse) matrix of r-Hankel and r -Toeplitz whose entries are geometric sequence \((1,q,q^{2},\cdots ,q^{n-1})\) and \((g,gq,gq^{2},\cdots ,gq^{n-1})\), such matrices are nonsingular if and only if \(r\ne 1\). Based on this analysis, we present the methods to study the inverse and generalized inverse matrix, then we get some prominent results. Then, we will give briefly an excellent explicit expression for determinant, inverse matrix, generalized inverse matrix, and group inverse of r-Toeplitz matrix and r-Hankel matrix whose entries are geometric sequence. Moreover, we will give lower and upper bounds for the spectral norms of r-Toeplitz matrix whose entries are geometric sequence.

Considering aforementioned literature, in this paper, we shall focus on the determinants, inverse matrices and some norms of r-Hankel and r-Toeplitz matrices involving geometric sequence \((1,q,q^{2},\cdots ,q^{n-1})\), (and \( (g,gq,gq^{2},\cdots ,gq^{n-1})\)), which can be briefly expressed \(\textbf{H} =H_{r}(1,q,q^{2},\cdots ,q^{n-1}),\) \(\textbf{T}=T_{r}(1,q,q^{2},\cdots ,q^{n-1}),\) in special case, \(\textbf{H},\) \(\textbf{T}\) are singular if and only if \(r=1,\) so we compute the Moore-Penrose inverse matrices of \(\textbf{H },\) \(\textbf{T}.\)

2 Preliminaries

To easily understand this study, this section reviews some definitions and provides Lemmas which can be the theoretical basis for subsequent sections. For convenience, the following symbols will be used:

  • the determinants of \(\textbf{H}\) and \(\textbf{T}\) will be denoted by \(| \textbf{H}|\) and \(|\textbf{T}|,\)

  • the spectral norms of \(\textbf{H}\) and \(\textbf{T}\) will be denoted by \(\Vert \textbf{H}\Vert _{2}\) and \(\Vert \textbf{T}\Vert _{2},\)

  • the inverse matrices of \(\textbf{H}\) and \(\textbf{T}\) will be denoted by \(\textbf{H}^{-1}\) and \(\textbf{T}^{-1},\)

  • the Moore-Penrose inverses of \(\textbf{H}\) and \(\textbf{T}\) will be denoted by \(\textbf{H}^{\dag }\) and \(\textbf{T}^{\dag },\)

  • the group inverses of \(\textbf{H}\) and \(\textbf{T}\) will be denoted by \(\textbf{H}^{\sharp }\) and \(\textbf{T}^{\sharp }.\)

Definition 1

(Horn and Johnson [21]) The spectral norm is defined by

$$\begin{aligned} \Vert A\Vert _{2}=\sqrt{\displaystyle \max _{1\le i\le n}\lambda _{i}(A^{*}A)}, \end{aligned}$$

where \(\lambda _{i}(A^{*}A)\) is the eigenvalue of \(A^{*}A\) and  \( A^{*}\) is the conjugate transpose of matrix A.

Remark 1

The spectral norm of normal matrix is equal to spectral radius.

Definition 2

(Ben-Israel and Greville [30]) Let A be \(m\times n\) complex matrix, when A is singular, the generalized inverse matrix (the Moore-Penrose inverse) \(n\times m\) matrix X satisfy the equations:

$$\begin{aligned} AXA= & {} A, \\ XAX= & {} X, \\ (AX)^{*}= & {} AX, \\ (XA)^{*}= & {} XA. \end{aligned}$$

To deal with the inverse matrix and Moore-Penrose inverse matrix of \(r-\) Hankel and \(r-\)Toeplitz matrices, some lemmas are presented as following.

Lemma 1

(Ben-Israel and Greville [30]) Let the rank of a matrix \( A_{m\times n}\) be positive natural number r. Then a full-rank factorization of A can be denoted by \(A=MN,\) where the rank of \(M_{m\times r}\) and \(N_{r\times n}\) are equal to r.

Lemma 2

(Ben-Israel and Greville [30]) If A satisfies Lemma 1, then the generalized inverse matrix (the Moore-Penrose inverse) of A is following formula

$$\begin{aligned} A^{\dag }=N^{*}(M^{*}MNN^{*})^{-1}M^{*}. \end{aligned}$$

Lemma 3

(Ben-Israel and Greville [30]) If A satisfies Lemma 1, then the group inverse matrix \((A^{\sharp })\) of A exists if and only if NM is nonsingular. In this case,

$$\begin{aligned} A^{\sharp }=M(NM)^{-2}N. \end{aligned}$$

Lemma 4

For r-Toeplitz matrix \(T_{r},\) notice that

$$\begin{aligned} T_{r}=t_{0}I_{n}+\sum _{j=1}^{n-1}rt_{j}P^{j}+\sum _{j=1}^{n-1}t_{-j}Q^{j}, \end{aligned}$$

where \(I_{n}\) is the \(n\times n\) identity matrix and

$$\begin{aligned} P=\left( \begin{array}{cc} 0 &{} 0 \\ &{} \\ I_{n-1} &{} 0 \end{array} \right) ,Q=\left( \begin{array}{cc} 0 &{} I_{n-1} \\ &{} \\ 0 &{} 0 \end{array} \right) . \end{aligned}$$

Lemma 5

((Merikoski, Haukkanen, Mattila, and Tossavainen [31]) If \(K\in C^{m\times n},\) then

$$\begin{aligned} \Vert K\Vert _{2}\ge \frac{1}{\sqrt{q}}\Vert K\Vert _{E},q=\min (m,n),~~\Vert K\Vert _{2}\ge \frac{2|r_{1}+r_{2}+\cdots +r_{m}|}{m+n}. \end{aligned}$$

Particularly, if K is a square matrix,

$$\begin{aligned} \Vert K\Vert _{2}\ge \frac{|r_{1}+r_{2}+\cdots +r_{n}|}{n},\Vert K\Vert _{2}\ge \sqrt{\frac{|r_{1}|^{2}+|r_{2}|^{2}+\cdots +|r_{n}|^{2}}{n}}, \end{aligned}$$

with \(r_{i}\) denote the i-th row sum of K.

3 Main results

Let us recall that, throughout the paper, \(r\in \mathbb {C} {\setminus } \left\{ 0\right\} .\) For r-Hankel and r-Toeplitz matrix \( \textbf{H},\) \(\textbf{T}\) with geometric sequence are defined by

$$\begin{aligned} \textbf{H}=\left( \begin{array}{ccccc} r &{} rq &{} \cdots &{} rq^{n-2} &{} q^{n-1} \\ rq &{} rq^{2} &{} \cdots &{} q^{n-1} &{} q^{n} \\ \vdots &{} \vdots &{} \cdots &{} \vdots &{} \vdots \\ rq^{n-2} &{} q^{n-1} &{} \cdots &{} q^{2n-4} &{} q^{2n-3} \\ q^{n-1} &{} q^{n} &{} \cdots &{} q^{2n-3} &{} q^{2n-2} \end{array} \right) , \end{aligned}$$
(3.1)

or

$$\begin{aligned} \mathbb {H}=\left( \begin{array}{ccccc} rg &{} rgq &{} \cdots &{} rgq^{n-2} &{} gq^{n-1} \\ rgq &{} rgq^{2} &{} \cdots &{} gq^{n-1} &{} gq^{n} \\ \vdots &{} \vdots &{} \cdots &{} \vdots &{} \vdots \\ rgq^{n-2} &{} gq^{n-1} &{} \cdots &{} gq^{2n-4} &{} gq^{2n-3} \\ gq^{n-1} &{} gq^{n} &{} \cdots &{} gq^{2n-3} &{} gq^{2n-2} \end{array} \right) , \end{aligned}$$

and

$$\begin{aligned} \textbf{T}=\left( \begin{array}{ccccc} 1 &{} q^{-1} &{} \cdots &{} q^{2-n} &{} q^{1-n} \\ rq &{} 1 &{} \cdots &{} q^{3-n} &{} q^{2-n} \\ \vdots &{} \vdots &{} &{} \vdots &{} \vdots \\ rq^{n-2} &{} rq^{n-3} &{} \cdots &{} 1 &{} q^{-1} \\ rq^{n-1} &{} rq^{n-2} &{} \cdots &{} rq &{} 1 \end{array} \right) , \end{aligned}$$
(3.2)

and

$$\begin{aligned} \mathbb {T}=\left( \begin{array}{ccccc} g &{} gq^{-1} &{} \cdots &{} gq^{2-n} &{} gq^{1-n} \\ rgq &{} g &{} \cdots &{} gq^{3-n} &{} gq^{2-n} \\ \vdots &{} \vdots &{} &{} \vdots &{} \vdots \\ rgq^{n-2} &{} rgq^{n-3} &{} \cdots &{} g &{} gq^{-1} \\ rgq^{n-1} &{} rgq^{n-2} &{} \cdots &{} rgq &{} g \end{array} \right) . \end{aligned}$$

The identities \(\mathbb {H}=g\textbf{H}\) and \(\mathbb {T}=g\textbf{T}\) are always held. Hence we just discuss the properties of \(\textbf{H}\) and \( \textbf{T}\) in this paper. At the beginning of this section, we obtain the determinant of (3.1).

Theorem 1

Let \(\textbf{H}\) be as matrix (3.1), then the determinant of \(\textbf{H}\) is shown as below:

$$\begin{aligned} |\textbf{H}|=\left( -1\right) ^{\genfrac(){0.0pt}1{n-1}{2}}q^{n\left( n-1\right) }\left( 1-r\right) ^{\left( n-1\right) }. \end{aligned}$$
(3.3)

Proof

Using elementary row and column operations, we have

$$\begin{aligned} |\textbf{H}|= & {} \left| \begin{array}{ccccc} r &{} rq &{} \cdots &{} rq^{n-2} &{} q^{n-1} \\ rq &{} rq^{2} &{} \cdots &{} q^{n-1} &{} q^{n} \\ \vdots &{} \vdots &{} \cdots &{} \vdots &{} \vdots \\ rq^{n-2} &{} q^{n-1} &{} \cdots &{} q^{2n-4} &{} q^{2n-3} \\ q^{n-1} &{} q^{n} &{} \cdots &{} q^{2n-3} &{} q^{2n-2} \end{array} \right| \\ |\textbf{H}|= & {} \left| \begin{array}{ccccc} r &{} rq &{} \cdots &{} rq^{n-2} &{} q^{n-1} \\ 0 &{} 0 &{} \cdots &{} q^{n-1}(1-r) &{} 0 \\ \vdots &{} \vdots &{} \cdots &{} \vdots &{} \vdots \\ 0 &{} q^{n-1}(1-r) &{} \cdots &{} q^{2n-4}(1-r) &{} 0 \\ q^{n-1}(1-r) &{} q^{n-2}(1-r) &{} \cdots &{} q^{2n-3}(1-r) &{} 0 \end{array} \right| \\ |\textbf{H}|= & {} \left( -1\right) ^{\genfrac(){0.0pt}1{n-1}{2}}q^{n\left( n-1\right) }\left( 1-r\right) ^{\left( n-1\right) }. \end{aligned}$$

\(\square \)

Remark 2

By virtue of Theorem 1, \(\textbf{H}\) is nonsingular if and only if \( r\ne 1.\)

Theorem 2

Let \(\mathbf {H=}\left( \textbf{h}_{ij}\right) _{i,j=0}^{n-1}\) be nonsingular \((r\ne 1),\) we can obtain the unique inverse matrix \(\textbf{H}^{-1}=\left( \mathfrak {h}_{ij}\right) _{i,j=0}^{n-1}\)

(3.4)

or equivalently

$$\begin{aligned} \mathfrak {h}_{ij}=\left\{ \begin{array}{ccc} \frac{1}{\left( r-1\right) q^{n-2}}, &{} &{} i+j=n-2 \\ &{} &{} \\ \frac{-1}{\left( r-1\right) q^{n-1}}, &{} &{} i+j=n-1 \\ &{} &{} \\ \frac{r}{\left( r-1\right) q^{2n-2}}, &{} &{} i=j=n-1 \end{array} \right. . \end{aligned}$$

Proof

For \(j<\left( n-1\right) \), only \(\left( n-j-2\right) j\) and \(\left( n-j-1\right) j\) entries are non-zero of \(\textbf{H}^{-1}\) matrix. Therefore, the entries of \(\textbf{HH}^{-1}\) are

$$\begin{aligned} \begin{array}{lll} \textbf{h}_{i\left( n-j-2\right) }\mathfrak {h}_{\left( n-j-2\right) j}+ \textbf{h}_{i\left( n-j-1\right) }\mathfrak {h}_{\left( n-j-1\right) j}, &{} &{} j<n-1 \\ &{} &{} \\ \textbf{h}_{i0}\mathfrak {h}_{0\left( n-1\right) }+\textbf{h}_{i\left( n-1\right) }\mathfrak {h}_{\left( n-1\right) \left( n-1\right) }, &{} &{} j=n-1 \end{array} \text {.} \end{aligned}$$
(3.5)

  • Now we calculate the first case of (3.5), i.e \(\textbf{h} _{i\left( n-j-2\right) }\mathfrak {h}_{\left( n-j-2\right) j}+\textbf{h} _{i\left( n-j-1\right) }\mathfrak {h}_{\left( n-j-1\right) j}\). For \(i-j\ge 1,\) we write

    $$\begin{aligned}\Rightarrow & {} i+n-j-1>n-1 \\\Rightarrow & {} i+n-j-2\ge n-1 \\\Rightarrow & {} h_{i\left( n-j-1\right) }=q^{i+n-j-1}\text { and }h_{i\left( n-j-2\right) }=q^{i+n-j-2}. \end{aligned}$$

    Then, we get

    $$\begin{aligned}{} & {} \textbf{h}_{i\left( n-j-2\right) }\mathfrak {h}_{\left( n-j-2\right) j}+ \textbf{h}_{i\left( n-j-1\right) }\mathfrak {h}_{\left( n-j-1\right) j}\nonumber \\{} & {} \quad =q^{i+n-j-2}\frac{1}{\left( r-1\right) q^{n-2}}-q^{i+n-j-1}\frac{1}{\left( r-1\right) q^{n-1}} \nonumber \\{} & {} \quad =q^{i-j}\frac{1}{\left( r-1\right) }-q^{i-j}\frac{1}{\left( r-1\right) } \nonumber \\{} & {} \quad =0. \end{aligned}$$
    (3.6)

    For \(i-j=0,\) we write

    $$\begin{aligned}\Rightarrow & {} i+n-j-1=n-1 \\\Rightarrow & {} i+n-j-2<n-1 \\\Rightarrow & {} h_{i\left( n-j-1\right) }=q^{i+n-j-1}\text { and }h_{i\left( n-j-2\right) }=rq^{i+n-j-2}. \end{aligned}$$

    Then, we have

    $$\begin{aligned}{} & {} \textbf{h}_{i\left( n-j-2\right) }\mathfrak {h}_{(n-j-2)j}+\textbf{h} _{i\left( n-j-1\right) }\mathfrak {h}_{\left( n-j-1\right) j}\nonumber \\{} & {} \quad =rq^{i+n-j-2} \frac{1}{\left( r-1\right) q^{n-2}}-q^{i+n-j-1}\frac{1}{\left( r-1\right) q^{n-1}} \nonumber \\{} & {} \quad =rq^{n-2}\frac{1}{\left( r-1\right) q^{n-2}}-q^{n-1}\frac{1}{\left( r-1\right) q^{n-1}} \nonumber \\{} & {} \quad =\frac{r}{r-1}-\frac{1}{r-1} \nonumber \\{} & {} \quad =1. \end{aligned}$$
    (3.7)

    For \(i-j<0,\) we write

    $$\begin{aligned}\Rightarrow & {} i+n-j-1<n-1 \\\Rightarrow & {} i+n-j-2<n-1 \\\Rightarrow & {} h_{i\left( n-j-1\right) }=rq^{i+n-j-1}\text { and }h_{i\left( n-j-2\right) }=rq^{i+n-j-2}. \end{aligned}$$

    So, we get

    $$\begin{aligned}{} & {} \textbf{h}_{i\left( n-j-2\right) }\mathfrak {h}_{\left( n-j-2\right) j}+ \textbf{h}_{i\left( n-j-1\right) }\mathfrak {h}_{\left( n-j-1\right) j}\nonumber \\{} & {} \quad =rq^{i+n-j-2}\frac{1}{\left( r-1\right) q^{n-2}}-rq^{i+n-j-1}\frac{1}{ \left( r-1\right) q^{n-1}} \nonumber \\{} & {} \quad =rq^{i-j}\frac{1}{\left( r-1\right) }-rq^{i-j}\frac{1}{\left( r-1\right) } \nonumber \\{} & {} \quad =0. \end{aligned}$$
    (3.8)

    Hence, from (3.6),(3.7) and (3.8), we get the first case.

  • We calculate the second case of (3.5), i.e \(\textbf{h} _{i\left( n-j-2\right) }\mathfrak {h}_{\left( n-j-2\right) j}+\textbf{h} _{i\left( n-j-1\right) }\mathfrak {h}_{\left( n-j-1\right) j}\). From matrix multiplication, we have

    $$\begin{aligned}{} & {} \textbf{h}_{i\left( n-j-2\right) }\mathfrak {h}_{\left( n-j-2\right) j}+ \textbf{h}_{i\left( n-j-1\right) }\mathfrak {h}_{\left( n-j-1\right) j}\\{} & {} \quad =\left\{ \begin{array}{lll} rq^{i}\frac{-1}{\left( r-1\right) q^{n-1}}+q^{i+n-1}\frac{r}{\left( r-1\right) q^{2n-2}}, &{} &{} i<j=n-1 \\ &{} &{} \\ q^{n-1}\frac{-1}{\left( r-1\right) q^{n-1}}+q^{2n-2}\frac{r}{\left( r-1\right) q^{2n-2}}, &{} &{} i=j=n-1 \end{array} \right. \\{} & {} \quad =\left\{ \begin{array}{lll} 0, &{} &{} i<j=n-1 \\ &{} &{} \\ 1, &{} &{} i=j=n-1 \end{array} \right. . \end{aligned}$$

\(\square \)

For example, if we set \(r=\sqrt{7}\) and \(n=5\) in (3.1), we have

$$\begin{aligned} \mathbf {H=}\left( \begin{array}{ccccc} \sqrt{7} &{} \sqrt{7}q &{} \sqrt{7}q^{2} &{} \sqrt{7}q^{3} &{} q^{4} \\ \sqrt{7}q &{} \sqrt{7}q^{2} &{} \sqrt{7}q^{3} &{} q^{4} &{} q^{5} \\ \sqrt{7}q^{2} &{} \sqrt{7}q^{3} &{} q^{4} &{} q^{5} &{} q^{6} \\ \sqrt{7}q^{3} &{} q^{4} &{} q^{5} &{} q^{6} &{} q^{7} \\ q^{4} &{} q^{5} &{} q^{6} &{} q^{7} &{} q^{8} \end{array} \right) . \end{aligned}$$

Using (3.3), the determinant of \(\textbf{H}\) is calculated as follows:

$$\begin{aligned} |\textbf{H}|= & {} q^{20}\left( 1-\sqrt{7}\right) ^{4} \\= & {} \left( -32\sqrt{7}+92\right) q^{20}. \end{aligned}$$

From (3.4), the inverse of \(\textbf{H}\) is calculated as follows:

$$\begin{aligned} \textbf{H}^{-1}=\left( \begin{array}{ccccc} 0 &{} 0 &{} 0 &{} \frac{1}{(\sqrt{7}-1)q^{3}} &{} -\frac{1}{(\sqrt{7}-1)q^{4}} \\ 0 &{} 0 &{} \frac{1}{(\sqrt{7}-1)q^{3}} &{} -\frac{1}{(\sqrt{7}-1)q^{4}} &{} 0 \\ 0 &{} \frac{1}{(\sqrt{7}-1)q^{3}} &{} -\frac{1}{(\sqrt{7}-1)q^{4}} &{} 0 &{} 0 \\ \frac{1}{(\sqrt{7}-1)q^{3}} &{} -\frac{1}{(\sqrt{7}-1)q^{4}} &{} 0 &{} 0 &{} 0 \\ -\frac{1}{(\sqrt{7}-1)q^{4}} &{} 0 &{} 0 &{} 0 &{} \frac{\sqrt{7}}{(\sqrt{7}-1)q^{8}} \end{array} \right) . \end{aligned}$$

Theorem 3

Let \(n\ge 1\) be any natural numbers and \(\textbf{H} =H_{r}(1,q,q^{2},\cdots ,q^{n-1})\) is a matrix defined by (3.1). For \(r=1\), the Moore-Penrose inverse matrix is as follows:

$$\begin{aligned} \textbf{H}^{\dag }=\left( \frac{1-q^{2}}{1-q^{2n}}\right) ^{2}\textbf{H} ^{*},~~~q\in R\setminus \{-1,1\}. \end{aligned}$$

Proof

Using elementary row operation, we find that

$$\begin{aligned} \textbf{H}=\left( \begin{array}{ccccc} 1 &{} q &{} \cdots &{} q^{n-2} &{} q^{n-1} \\ q &{} q^{2} &{} \cdots &{} q^{n-1} &{} q^{n} \\ \vdots &{} \vdots &{} \cdots &{} \vdots &{} \vdots \\ q^{n-2} &{} q^{n-1} &{} \cdots &{} q^{2n-4} &{} q^{2n-3} \\ q^{n-1} &{} q^{n} &{} \cdots &{} q^{2n-3} &{} q^{2n-2} \end{array} \right) \longrightarrow \left( \begin{array}{ccccc} 1 &{} q &{} \cdots &{} q^{n-2} &{} q^{n-1} \\ 0 &{} 0 &{} \cdots &{} 0 &{} 0 \\ \vdots &{} \vdots &{} \cdots &{} \vdots &{} \vdots \\ 0 &{} 0 &{} \cdots &{} 0 &{} 0 \\ 0 &{} 0 &{} \cdots &{} 0 &{} 0 \end{array} \right) . \end{aligned}$$

Thus rank \(r(\textbf{H})=1,\) and the first column of \(\textbf{H}\) is \(c_{1}( \textbf{H})\ne 0.\) Let \(\textbf{H}=[\textbf{H}_{1}|\textbf{H}_{2}],\) \( \textbf{H}_{1}\in R^{n\times 1},\) \(\textbf{H}_{2}\in R^{n\times (n-1)},\) by Lemma 1, the full-rank factorization is showed that \(\textbf{H}= \textbf{H}_{1}\textbf{W},\) \(\textbf{H}_{1}=(1,q,q^{2},\cdots ,q^{n-2},q^{n-1})^{T},\) \(\textbf{W}=(1,q,q^{2},\cdots ,q^{n-2},q^{n-1}),\)

$$\begin{aligned} \textbf{H}_{1}^{*}\textbf{H}_{1}\textbf{WW}^{*}=\left( \displaystyle \sum _{k=0}^{n-1}q^{2k}\right) ^{2}=\left( \frac{1-q^{2n}}{1-q^{2}}\right) ^{2}, \end{aligned}$$

meanwhile, by Lemma 2, we obtain

$$\begin{aligned} \textbf{H}^{\dag }=\textbf{W}^{*}(\textbf{H}_{1}^{*}\textbf{H}_{1} \textbf{WW}^{*})^{-1}\textbf{H}_{1}^{*}=\left( \frac{1-q^{2}}{ 1-q^{2n}}\right) ^{2}\textbf{H}^{*},~~~q\in R\setminus \{-1,1\}. \end{aligned}$$

\(\square \)

We can discuss individually the case when \(q=-1\) and 1,  then we give the generalized inverse matrices.

Theorem 4

For Hankel matrix \(\textbf{H}\), \(r=1,\) and \(q=-1,1,\) we have

$$\begin{aligned} \textbf{H}^{\dag }=\frac{1}{n^{2}}\textbf{H}^{*}. \end{aligned}$$

Proof

When \(r=q=1,\)

$$\begin{aligned} \textbf{H}=\left( \begin{array}{ccccc} 1 &{} 1 &{} \cdots &{} 1 &{} 1 \\ 1 &{} 1 &{} \cdots &{} 1 &{} 1 \\ \vdots &{} \vdots &{} \cdots &{} \vdots &{} \vdots \\ 1 &{} 1 &{} \cdots &{} 1 &{} 1 \\ 1 &{} 1 &{} \cdots &{} 1 &{} 1 \end{array} \right) \longrightarrow \left( \begin{array}{ccccc} 1 &{} 1 &{} \cdots &{} 1 &{} 1 \\ 0 &{} 0 &{} \cdots &{} 0 &{} 0 \\ \vdots &{} \vdots &{} \cdots &{} \vdots &{} \vdots \\ 0 &{} 0 &{} \cdots &{} 0 &{} 0 \\ 0 &{} 0 &{} \cdots &{} 0 &{} 0 \end{array} \right) , \end{aligned}$$

notice that \(r(\textbf{H})=1,\) by Theorem 3 and Lemma 2, matrix \(\textbf{H}\) has the following full-rank factorization \( \textbf{H}=\textbf{H}_{3}\textbf{W}_{1},\) where \(\textbf{H} _{3}=(1,1,1,\cdots ,1,1)^{T},\) \(\textbf{W}_{1}=(1,1,1,\cdots ,1,1),\) and \( \textbf{H}_{3}^{*}\textbf{H}_{3}\textbf{W}_{1}\textbf{W}_{1}^{*}=n^{2},\) thus

$$\begin{aligned} \textbf{H}^{\dag }=\textbf{W}_{1}^{*}(\textbf{H}_{3}^{*}\textbf{H} _{3}\textbf{W}_{1}\textbf{W}_{1}^{*})^{-1}\textbf{H}_{3}^{*}=\frac{1 }{n^{2}}\textbf{H}^{*}. \end{aligned}$$

When \(r=1,q=-1,\) \(\textbf{H}=H(1,-1,\cdots ,-1,1)\) or \(\textbf{H} =H(1,-1,\cdots ,-1),\) using similar methods as mentioned above, the generalized inverse of this case is obtained

$$\begin{aligned} \textbf{H}^{\dag }=\frac{1}{n^{2}}\textbf{H}^{*}. \end{aligned}$$

\(\square \)

Theorem 5

The group inverse matrix of \(\textbf{H}\) can be denoted by the form

$$\begin{aligned} \textbf{H}^{\sharp }=\left( \frac{1-q^{2n}}{1-q^{2}}\right) ^{-2}\textbf{H} ,~~~q\in R\setminus \{-1,1\}. \end{aligned}$$

Proof

By Theorem 3 and Lemma 3, \(\textbf{H}\) has full-rank factorization \(\textbf{H}=\textbf{H}_{1}\textbf{W},\) \(\textbf{H} _{1}=(1,q,q^{2},\cdots ,q^{n-2},q^{n-1})^{T},\) \(\textbf{W}=(1,q,q^{2},\cdots ,q^{n-2},q^{n-1}),\) so

$$\begin{aligned} \textbf{WH}_{1}=\displaystyle \sum _{k=0}^{n-1}q^{2k}=\frac{1-q^{2n}}{1-q^{2}}. \end{aligned}$$

then we have

$$\begin{aligned} \textbf{H}^{\sharp }=\textbf{H}_{1}(\textbf{WH}_{1})^{-2}\textbf{W}=\left( \frac{1-q^{2n}}{1-q^{2}}\right) ^{-2}\textbf{H},~~~q\in R\setminus \{-1,1\}. \end{aligned}$$

\(\square \)

Theorem 6

Let \(\textbf{H}\) be singular r-Hankel matrix with geometric sequence. Then \(\textbf{H}^{\sharp }\) exists if and only if q does not lie on the unit circle.

Proof

By Theorem 5 and Lemma 3, \(\textbf{H} ^{\sharp }\) exists, namely, \(\frac{1-q^{2n}}{1-q^{2}}\ne 0\) and q does not lie on the unit circle. \(\square \)

Theorem 7

Let \(\textbf{T}\) be a r-Toeplitz matrix satisfying \( \textbf{T}=T_{r}\left( 1,q,\cdots ,q^{n-1}\right) ,\) we have

$$\begin{aligned} |\textbf{T}|=(1-r)^{n-1}. \end{aligned}$$
(3.9)

Proof

Using elementary row operations, we have

$$\begin{aligned} |\textbf{T}|= & {} \left| \begin{array}{ccccc} 1 &{} q^{-1} &{} \cdots &{} q^{2-n} &{} q^{1-n} \\ rq &{} 1 &{} \cdots &{} q^{3-n} &{} q^{2-n} \\ \vdots &{} \vdots &{} &{} \vdots &{} \vdots \\ rq^{n-2} &{} rq^{n-3} &{} \cdots &{} 1 &{} q^{-1} \\ rq^{n-1} &{} rq^{n-2} &{} \cdots &{} rq &{} 1 \end{array} \right| \\= & {} \left| \begin{array}{ccccc} 1 &{} q^{-1} &{} \cdots &{} q^{2-n} &{} q^{1-n} \\ 0 &{} 1-r &{} \cdots &{} q^{3-n}(1-r) &{} q^{2-n}(1-r) \\ \vdots &{} \vdots &{} &{} \vdots &{} \vdots \\ 0 &{} 0 &{} \cdots &{} 1-r &{} q^{-1}(1-r) \\ 0 &{} 0 &{} \cdots &{} 0 &{} 1-r \end{array} \right| \\ |\textbf{T}|= & {} (1-r)^{n-1}. \end{aligned}$$

\(\square \)

Remark 3

\(\textbf{T}\) is nonsingular if and only if \(r\ne 1\).

Theorem 8

Assume that \(\textbf{T}\) is defined as matrix (3.2), \(\textbf{T} =T_{r}(1,q,q^{2},\cdots ,q^{n-1})\), for \(r\ne 1,\)

$$\begin{aligned} \textbf{T}_{n}=\left( \begin{array}{cc} \textbf{T}_{n-1} &{} ~~E \\ &{} \\ rF^{T} &{} ~~1 \end{array} \right) , \end{aligned}$$

with \((n-1)\times 1\) matrix \(E=\left( \frac{1}{q^{n-1}},\frac{1}{q^{n-2}} ,\cdots ,\frac{1}{q}\right) ^{T},\) matrix \(F=(rq^{n-1},rq^{n-2},\cdots ,rq)^{T}.\) For \(r\ne 1,\) \(\textbf{T}\) is nonsingular, then the inverse matrix of \(\textbf{T}\) is

$$\begin{aligned} \textbf{T}_{n}^{-1}=\left( \begin{array}{cc} \textbf{T}_{n-1}^{-1}+ru\textbf{T}_{n-1}^{-1}EF^{T}\textbf{T}_{n-1}^{-1} &{} ~~~-u\textbf{T}_{n-1}^{-1}E \\ &{} \\ -ruF^{T}\textbf{T}_{n-1}^{-1} &{} ~~~u \end{array} \right) , \end{aligned}$$

with \(u=\frac{1}{1-rF^{T}\textbf{T}_{n-1}^{-1}E}.\)

Proof

We use the mathematical induction on n to complete the proof. For \(n=2,\) we have

$$\begin{aligned} \textbf{T}_{2}= & {} \left( \begin{array}{cc} 1 &{} ~~\frac{1}{q} \\ &{} \\ rq &{} ~~1 \end{array} \right) ,\\ \textbf{T}_{2}^{-1}= & {} \left( \begin{array}{cc} \frac{1}{1-r} &{} ~~~-\frac{1}{q(1-r)} \\ &{} \\ -\frac{rq}{1-r} &{} ~~~\frac{1}{1-r} \end{array} \right) . \end{aligned}$$

So our assertion is true. Now we assume that the assertion is true for \(n-1,\) \(\textbf{T}_{n-1}^{-1}\textbf{T}_{n-1}=I_{n-1},\) so \(\textbf{T} _{n-1}^{-1}E=e_{n-1},\) where \(e_{n-1}=(0,0,\cdots ,1).\) Next, for n,  by the identity \(\textbf{T}_{n}^{-1}\textbf{T}_{n}=I_{n},\)

$$\begin{aligned} \textbf{T}_{n}^{-1}=\left( \begin{array}{cc} \textbf{T}_{n-1}^{-1}+ru\textbf{T}_{n-1}^{-1}EF^{T}\textbf{T}_{n-1}^{-1} &{} ~~~~~-u\textbf{T}_{n-1}^{-1}E \\ &{} \\ -ruF^{T}\textbf{T}_{n-1}^{-1} &{} ~~~~~u \end{array} \right) \left( \begin{array}{cc} \textbf{T}_{n-1} &{} ~~E \\ &{} \\ rF^{T} &{} ~~1 \end{array} \right) =\left( \begin{array}{cc} I_{n-1} &{} 0 \\ &{} \\ 0 &{} 1 \end{array} \right) . \end{aligned}$$

So the proof is completed. \(\square \)

Theorem 9

For \(r\ne 1,\) the inverse matrix of \(\mathbf {T=}\left( \textbf{t} _{ij}\right) _{i,j=0}^{n-1}\) is \(\textbf{T}^{-1}=\left( \mathfrak {t} _{ij}\right) _{i,j=0}^{n-1}\)

$$\begin{aligned} \textbf{T}^{-1}=\frac{1}{1-r}\left( \begin{array}{ccccc} 1 &{} -\frac{1}{q} &{} &{} &{} \\ &{} 1 &{} -\frac{1}{q} &{} &{} \\ &{} &{} \ddots &{} \ddots &{} \\ &{} &{} &{} 1 &{} -\frac{1}{q} \\ -rq^{n-1} &{} &{} &{} &{} 1 \end{array} \right) _{n\times n}, \end{aligned}$$
(3.10)

or equivalently

$$\begin{aligned} \mathfrak {t}_{ij}=\left\{ \begin{array}{ccc} -r\frac{q^{i-j}}{1-r}, &{} &{} i=n-1\text { and }j=0 \\ &{} &{} \\ -\frac{q^{i-j}}{\left( r-1\right) }, &{} &{} 0\le j-i\le 1 \\ &{} &{} \\ 0, &{} &{} \text {otherwise} \end{array} \right. . \end{aligned}$$

Proof

For \(j>0\), only jj ve \(\left( j-1\right) j\) entries are non-zero of \( \textbf{T}^{-1}\) matrix. Therefore, entries of multiplication are

$$\begin{aligned} \begin{array}{lll} \textbf{t}_{ij}\mathfrak {t}_{jj}+\textbf{t}_{i\left( j-1\right) }\mathfrak {t} _{\left( j-1\right) j}, &{} &{} j>0 \\ &{} &{} \\ \textbf{t}_{i0}\mathfrak {t}_{00}+\textbf{t}_{i\left( n-1\right) }\mathfrak {t} _{\left( n-1\right) 0}, &{} &{} j=0 \end{array} \text {.} \end{aligned}$$
(3.11)

Now we calculate the first case of (3.11), i.e \(\textbf{t}_{ij} \mathfrak {t}_{jj}+\textbf{t}_{i\left( j-1\right) }\mathfrak {t}_{\left( j-1\right) j}.\) For \(j>0,\) we have

$$\begin{aligned}= & {} \left\{ \begin{array}{ccc} rq^{i-j}\frac{1}{\left( 1-r\right) }-rq^{i-j+1}\frac{1}{\left( 1-r\right) q}, &{} &{} i>j \\ &{} &{} \\ q^{i-j}\frac{1}{\left( 1-r\right) }-q^{i-j+1}\frac{1}{\left( 1-r\right) q}, &{} &{} i<j \\ &{} &{} \\ q^{i-j}\frac{1}{\left( 1-r\right) }-rq^{i-j+1}\frac{1}{\left( 1-r\right) q}, &{} &{} i=j \end{array} \right. \\= & {} \left\{ \begin{array}{lll} 0, &{} &{} i\ne j \\ &{} &{} \\ 1, &{} &{} i=j \end{array} \right. . \end{aligned}$$

For \(j=0,\) i.e \(\textbf{t}_{i0}\mathfrak {t}_{00}+\textbf{t}_{i\left( n-1\right) }\mathfrak {t}_{\left( n-1\right) 0},\) we have

$$\begin{aligned}= & {} \left\{ \begin{array}{ccc} rq^{i}\frac{1}{\left( 1-r\right) }-q^{i-n+1}\frac{rq^{n-1}}{\left( 1-r\right) }, &{} &{} i>j=0 \\ &{} &{} \\ q^{0}\frac{1}{\left( 1-r\right) }-q^{1-n}\frac{rq^{n-1}}{\left( 1-r\right) }, &{} &{} i=j=0 \end{array} \right. \\= & {} \left\{ \begin{array}{lll} 0, &{} &{} i>j=0 \\ &{} &{} \\ 1, &{} &{} i=j=0 \end{array} \right. . \end{aligned}$$

So the proof is completed. \(\square \)

For example, \(r=\frac{1}{3}\) and \(n=5\) in (3.2), we have

$$\begin{aligned} \mathbf {T=}\left( \begin{array}{ccccc} 1 &{} q^{-1} &{} q^{-2} &{} q^{-3} &{} q^{-4} \\ \frac{1}{3}q &{} 1 &{} q^{-1} &{} q^{-2} &{} q^{-3} \\ \frac{1}{3}q^{2} &{} \frac{1}{3}q &{} 1 &{} q^{-1} &{} q^{-2} \\ \frac{1}{3}q^{3} &{} \frac{1}{3}q^{2} &{} \frac{1}{3}q &{} 1 &{} q^{-1} \\ \frac{1}{3}q^{4} &{} \frac{1}{3}q^{3} &{} \frac{1}{3}q^{2} &{} \frac{1}{3}q &{} 1 \end{array} \right) . \end{aligned}$$

Using (3.9), the determinant of \(\textbf{T}\) is calculated as follows:

$$\begin{aligned} |\textbf{T}|=\frac{16}{81}. \end{aligned}$$

From (3.10), the inverse of \(\textbf{T}\) is calculated as follows:

$$\begin{aligned} \textbf{T}^{-1}=\left( \begin{array}{ccccc} \frac{3}{2} &{} -\frac{3}{2q} &{} 0 &{} 0 &{} 0 \\ 0 &{} \frac{3}{2} &{} -\frac{3}{2q} &{} 0 &{} 0 \\ 0 &{} 0 &{} \frac{3}{2} &{} -\frac{3}{2q} &{} 0 \\ 0 &{} 0 &{} 0 &{} \frac{3}{2} &{} -\frac{3}{2q} \\ -\frac{1}{2}q^{4} &{} 0 &{} 0 &{} 0 &{} \frac{3}{2} \end{array} \right) . \end{aligned}$$

Remark 4

When \(r=1\), \(\textbf{T}\) is singular, so we compute the generalized inverse matrix by using full-rank factorization.

Theorem 10

Let \(\textbf{T}\) be defined by matrix (3.2) and \(r=1.\) Then, we get

$$\begin{aligned} \textbf{T}^{\dag }=q^{2n-2}\frac{\left( q^{2}-1\right) ^{2}}{\left( q^{2n}-1\right) ^{2}}\textbf{T}^{*},~~~q\in R\setminus \{-1,1\}. \end{aligned}$$

Proof

When \(r=1,\) Toeplitz matrix with geometric sequence is always singular by Theorem 7, and rank \(r(\textbf{T})=1,\) so we write \(\textbf{T }=[\textbf{T}_{1}|\textbf{T}_{2}],\) \(r(\textbf{T}_{1})=1,\) then the full-rank factorization is \(\textbf{T}=\textbf{T}_{1}\textbf{G},\) with \( \textbf{T}_{1}=(1,q,q^{2},\cdots ,q^{n-1})^{T},\) \(\textbf{G}=(1,\frac{1}{q},\cdots ,\frac{1}{q^{n-1}}),\) thus

$$\begin{aligned} \textbf{T}_{1}^{*}\textbf{T}_{1}\textbf{GG}^{*}=\displaystyle \sum _{k=0}^{n-1}q^{2k}\sum _{k=0}^{n-1}q^{-2k}=\frac{q^{-2n+2}}{\left( q^{2}-1\right) ^{2}}\left( q^{2n}-1\right) ^{2}. \end{aligned}$$

By Lemma 2,

$$\begin{aligned} \textbf{T}^{\dag }=\textbf{G}^{*}(\textbf{T}_{1}^{*}\textbf{T}_{1} \textbf{GG}^{*})^{-1}\textbf{T}_{1}^{*}=q^{2n-2}\frac{\left( q^{2}-1\right) ^{2}}{\left( q^{2n}-1\right) ^{2}}\textbf{T}^{*}. \end{aligned}$$

\(\square \)

Theorem 11

The group inverse matrix of \(\textbf{T}\) always exist and can be denoted by

$$\begin{aligned} \textbf{T}^{\sharp }=\frac{1}{n^{2}}\textbf{T}. \end{aligned}$$

Proof

By Theorem 10 and Lemma 3,

$$\begin{aligned} \textbf{T}^{\sharp }=\textbf{T}_{1}(\textbf{GT}_{1})^{-2}\textbf{G}=\frac{1}{ n^{2}}\textbf{T}. \end{aligned}$$

Theorem 12

Let \(\textbf{T}\) be a r-Toeplitz matrix satisfying \(\textbf{T}=T_{r}\left( 1,q,\cdots ,q^{n-1}\right) .\) We have

$$\begin{aligned} \Vert \textbf{T}\Vert _{2}\le \left\{ \begin{array}{ccc} -\frac{1}{\left| q\right| ^{n}\left( \left| q\right| -1\right) }\left( \left| q\right| -\left| q\right| ^{n+1}-\left| q\right| ^{2n}\left| r\right| +\left| q\right| ^{n+1}\left| r\right| \right) , &{} &{} q\in \mathbb {R} -\left\{ -1,1\right\} \\ &{} &{} \\ n+\left| r\right| \left( n-1\right) , &{} &{} q=-1,1 \end{array} \right. . \end{aligned}$$

Proof

$$\begin{aligned} \textbf{T}=\left( \begin{array}{ccccc} 1 &{} q^{-1} &{} \cdots &{} q^{2-n} &{} q^{1-n} \\ rq &{} 1 &{} \cdots &{} q^{3-n} &{} q^{2-n} \\ \vdots &{} \vdots &{} &{} \vdots &{} \vdots \\ rq^{n-2} &{} rq^{n-3} &{} \cdots &{} 1 &{} q^{-1} \\ rq^{n-1} &{} rq^{n-2} &{} \cdots &{} rq &{} 1 \end{array} \right) _{n\times n}. \end{aligned}$$

By Lemma 5, we have the new lower bounds of \(\Vert \textbf{T} \Vert _{2},\)

$$\begin{aligned} \Vert \textbf{T}\Vert _{2}\ge \frac{|r_{1}+r_{2}+\cdots +r_{n}|}{n}. \end{aligned}$$

On the other hand, let P and Q be the following form

$$\begin{aligned} P=\left( \begin{array}{cc} 0 &{} ~~0 \\ I_{n-1} &{} ~~0 \end{array} \right) _{n\times n},Q=\left( \begin{array}{cc} 0 &{} ~~I_{n-1} \\ 0 &{} ~~0 \end{array} \right) _{n\times n}, \end{aligned}$$

so, we have \(\Vert P\Vert _{2}=\Vert Q\Vert _{2}=1\),

$$\begin{aligned} \textbf{T}= & {} I_{n}+\displaystyle \sum _{j=1}^{n-1}rq^{j}P^{j}+\displaystyle \sum _{j=1}^{n-1}q^{-j}Q^{j}.\\ \Vert \textbf{T}\Vert _{2}= & {} \Vert I_{n}+\displaystyle \sum _{j=1}^{n-1}rq^{j}P^{j}+\displaystyle \sum _{j=1}^{n-1}q^{-j}Q^{j}\Vert _{2} \\\le & {} 1+|r|\displaystyle \sum _{j=1}^{n-1}|q|^{j}\Vert P\Vert _{2}^{j}+ \displaystyle \sum _{j=1}^{n-1}|q|^{-j}\Vert Q\Vert _{2}^{j} \\= & {} \left\{ \begin{array}{ccc} -\frac{1}{\left| q\right| ^{n}\left( \left| q\right| -1\right) }\left( \left| q\right| -\left| q\right| ^{n+1}-\left| q\right| ^{2n}\left| r\right| +\left| q\right| ^{n+1}\left| r\right| \right) , &{} &{} q\in \mathbb {R} -\left\{ -1,1\right\} \\ &{} &{} \\ n+\left| r\right| \left( n-1\right) , &{} &{} q=-1,1 \end{array} \right. . \end{aligned}$$

\(\square \)

Remark 5

For Hadamard inverse of r-Hankel and r-Toeplitz matrices, the reader can continue to study same properties.

4 Conclusion

In recent years, the study of matrices has emerged as a prominent and intriguing research area in both applied and computational mathematics. Numerous researchers have focused on various applications of matrices and continue to explore this field. In this paper, we investigated the characteristics of r-Hankel and r-Toeplitz matrices, where the entries are geometric sequences. We subsequently derived the determinants, inverse matrix, generalized inverse matrix, and spectral norms of these matrices. Furthermore, we propose a new approach to enhance the bounds of matrix norms for r-Toeplitz matrices, which surpasses the results presented in [22] . For \(e(x)=e^{2\pi ix}\), we chose the elements \(e(\frac{k}{n})\) which is a special geometric sequence and common ratio is \(e(\frac{1}{n}).\) Therefore, we obtained the results showed in the paper [22]. Particularly, common ratio \(e(\frac{1}{n})\) is complex number which lies on unit circle. Maybe we can get some interesting results later.