# On the Norms ofr-Hankel and r-Toeplitz Matrices

Authors: Baijuan Shi
Journal: Mathematical Problems in Engineering
 (2019)

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## Abstract

In this paper, using the properties ofr-Hankel and r-Toeplitz matrices, combining the properties of exponential form, we shall study the spectral norms of r-Hankel and r–Toeplitz matrices involving exponential form e(x).


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## Body

## 1. Introduction

Circulant matrix is first proposed by Professor T. Muir in 1885, and he has carried on the preliminary study. Until 1950-1955, Good et al. began to study the determinants and inverses of circulant matrices, which have opened the curtain of the research on circulant matrices. Recently, circulant matrices have been one of the most important research subjects in the field of the computation and pure mathematics, which has been widely applied in the field of modern science and technology engineering, for example, signal processing, coding theory, image processing and electrodynamics, etc. Particularly, circulant,r-circulant, geometric circulant matrices, r-Hankel, and r-Toeplitz matrices occupy very important position in matrices theory. As we all know, circulant matrix is a special form of Toeplitz matrix. So studying the Toeplitz matrix is a very meaningful job. Hence, based on the special properties and structures of circulant matrices, many scholars at home and abroad have studied the determinants, norms, and inverses of above matrices with well-known number sequences and made great achievements [1–11]. For example, Solak, S has studied the norms of circulant matrices with the Fibonacci and Lucas numbers. C. Köme and Y. Yazlik [5, 10] have studied the determinants, inverses, and spectral norms of of r-circulant matrices with biperiodic Fibonacci and Lucas numbers. H. Gökbas and R. Türkmen have studied the norms of r-Toeplitz matrices involving Fibonacci and Lucas numbers. The authors Y. Yazlik, N. Yilmaz, and N. Taskara [11] have studied the norms of Hankel matrices with the k-Jacobsthal and k-Jacobsthal Lucas numbers. In 2017, H. Gökbas and H. Köse have studied the norms of r-Hankel matrices involving Pell and Pell-Lucas numbers. Considering all of references, we shall study some new problems. As far as we know, it actually seems that no one has studied the upper and lower estimate problems for the spectral norms involving exponential form e(x) yet, where we define e(x)=e2πix.Fore(x), ex=1, by eiθ=cos⁡θ+isin⁡θ, note thate ( 0 ) = e ( 1 ) = e ( - 1 ) = e ( n ) = 1, and by the trigonometric sums, we have (1)∑k=1nekmn=n,n∣m0,otherwiseA n×nr-Hankel matrix Hr is defined by [9] (2)Hr=rh0rh1⋯rhn-2hn-1rh1rh2⋯hn-1hn⋮⋮⋯⋮⋮rhn-2hn-1⋯h2n-4h2n-3hn-1hn⋯h2n-3h2n-2n×n.Namely, (3)Hr=rhi+j,ifi≤jhi+j,ifi>j.A matrix Tr=tij is called a r-Toeplitz matrix (4)Tr=rti-j,ifi<jti-j,ifi≥jTr=t0t-1⋯t2-nt1-nrt1t0⋯t3-nt2-n⋮⋮⋮⋮rtn-2rtn-3⋯t0t-1rtn-1rtn-2⋯rt1t0n×n.Obviously, when the parameter satisfiesr=1, we can get the classical Hankel and Toeplitz matrices. In this paper, we shall study the norms of A=Hr and P=Tr whose entries are exponential form e(x).(5)A=Hre0n,e1n,e2n,…,en-1n,P=Tre0n,e-1n,e-2n,…,e1-nn.Then we get some interesting and concise results which are stated by the following theorems.Theorem 1.
For any positive numbern>2, let A=Hre(0/n),e(1/n),e(2/n),…,e(n-1/n) be an r-Hankel matrix, then we have(6)r≥1,n-1n+2n≤A2≤nn-1r2+1;r<1,n-1n+2nr≤A2≤n.Theorem 2.
For any positive numbern>2, let P=Tre(0/n),e(-1/n),e(-2/n),…,e(1-n/n) be an r-Toeplitz matrix, then we have(7)r≥1,n≤P2≤rnn-1;r<1,nr≤P2≤nn-1.


## 2. Preliminaries

Definition 3 (see [3]).
Let any matrixA=(aij)∈Mm×n(C), the spectral norm is defined by A2=max1≤i≤n⁡λi(AHA),
and the Euclidean norm of matrixA is given by AE=∑i=1m∑j=1naij21/2,
whereλi(AHA) is the eigenvalues of matrices AHA and AH is the conjugate transpose of A.
The following important inequalities hold between the Euclidean norm and spectral norm:(8)1nAE≤A2≤AE.Definition 4 (see [3]).
LetA=(aij) and B=(bij) be m×n matrices, then the Hadamard product of A and B is A∘B=(aijbij).
Then we have the following inequalities:(9)A∘B2≤r1AC1B,r1(A)=max1≤i≤m⁡∑j=1naij2, C1(B)=max1≤j≤n⁡∑i=1mbij2.Lemma 5.
For any positive integern>2, we have (10)∑k=0n-1e2kn=0,∑k=0n-1ke2kn=ne2/n-1.Proof.
Takinge(x)=e2πix and eiθ=cos⁡θ+isin⁡θ, by the trigonometric sums, we have (11)∑k=0n-1e2kn=∑k=0n-1e2kn=0.Taking S=∑k=0n-1ke2k/n=∑k=0n-1ke2k/n, we have (12)S=e2n+2e4n+3e6n+⋯+n-2e2n-2n+n-1e2n-1n,e2nS=e4n+2e6n+⋯+n-2e2n-1n+n-1e2nn,and therefore,(13)1-e2nS=e2n+e4n+⋯+e2n-1n-n-1=∑k=1n-1e2kn-n-1=-n.That is, S=n/e2/n-1.


## 3. Proofs of Theorems

Proof of Theorem1.Proof.
For the matrixA=Hre(0/n),e(1/n),e(2/n),…,e(n-1/n)(14)A=re0nre1n⋯ren-2nen-1nre1nre2n⋯en-1nenn⋮⋮⋯⋮⋮ren-2nen-1n⋯e2n-4ne2n-3nen-1nenn⋯e2n-3ne2n-2nn×n.(i) From r≥1, using the definition of Euclidean norm and properties of e(x), we have (15)AE2=∑k=0n-2r2k+1ekn2+∑k=n-12n-22n-1-kekn2≥∑k=0n-2k+1+∑k=n-12n-22n-1-k=n2.by (8); that is to say, (16)A2≥1nAE≥n.
On the other hand, let the matricesB and C be defined by (17)B=rr⋯r1rr⋯11⋮⋮⋮⋮r1⋯1111⋯11n×n,and (18)C=e0ne1n⋯en-2nen-1ne1ne2n⋯en-1nenn⋮⋮⋯⋮⋮en-2nen-1n⋯e2n-4ne2n-3nen-1nenn⋯e2n-3ne2n-2nn×n,then A=B∘C. So A2=B∘C2≤r1(B)C1(C), (19)r1B=max1≤i≤n⁡∑j=1nbij2=n-1r2+1;c1C=max1≤j≤n⁡∑i=1ncij2=∑k=0n-1ekn2=n,Therefore, we have (20)A2≤nn-1r2+1.Thus, we can obtain the inequality (21)n-1n+2n≤A2≤nn-1r2+1.(ii) From |r|<1, (22)AE2=∑k=0n-2r2k+1ekn2+∑k=n-12n-22n-1-kekn2≥∑k=0n-2r2k+1+∑k=n-12n-2r22n-1-k=r2n2,we can get(23)A2≥1nAE≥rn.For another, let the matrices B and C be as mentioned above, A=B∘C.
SoA2=B∘C2≤r1(B)C1(C)=n.
Therefore, we havern≤A2≤n.
This proofs Theorem1.Now we proof Theorem2.Proof.
ForP=Tre(0/n),e(-1/n),e(-2/n),…,e(1-n/n), pij=e(i-j/n), (24)P=e0ne-1n⋯e2-nne1-nnre1ne0n⋯e3-nne2-nn⋮⋮⋮⋮ren-2nren-3n⋯e0ne-1nren-1nren-2n⋯re1ne0nn×n,
(i) Fromr≥1, by ex=1, we have (25)PE2=ne0n2+∑k=1n-1r2n-kekn2+∑k=1n-1n-ke-kn2≥n+2∑k=1n-1n-k=n2;that is, P2≥1/nPE≥n.
On the other hand, let the matricesR and S be given by (26)R=e0n1⋯11re0n⋯11⋮⋮⋮⋮rr⋯e0n1rr⋯re0nn×n,and (27)S=e0ne-1n⋯e2-nne1-nne1ne0n⋯e3-nne2-nn⋮⋮⋯⋮⋮en-2nen-3n⋯e0ne-1nen-1nen-2n⋯e1ne0nn×n,then P=R∘S. So P2=R∘S2≤r1(R)C1(S), (28)r1R=max1≤i≤n⁡∑j=1nrij2=n-1r2+1;c1S=max1≤j≤n⁡∑i=1nsij2=∑k=0n-1ekn2=n,Therefore, we have (29)P2≤nn-1r2+1.Thus, we can obtain (30)n≤P2≤nn-1r2+1.(ii) From r<1, (31)PE2=ne0n2+∑k=1n-1r2n-kekn2+∑k=1n-1n-ke-kn2≥nr2+2r2∑k=1n-1n-k=r2n2,we can get (32)P2≥1nPE≥nr.
On the other hand, for the matricesR and S as mentioned above, P=R∘S.
SoP2=R∘S2≤r1(R)C1(S)=n.
Therefore, we havenr≤P2≤n.
This proofs Theorem2.This completes all of the theorems.



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