2015+An eigenvalue problem for even order tensors

See discussions, stats, and author profiles for this publication at:https:///publication/281367651An eigenvalue problem for even order tensors with its applicationsARTICLE in LINEAR AND MULTILINEAR ALGEBRA · AUGUST 2015Impact Factor: 0.74 · DOI: 10.1080/03081087.2015.1071311READS394 AUTHORS, INCLUDING:Chuan ChenHong Kong Baptist University5 PUBLICATIONS 1 CITATIONSEE PROFILEWen LiSouth China Normal University107 PUBLICATIONS 510 CITATIONSSEE PROFILEAvailable from: Chuan ChenRetrieved on: 23 February 2016This article was downloaded by: [University of Hong Kong Libraries]On: 07 August 2015, At: 03:04Publisher: Taylor & FrancisInforma Ltd Registered in England and Wales Registered Number: 1072954 Registeredoffice: 5 Howick Place, London, SW1P 1WGClick for updatesLinear and Multilinear AlgebraPublication details, including instructions for authors and subscription information:/loi/glma20An eigenvalue problem for even order tensors with its applicationsLu-Bin Cui a, Chuan Chen b, Wen Li c& Michael K. NgbaHenan Engineering Laboratory for Big Data Statistical Analysis and Optimal Control, School of Mathematics and Information Sciences, Henan Normal University , XinXiang, P .R. China.bDepartment of Mathematics, Hong Kong Baptist University , Hong Kong, China.cSchool of Mathematical Sciences, South China Normal University ,Guangzhou, China.Published online: 06 Aug 2015.PLEASE SCROLL DOWN FOR ARTICLETaylor & Francis makes every effort to ensure the accuracy of all the information (the “Content”) contained in the publications on our platform. However , Taylor & Francis,our agents, and our licensors make no representations or warranties whatsoever as to the accuracy, completeness, or suitability for any purpose of the Content. Any opinions and views expressed in this publication are the opinions and views of the authors,and are not the views of or endorsed by Taylor & Francis. 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Terms &D o w n l o a d e d b y [U n i v e r s i t y o f H o n g K o n g L i b r a r i e s ] a t 03:04 07 A u g u s t 2015Linear and Multilinear Algebra ,2015/10.1080/03081087.2015.1071311An eigenvalue problem for even order tensors with its applicationsLu-Bin Cui a ,Chuan Chen b ,Wen Li c ∗and Michael K.Ng ba Henan Engineering Laboratory for Big Data Statistical Analysis and Optimal Control,School of Mathematics and Information Sciences,Henan Normal University,XinXiang,P .R.China;b Department of Mathematics,Hong Kong Baptist University,Hong Kong,China;c School ofMathematical Sciences,South China Normal University,Guangzhou,ChinaCommunicated by R.-C.Li(Received 6August 2014;accepted 6July 2015)In this paper,we study an eigenvalue problem for even order ing the matrix unfolding of even order tensors,we can establish the relationship between a tensor eigenvalue problem and a multilevel matrix eigenvalue problem.By considering a higher order singular value decomposition of a tensor,we show that higher order singular values are the square root of the eigenvalues of the product of the tensor and its conjugate transpose.This result is similar to that in matrix case.Also we study an eigenvalue problem for Toeplitz/circulant tensors,and give the lower and upper bounds of eigenvalues of Toeplitz tensors.An application in image restoration is also discussed.Keywords:tensors;eigenvalues;eigenvectors;higher order singular value decomposition;multilevel matrices;Toeplitz tensors;circulant tensors AMS Subject Classifications:15A18;15A691.IntroductionAtensor is a multidimensional array.Let C be the complex field.An m th-order n -dimensional tensor A consisting of n m entries in C is denoted by:A =(a i 1,i 2,...,i m ),a i 1,i 2,...,i m ∈C ,1≤i k ≤n ,k =1,2,...,m .(1)In the following discussion,we also use A (i 1,i 2,...,i m )to denote (i 1,i 2,...,i m )th entry of A .The tensor eigenpair was introduced by Lim [1]and Qi [2]independently in 2005.At present,the tensor eigenvalue problem becomes a hot topic because of its applications in diffusion tensor imaging,higher order Markov chains and data mining et al.,see e.g.[1–13].Below is the definition of eigenvalues of tensors.[2]Definition 1.1Let A be a real m th-order n -dimensional tensor.If x ∈C n \{0},λ∈C ,xand λsatisfy∗Corresponding author.Email:liwen@©2015Taylor &FrancisD o w n l o a d e d b y [U n i v e r s i t y o f H o n g K o n g L i b r a r i e s ] a t 03:04 07 A u g u s t 20152L.-B.Cui et al.A x m −1=λx [m −1],(2)then we call λan eigenvalue of A ,and x its corresponding eigenvector.In particular,if x is real,then λis also real.Here x T is the transpose of x ,A x m −1:=⎛⎝ni 2,...,i m =1a i 1,i 2,...,i m x i 2...x i m ⎞⎠1≤i 1≤nand x [m −1]= x m −1i1≤i ≤nwith x =(x 1,x 2,...,x n )T .It is clear when m =2,the above definition is same as that of eigenvalues and eigen-vectors of real matrices.Hence,the tensor eigenvalue can be regarded as a generalization of matrix eigenvalues.According to Definition 1.1,we see that a tensor eigenvalue problem is equivalent to solving a set of multivariate polynomials of variables x 1,x 2,...,x n and an unknown λ.In general,the tensor eigenvalue problem given by Definition 1.1is NP-hard.[4]However,it is interesting to note that there are some other ways to define the tensor eigenvalue.For example,in computational mechanics [14–16]and signal processing[17],the eigenvalue of a fourth-order symmetric tensor C was introduced as follows when they studied the elasticity of isotropic materials,C ·E =⎛⎝nk ,l =1C i jkl E kl ⎞⎠=λE ,(3)where C is a fourth-order n -dimension symmetric tensor and E is an n ×n square matrix.Here,the symmetry means that C i jkl =C kli j for all i ,j ,k ,l ∈{1,2,...,n }holds.Obviously,the eigenvalue problem given by (3)can be considered as a linear transfor-mation as an fourth-order square tensor on a second-order square tensor (matrix),which generalizes the idea of the matrix case.Recently,Qi [18]extended (3)to a (2m )th-order square tensor,i.e.Definition 1.2[17,18]Let A be a complex 2m th-order n -dimensional tensor.If X is a nonzero complex m th-order n -dimensional tensor,λ∈C ,and X and λsatisfyA ·X =λX ,(4)where(A ·X )i 1,...,i m =n j 1=1n j 2=1···n j m =1a i 1,...,i m ,j 1,...,j m x j 1,...,j m ,1≤i k ≤n ,1≤k ≤m .(5)We call λand X are eigenvalue and eigentensor of A .It is noted that if m =1,then A is a square matrix and Definition 1.2reduces to thematrix eigenvalue.It is also noted that Definition 1.2is very different from the one in (2).From now on,if no other special illustration,we say the eigenvalue means the eigenvalue given by Definition 1.2.D o w n l o a d e d b y [U n i v e r s i t y o f H o n g K o n g L i b r a r i e s ] a t 03:04 07 A u g u s t 2015Linear and Multilinear Algebra3Figure 1.A stack of brain MRI images.The first and second images in the first row are different,but the parts in the red circles in these two images are similar.A natural question is that why we study the tensor eigenvalue problem?In fact,for high-dimensional problems,the data have inherent tensor structure.Firstly,the tensor structure means that the data is high dimensional.The difference slices of the data may have some relationship.For example,human brain is a three-dimensional structure,any point in the brain can be localized on the x ,y and z planes.The brain can be cut on any of these planes and are named the coronal plane,the horizontal plane or the sagittal plane.Figure 1shows a stack of brain MRI images on horizontal plane.The data of the stack of brain MRI images on horizontal plane is high dimensional.But the slices of brain are not independent.As we know,brain is made up of many specialized regions.For example,thinking and voluntary movements are controlled by telencephalon.If we want to reconstruct a part of telencephalon,only one slice of brain MRI images on horizontal is not enough,many slices are needed.These slices contain information of the part of telencephalon.So they have a relationship.If we just process these MRI images slice by slice,we may lose some information of the tensor structure of MRI images.So it is necessary to study this tensor eigenvalue problem.As we know,the matrix unfolding of a tensor is a useful tool for studying tensor problems.For example,in [19],the authors used the matrix unfolding to solve multilinear systems in quantum mechanical models and high-dimensional PDEs.In this paper,we will apply the matrix unfolding technique to study the eigenvalue problem.The singular value of a matrix A is the square root of the eigenvalue of A ∗A .In multilinear algebra,the higher order singular value decomposition (HOSVD)[20]was a generalization of SVD of a matrix.Naturally,one may ask:what is the relationship between the eigenvalue problem in Definition 1.2and HOSVD?One contribution of this paper is to establish the relationship between the tensor eigenvalue problem in Definition 1.2and HOSVD.More precisely,we show that the singular values and the associated singular vectors of A are just the eigenvalues and the associated eigentensors of A ∗ A .Here A ∗represents the conjugate transpose of A ,and refers to the multiplication of two tensors.Their definitions will be given in Section 2.Another contribution of this paper is to study eigenvalue problems for Toeplitz/circulant tensors,which can be applied to image processing.[21,22]In particular,we construct eigentensors to diagonalize circulant tensors to obtain the eigenvalues.For Toeplitz tensors,we present the lower and upper bounds of the eigenvalues based on generating functions.The remaining of the paper is organized as follows.In Section 2,we give the basic properties of the tensor eigenvalue problem and demonstrate the relationship between tensorD o w n l o a d e d b y [U n i v e r s i t y o f H o n g K o n g L i b r a r i e s ] a t 03:04 07 A u g u s t 20154L.-B.Cui et al.eigenvalue problem and HOSVD.In Section 3,we study Toepltiz and circulant tensors and analyse their eigenvalues,and give an application in image restoration.The concluding remarks are given in Section 4.2.Properties of tensor eigenvalues 2.1.Unfolding operationsSuppose that A is a 2m th-order n -dimensional tensor.We can reorder A as a square matrix using the square matrix unfolding of tensors.In [23],Kofidis et al.employed the square matrix unfolding of tensors,and their aim is to study the problem of the best rank-one approximation of a super-symmetric tensor.Here,the super-symmetric meansthat A i 1,i 2,...,i m =A i 1,i 2...,i m ,where (i 1,i 2...,i m )is any permutation of (i 1,i 2,...,i m ),i k ∈{1,2,...,n },k ∈{1,2,...,m }.Definition 2.1[23]Let A be a (2m )th-order n -dimensional tensor.The square matrix unfolding of A with an ordering P is an n m -by-n m matrix A P where its (k ,h )th entry is given byA P (k ,h )=A (i 1,i 2,...,i m ,j 1,j 2,...,j m ),withk =n m −1(i 1−1)+n m −2(i 2−1)+···+n (i m −1−1)+i m ,1≤ik ≤n ,1≤k ≤m ,h =n m −1(j 1−1)+n m −2(j 2−1)+···+n (j m −1−1)+j m ,1≤j k ≤n ,1≤k ≤m ,and P is the permutation matrix corresponding to the ordering P :(i 1,i 2,...,im )=(i 1,i 2,...,i m )P ,(j 1,j 2,...,jm )=(j 1,j 2,...,j m )P .Let us consider a simple example of the square matrix unfolding with the natural orderingI .The permutation matrix P is just the identity matrix I .Suppose that A =(a i ,j ,k ,l )is a 4th-order three-dimensional tensor,the square matrix unfolding of A with the natural ordering I is a 32-by-32matrix givenbySince A is a 4th-order tensor,we have the other square matrix unfolding and the permutation matrix is given as follows:P = 0110.D o w n l o a d e d b y [U n i v e r s i t y o f H o n g K o n g L i b r a r i e s ] a t 03:04 07 A u g u s t 2015Linear and Multilinear Algebra 5We note that (i 1,i 2)=(i 1,i 2)P =(i 2,i 1).The corresponding square matrix unfolding of A is equal to a 32-by-32matrix:Remark 1In Definition 1.1,the tensor is a (2m )th-order n -dimensional tensor.We maygeneralize this definition to a (2m )th-order n 1×n 2×···×n m ×n 1×n 2×···×n m -dimensional tensor,where n i =n i,i =1,...,m .In this paper,we study the eigenvalue problem of a (2m )th-order n -dimensional tensor.But those results can be generalized to a(2m )th-order n 1×n 2×···×n m ×n 1×n 2×···×n m -dimensional tensor,where n i =n i ,i =1,...,m ,easily.Given two different orderings P and P ,it is interesting to note that A P and A P are similar via a permutation matrix.The permutation matrix is called a perfect shuffle permutation.[24]Proposition 2.2Suppose P and P are two different orderings.Then there exist a permutation matrix P ,P such thatP ,P A P T P ,P =A P .(8)For example,(6)and (7)are the two square matrix unfoldings of A with two differentorderings.It is easy to check that the perfect shuffle permutation matrix P ,I is givenbyand P ,I A P P ,I =A I .Without loss of generality,we can assume that the natural ordering I is used in the square matrix unfolding of tensors in the following discussion.For simplicity,we denote A I by A .We remark that the square matrix unfolding of an (2m )th-order n -dimensional tensor is a multilevel matrix with m levels.More precisely,at the first level,A is an n -by-n block matrix with n m −1-by-n m −1blocks:D o w n l o a d e d b y [U n i v e r s i t y o f H o n g K o n g L i b r a r i e s ] a t 03:04 07 A u g u s t 20156L.-B.Cui et al.A =A (1)=⎛⎜⎜⎜⎜⎝A (2)(1,1)A (2)(1,2)···A (2)(1,n )A (2)(2,1)A (2)(2,2)···A (2)(2,n )............A (2)(n ,1)A (2)(n ,2)···A (2)(n ,n )⎞⎟⎟⎟⎟⎠,(9)where the (i ,j )th-block A (2)(i ,j )is a second-level matrix given by a n -by-n block with nm −2-by-n m −2block matrix:A (2)(i ,j )=⎛⎜⎜⎜⎜⎝A (3)(i ,1,j ,1)A (3)(i ,1,j ,2)···A (3)(i ,1,j ,n )A (3)(i ,2,j ,1)A (3)(i ,2,j ,2)···A (3)(i ,2,j ,n )............A (3)(i ,n ,j ,1)A (3)(i ,n ,j ,2)···A (3)(i ,n ,j ,n )⎞⎟⎟⎟⎟⎠,1≤i ,j ≤n .In general,the th-level matrix is an n -by-n block matrix with n m − -by-n m − blocks given byA ( )(i 1,...,i −1,j 1,...,j −1)=⎛⎜⎜⎜⎜⎝A ( +1)(i 1,...,i −1,1,j 1,...,j −1,1)A ( +1)(i 1,...,i −1,1,j 1,...,j −1,2)···A ( +1)(i 1,...,i −1,1,j 1,...,j −1,n )A ( +1)(i 1,...,i −1,2,j 1,...,j −1,1)A ( +1)(i 1,...,i −1,2,j 1,...,j −1,2)···A ( +1)(i 1,...,i −1,2,j 1,...,j −1,n )............A ( +1)(i 1,...,i −1,n ,j 1,...,j −1,1)A ( +1)(i 1,...,i −1,n ,j 1,...,j −1,2)···A ( +1)(i 1,...,i −1,n ,j 1,...,j −1,n )⎞⎟⎟⎟⎟⎠,for 1≤i ,j ≤n and 2≤ ≤m .It is clear that when =m ,A (m +1)(i 1,...,i m −1,j 1,...,j m −1)is an n -by-n matrix.We will discuss multilevel Toeplitz and circulant matrices in Section 3.To change the new tensor eigenvalue problem into a multilevel matrix eigenvalue problem,we also need to change the eigentensor into a column vector.Definition 2.3Let X be an m th-order n -dimensional tensor.The vectorization of X with an ordering P is an n m -vector x P where its i th entry x P (j )is given byx P (j )=X i 1,i 2,...,i m ,1≤i k ≤n ,1≤k ≤m ,with j = m −1k =1nm −k (ik −1)+i m and P is the permutation matrix corresponding to the ordering P :(i 1,i 2,...,i m )=(i 1,i 2,...,i m )P .Using the same ordering P on A and X ,we have the following characterization for thetensor eigenvalue problem.Proposition 2.4The tensor eigenvalue problem in (4)is equivalent to the following the matrix eigenvalue system:A P x P =λx P .D o w n l o a d e d b y [U n i v e r s i t y o f H o n g K o n g L i b r a r i e s ] a t 03:04 07 A u g u s t 2015Linear and Multilinear Algebra 7Remark 2According to Proposition 2.4,we can calculate the eigenvalues of A and its associated eigentensor by solving the eigenvalue problem of the corresponding matrix A P based on the ordering P .Indeed we see by Proposition 2.2that both the eigenvalue and eigentensor are unique up to the ordering P .Proposition 2.5Let A be a (2m )th-order n-dimensional tensor.Suppose P and P are two different orderings,λis an eigenvalue of A P and y is the associated eigenvector.Then λis also eigenvalue of A P and P ,P y is an eigenvector of A P ,where P ,P is perfect shuffle permutation matrix in Proposition 2.2.Proof We know that A P y =λy and P ,P A P T P ,P=A P .It implies thatP ,P A P T P ,P P ,P y =λ P ,P y .The result follows.It is noted that different from the tensor eigenvalue problem in Definition 1.1,the new tensor eigenvalue problem in Definition 1.2is solvable and computable.2.2.Relationship with HOSVDIn this subsection,we will establish the relationship between the proposed tensor eigen-value problem and HOSVD [20]of a tensor.In numerical multilinear algebra,there are many applications for HOSVD [20].The computational procedure of HOSVD involves the calculation of singular value decomposition of matrices with respect to tensor unfolding at different indices.The HOSVD of A is given byA =S ×1U 1×2U 2×3···×2m U 2m ,(10)where S is a (2m )th-order n -dimensional all-orthogonal and ordering tensor and U k are n -by-n unitary matrices.Here,the multiplication ×k of a tensor A with a matrix U k is a (2m )th-order n -dimensional tensor given by(A ×k U k )i 1,...,i k −1,j k ,i k +1,...,i 2m =n i k =1a i 1,...,i k −1,i k ,i k +1,...,i 2m U k (j k ,i k ),for 1≤i l ,j k ≤n and 1≤l ≤2m .Next,we will show that the HOSVD of A can provide the information for the eigenvalues and their associated eigentensors for the multiplication of A and its conjugate transpose.Let us first define the conjugate transpose of a tensor.Definition 2.6Let A be a (2m )th-order n -dimensional tensor.A ∗is called the conjugate transpose of A where its entry is given by a j 1,...,j m ,i 1,...,i m for 1≤i k ,j k ≤n and 1≤k ≤m .A is called Hermitian ifa i 1,...,i m ,j 1,...,j m =a j 1,...,j m ,i 1,...,i m ,1≤i k ,j k ≤n ,1≤k ≤m ,i.e.A =A ∗.D o w n l o a d e d b y [U n i v e r s i t y o f H o n g K o n g L i b r a r i e s ] a t 03:04 07 A u g u s t 2015The contraction product of two (2m )th-order n -dimensional tensors (the multiplication of two square tensors)can be defined as follows:(A B )i 1,...,i m ,j 1,...,j m =n k 1,...,k m =1a i 1,...,i m ,k 1,...,k mb k 1,...,k m ,j 1,...,j m ,(11)for 1≤i l ,j l ≤n and 1≤l ≤m .Indeed,the contraction product of two square tensorscan be expressed in terms of the multiplication of two multilevel matrices.Also we have (A B )(:,...,:,j 1,...,j m )=A ·B (:,...,:,j 1,...,j m ),1≤j k ≤n ,1≤k ≤m .(12)Proposition 2.7Let A and B be (2m )th-order n-dimensional tensors.If their corre-sponding matrix unfolding are A P and B P under the ordering P,then the multilevel matrix of A B under the ordering P is equal to A P B P .Proof Using Proposition 2.2,it is sufficient to consider the natural ordering.The (i ,j )entry of AB isn mk =1A (i ,k )B (k ,j )=n k 1,...,k m =1a i 1,...,i m ,k 1,...,k mb k 1,...,k m ,j 1,...,j m .Therefore,the multilevel matrix AB is the square matrix unfolding of A B under thenatural ordering. Because A has a HOSVD given in (10),we can make use of the decomposition to express the multilevel matrix A corresponding to A according to the natural ordering:A =(U 1⊗U 2⊗···⊗U m )SU T m +1⊗U T m +2⊗···⊗U T2m ,where S is the square matrix unfolding of S under the natural ordering.This implies thatA ∗A =(¯U m +1⊗¯U m +2⊗···⊗¯U 2m )S ∗SU T m +1⊗U T m +2⊗···⊗U T 2m ,(13)where ¯Uk is a matrix where its entry is the complex conjugate of the entry of U k .We make use of the singular value decomposition of S ,i.e.S =QDW ∗where Q and W are unitary matrices and D is a real non-negative diagonal matrix.Therefore,we haveS ∗S =WD 2W ∗.(14)By substituting (14)into (13),we obtainA ∗A =(¯U m +1⊗¯U m +2⊗···⊗¯U 2m )WD 2W ∗ U T m +1⊗U T m +2⊗···⊗U T 2m .(15)It is clear that the eigenvectors of A ∗A are given by the unitary matrix:V =(¯Um +1⊗¯U m +2⊗···⊗¯U 2m )W or the eigentensors of A ∗A are just given byV =W ×1¯Um +1×2¯U m +2×3···×m ¯U 2m ,D o w n l o a d e d b y [U n i v e r s i t y o f H o n g K o n g L i b r a r i e s ] a t 03:04 07 A u g u s t 2015where V and W are (2m )th-order n -dimensional tensors and their square matrix unfoldings are equal to V and W ,respectively.Correspondingly,the eigenvalues of A ∗A are given by the eigenvalues D 2of S ∗S or S ∗ S ,i.e.the square of singular values D of S .In other words,the square root of the eigenvalue of A ∗A is just equal to the singular value of S corresponding to square matrix unfolding of S which is the core tensor of A .In other words,we obtain the eigen-decomposition of A ∗ A :(A ∗ A ) V =V D ,where D is a (2m )th-order n -dimensional diagonal tensor with its entries given byD i 1,i 2,...,i m ,j 1,j 2,...,j m =D 2k ,k ,k =n l =1(i l −1)n m −l +i m ,i l =j l ,1≤l ≤m ,0,otherwise .Example 1Let A be a 4th-order three-dimensional withA (1,:,1,:)=⎛⎝−0.1050−0.14470.14170.05920.08210.08880.14140.12620.0567⎞⎠,A (1,:,2,:)=⎛⎝−0.2156−0.03060.21270.22080.08320.02240.39450.2600−0.0458⎞⎠,A (1,:,3,:)=⎛⎝0.2830−0.8334−0.0230−0.61530.17100.4950−0.8596−0.34390.5920⎞⎠,A (2,:,1,:)=⎛⎝0.2631−0.4137−0.48890.89700.15070.2317−0.01130.32170.5491⎞⎠,A (2,:,2,:)=⎛⎝0.2353−0.6650−0.71351.30660.45970.4064−0.12150.56020.8921⎞⎠,A (2,:,3,:)=⎛⎝0.69100.2683−0.01540.0391−1.0960−0.30630.4814−0.4065−0.3998⎞⎠,A (3,:,1,:)=⎛⎝−0.1438−0.22050.0344−0.06740.27190.31310.03300.25640.2619⎞⎠,A (3,:,2,:)=⎛⎝−0.3521−0.20770.0935−0.08860.30470.49610.16730.44240.3168⎞⎠,A (3,:,3,:)=⎛⎝0.6487−0.5308−0.1976−0.04710.4314−0.1700−0.5455−0.30520.3088⎞⎠.D o w n l o a d e d b y [U n i v e r s i t y o f H o n g K o n g L i b r a r i e s ] a t 03:04 07 A u g u s t 2015The unitary matrices in the HOSVD of A are given byU 1=⎛⎝−0.58750.4445−0.6762−0.4185−0.8821−0.2163−0.69260.15590.7043⎞⎠,U 2=⎛⎝−0.2879−0.84420.4522−0.92950.36010.0804−0.2307−0.3971−0.8883⎞⎠,U 3=⎛⎝−0.5579−0.1280−0.8200−0.7333−0.38660.5593−0.38850.91330.1218⎞⎠,U 4=⎛⎝−0.5798−0.55210.5992−0.28810.82680.4830−0.76210.1074−0.6385⎞⎠.The core tensor of A is S ,and its entries are given byS (1,:,1,:)=⎛⎝0.9172−0.0540−0.0119−0.28580.53080.33710.7572−0.77920.1622⎞⎠,S (1,:,2,:)=⎛⎝0.6020−0.2290−0.4427−0.2630−0.91330.10670.65410.1524−0.9619⎞⎠,S (2,:,1,:)=⎛⎝0.75370.9340−0.7943−0.38040.1299−0.31120.5678−0.56880.5285⎞⎠,S (2,:,2,:)=⎛⎝0.6892−0.82580.0046−0.74820.53830.77490.4505−0.99610.8173⎞⎠,S (3,:,1,:)=⎛⎝0.07590.4694−0.1656000000⎞⎠,S (3,:,2,:)=⎛⎝0.08380.0782−0.86870000⎞⎠,S (1,:,3,:)=S (2,:,3,:)=S (3,:,3,:)=⎛⎝000000000⎞⎠.The square matrix unfolding of S with the natural ordering is given byS =⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝0.9172−0.0540−0.01190.6020−0.2290−0.4427000−0.28580.53080.3371−0.2630−0.91330.10670000.7572−0.77920.16220.65410.1524−0.96190000.75370.9340−0.79430.6892−0.82580.0046000−0.38040.1299−0.3112−0.74820.53830.77490000.5678−0.56880.52850.4505−0.99610.81730000.07590.4694−0.16560.08380.0782−0.868700000000000000000000⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠.D o w n l o a d e d b y [U n i v e r s i t y o f H o n g K o n g L i b r a r i e s ] a t 03:04 07 A u g u s t 2015Then we obtainA =(U 1⊗U 2)S (U T 3⊗U T 4).Next,we will find an unitary matrix W such that W ∗S ∗SW = 2is a diagonal matrix,whereW =⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝−0.61300.02140.0793−0.41620.66460.05050000.05980.3451−0.77590.11600.17260.48170000.0105−0.00740.50600.57470.26500.5858000−0.5777−0.00020.0490−0.1615−0.67250.43070000.3679−0.6915−0.0601−0.43540.06300.43520000.38920.63430.3599−0.5170−0.04470.2177000000000100000000010000000001⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠.The matrix V is given byV =(U 3⊗U 4)W=⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝−0.23010.0182−0.4107−0.29410.13750.0148−0.49130.45270.4754−0.2132−0.11980.2202−0.2032−0.0729−0.4156−0.3961−0.67800.2362−0.29010.04710.2955−0.03120.29650.25510.5235−0.08810.6249−0.3821−0.1425−0.5880−0.39190.10910.09800.3351−0.3088−0.3243−0.4241−0.09950.2655−0.1442−0.1448−0.62070.27010.4624−0.1611−0.43180.16650.4481−0.12980.27010.4271−0.35710.0601−0.42620.20570.7764−0.0650−0.18060.4249−0.35010.0730−0.0672−0.07060.5118−0.34940.2637−0.70610.17400.05230.05880.1007−0.03510.0298−0.4475−0.06850.38570.7557−0.2439−0.07780.0131−0.0928⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠,and2=diag (6.0484,3.8184,3.0304,1.1585,0.0968,0.0240,0,0,0).Therefore,the eigenvalues of A ∗ A are6.0484,3.8184,3.0304,1.1585,0.0968,0.0240,0,0,0and the eigentensors are V .The entries of the tensor V can be constructed correspondingly.3.Toeplitz and circulant tensorsIn this section,we will study the eigenvalue problem for special structured tensors:Toeplitz and circulant tensors.Definition 3.1A (2m )th-order n -dimensional tensor T =(t i 1,...,i m ,j 1,...,j m )is called a Toeplitz tensor ift i 1,...,i m ,j 1,...,j m =r j 1−i 1,...,j m −i m ,1≤i k ,j k ≤n ,1≤k ≤m ,(16)where R =(r i 1,...,i m )is an m th-order (2n −1)-dimensional tensor.D o w n l o a d e d b y [U n i v e r s i t y o f H o n g K o n g L i b r a r i e s ] a t 03:04 07 A u g u s t 2015Figure 2.A 4th-order Toeplitz Tensor.It is clear when m =1,the above definition is the same as that of Toeplitz matrix.Because of its special structure,we only require m 2n −1entries to construct a Toeplitz tensor.This Toeplitz tensor has been studied in [17].In Figure 2,we give an example to show the structure of a 4th-order n -dimensional Toeplitz tensor.Let us fix two indices i 2and j 2.For example,when i 2=3and j 2=1,T (:,3,:,1)is a Toeplitz matrix (the green one as shown in the figure).Indeed,all front slices of each cuboid,T (:,i 2,:,j 2),are Toeplitz matrices.Similarly,all the slices T (i 1,:,j 1,:)are also Toeplitz matrices.It is worth to noting that if j 2−i 2(or j 1−i 1)are fixed,then the corresponding matrices T (:,i 2,:,j 2)(or T (i 1,:,j 1,:))are the same.For example,when j 2−i 2=0,these matrices shown in red are the same in different cuboids in Figure 2.Similar to (9),we can use the square matrix unfolding with natural ordering to present a Toeplitz tensor.It is an n m ×n m multilevel Toeplitz matrix T .Indeed,T can be regarded as the first-level matrix,i.e.T =T (1)=⎛⎜⎜⎜⎜⎝T (2)(0)T (2)(1)···T (2)(n −1)T (2)(−1)T (2)(0)···T (2)(n −2)............T (2)(1−n )T (2)(2−n )···T (2)(0)⎞⎟⎟⎟⎟⎠,where (i ,j )th-block T (2)(j −i )is a second-level matrix.T (2)(k 1)is given by:T (2)(k 1)=⎛⎜⎜⎜⎜⎝T (3)(k 1,0)T (3)(k 1,1)···T (3)(k 1,n −1)T (3)(k 1,−1)T (3)(k 1,0)···T (3)(k 1,n −2)............T (3)(k 1,1−n )T (3)(k 1,2−n )···T (3)(k 1,0)⎞⎟⎟⎟⎟⎠,1−n ≤k 1≤n −1.In general,the th-level matrix is an n -by-n block matrix with n m − -by-n m − blocks given byT ( )(k 1,k 2,...,k −1)=⎛⎜⎜⎜⎜⎝T ( +1)(k 1,k 2,...,k −1,0)T ( +1)(k 1,k 2,...,k −1,1)···T ( +1)(k 1,k 2,...,k −1,n −1)T ( +1)(k 1,k 2,...,k −1,−1)T ( +1)(k 1,k 2,...,k −1,0)···T ( +1)(k 1,k 2,...,k −1,n −2)............T ( +1)(k 1,k 2,...,k −1,1−n )T ( +1)(k 1,k 2,...,k −1,2−n )···T ( +1)(k 1,k 2,...,k −1,0)⎞⎟⎟⎟⎟⎠,D o w n l o a d e d b y [U n i v e r s i t y o f H o n g K o n g L i b r a r i e s ] a t 03:04 07 A u g u s t 2015。