In this paper we concern the spectral properties of hermitian Toeplitz
matrices. Based on the fact that every centrohermitian matrix can be reduced to a real matrix by a
simple similarity transformation, we first consider the eigenstructure of hermitian Toeplitz matrices
and then discuss a related inverse eigenproblem. We show that the dimension of the subspace of
hermitian Toeplitz matrices with two given eigenvectors is at least two and independent of the size
of the matrix, the solution of the inverse hermitian Toeplitz eigenproblem with two given eigenpairs
is unique.FC

Li Jing

Liu Zhongyun

Zhang Yulin

Universidade do Minho: RepositoriUM

On the Eigenstructure of Hermitian ToeplitzMatrices with Prescribed EigenpairsZhongyun Liu1 Jing Li1 Yulin Zhang21School of Mathematics and Computing Science, Changsha University of Science and Technology,Changsha, Hunan, 410076, P. R. China.2Department of Mathematics, University of Minho, Campus de Gualtar, 4710-057 Braga, Portugal.Abstract Toeplitz matrices have found important applications in bioinformatics and computa-tional biology [5, 6, 11, 12]. In this paper we concern the spectral properties of hermitian Toeplitzmatrices. Based on the fact that every centrohermitian matrix can be reduced to a real matrix by asimple similarity transformation, we first consider the eigenstructure of hermitian Toeplitz matricesand then discuss a related inverse eigenproblem. We show that the dimension of the subspace ofhermitian Toeplitz matrices with two given eigenvectors is at least two and independent of the sizeof the matrix, the solution of the inverse hermitian Toeplitz eigenproblem with two given eigenpairsis unique.Keywords Centrohermitian matrix, hermitian Toeplitz matrix, eigenstructure, inverse eigenprob-lems1 IntroductionHermitian Toeplitz matrices play an important role in the trigonometric moment prob-lem, the szego¨ theory, the stochastic filtering, the signal processing, the biological infor-mation processing and other engineering problems [1, 3, 5, 6, 11, 12]. Many propertiesof hermitian Toeplitz matrices have been studied for decades.Recall that a matrix A ∈ Cn×n is said to be centrohermitian [10], if JAJ = A, where Adenotes the element-wise conjugate of the matrix and J is the exchange matrix with oneson the cross diagonal (lower left to upper right) and zeros elsewhere. Hermitian Toeplitzmatrices are an important subclass of centrohermitian matrices and have the followingformH =h0 h1 · · · hn−1h¯1 h0. . ........ . . . . . h1h¯n−1 · · · h¯1 h0 . (1)A vector x ∈ Cn is said to be hermitian if Jx = x¯. Let A ∈ Cn×n be a hermitiancentrohermitian matrix and x ∈ Cn be an eigenvector of A associated with the eigenvalueλ , then Ax = λx implies AJx¯ = λJx¯, which means that x+ Jx¯ is also an eigenvector of Aassociated with the eigenvalue λ , and x+ Jx¯ is hermitian. So we claim that an hermitiancentrohermitian matrix A has an orthonormal basis consisting of n hermitian eigenvectors.Naturally, an hermitian Toeplitz matrix also has an orthonormal basis consisting of nhermitian eigenvectors. In this paper we consider the following problem:Problem A Given a set of hermitian unitary vectors {x( j)}kj=1 ∈ Cn (k < n), and a set ofscalars {λ j}kj=1 ∈ R, find an n×n hermitian Toeplitz matrix H such thatHx( j) = λ jx( j), for j = 1, . . . ,k,where R and C denote the field of real and complex numbers respectively.Since H is required to have a given structure, in our case, it is a hermitian Toeplitzmatrix, so the eigenvalues cannot be totally arbitrary.We remark here that Problem A is actually one of the partially described inverse eigen-value problems (PDIEPs) [1], which is of interests in biomedical engineering [7, 8, 9].On this topic of PDIEPs, the earlier study can be found for real symmetric Toeplitz ma-trices and Jacobi matrices in [3], and some of the recent works can be found for anti-symmetric matrices, anti-persymmetric matrices, centrosymmetric matrices, symmetricanti-bidiagonal matrices, K-symmetric matrices and K-centrohermitian matrices, see [10]and references therein.2 PreliminariesWe begin with a brief overview on the reducibility of centrohermitian matrices. Allthe formulas become slightly more complicated when n is odd. For simplicity, we restrictour attention to the case of even n = 2m throughout this paper.A centrohermitian matrix of order n can be partitioned as follows:A =[B JCJC JBJ], n = 2m. (2)We defineQ =√22[I iIJ −iJ], n = 2m. (3)We then have the following well known theorems (cf. [10]).Theorem 1. Let Q be defined as in (3). Then A ∈ Cn×n is centrohermitian if and only ifQHAQ ∈ Rn×n (denoted by RA), whereRA =[Re(B+ JC) −Im(B+ JC)Im(B− JC) Re(B− JC)], n = 2m.Corollary 2. Let Q be defined as in (3). Then x∈Cn is hermitian if and only if QHx∈Rn.We note that an n×n hermitian Toeplitz matrix H can be completely characterized bythe real and imaginary parts of its first row (or column). Let h = (h0, Re(h1), Im(h1), . . . ,Re(hn−1), Im(hn−1))T , which is a (2n−1)-dimensional vector; and letS j =[0n− j, j In− j0 j,n− j 0 j, j], j = 0,1, · · · ,n−1,where 0p,q denotes the p×q zero matrix. Then H in (1) can be parameterized as follows:H = φ0I+2n−2∑j=1φ jH j, (denoted by H(h)), (4)where φ0 = h0, φ2p−1 = Re(hp), φ2p = Im(hp), H2p−1 = Sp +STp , and H2p = i(Sp−STp ),for p = 1, . . . ,n−1.Eq. (4) gives an one-to-one correspondence between complex hermitian Toeplitz n×nmatrices and real (2n−1)-vectors.Applying Theorem 1 to (4) givesRH(h) = φ0I+2n−2∑j=1φ jRH j (5)where all RH j , for j = 1, . . . ,2n−2, are real symmetric, and their matrix structures for thecase n = 2m are given as follows:(i) For 1≤ j ≤ m−1,RH2 j−1 =[Tˆ2 j−1T˜2 j−1], RH2 j =[Tˇ2 jTˇ T2 j], (6)whereTˆ2 j−1 = T (e j+1)+[0 00 J j], T˜2 j−1 = T (e j+1)+[0 00 −J j], andTˇ2 j =[0 −Im− jIm− j J j].(ii) For m≤ j ≤ n−1,RH2 j−1 =J2m− j 00 0−J2m− j 00 0 , RH2 j =J2m− j 00 0J2m− j 00 0 . (7)Here T (e j+1) denotes the Toeplitz matrix generated by the m-dimensional unit vectore j+1; Is and Js denote the identity matrix and exchange matrix of order s, respectively.Based on the above analysis, Problem A can be restated as follows:Problem B Given a set of orthonormal vectors {y( j)}kj=1 ∈Rn (n> k) and a set of scalars{λ j}kj=1 ∈ R, find a symmetric matrix RH(h) ∈ Rn×n in the form (5) such thatRH(h)y( j) = λ jy( j), for j = 1, . . . ,k.In this paper, we concentrate our study on the eigenpairs for the cases k = 1 and k = 2.3 Hermitian Toeplitz matrices with a given eigenvectorSuppose that x is an eigenvector of two matrices A and B, with associated eigenval-ues λ and µ , respectively, then x is also eigenvector of matrix A+ B with associatedeigenvalue λ + µ . Hence, given any vector, the space of matrices with that vector as aneigenvector is a linear subspace. Since there is an one-to-one correspondence betweencomplex hermitian Toeplitz n×n matrices H and real (2n−1)-vectors h, then the collec-tion of these (2n−1)-vectors form a linear subspace of R(2n−1).Assume that x ∈ Cn is an arbitrary hermitian vector. LetS(x) = {h ∈ R(2n−1)|H(h)x = λx, for some λ ∈ R}be this linear subspace. It is evident that S(x) is nonempty. In fact, the standard basis(2n−1)-vector e1 = (1,0, . . . ,0)T ∈ S(x) for all x. This means that the dimension of S(x)is at least 1. Furthermore, letS0(x) = {h ∈ R(2n−1)|H(h)x = 0}denote the linear subspace consisting of all hermitian Toeplitz matrices for which x is aneigenvector associated with eigenvalue 0. Clearly, H(h)x = λx if and only if h−λe1 ∈S0(x). SoS(x) =< e1 >⊕S0(x).The following result gives the precise dimension of S0(x) for general hermitian vector x.Lemma 3. Let x ∈ Cn be hermitian. Thendimension(S0(x)) = n−1.Proof. From the hypothesis, we know that H(h) is centrohermitian and x is hermitian.By Theorem 1 and Corollary 2, we have thatH(h)x = 0is equivalent toRH(h)z = 0,where RH(h) ∈ Rn×n is defined as in (5) and z = QHx ∈ Rn. Note that RH(h)z is a linearfunction of both entries of z and h. So we can writeRH(h)z = A(z)h, (8)where A(z) is an n× (2n− 1) matrix whose entries depend linearly on the n-vector z =(z1, . . . ,zn)T . Thus the dimension of S0(x) is the nullity of A(z). Note that A(z)h = 0 is ahomogeneous linear system of n equations in 2n− 1 unknowns, so the nullity of A(z) isat least n−1.We now show that the nullity of A(z) is exactly n−1. Note thatA(z) = [ z RH 1z . . . RH jz . . . RH 2n−2z ],where RH j, j = 1, · · · ,2n− 2, are defined as in (6) and (7), respectively. Note also thatthe RH j, j = 1, · · · ,2n−2, are direct sum ( j is odd) or anti-direct sum ( j is even) of twomatrices with the same structure, so the first m rows and the last m rows of A(z) has alsothe same structure. Now we exchange the order of rows, we put together the rows whoseright side has the same number of zeros, then we will get a block echelon matrix like,z1 · · · · · z2 zm+2 z1 zm+1zm+1 · · · · · −zm+2 z2 −zm+1 z1· · · · z2 zm+2 z1 zm+1 0 0· · · · −zm+2 z2 −zm+1 z1 0 0...... · · · · · · 0 0...... · · · · · ·...... z1 zm+1 0 0zn... −zm+1 z1 0 0In case z21+z2m+1 6= 0, the rank of this matrix is n. When both z1 and zm+1 are zero, weget another block echelon form, which is[z2 zm+2−zm+2 z2]. In case z2 and zm+2 are notboth zero, then the rank of that matrix is n, when they are both zero, we go on to anotherblock, go on with this process, since z is a nonzero vector, so at least one of the block isnonsingular, which guarantees the rank of this matrix is n. So the nullity of A(z) is n−1,which means dimension(S0(x)) = n−1. ¤It is useful to illustrate the structure used above with a 6×6 example. Assume that z=(z1, . . . ,z6)T ∈ R6 is a nonzero vector and h = (φ0,φ1, . . . ,φ10)T . Then RH(h)z = A(z)hwith A(z) taking the following formz1 z2 −z5 z3 −z6 z3 z6 z2 −z5 z1 z4z2 z1+ z3 z4− z6 z3 z6 z2 z5 z1 −z4 0 0z3 z2+ z3 z5+ z6 z1+ z2 z4+ z5 z1 z4 0 0 0 0z4 z5 z2 z6 z3 −z6 z3 −z5 −z2 −z4 z1z5 z4+ z6 z3− z1 −z6 z3 −z5 z2 −z4 −z1 0 0z6 z5− z6 z3− z2 z4− z5 z2− z1 −z4 z1 0 0 0 0 .After exchange of the rows, we can get a matrix likez1 z2 −z5 z3 −z6 z3 z6 z2 z5 z1 z4z4 z5 z2 z6 z3 −z6 z3 −z5 z2 −z4 z1z2 z1+ z3 z4− z6 z3 z6 z2 z5 z1 z4 0 0z5 z4+ z6 −z1+ z3 −z6 z3 −z5 z2 −z4 z1 0 0z3 z2+ z3 z5+ z6 z1+ z2 z4+ z5 z1 z4 0 0 0 0z6 z5− z6 −z2+ z3 z4− z5 −z1+ z2 −z4 z1 0 0 0 0 .The elements z1,z4 form a block echelon form and the rank of this matrix is 6 if z1and z4 are not both zero. If z1 and z4 are both zero, then the elements z2,z5 form anotherblock echelon form, if z2 and z5 are both zero, then the block echelon form go to elementsz3,z6. Because we assume that z is a nonzero vector, so at least one of zi cannot be zero,that is rank(A(z)) = 6.The following theorem gives the precise dimension of S(x) and shows that for anyhermitian x ∈ Cn, there is a large collection of Hermitian Toeplitz matrices with x as aneigenvector..Theorem 4. Let x ∈ Cn be hermitian. Then dimension(S(x)) = n.4 Hermitian Toeplitz matrices with two given eigenvec-torsIn this section, we consider the case that two eigenvectors are given. Assume thatx,y ∈ Cn (xTy = 0) are arbitrary hermitian vectors. LetS(x) = {h ∈ R(2n−1)|H(h)x = λx, for some λ ∈ R}S(y) = {h ∈ R(2n−1)|H(h)y = γy, for some γ ∈ R}Our objective is find the dimension of S(x)∩S(y). Since the standard basis (2n−1)-vector e1 = (1,0, . . . ,0)T ∈ S(x)∩S(y), so S(x)∩S(y) is nonempty. That means that thedimension of S(x)∩S(y) is at least 1.As we did in previous section, we first transform our equations into real equations,that is rewrite them asRH(h)z = λz,RH(h)w = γw.(9)or(RH(h)−λ I)z = 0(RH(h)− γI)w = 0. (10)where RH(h) ∈ Rn×n is defined as in (5) and z = QHx, w = QHy ∈ Rn.Denoting t = (φ0 − γ,φ0 − λ ,φ1, . . . ,φ2n−2)T , we then have that the system (10) isequivalent toMt = 0 (11)whereM is the 2n×2n matrix defined byM=[0 z RH1z · · · RH j z · · · RH2n−2zw 0 RH1w · · · RH j w · · · RH2n−2w].Suppose that s = (s′0,s0,s1,s2, ...,s2n−2)T is a solution of (11). For arbitrary λ and α ,defineφ0 := αs0+λ ,φi := αsi, i = 1, ...,2n−2. (12)andγ := α(s0− s′0)+λ . (13)Then z,w are eigenvectors of RH(h), or we say that S(x)∩ S(y) is the direct sum ofthe subspace spanned by e1 = (1,0, ...,0)T and the subspace obtained by deleting the firstcomponent from ker(M) ( see h = α s¯+ λe1, s¯ = (s0,s1,s2, ...,s2n−2)T ). On the otherhand, suppose the two eigenvalues λ ,γ are given, then by (13), the α in (12) must beα =λ − γs′0− s0provided s′0 6= s0. This gives us the following lemmaLemma 5. Suppose that s = (s′0,s0,s1,s2, ...,s2n−2) is a solution of (11) with s′0 6= s0.Then corresponding to the direction of s, there is only one solution to (9)Now we determine the null space of M. We multiply on the left of M a nonsingularmatrix T defined byT= [T1, T2, ..., T2n]T ,whereT1 = (−w1,−w2, ...,−wn,z1,z2, ...,zn) = (−wT, zT),Ti = eTi , i = 2, ...,2n.Then we can see that the first row of of TM is identically zero because of the orthogo-nality condition zTw = 0,wTz = 0 and the symmetries of RHi ′s, i = 1,2, ...,2n−2. So therank ofM is at most 2n−1. On the other side, by the proof of lemma 3, we know that thelast n row of TM can form an echelon block with rank n, so the rank ofM is at least n.In conclusion, we give the following theoremTheorem 6. Suppose that n is even, and x and y are two hermitian orthogonal vectors.Then2≤ dim(S(x)∩S(y))≤ n+1.5 ConclusionIn this paper we have exploited the facts that every centrohermitian matrix can bereduced to be a real matrix by a simple similarity transformation and that every n× nhermitian Toeplitz matrix H can be completely characterized by the real and imaginaryparts of its first row (or column) (viz. there exists a one-to-one corresponding betweencomplex hermitian Toeplitz n×n matrices and real (2n−1)-vectors) to show some theo-retical results, which can be thought of extensions of the works in [4] and [2], from realsymmetric Toeplitz matrices to complex hermitian Toeplitz matrices.We conclude that for Problem A in the cases k = 1,2, being hermitian is sufficientfor a single vector to be an eigenvector of a hermitian Toeplitz matrix, and the collectionof hermitian Toeplitz matrices with one given eigenvector is quite large. Also, the setS(x)∩S(y) contains all the hermitian Toeplitz matrices with two prescribed eigenvectorsand its dimension is at least 2 independently of the size of the problem. Moreover, foreach direction of kerM, there is only one hermitian Toeplitz matrix with two prescribedeigenpairs.In general case, suppose that {x(1),x(2), ...,x(k)},k ≥ 3 is a set of hermitian orthonor-mal vectors. Thenk⋂i=1S(x(i)) contains all hermitian Toeplitz matrices for which each x(i)is an eigenvector. Evidently, (2n− 1)-vector e1 = (1,0, . . . ,0)T ∈k⋂i=1S(x(i)) for all i. Sok⋂i=1S(x(i)) is at least of dimension 1. Analogously, we may transferk⋂i=1S(x(i)) into linearequations. The solvability of Problem A depends on if the system is consistent.6 BibiliographyAcknowledgesThis work is supported by the National Natural Science Foundation of China, No.10771022 and 10571012, Scientific Research Foundation for the Returned Overseas Chi-nese Scholars, State Education Ministry of China, No. 890 [2008], and Major Foundationof Educational Committee of Hunan Province [2009]; Portuguese Foundation for Scienceand Technology (FCT) through the Research Programme POCTI, respectively.References[1] M. Chu, Inverse eigenvalue problems, SIAM Rev., Vol. 40 (1998), pp. 1–39.[2] M. T. Chu and M. A. Erbrecht, Symmetric Toeplitz matrices with two prescribedeigenpairs, SIAM J. Matrix Anal. Appl., Vol. 15 (1994), pp.623–635.[3] M. Chu, and G. H. Golub, Structured inverse eigenvalue problems, Acta Numer.,Vol. 11 (2002), pp. 1–71.[4] G. Cybenko, On the eigenstructure of Toeplitz matrices, IEEE Trans. Acoust. SpeechSignal Process., Vol. 32 (1984), pp. 918-920.[5] W. J. Ewens and G. R. Grant, Statistical Methods in Bioinformatics, Springer, NY,2001.[6] M. G. Grigorov, Global dynamics of biological systems from time-resolved omicsexperiments, Bioinformatics 2006, 22(12):1424-1430.[7] J. Hu and S. Ye, Modern signal processing theory in MEG research, Chinese J.Medical Physics No. 4 (2003).[8] Jun Li, Application and Comparison of Some Optimization Methods on MEG In-verse Problem, Acta Electronica Sinica, No. 1 (2001).[9] Jun Li, MEG Inverse Solution Using Gauss-Newton Algorithm Modified by Moore-Penrose Inversion, J. Biomedical Engineering, No. 2 (2001).[10] Z. Liu and H. Faßbender, An inverse eigenvalue problem and an associated ap-proximation problem for generalized K-centrohermitian matrices, J. Comput. Appl.Math., 206/1 (2007) pp. 578-585.[11] W. Miller, Comparison of genomic DNA sequences: solved and unsolved problems,Bioinformatics 17 (2001) 391-397.[12] T. D. Pham, Spectral distortion measures for biological sequence comparisons anddatabase searching, Pattern Recognition, 40 (2007) 516 - 529.

On the eigenstructure of hermitian Toeplitz matrices with prescribed eigenpairs

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On the eigenstructure of hermitian Toeplitz matrices with prescribed eigenpairs

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