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Applied Mathematics Letters 20 (2007) 1189–1193 www.elsevier.com/locate/aml

A note on inversion of Toeplitz matrices$

Xiao-Guang Lv, Ting-Zhu Huang∗

School of Applied Mathematics, University of Electronic Science and Technology of China, Chengdu, Sichuan, 610054, PR China

Received 18 October 2006; accepted 30 October 2006

Abstract

It is shown that the invertibility of a Toeplitz can be determined through the solvability of two standard equations. The inverse matrix can be denoted as a sum of products of circulant matrices and upper triangular Toeplitz matrices. The stability of the inversion formula for a Toeplitz matrix is also considered. c 2007 Elsevier Ltd. All rights reserved.

Keywords: Toeplitz matrix; ; Inversion; Algorithm

1. Introduction

Let T be an n-by-n Toeplitz matrix:

a0 a−1 a−2 ··· a1−n ··· a1 a0 a−1 a2−n  ···  T = a2 a1 a0 a3−n , . . . . .  ......  an−1 an−2 an−3 ··· a0 where a−(n−1),..., an−1 are complex numbers. We use the shorthand n T = (ap−q )p,q=1 for a Toeplitz matrix. The inversion of a Toeplitz matrix is usually not a Toeplitz matrix. A very important step is to answer the question of how to reconstruct the inversion of a Toeplitz matrix by a low number of its columns and the entries of the original Toeplitz matrix. It was first observed by Trench [1] and rediscovered by Gohberg and Semencul [2] that T −1 can be reconstructed from its first and last columns provided that the first component of the first column does not vanish. Gohberg and Krupnik [3] observed that T −1 can be recovered from its first and second columns if the last component of the first column does not vanish.

$ Supported by NCET-04-0893 and applied basic research foundations of Sichuan province (05JY029-068-2). ∗ Corresponding author. E-mail address: [email protected] (T.-Z. Huang).

0893-9659/$ - see front matter c 2007 Elsevier Ltd. All rights reserved. doi:10.1016/j.aml.2006.10.008 1190 X.-G. Lv, T.-Z. Huang / Applied Mathematics Letters 20 (2007) 1189–1193

In Heinig and Rost [4], an inversion formula was exhibited for every nonsingular Toeplitz matrix. The method requires the solution of linear systems of equations (the so-called fundamental equations), where the right-hand side of one of them is a shifted column of the Toeplitz matrix T . In [5], Ben-Artzi and Shalom proved that three columns of the inverse of a Toeplitz matrix, when properly chosen, are always enough to reconstruct the inverse. Labahn and Shalom [6], and Ng, Rost and Wen [7] presented modifications of this result. In [8], Georg Heinig discussed the problem of the reconstruction of Toeplitz matrix inverses from columns. In this work, we give a new Toeplitz matrix inversion formula. The inverse matrix T −1 can be denoted as a sum of products of circulant matrices and upper triangular Toeplitz matrices. The results obtained show that this formula is numerically forward stable.

2. Toeplitz inversion formula

= n × Lemma 1. Let T (ap−q )p,q=1 be a n n Toeplitz matrix; then it satisfies the formula

T T KT − TK = f en − e1 f J, where 0 1  1  ..  1 .  . K =   , J =  ..  ,  .. ..     . .  1 1 0 0  1 0 a − a 0 .  n−1 −1   . .  e1 = . , en =   , f = .  . . 0       a − a  0 1 2 −n+2 a1 − a−n+1

= n = = Theorem 1. Let T (ap−q )p,q=1 be a Toeplitz matrix. If each of the systems of equations T x f, T y e1 is T T solvable, x = (x1, x2,..., xn) , y = (y1, y2,... yn) , then

(a) T is invertible; −1 (b) T = T1U1 + T2U2, where     y1 yn ··· y2 1 −xn · · · −x2  ..   .. .  y2 y1 .   1 . .  T1 =   , U1 =   , . .. ..   ..  . . . yn  . −xn yn ··· y2 y1 1     x1 xn ··· x2 o yn ··· y2  .. .   .. .  x2 x1 . .   o . .  T2 =   and U2 =   . . .. ..   ..  . . . xn  . yn xn ··· x2 x1 o

Proof. From Lemma 1 and T x = f, T y = e1, we have T T KT = TK + f en − e1 f J T T = T [K + xen − y f J], X.-G. Lv, T.-Z. Huang / Applied Mathematics Letters 20 (2007) 1189–1193 1191 then,

i i−1 T T K T = K T [K + xen − y f J] = · · · T T i = T [K + xen − y f J] . Therefore,

i i T T i K e1 = K T y = T [K + xen − y f J] y. Let T T i−1 ˆ ti = [K + xen − y f J] y and T = (t1, t2,..., tn). Then

T T i−1 i−1 T ti = T [K + xen − y f J] y = K e1 = ei , ˆ T T = T (t1, t2,..., tn) = (e1, e2,..., en) = In. So the matrix T is invertible, and the inverse of T is the matrix Tˆ = T −1. For (b): First of all, it is easy to see that

T T t1 = y, ti = [K + xen − y f J]ti−1 (i = 1, 2,..., n), −1 ti = T ei , Jei = en−i+1, JTJ = T T, JJ = I, J T = J.

Then, for i > 1

T T ti = K ti−1 + xen ti−1 − y f Jti−1 T −1 T −1 = K ti−1 + xen T ei−1 − y f JT ei−1 T −1 T −T = K ti−1 + xen JJT Jen−i+2 − y f T Jei−1 T −T T −T = K ti−1 + xe1 T en−i+2 − y f T en−i+2 T T = K ti−1 + xy en−i+2 − yx en−i+2 = K ti−1 + yn−i+2x − xn−i+2 y. So we have

t1 = y, t2 = K y + yn x − xn y, ..., n−1 n−2 n−2 tn = K y + K xyn − K yxn + · · · + xy2 − yx2 −1 T = (t1, t2,... tn)     1 −xn · · · −x2 o yn ··· y2  .. .   .. .  n−  1 . .  n−  o . .  = (y, K y,..., K 1 y)   + (x, K x,..., K 1x)    ..   ..   . −xn  . yn 1 o         y1 yn ··· y2 1 −xn · · · −x2 x1 xn ··· x2 o yn ··· y2  ..   .. .   .. .   .. .  y2 y1 .   1 . .  x2 x1 . .   o . .  =     +     .  . .. ..   ..  . .. ..   ..  . . . yn  . −xn . . . xn  . yn yn ··· y2 y1 1 xn ··· x2 x1 o 1192 X.-G. Lv, T.-Z. Huang / Applied Mathematics Letters 20 (2007) 1189–1193

= n Remark. In Theorem 1, let the Toeplitz matrix T (ap−q )p,q=1 be a circulant Toeplitz matrix. That is to say, the = n = = − = elements of the matrix T (ap−q )p,q=1 satisfy ai ai−n for all i 1,..., n 1. It is easy to see that f 0. Thus, x = T −1 f = 0. From (b) of Theorem 1, we get   y1 yn ··· y2  ..  − y2 y1 .  T 1 =   . . .. ..  . . . yn yn ··· y2 y1 We conclude: The inverses of the circulant Toeplitz matrices are circulant Toeplitz matrices.

3. Stability analysis

In this section, we will show that the Toeplitz inversion formula presented in Section 2 is evaluation forward stable. An algorithm is called forward stable if for all well conditioned problems, the computed solution x˜ is close to the true solution x in the sense that the relative error kx −x ˜k2 / kxk2 is small. In the matrix computation, round-off errors occur. Let A, B ∈ Cn,n and α ∈ C. If we neglect the O(ε2) terms, then for any floating-point arithmetic with machine ε, then (cf. [9]) √ fl(α A) = α A + E, kEk ≤ ε |α| kAk ≤ ε n |α| kAk , F F √ 2 fl(A + B) = A + B + E, kEkF ≤ ε kA + BkF ≤ ε n kA + Bk2 , fl(AB) = AB + E, kEkF ≤ εn kAkF kBkF . According to the floating-point arithmetic, we have the following bound.

= n Theorem 2. Let T (ap−q )p,q=1 be a nonsingular Toeplitz matrix and be well conditioned; then the formula in Theorem 1 is forward stable.

Proof. Assume that we have computed the solutions x˜, y˜ in Theorem 1 which are perturbed by the normwise relative errors bounded by ε˜,

kx˜k2 ≤ kxk2 (1 +ε), ˜ ky˜k2 ≤ kyk2 (1 +ε). ˜ Therefore, we have √ √ kT1kF = n kyk2 , kT2kF = n kxk2 , √ q √ k k ≤ + k k2 k k ≤ k k U1 F n 1 x 2, U2 F n y 2 . Using the perturbed solutions x˜, y˜, the inversion formula in Theorem 1 can be expressed as

˜ −1  ˜ ˜ ˜ ˜  T = f l T1U1 + T2U2

= f l ((T1 + ∆T1)(U1 + ∆U1) + (T2 + ∆T2)(U2 + ∆U2)) −1 = T + ∆T1U1 + T1∆U1 + ∆T2U2 + T2∆U2 + E + F. Here, E is the matrix containing the error which results from computing the matrix products, and F contains the error from subtracting the matrices. For the error matrices ∆T1, ∆U1, ∆T2 and ∆U2, we have √ k∆T k ≤ε ˜ kT k =ε ˜ n kyk , 1 F 1 F √ 2 k∆T2kF ≤ε ˜ kT2kF =ε ˜ n kxk2 , √ q 2 k∆U1kF ≤ε ˜ kU1kF ≤ε ˜ n 1 + kxk , √ 2 k∆U2kF ≤ε ˜ kU2kF ≤ε ˜ n kyk2 . X.-G. Lv, T.-Z. Huang / Applied Mathematics Letters 20 (2007) 1189–1193 1193

It follows that

kEk2 ≤ kEkF ≤ εn(kT1kF kU1kF + kT2kF kU2kF ) q  ≤ 2 k k + k k2 + k k εn y 2 1 x 2 x 2 √ 2 −1 ≤ εn kyk2 (1 + 2 kxk2), kFk2 ≤ ε n T . 2 Adding all these error bounds, we have √ ˜ −1 −1 −1 T − T ≤ n(2ε˜ + nε) kyk2 (1 + 2 kxk2) + ε n T . 2 2

Note that T y = e1 and T x = f ; then −1 −1 kyk2 ≤ T and kxk2 ≤ T k f k2 . 2 2 Thus, the relative error is

˜ −1 −1 T − T √ 2  −1  ≤ n(2ε˜ + nε) 1 + 2 T k f k2 + ε n. −1 2 T 2 −1 k k As T is well conditioned, thus, T 2 is finite. Obviously, f 2 is finite. Therefore, the formula presented in Theorem 1 is forward stable.  References

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