MATRIX INVERSE EIGENVALUE PROBLEM

Transcript

1 English NUMERICAL MATHEMATICS Vol.14, No.2 Series A Journal of Chinese Universities May 2005 A STABILITY ANALYSIS OF THE (k) JACOBI MATRIX INVERSE EIGENVALUE PROBLEM Hou Wenyuan ( ΛΠ) Jiang Erxiong( Ξ) Abstract In this paper we will analyze the perturbation quality for a new algorithm of the (k) Jacobi matrix inverse eigenvalue problem. Key words eigenvalue, Jacobi matrix, (k) inverse problem. AMS(2000)subject classifications 65F 18 1 Introduction Let T n = α 1 β 1 0 β 1 α 2 β 2 β β 0 β α n be an n n unreduced symmetric tridiagonal matrix, and denote its submatrix T p,q, (p <q)as follows α p β p 0 β p α p+1 β p+1. T p,q = β..... p+1 p<q βq 1 0 β q 1 α q We call an unreduced symmetric tridiagonal matrix with β i > 0asaJacobimatrix. Consider T 1,n and T p,q to be Jacobi matrices. The matrix ( ) T 1, 0 W k = 0 T k+1,n is obtained by deleting the k th row and the k th column (k =1, 2,..., n) fromt n. Received: Mar. 15, 2003.

2 116 Hou Wenyuan Jiang Erxiong Problem If we don t know the matrix T 1,n, but we know all eigenvalues of matrix T 1,, all eigenvalues of matrix T k+1,n, and all eigenvalues of matrix T 1,n,could we construct the matrix T 1,n. Let S1 =(µ 1,µ 2,,µ ), S2 =(µ k,µ k+1,,µ ), λ =(λ 1,λ 2,,λ n ) are the eigenvalues of matrices T 1,,T k+1,n and T 1,n respectively. The problem is that from above 2n-1 data to find other 2n-1 data: α 1,α 2,,α n, and β 1,β 2,,β. Obviously, when k =1ork = n this problem has been solved and there are many algorithms to construct T 1,n [3][6], and its stability analysis can be founded in [2]. While k =2, 3...n 1, a new algorithm has been put forward to construct T 1,n [1]. In this paper we will give some stability properties of the new algorithm in case k =2, 3...n 1. 2 Basic Theorem Theorem 2.1 [1] If there is no common number between µ 1,µ 2,,µ and µ k,µ k+1,,µ, then the necessary and sufficient condition for the (k) problem having a solution is: λ 1 <µ j1 <λ 2 <µ j2 < <µ j <λ n, (2.1) where µ =(µ j1,µ j2...µ j ), and µ i, (i =1, 2...n 1) are recorded as µ ji, (i =1, 2...n 1) such that µ j1 <µ j2 < <µ j. (2.2) Furthermore, if a given (k) problem has a solution, then the solution is unique. Algorithm 2.2 [1] Given three vectors λ =(λ 1,λ 2 λ n ) T,S1 =(µ 1,µ 2 µ ) T and S2 =(µ k, µ k+1 µ ) T which are satisfied with (2.1), then we can solve (k) problem by following algorithm: Step 1 Find α k as α k =trace(t 1,n ) trace(w k )= n λ i µ i. (2.3)

3 A Stability Analysis of the (k)-jacobi Matrix Inverse Eigenvalue Problem 117 Step 2 Find x =(x 1,x 2,,x )as x j = n (µ j λ i ), j > 0,j =1, 2,n 1. (2.4) (µ j µ i ) Step 3 Compute β and β k as β = x i, (2.5) β k = x i. (2.6) Step 4 Compute S (1),i,, 2,,k 1andS(2) 1,i i = k, k +1,,n 1. as: S (1),j = x j /β,j =1, 2,,k 1, (2.7) S (2) 1,i = x i /β k,i= k, k +1,,n 1, (2.8) where S (1),j is the last element of the unit eigenvector of T 1,, corresponding to the eigenvalue µ j,and S (2) 1,i is the first element of the unit eigenvector of T k+1,n, corresponding to the eigenvalue µ i. Step 5 Compute T 1, from S (1),i,i =1, 2,,k 1andµ 1,µ 2,,µ by Lanczos Process or Givens Orthogonal Reduction Process [3],[6],[7]. Compute T k+1,n from S (2) 1,i,i= k, k+1,,andµ k,µ k+1,,µ by Lanczos Process or Givens Orthogonal Reduction Process [3],[6],[7]. 3 Stability Analysis Let λ =(λ 1,λ 2 λ n ) T,S1=(µ 1,µ 2 µ ) T,S2=(µ k,µ k+1 µ ) T (3.1) and λ =( λ 1, λ 2 λ n ) T, S1 =( µ 1, µ 2 µ ) T, S2 =( µ k, µ k+1 µ ) T (3.2) are satisfied with (2.1) to be that λ 1 <µ j1 <λ 2 <µ j2 < <µ j <λ n,

4 118 Hou Wenyuan Jiang Erxiong and λ 1 < µ j1 < λ 2 < µ j2 < < µ j < λ n. By using the Algorithm 2.2, we can get a matrix T 1,n by the date (3.1) and a matrix T 1,n by the data (3.2), now we will analyze the bound of T 1,n T 1,n F. is Lemma 3.1 The derivative of the function f(t) defined as f (t) =f(t)( f(t) = n n (a i + b i t), j (c i + d i t) b i a i + b i t, j d i c i + d i t ). Proof Start by noticing that f (t) = = ( n (a k + b k t)) k=1 k=1, j n = f(t)( (c k + d k t) b i n k=1 (a i + b i t) n Defined a function f j (t) as f j (t) = +( n (a k + b k t))( k=1 (a k + b k t) k=1, j (c k + d k t) b i a i + b i t, j d i c i + d i t )., j k=1, j n [µ j λ i + t( µ j µ j + λ i λ i )], j By the Lemma 3.1, we get that f j(t) =f j (t)( [µ j µ i + t( µ j µ j + µ i µ i )], j n 1 (c k + d k t) d i (c i + d i t) n ) (a k + b k t) k=1 µ j µ j + λ i λ i µ j λ i + t( µ j µ j + λ i λ i ) k=1, j (c k + d k t), t [0, 1]. (3.3) µ j µ j + µ i µ i ). (3.4) µ j µ i + t( µ j µ j + µ i µ i )

5 A Stability Analysis of the (k)-jacobi Matrix Inverse Eigenvalue Problem 119 Using the mean value formula for differential calculus, we have f j (1) f j (0) = f j(τ),τ (0, 1). (3.5) Symbolized A j = B j = n, j F j = max t (0,1) f j(t), (3.6) 1 min{ µ j λ i, µ j λ i }, (3.7) 1 min{ µ j µ i, µ j µ i }, (3.8) θ j =max{a j,b j }, (3.9) thus from (3.4) we get f j (t) F j + + n, j n, j A j µ j µ j + λ i λ i µ j λ i + t( µ j µ j + λ i λ i ) µ j µ j + µ i µ i µ j µ i + t( µ j µ j + µ i µ i ) ) µ j µ j + λ i λ i min{ µ j λ i, µ j λ i } n µ j µ j + µ i µ i min{ µ j µ i, µ j µ i } µ j µ j + λ i λ i +B j, j n n A j ( µ j µ j + λ i λ i ) + B j (, j µ j µ j +, j µ i µ i ) θ j [2(n 1) µ j µ j + µ µ 1 + λ λ 1 ] µ j µ j + µ i µ i (2n 1)θ j ( µ µ 1 + λ λ 1 ). (3.10) Theorem 3.2 Let x j, β, β k, x j, β and β k are the matrix element defined by (2.4),(2.5) and (2.6) from the data (3.1) and (3.2) respectively, then we have x j x j ϱ j ( µ µ 1 + λ λ 1 ) (3.11) β β ρ 1 ( µ µ 1 + λ λ 1 ) (3.12) β k β k ρ 2 ( µ µ 1 + λ λ 1 ), (3.13)

6 120 Hou Wenyuan Jiang Erxiong where ϱ j,ρ 1 and ρ 2 are defined as ϱ j = (2n 1)F jθ j xj + x j, (3.14) ρ 1 = (2n 1) F i θ i β + β, (3.15) ρ 2 = (2n 1) F i θ i β k + β k. (3.16) Proof Start by observing that x j x j = x j x j xj + x j = f j(1) f j (0) xj + x j (2n 1)F jθ j xj + x j ( µ µ 1 + λ λ 1 ), so it is easy to prove that β β = x x i i x i x i x i + x i (2n 1) F i θ i β + β ( µ µ 1 + λ λ 1 ), β k β k = x x i i x i x i x i + x i (2n 1) F i θ i β k + β k ( µ µ 1 + λ λ 1 ). Theorem 3.3 Let q 1 =(S (1),1,S(1),2 S(1), )T and q 1 =( S (1) (1),1, S S (1), )T, where S (1) (1),j and S,j,j =1, 2...k 1 are defined by (2.7), then,2 q 1 q 1 2 η 1 ( µ µ 1 + λ λ 1 ), (3.17)

7 A Stability Analysis of the (k)-jacobi Matrix Inverse Eigenvalue Problem 121 where η 1 is defined as ϱ 2 i + ρ 1 η 1 = β (3.18) and ϱ i, ρ 1 are defined as (3.14),(3.15). Proof q 1 q 1 2 =( =( =( ( S (1) (1),i S,i 2 ) 1 2 ϱ 2 i xi β xi β xi β 2 ) 1 2 xi β + xi β xi x i 2 ) 1 2 +( β x i ρ 1 xi β 2 ) 1 2 xi (β β ) β β 2 ) 1 2 ( + )( µ µ 1 + λ β λ 1 ) β β ϱ 2 i + ρ 1 = β ( µ µ 1 + λ λ 1 ). So by the same method, we can get the theorem below immediately. Theorem 3.4 Let q 2 =(S (2) 1,k,S(2) 1,k+1 S(2) S (2) 1, )T, where S (2) 1,j where η 2 is defined as and S (2) 1,j and ϱ i, ρ 2 are defined as (3.14),(3.16). 1, )T and q 2 =( S (2) (2) 1,k, S,j = k, k +1...n 1 are defined by (2.8), then 1,k+1 q 2 q 2 2 η 2 ( µ µ 1 + λ λ 1 ), (3.19) η 2 = ϱ 2 i + ρ 2 β k (3.20) Lemma 3.5 [2] Let q 1 =(S (1),1,S(1),2 S(1), )T and q 2 =(S (2) 1,k,S(2) 1,k+1 S (2) 1, )T with S (1),i > 0(i =1, 2 k 1), S(2) 1,i > 0(i = k, k +1 n 1) and q 1 2 =1, q 2 2 =1,letΛ 1 =diag(µ 1,µ 2 µ )andλ 2 =diag(µ k,µ k+1 µ ). Then the matrices K 1 =[Λ 1 q 1, Λ k 3 1 q 1, q 1 ] R () (), (3.21)

8 122 Hou Wenyuan Jiang Erxiong K 2 =[q 2, Λ 2 q 2, Λ n 1 q 2 ] R (n k) (n k) (3.22) are nonsingular, and K α 1, K α 2, where Lemma 3.6 [4] E R n n satisfy α 1 = k 1 α 2 = n k max 1i,i j max ki,i j 1+ λ i λ j λ i / min 1i S(1),i, (3.23) 1+ λ i λ j λ i / min ki S(2) 1,i. (3.24) Let A = QR be the QR factorization of A R n n with rank(a) =n. Let A 1 2 E 2 < 1. (3.25) Then there is a unique QR factorization A + E =(Q + W )(R + F ) (3.26) and W F (1 + 2) A A 1 2 E 2 E F. (3.27) By the same way, we can get the result on the condition that A = QL be the QL factorization of A R n n with rank(a)=n. Let E R n n satisfy A 1 2 E 2 < 1. (3.28) Then there is a unique QL factorization A + E =(Q + W )(L + F ) (3.29) and W F (1 + 2) A A 1 2 E 2 E F. (3.30) Theorem 3.7 Let T 1,,and T 1, are the solution by the data (3.1) and (3.2) respectively. If α 1 ζ 1 < 1, then where α 1 are defined as (3.23) and T 1, T 1, F l 1 ( µ µ 1 + λ λ 1 ), (3.31) ζ 1 =( k 1δ 1 η 1 + δ k 3 1 )( µ µ 1 + λ λ 1 ), (3.32)

9 A Stability Analysis of the (k)-jacobi Matrix Inverse Eigenvalue Problem 123 l 1 =1+2 (δ1η i 1 + iδ1 i 1 (1 + 2)α 1 ) S1 2, (3.33) 1 α 1 ζ 1 δ 1 =max{ S1, S1, 1}. (3.34) Proof Start by observing that T 1, = Q 1 Λ 1 Q T 1, T 1, = Q 1 Λ1 QT 1 and Q T 1 e = q 1, Q T 1 ẽ = q 1. (3.35) Let W 1 = Q 1 Q 1, Ω 1 =Λ 1 Λ 1. (3.36) Thus T 1, T 1, = Q 1 Λ 1 Q T 1 Q 1 Λ1 QT 1 = Q 1 Λ 1 W T 1 +(Q 1 Λ 1 Q 1 Λ1 ) Q T 1 = Q 1 Λ 1 W T 1 +(Q 1 Λ 1 Q 1 Λ 1 + Q 1 Λ 1 Q 1 Λ1 ) Q T 1 = Q 1 Λ 1 W T 1 + W 1Λ 1 QT 1 + Q 1 Ω 1 QT 1. (3.37) Hence we have T 1, T 1, F Ω 1 F +2 Λ 1 F W 1 F. (3.38) As K 1 =[Λ 1 q 1, Λ k 3 1 q 1, q 1 ] R () (), it is easy to get that Q 1 K 1 =[T 1, e,t k 3 1, e, e ]=L 1, (3.39) where L 1 is an lower triangular matrix with positive diagonal elements, and K 1 = Q T 1 L 1 is the QL factorization of K 1. From the same reason, we can get the QL factorization of K 1 as K 1 = Q T 1 L 1. Let E 1 = K 1 K 1. (3.40) Then E 1 2 = K 1 K 1 2

10 124 Hou Wenyuan Jiang Erxiong = [Λ 1 q 1 Λ 1 q 1, Λ k 3 k 3 1 q 1 Λ 1 q 1,,q 1 q 1 ] 2 On the other hand we notice that max 0i Λi 1q 1 Λ i 1 q 1 1 Thus by Lemma 3.5 and 3.6 we know if then max ( 0i Λi 1 1 q 1 q Λ i 1 Λ i 1 1) δ1 q 1 q 1 1 +δ1 k 3 Λ 1 Λ 1 1 ( k 1δ1 η 1 + δ1 k 3 )( µ µ 1 + λ λ 1 ) = ζ 1. (3.41) K 1 +( E 1 )=(Q T 1 +( W 1 ) T )(L 1 + F 1 ). K E 1 2 α 1 ζ 1 < 1. W 1 F (1 + 2) K K E 1 2 E 1 F (1 + 2)α 1 E 1 F. (3.42) 1 α 1 ζ 1 So by the Lemma 3.3 we can get that E 1 F = K 1 K 1 F = [Λ 1 q 1 Λ 1 q 1, Λ k 3 1 q 1 Λ k 3 1 q 1,,q 1 q 1 ] F Λ i 1 q 1 Λ i 1 q 1 2 Λ i 1 2 q 1 q δ1 i q 1 q Λ i 1 Λ i 1 2 iδ i 1 1 Λ 1 Λ 1 2 (δ1 i η 1 + iδ1 i 1 )( µ µ 1 + λ λ 1 ). (3.43) Thus substituting (3.42) and (3.43) into (3.38) we get the inequality that T 1, T 1, F Ω 1 F +2 Λ 1 F W 1 F [1 + 2 (δ1η i 1 + iδ1 i 1 (1 + 2)α 1 ) S1 2 ] 1 α 1 ζ 1 ( µ µ 1 + λ λ 1 ) = l 1 ( µ µ 1 + λ λ 1 ). (3.44)

11 A Stability Analysis of the (k)-jacobi Matrix Inverse Eigenvalue Problem 125 From this we have proved the Theorem 3.7. Theorem 3.8 Let T k,,and T k, are the solution by the data (3.1) and (3.2) respectively. If α 2 ζ 2 < 1, then where α 2 are defined as (3.24) and T k, T k, l 2 ( µ µ 1 + λ λ 1 ), (3.45) ζ 2 =( n kδ1 n η 2 + δ2 n )( µ µ 1 + λ λ 1 ) (3.46) n l 2 =1+2 (δ2η i 2 + iδ2 i 1 (1 + 2)α 2 ) S2 2 1 α 2 ζ 2 (3.47) δ 2 =max{ S2, S2, 1}. (3.48) Proof The proof of this theorem are absolutely similar to the theorem above. The only difference we need to notice is the (3.39) change to Q 2 K 2 =[e 1,T k, e 1 T n k, e 1]=R 2, where R 2 is an upper triangular matrix and K 2 = Q T 2 R 2 is the QR factorization of K 2. Theorem 3.9 Let T 1,n,and T 1,n are the solution by the data (3.1) and (3.2) respectively. If max{α 1 ζ 1,α 2 ζ 2 } < 1, then T 1,n T 1,n F L( µ µ 1 + λ λ 1 ) (3.49) where L =1+ρ 1 + ρ 2 + l 1 + l 2 (3.50) and ρ 1, ρ 2, l 1, l 2 are defined by (3.15)(3.16)(3.33)(3.47) respectively. Proof By the theorem (3.2)(3.7) and (3.8) we obtain that T 1,n T 1,n F T 1, T 1, F + T k, T k, F + β β + β k β k + a k ã k (1 + ρ 1 + ρ 2 + l 1 + l 2 )( µ µ 1 + λ λ 1 ). Theorem 3.10 The algorithm 2.2 is Lipschitz continuous. Proof By the Cauchy-Schwarz inequality n n x i y i ( x 2 i ) n 1 2 ( yi 2 ) 1 2 we get from the Theorem 3.9 immediately that T 1,n T 1,n F 2n 1L( µ µ λ λ 2 2 ) 1 2. Thus we obtain our result.

12 126 Hou Wenyuan Jiang Erxiong 4 Numerical Example Now we will solve a practical example by using the algorithm 2.2 and look its stability. Let T 1,9 = (4.1) Its eigenvalues are λ = (4.2) Pick k = 5, and delete 5th row and 5th column from T 1,9. There are two submatrices T 1,4 and T 6,9. The eigenvalues of them are S1 = , (4.3) respectively. S2 = Thus we can construct a Jacobi matrix T (1) 1, (4.4) by algorithm 2.2 from the data λ,s1 ands2. Note ε 1 = (1, 1, 1, 1, 1, 1, 1, 1, 1), ε 2 = (1, 1, 1, 1) and ε 3 = (1, 1, 1, 1), so we can construct T (2) 1,9 by the data λ + ε 1,S1 +ε 2, S2+ε 2 and T (3) 1,9 by the data λ + ε 1, S1+ε 3, S2+ε 3 respectively. Now we get a table to compare the difference between T (1) (2) (3) 1,9,T 1,9 and T 1,9.

13 A Stability Analysis of the (k)-jacobi Matrix Inverse Eigenvalue Problem 127 α i T 1,9 T (1) 1,9 T (2) 1,9 T (3) 1, β i T 1,9 T (1) 1,9 T (2) 1,9 T (3) 1, References 1 Jiang E X. An inverse eigenvalue problem for Jacobi Matrices. J. Comput. Math., Vol. 21, No. 3, Sept. 2003, 21: Xu S F. A stability analysis of the Jacobi matrix inverse eigenvalue problem. BIT, 1993, 33: Hald O H. Inverse eigenvalue problems for Jacobi matrices. Linear Algebra Appl., 1976, 14: Sun J G. Perturbation bounds for the Cholesky and QR factorizations. BIT, 1991, 33: Gautschi W. Norm estimate for inverse of Vandermonde matrices. Numer. Math., 1975, 23: Jiang E X. Symmetrical Matrix Computing (Chinese). Shanghai: Shanghai Science and Technology Press, Wilkinson J H. The algebraic eigenvalue problem (Chinese). Beijing: Science Press, 2001 Hou Wenyuan Dept. of Math., Shanghai University, Shanghai , PRC. Jiang Erxiong Dept. of Math., Shanghai University, Shanghai , PRC.