Polynomial Jacobi Davidson Method for Large/Sparse Eigenvalue Problems

Transcription

1 Polynomial Jacobi Davidson Method for Large/Sparse Eigenvalue Problems Tsung-Ming Huang Department of Mathematics National Taiwan Normal University, Taiwan April 28, 2011 T.M. Huang (Taiwan Normal Univ.) Poly. JD method for PEP April 28, / 42

2 Outline 1 Jacobi s orthogonal component correction 2 Davidson s method 3 Jacobi-Davidson method 4 Polynomial Jacobi-Davidson method 5 Non-equivalence deflation T.M. Huang (Taiwan Normal Univ.) Poly. JD method for PEP April 28, / 42

3 References SIAM J. Matrix Anal. Vol. 17, No. 2, PP , 1996, Van der Vorst etc. BIT. Vol. 36, No. 3, PP , 1996, Van der Vorst etc. T.-M. Hwang, W.-W. Lin and V. Mehrmann, SIAM J. Sci. Comput. T.M. Huang (Taiwan Normal Univ.) Poly. JD method for PEP April 28, / 42

4 Some basic theorems Theorem 1 Let X be an eigenspace of A and let X be a basis for X. Then there is a unique matrix L such that The matrix L is given by AX = XL. L = X I AX, where X I is a matrix satisfying X I X = I. If (λ, x) is an eigenpair of A with x X, then (λ, X I x) is an eigenpair of L. Conversely, if (λ, u) is an eigenpair of L, then (λ, Xu) is an eigenpair of A. T.M. Huang (Taiwan Normal Univ.) Poly. JD method for PEP April 28, / 42

5 Proof: Let X = [x 1 x k ] and Y = AX = [y 1 y k ]. Since y i X and X is a basis for X, there is a unique vector l i such that y i = Xl i. If we set L = [l 1 l k ], then AX = XL and L = X I XL = X I AX. Now let (λ, x) be an eigenpair of A with x X. Then there is a unique vector u such that x = Xu. However, u = X I x. Hence λx = Ax = AXu = XLu λu = λx I x = Lu. Conversely, if Lu = λu, then A(Xu) = (AX)u = (XL)u = X(Lu) = λ(xu), so that (λ, Xu) is an eigenpair of A. T.M. Huang (Taiwan Normal Univ.) Poly. JD method for PEP April 28, / 42

6 Theorem 2 (Optimal residuals) Let [X X ] be unitary. Let R = AX XL and S H = X H A LX H. Then R and S are minimized when L = X H AX, in which case (a) R = X H AX, (b) S = X H AX, (c) X H R = 0. T.M. Huang (Taiwan Normal Univ.) Poly. JD method for PEP April 28, / 42

7 Proof: Set Then [ X H X H [ X H X H ] ] [ ˆL H R = G M A [ [ ] ˆL H X X = G M ] [ X H X H X H ] [ X H X X H It implies that [ ] [ ] R = X H R ˆL L =, G which is minimized when L = ˆL = X H AX and min R = G = X H AX. The proof for S is similar. If L = X H AX, then ]. ] [ ˆL L XL = G ]. X H R = X H AX X H XL = X H AX L = 0. T.M. Huang (Taiwan Normal Univ.) Poly. JD method for PEP April 28, / 42

8 Definition 3 Let X be of full column rank and let X I be a left inverse of X. Then X I AX is a Rayleigh quotient of A. Theorem 4 Let X be orthonormal, A be Hermitian and R = AX XL. If l 1,..., l k are the eigenvalues of L, then there are eigenvalues λ j1,..., λ jk of A such that l i λ ji R 2 and k (l i λ ji ) 2 2 R F. i=1 T.M. Huang (Taiwan Normal Univ.) Poly. JD method for PEP April 28, / 42

9 Jacobi s orthogonal component correction, 1846 Consider the eigenvalue problem [ ] [ 1 α c T A z b F ] [ 1 z ] [ 1 = λ z ], (1) where A is diagonal dominant and α is the largest diagonal element. (1) is equivalent to { λ = α + c T z, (F λi)z = b. Jacobi iteration : (with z 1 = 0) { θk = α + c T z k, (D θ k I)z k+1 = (D F )z k b (2) where D = diag(f ). T.M. Huang (Taiwan Normal Univ.) Poly. JD method for PEP April 28, / 42

10 Davidson s method (1975) Algorithm 1 (Davidson s method) Given unit vector v, set V = [v] Iterate until convergence Compute desired eigenpair (θ, s) of V T AV. Compute u = V s and r = Au θu. If ( r 2 < ε), stop. Solve (D A θi)t = r. Orthog. t V v, V = [V, v] end T.M. Huang (Taiwan Normal Univ.) Poly. JD method for PEP April 28, / 42

11 Let u k = (1, z T k )T. Then r k = (A θ k I)u k = [ α θk + c T z k (F θ k I)z k + b Substituting the residual vector r k into linear systems [ α 0 (D A θ k I)t k = r k, where D A = 0 D we get (D θ k I)y k = (F θ k I)z k b From (2) and above equality, we see that ] ], = (D F )z k (D θ k I)z k b (D θ k I)(z k + y k ) = (D F )z k b = (D θ k I)z k+1. This implies that z k+1 = z k + y k as one step of JOCC starting with z k. T.M. Huang (Taiwan Normal Univ.) Poly. JD method for PEP April 28, / 42

12 Jacobi-Davidson method (1996) (θ k, u k ): approx. eigenpair of A, θ k λ, with u k = V k s k, V T k AV ks k = θ k s k and s k 2 = 1. Definition 5 (θ k, u k ) is called a Ritz pair of A. θ k is called a Ritz value and u k is a Ritz vector. T.M. Huang (Taiwan Normal Univ.) Poly. JD method for PEP April 28, / 42

13 Jacobi-Davidson method (1996) (θ k, u k ): approx. eigenpair of A, θ k λ, with Then u k = V k s k, V T k AV ks k = θ k s k and s k 2 = 1. u T k r k u T k (A θ ki)u k = s T k V T k AV ks k θ k s T k V T k V ks k = 0 r k u k Find the correction t u k such that That is A(u k + t) = λ(u k + t). (A λi)t = λu k Au k = r k + (λ θ k )u k. T.M. Huang (Taiwan Normal Univ.) Poly. JD method for PEP April 28, / 42

14 Since (I u k u T k )r k = r k, we have ( I uk u T ) k (A λi)t = rk. Correction equation (I u k u T k )(A θ ki) ( I u k u T k ) t = rk and t u k, T.M. Huang (Taiwan Normal Univ.) Poly. JD method for PEP April 28, / 42

15 Solving correction vector t Correction equation: ( I uk u T ) k (A θk I)(I u k u T k )t = r k, t u k. (3) Scheme S OneLS : Use preconditioning iterative approximations, e.g., GMRES, to solve (3). Use a preconditioner M p ( I u k u T k ) M ( I uk u T k ), where M is an approximation of A θ k I. In each of the iterative steps, it needs to solve M p y = b, y u k (4) for a given b. T.M. Huang (Taiwan Normal Univ.) Poly. JD method for PEP April 28, / 42

16 Since y u k, Eq. (4) can be rewritten as ( I uk u T ) ( k My = b My = u T k My ) u k + b η k u k + b. Hence where y = M 1 b + η k M 1 u k, η k = ut k M 1 b u T k M 1 u k. SSOR preconditioner: Let A θ k I = L + D + U. Then M = (D + ωl)d 1 (D + ωu). T.M. Huang (Taiwan Normal Univ.) Poly. JD method for PEP April 28, / 42

17 Scheme S T wols : Since t u k, Eq. (3) can be rewritten as (A θ k I)t = ( u T k (A θ ki)t ) u k r k εu k r k. (5) Let t 1 and t 2 be approximated solutions of the following linear systems: (A θ k I)t = r k and (A θ k I)t = u k, respectively. Then the approximated solution t for (5) is t = t 1 + εt 2 for ε = ut k t 1 u T k t. 2 Scheme S OneStep : The approximated solution t for (5) is t = M 1 r k + εm 1 u k for ε = u k M 1 r k u k M 1 u k, where M is an approximation of A θ k I. T.M. Huang (Taiwan Normal Univ.) Poly. JD method for PEP April 28, / 42

18 Algorithm 2 (Jacobi-Davidson Method) Choose an n-by-m orthonormal matrix V 0 Do k = 0, 1, 2, Compute all the eigenpairs of Vk T AV ks = λs. Select the desired (target) eigenpair (θ k, s k ) with s k 2 = 1. Compute u k = V k s k and r k = (A θ k I)u k. If ( r k 2 < ε), λ = θ k, x = u k, Stop Solve (approximately) a t k u k from (I u k u T k )(A θ ki)(i u k u T k )t = r k. Orthogonalize t k V k v k+1, V k+1 = [V k, v k+1 ] T.M. Huang (Taiwan Normal Univ.) Poly. JD method for PEP April 28, / 42

19 Locking: V k with V k V = I k are convergent Schur vectors, i.e., AV k = V k T k for some upper triangular T k. Set V = [V k, V q ] with V V = I k+q in k + 1-th iteration of Jacobi-Davidson Algorithm. Then [ ] [ ] V V AV = = k AV k Vq AV k [ Tk V k AV q 0 V q AV q Vk AV q Vq AV q ]. = T k V q V k T k Vk AV q V q AV q Restarting: Keep the locked Schur vectors as well as the Schur vectors of interest in the subspace and throw away those we are not interested. T.M. Huang (Taiwan Normal Univ.) Poly. JD method for PEP April 28, / 42

20 Remark 1 If ε = 0, we obtain Davidson method with ( t is NOT orthogonal to u k ) t 1 = (D A θ k I) 1 r. If the linear systems in (6) are exactly solved, then the vector t becomes t = ε(a θ k I) 1 u k u k. (6) Since t is made orthogonal to u k, (6) is equivalent to t = (A θ k I) 1 u k which is equivalent to shift-invert power iteration. Hence it is quadratic convergence. T.M. Huang (Taiwan Normal Univ.) Poly. JD method for PEP April 28, / 42

21 Consider Ax = λx and assume that λ is simple. Lemma 6 Consider w with w T x 0. Then the map ) ) F p (I xwt w T (A λi) (I xwt x w T x is a bijection from w to w. Proof: Suppose y w and F p y = 0. That is Then it holds that ) ) (I xwt w T (A λi) (I xwt x w T y = 0. x (A λi)y = εx. T.M. Huang (Taiwan Normal Univ.) Poly. JD method for PEP April 28, / 42

22 Therefore, both y and x belong to the kernel of (A λi) 2. The simplicity of λ implies that y is a scale multiple of x. The fact that y w and x T w 0 implies y = 0, which proves the injectivity of F p. An obvious dimension argument implies bijectivity. Extension F p t Then ) ) (I uut u T (A θi) (I uut u u T t = r, t u, r u. u t u Fp r u. T.M. Huang (Taiwan Normal Univ.) Poly. JD method for PEP April 28, / 42

23 Theorem 7 Assume that the correction equation ( I uu T ) (A θi) ( I uu T ) t = r, t u (7) is solved exactly in each step of Jacobi-Davidson Algorithm. Assume u k = u x and u T k x has non-trivial limit. Then if u k is sufficiently chosen to x, then u k x with locally quadratical convergence and θ k = u T k Au k λ. T.M. Huang (Taiwan Normal Univ.) Poly. JD method for PEP April 28, / 42

24 Proof: Suppose Ax = λx with x such that x = u + z for z u. Then (A θi) z = (A θi) u + (λ θ) x = r + (λ θ) x. (8) Consider the exact solution z 1 u of (7): (I P ) (A θi) z 1 = (I P ) r, (9) where P = uu T. Note that (I P ) r = r since u r. Since x (u + z 1 ) = z z 1 and z = x u, for quadratic convergence, it suffices to show that x (u + z 1 ) = z z 1 = O ( z 2). (10) Multiplying (8) by (I P ) and subtracting the result from (9) yields (I P ) (A θi) (z z 1 ) = (λ θ) (I P ) z + (λ θ) (I P ) u. (11) T.M. Huang (Taiwan Normal Univ.) Poly. JD method for PEP April 28, / 42

25 Multiplying (8) by u T and using r u leads to λ θ = ut (A θi) z u T. (12) x Since u T k x has non-trivial limit, we obtain (λ θ) (I P ) z = u T (A θi) z u T x (I P ) z. (13) From (11), Lemma 6 and (I P ) u = 0, we have [ ] 1 z z 1 = (I P ) (A θ k I) u (λ θ) (I P ) z k [ ] 1 = u T k (I P ) (A θ k I) (A θ ki) z u k u T k x (I P ) z = O ( z 2). T.M. Huang (Taiwan Normal Univ.) Poly. JD method for PEP April 28, / 42

26 Jacobi Davidson method as on accelerated Newton Scheme Consider Ax = λx and assume that λ is simple. Choose w T x = 1. Consider nonlinear equation F (u) = Au θ(u)u = 0 with θ (u) = wt Au w T u, where u = 1 or w T u = 1. Then F : {u w T u = 1} w. In particular, r F (u) = Au θ (u) u w. Suppose u k x and the next Newton approximation u k+1 : ( 1 F u k+1 = u k u F (u k ) u=uk) is given by u k+1 = u k + t, i.e., t satisfies that ( ) F t = F (u k ) = r. u u=uk T.M. Huang (Taiwan Normal Univ.) Poly. JD method for PEP April 28, / 42

27 Since 1 = u T k+1 w = (u k + t) T w = 1 + t T w, it implies that w T t = 0. By the definition of F, we have F u = A θ (u) I ( w T Au ) uw T + ( w T u ) uw T A (w T u) 2 = A θi + wt Au (w T u) 2 uwt uwt A (I w T u = u kw T ) w T (A θ k I). u k Consequently, the Jacobian of F acts on w and is given by ( ) F t = (I u kw T ) u w T (A θ k I) t, t w. u k u=uk Hence the correction equation of Newton method read as t w, (I u kw T ) w T (A θ k I) t = r, u k which is the correction equation of Jacobi-Davidson method in (7) with w = u. T.M. Huang (Taiwan Normal Univ.) Poly. JD method for PEP April 28, / 42

28 Polynomial Jacobi-Davidson method (θ k, u k ): approx. eigenpair of A(λ) τ k=0 λk A k, θ k λ, with Let Then u k = V k s k, V k A(λ)V ks k = 0 and s k 2 = 1. r k = A(θ k )u k. u k r k = u k A(θ k)u k = s k V k A(θ k)v k s k = 0 r k u k Find the correction t such that That is A(λ)(u k + t) = 0. A(λ)t = A(λ)u k = r k + (A(θ k ) A(λ))u k. T.M. Huang (Taiwan Normal Univ.) Poly. JD method for PEP April 28, / 42

29 Let p k = A (θ k )u k ( τ i=1 iθ i 1 k A i ) u k. A(λ) = A λi: A(λ) = A λb: p k = u k, (A(θ k ) A(λ))u k = (λ θ k )u k = (θ k λ k )p k p k = Bu k, (A(θ k ) A(λ))u k = (λ θ k )Bu k = (θ k λ)p k A(λ) = τ i=0 λi A i with τ 2: [ (A(θ k ) A(λ))u k = (θ k λ)a (θ k ) 1 ] 2 (θ k λ) 2 A (ξ k ) u k = (θ k λ)p k 1 2 (θ k λ) 2 A (ξ k )u k T.M. Huang (Taiwan Normal Univ.) Poly. JD method for PEP April 28, / 42

30 Hence A(λ)t = r k + (θ k λ)p k 1 2 (θ k λ) 2 A (ξ k )u k Since r k u k, we have ( I p ku ) k u k p A(λ)t = r k 12 ( (θ k λ) 2 I p ku ) k k u k p A (ξ k )u k. k Correction equation: ( I p ku ) k u k p A(θ k )(I u k u k )t = r k and t u k, k or ( I p ku ) ( k u k p (A θ k B) I u kp ) k k p k u t = r k and t B u k, k with symmetric positive definite matrix B. T.M. Huang (Taiwan Normal Univ.) Poly. JD method for PEP April 28, / 42

31 Algorithm 3 (Jacobi-Davidson Algorithm for solving A(λ)x = 0) Choose an n-by-m orthonormal matrix V 0 Do k = 0, 1, 2, Compute all the eigenpairs of Vk A(λ)V k = 0. Select the desired (target) eigenpair (θ k, s k ) with s k 2 = 1. Compute u k = V k s k, r k = A(θ k )u k and p k = A (θ k )u k. If ( r k 2 < ε), λ = θ k, x = u k, Stop Solve (approximately) a t k u k from (I p ku T k u T k p )A(θ k )(I u k u T k k )t = r k. Orthogonalize t k V k v k+1, V k+1 = [V k, v k+1 ] p k = A (θ k )u k ( τ i=1 iθ i 1 k A i ) u k. T.M. Huang (Taiwan Normal Univ.) Poly. JD method for PEP April 28, / 42

32 Non-equivalence deflation of quadratic eigenproblems Let λ 1 be a real eigenvalue of Q(λ) λ 2 M + λc + K and x 1, z 1 be the associated right and left eigenvectors, respectively, with z T 1 Kx 1 = 1. Let θ 1 = (z T 1 Mx 1 ) 1. We introduce a deflated quadratic eigenproblem [ ] Q(λ)x λ 2 M + λ C + K x = 0, where M = M θ 1 Mx 1 z T 1 M, C = C + θ 1 λ 1 (Mx 1 z T 1 K + Kx 1 z T 1 M), K = K θ 1 λ 2 Kx 1 z1 T K. 1 T.M. Huang (Taiwan Normal Univ.) Poly. JD method for PEP April 28, / 42

33 Complex deflation Let λ 1 = α 1 + iβ 1 be a complex eigenvalue of Q(λ) and x 1 = x 1R + ix 1I, z 1 = z 1R + iz 1I be the associated right and left eigenvectors, respectively, such that Z T 1 KX 1 = I 2, where X 1 = [x 1R, x 1I ] and Z 1 = [z 1R, z 1I ]. Let Θ 1 = (Z T 1 MX 1 ) 1. Then we introduce a deflated quadratic eigenproblem with M = M MX 1 Θ 1 Z1 T M, C = C + MX 1 Θ 1 Λ T 1 Z1 T K + KX 1 Λ 1 1 ΘT 1 Z1 T M, K = K KX 1 Λ 1 1 Θ 1Λ T 1 Z1 T K [ ] α1 β in which Λ 1 = 1. β 1 α 1 T.M. Huang (Taiwan Normal Univ.) Poly. JD method for PEP April 28, / 42

34 Theorem 8 (i) Let λ 1 be a simple real eigenvalue of Q(λ). Then the spectrum of Q(λ) is given by provided that λ 2 1 θ 1. (σ(q(λ)) {λ 1 }) { } (ii) Let λ 1 be a simple complex eigenvalue of Q(λ). Then the spectrum of Q(λ) is given by ( σ(q(λ)) {λ1, λ 1 } ) {, } provided that Λ 1 Λ T 1 Θ 1. Furthermore, in both cases (i) and (ii), if λ 2 λ 1 and (λ 2, x 2 ) is an eigenpair of Q(λ) then the pair (λ 2, x 2 ) is also an eigenpair of Q(λ). T.M. Huang (Taiwan Normal Univ.) Poly. JD method for PEP April 28, / 42

35 Suppose that M, C, K are symmetric. Given an eigenmatrix pair (Λ 1, X 1 ) R r r R n r of Q(λ), where Λ 1 is nonsingular and X 1 satisfies X T 1 KX 1 = I r, Θ 1 := (X T 1 MX 1 ) 1. We define Q(λ) := λ 2 M + λ C + K, where M := M MX 1 Θ 1 X1 T M, C := C + MX 1 Θ 1 Λ T 1 X1 T K + KX 1 Λ 1 1 Θ 1X1 T M, K := K KX 1 Λ 1 1 Θ 1Λ T 1 X1 T K. Theorem 9 Suppose that Θ 1 Λ 1 Λ T 1 is nonsingular. Then the eigenvalues of the real symmetric quadratic pencil Q(λ) are the same as those of Q(λ) except that the eigenvalues of Λ 1, which are closed under complex conjugation, are replaced by r infinities. T.M. Huang (Taiwan Normal Univ.) Poly. JD method for PEP April 28, / 42

36 Proof: Since (Λ 1, X 1 ) is an eigenmatrix pair of Q(λ), i.e., we have MX 1 Λ CX 1 Λ 1 + KX 1 = 0, Q(λ) = Q(λ) + [MX 1 (λi r + Λ 1 ) + CX 1 ] Θ 1 Λ T 1 (X T 1 K λλ T 1 X T 1 M) = Q(λ) + Q(λ)X 1 (λi r Λ 1 ) 1 Θ 1 Λ T 1 (X T 1 K λλ T 1 X T 1 M). By using the identity where R, S T R n m, we have det[ Q(λ)] det(i n + RS) = det(i m + SR), = det[q(λ)]det[i + X 1 (λi r Λ 1 ) 1 Θ 1 Λ T 1 (X T 1 K λλ T 1 X T 1 M)] = det[q(λ)]det[i r + (λi r Λ 1 ) 1 Θ 1 Λ T 1 (I r λλ T 1 Θ 1 1 )] det[q(λ)] = det(λi r Λ 1 ) det(θ 1Λ T 1 Λ 1 ). T.M. Huang (Taiwan Normal Univ.) Poly. JD method for PEP April 28, / 42

37 Since (Θ 1 Λ 1 Λ T 1 ) Rr r is nonsingular, we have det(θ 1 Λ T 1 Λ 1 ) 0. Therefore, Q(λ) has the same eigenvalues as Q(λ) except that r eigenvalues of Λ 1 are replaced by r infinities. T.M. Huang (Taiwan Normal Univ.) Poly. JD method for PEP April 28, / 42

38 Non-equiv. deflation for cubic poly. eigenproblems Let (Λ, V u ) R r r R n r be an eigenmatrix pair of A(λ) λ 3 A 3 + λ 2 A 2 + λa 1 + A 0 (14) with V T u V u = I r and 0 / σ(λ), i.e., A 3 V u Λ 3 + A 2 V u Λ 2 + A 1 V u Λ + A 0 V u = 0. (15) Define a new deflated cubic eigenvalue problem by Ã(λ)u = (λ 3 à 3 + λ 2 à 2 + λã1 + Ã0)u = 0, (16) where à 0 = A 0, à 1 = A 1 (A 1 V u V T u + A 2 V u ΛV T u + A 3 V u Λ 2 V T u ), à 2 = A 2 (A 2 V u V T u + A 3 V u ΛV T u ), à 3 = A 3 A 3 V u V T u. (17) T.M. Huang (Taiwan Normal Univ.) Poly. JD method for PEP April 28, / 42

39 Lemma 10 Let A(λ) and Ã(λ) be cubic pencils given by (14) and (16), respectively. Then it holds Ã(λ) = A(λ) ( I n λv u (λi r Λ) 1 V T u ). (18) Theorem 11 Let (Λ, V u ) be an eigenmatrix pair of A(λ) with V T u V u = I r. Then (i) (σ(a(λ)) σ(λ)) { } = σ(ã(λ)). (ii) Let (µ, z) be an eigenpair of A(λ) with z 2 = 1 and µ / σ(λ). Define z = (I n µv u Λ 1 V T u )z T (µ)z. (19) Then (µ, z) is an eigenpair of Ã(λ). T.M. Huang (Taiwan Normal Univ.) Poly. JD method for PEP April 28, / 42

40 Proof of Lemma: Using (17) and (15), and the fundamental matrix calculation, we have Ã(λ) = A(λ) λ ( λ 2 A 3 V F VF T + λa 2 V F VF T + λa 3 V F ΛVF T +A 1 V F VF T + A 2 V F ΛVF T + A 3 V F Λ 2 VF T ) = A(λ) λ ( A 3 V F (λi r Λ) 3 (λi r Λ) 1 VF T +3A 3 V F Λ(λI r Λ) 2 (λi r Λ) 1 VF T +3A 3 V F Λ 2 (λi r Λ)(λI r Λ) 1 VF T +A 2 V F (λi r Λ) 2 (λi r Λ) 1 VF T +2A 2 V F Λ(λI r Λ)(λI r Λ) 1 VF T +A 1 V F (λi r Λ)(λI r Λ) 1 VF T ) T.M. Huang (Taiwan Normal Univ.) Poly. JD method for PEP April 28, / 42

41 Ã(λ) = A(λ) λ {[ A 3 V F (λ 3 I r Λ 3 ) + A 2 V F (λ 2 I r Λ 2 ) +A 1 V F (λi r Λ) + A 0 V F A 0 V F ] (λi r Λ) 1 VF T } = A(λ) λ [ A(λ)V F (λi r Λ) 1 VF T ] = A(λ) [ I n λv F (λi r Λ) 1 VF T ]. T.M. Huang (Taiwan Normal Univ.) Poly. JD method for PEP April 28, / 42

42 Proof of Theorem: (i) Using the identity det(i n + RS) = det(i m + SR) and Lemma 10, we have det(ã(λ)) = det(a(λ)) det ( I n λv F (λi r Λ) 1 VF T ) = det(a(λ)) det ( I n λ(λi r Λ) 1) = det(a(λ)) det(λi r Λ) 1 det( Λ). Since 0 / σ(λ), det( Λ) 0. Thus, Ã(λ) and A(λ) have the same finite spectrum except the eigenvalues in σ(λ). Furthermore, dividing Eq. (16) by λ 3 and using the fact that à 3 V F = (A 3 A 3 V F V T F )V F = 0, we see that (diag r {,, }, V F ) is an eigenmatrix pair of Ã(λ) corresponding to infinite eigenvalues. T.M. Huang (Taiwan Normal Univ.) Poly. JD method for PEP April 28, / 42

43 (ii) Since µ / σ(λ), the matrix T (µ) = (I µv F Λ 1 VF T ) in (19) is invertible with the inverse T (µ) 1 = I n µv F (µi r Λ) 1 VF T. (20) From Lemma 10, we have Ã(µ) z = A(µ) [ I n µv F (µi r Λ) 1 VF T ] [ In µv F Λ 1 VF T ] z = 0. This completes the proof. T.M. Huang (Taiwan Normal Univ.) Poly. JD method for PEP April 28, / 42