Low Rank ADI Solution of Sylvester Equation via Exact Shifts

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)Z i+1 =((A − β i I) −1 G (A − α i I)(A − β i I) −1 Z i ,⎛⎞⎜D i+1 =(β i − α i )I ⎟⎝ ⎠ ,D i⎛⎜ F ∗ (B − α i I) −1 ⎟⎝⎠ .Yi+1 ∗ =Y ∗i (B − β i I)(B − α i I) −1 ⎞After renaming the parameters {α i } and {β i } as in [47] or [11], since Z 0 = 0and Y 0 = 0, we can write,)Z k =(Z (1) Z (2) · · · Z (k) ,⎧Z⎪⎨(1) = (A − β 1 I) −1 G,(2.3)with Z (i+1) = (A − α i I)(A − β i+1 I) −1 Z (i)and Y k =⎪⎩(Y (1) Y (2) · · · Y (k) ),= Z (i) + (β i+1 − α i )(A − β i+1 I) −1 Z (i) ,with⎧Y (1)∗ = F ∗ (B − α 1 I) −1 ,⎪⎨Y (i+1)∗ = Y (i)∗ (B − α i+1 I) −1 (B − β i I)(2.4)⎪⎩= Y (i)∗ + (α i+1 − β i )Y (i)∗ (B − α i+1 I) −1 .All together giveX k = Z k D k Y ∗k , D k = diag ((β 1 − α 1 )I, . . . , (β k − α k )I) ,ork∑X k = (β j − α j )Z (j) Y (j)∗ . (2.5)j=13 The diagonalizable caseIn this section we will present an upper bound for the norm of the solution ofSylvester equation (2.1) for diagonalizable matrices A and B, an error boundfor the kth LR-ADI solution with exact shifts, and a first order perturbationbound when A, B, G, and F are perturbed slightly.5
Suppose A and B are diagonalizable, i.e.,A = SΛS −1 , Λ = diag(λ 1 , . . . , λ m ), (3.1)B = T ΩT −1 , Ω = diag(µ 1 , . . . , µ n ). (3.2)Then the (i + 1)st block of Z k and Y ∗kcan be written asZ (i+1) = S(Λ − α i I)(Λ − β i+1 I) −1 S −1 Z (i) , (3.3)Y (i+1)∗ = Y (i)∗ T (Ω − α i+1 I) −1 (Ω − β i I)T −1 . (3.4)Define∆ i = (Λ − α i I)(Λ − β i I) −1 , (∆ i ) (j,j) = λ j − α iλ j − β i,Θ i = (Ω − β i I)(Ω − α i I) −1 , (Θ i ) (j,j) = µ j − β iµ j − α i.Equations (3.3) and (3.4) implyZ (i+1) = S(Λ − α i I)(Λ − β i+1 I) −1 (Λ − α i−1 I)(Λ − β i I) −1 · · ·· · · (Λ − α 1 I)(Λ − β 2 I) −1 (Λ − β 1 I) −1 S −1 G= S(Λ − β i+1 I) −1 ∆ i ∆ i−1 · · · ∆ 1 S −1 G,Y (i+1)∗ = F ∗ T (Ω − α 1 I) −1 (Ω − α 2 I) −1 (Ω − β 1 I) · · ·· · · (Ω − α i+1 I) −1 (Ω − β i I)T −1= F ∗ T Θ 1 Θ 2 · · · Θ i (Ω − α i+1 I) −1 T −1 .Notice (Λ − β i+1 I) −1 ∆ i ∆ i−1 · · · ∆ 1 is diagonal with its jth diagonal entry((Λ − βi+1 I) −1 ∆ i ∆ i−1 · · · ∆ 1)(j,j) = i∏l=1λ j − α l 1· ,λ j − β l λ j − β i+1and similarly Θ 1 Θ 2 · · · Θ i (Ω − α i+1 I) −1 is also diagonal with its jth diagonalentry(Θ1 Θ 2 · · · Θ i (Ω − α i+1 I) −1) (j,j) = i∏l=1µ j − β l 1· ,µ j − α l µ j − α i+1Our first theorem gives an upper bound for the norm of the solution ofSylvester equation (2.1). But before we state the first theorem, we note that(2.2) implies that if {α j } i j=0 contains all of A’s eigenvalues (multiple eigenvaluescounted as many times as their algebraic multiplicities) or if {β j } i j=0contains all of B’s eigenvalues, then X i+1 − X ≡ 0. This is because, bythe Cayley-Hamilton theorem, p(A) ≡ 0 for A’s characteristic polynomial6
Page 1: Low Rank ADI Solution ofSylvester E
Page 4 and 5: greatly influence the norm of X and
Page 8 and 9: p(λ) def = det(λI − A) and q(B)
Page 10 and 11: ThereforeX − X k =n 0 ∑j=k+1⎛
Page 12 and 13: Neglecting the second order terms a
Page 14 and 15: ∑where γ = n 0j=1|µ j − λ j
Page 16 and 17: ⎛⎞79.750 303.69 −3.0939 · 10
Page 18 and 19: Since in the application often eige
Page 20 and 21: η(i, j) = σ(i, j) for i = 1, . .
Page 22 and 23: (Ξ1 Ξ 2 . . . Ξ j−1 (J B −
Page 24 and 25: Similarly from (4.19), (4.20), (4.8
Page 26 and 27: Then A and B are given as⎛⎞9.34
Page 28 and 29: [6] R. H. Bartels and G. W. Stewart
Page 30: [40] D. Salkuyeh, F. Toutounian, Ne

Low Rank ADI Solution of Sylvester Equation via Exact Shifts

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