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Structure-preserving Krylov subspace methods for Hamiltonian and symplectic eigenvalue problems David S. Watkins watkins@math.wsu.edu Department of Mathematics Washington State University March 2007 p.1 Definitions March 2007 p.2

  1. Preservation of Structure � A = S − 1 AS March 2007 – p.9

  2. Preservation of Structure � A = S − 1 AS S symplectic structure preserved ⇒ March 2007 – p.9

  3. Preservation of Structure � A = S − 1 AS S symplectic structure preserved ⇒ Lanczos process is a partial similarity transformation. March 2007 – p.9

  4. Preservation of Structure � A = S − 1 AS S symplectic structure preserved ⇒ Lanczos process is a partial similarity transformation. Vectors produced are columns of transforming matrix. March 2007 – p.9

  5. Preservation of Structure � A = S − 1 AS S symplectic structure preserved ⇒ Lanczos process is a partial similarity transformation. Vectors produced are columns of transforming matrix. Need process that produces vectors that are columns of a symplectic matrix. March 2007 – p.9

  6. Preservation of Structure � A = S − 1 AS S symplectic structure preserved ⇒ Lanczos process is a partial similarity transformation. Vectors produced are columns of transforming matrix. Need process that produces vectors that are columns of a symplectic matrix. Isotropy! March 2007 – p.9

  7. Isotropy March 2007 – p.10

  8. Isotropy Def: U = R ( U ) is isotropic if x T Jy = 0 for all x , y ∈ U , i.e. March 2007 – p.10

  9. Isotropy Def: U = R ( U ) is isotropic if x T Jy = 0 for all x , y ∈ U , i.e. U T JU = 0 March 2007 – p.10

  10. Isotropy Def: U = R ( U ) is isotropic if x T Jy = 0 for all x , y ∈ U , i.e. U T JU = 0 Symplectic matrix S = [ U V ] satisfies S T JS = J , i.e. March 2007 – p.10

  11. Isotropy Def: U = R ( U ) is isotropic if x T Jy = 0 for all x , y ∈ U , i.e. U T JU = 0 Symplectic matrix S = [ U V ] satisfies S T JS = J , i.e. V T JV = 0 , U T JV = I. U T JU = 0 , March 2007 – p.10

  12. Isotropy Def: U = R ( U ) is isotropic if x T Jy = 0 for all x , y ∈ U , i.e. U T JU = 0 Symplectic matrix S = [ U V ] satisfies S T JS = J , i.e. V T JV = 0 , U T JV = I. U T JU = 0 , In particular, R ( U ) , R ( V ) are isotropic. March 2007 – p.10

  13. Theorem: If B is skew Hamiltonian, then every Krylov subspace � � x, Bx, B 2 x, . . . , B j − 1 x κ j ( B, x ) = span is isotropic. March 2007 – p.11

  14. Theorem: If B is skew Hamiltonian, then every Krylov subspace � � x, Bx, B 2 x, . . . , B j − 1 x κ j ( B, x ) = span is isotropic. Proof: Mehrmann / W (2001) March 2007 – p.11

  15. Theorem: If B is skew Hamiltonian, then every Krylov subspace � � x, Bx, B 2 x, . . . , B j − 1 x κ j ( B, x ) = span is isotropic. Proof: Mehrmann / W (2001) Corollary: Every Krylov subspace method automatically preserves skew-Hamiltonian structure. March 2007 – p.11

  16. Skew-Hamiltonian Lanczos Process March 2007 – p.12

  17. Skew-Hamiltonian Lanczos Process = Bu j − u j a j d j − u j − 1 b j − 1 d j u j +1 b j d j B T w j − w j d j a j − w j − 1 d j − 1 b j − 1 = w j +1 d j +1 b j March 2007 – p.12

  18. Skew-Hamiltonian Lanczos Process = Bu j − u j a j d j − u j − 1 b j − 1 d j u j +1 b j d j B T w j − w j d j a j − w j − 1 d j − 1 b j − 1 = w j +1 d j +1 b j U T W T U T j JU j = 0 , j JW j = 0 , j W j = I March 2007 – p.12

  19. Skew-Hamiltonian Lanczos Process = Bu j − u j a j d j − u j − 1 b j − 1 d j u j +1 b j d j B T w j − w j d j a j − w j − 1 d j − 1 b j − 1 = w j +1 d j +1 b j U T W T U T j JU j = 0 , j JW j = 0 , j W j = I Let v k = − Jw k (So W j = JV j ) March 2007 – p.12

  20. Skew-Hamiltonian Lanczos Process = Bu j − u j a j d j − u j − 1 b j − 1 d j u j +1 b j d j B T w j − w j d j a j − w j − 1 d j − 1 b j − 1 = w j +1 d j +1 b j U T W T U T j JU j = 0 , j JW j = 0 , j W j = I Let v k = − Jw k (So W j = JV j ) ( JB ) T = − JB − JB T = − BJ ⇒ March 2007 – p.12

  21. Skew-Hamiltonian Lanczos Process = Bu j − u j a j d j − u j − 1 b j − 1 d j u j +1 b j d j = Bv j − v j d j a j − v j − 1 d j − 1 b j − 1 v j +1 d j +1 b j March 2007 – p.13

  22. Skew-Hamiltonian Lanczos Process = Bu j − u j a j d j − u j − 1 b j − 1 d j u j +1 b j d j = Bv j − v j d j a j − v j − 1 d j − 1 b j − 1 v j +1 d j +1 b j Start with u T 1 Jv 1 = 1 . March 2007 – p.13

  23. Skew-Hamiltonian Lanczos Process = Bu j − u j a j d j − u j − 1 b j − 1 d j u j +1 b j d j = Bv j − v j d j a j − v j − 1 d j − 1 b j − 1 v j +1 d j +1 b j Start with u T 1 Jv 1 = 1 . U T V T U T j JU j = 0 , j JV j = 0 , j JV j = I March 2007 – p.13

  24. Skew-Hamiltonian Lanczos Process = Bu j − u j a j d j − u j − 1 b j − 1 d j u j +1 b j d j = Bv j − v j d j a j − v j − 1 d j − 1 b j − 1 v j +1 d j +1 b j Start with u T 1 Jv 1 = 1 . U T V T U T j JU j = 0 , j JV j = 0 , j JV j = I These are the columns of a symplectic matrix. March 2007 – p.13

  25. Hamiltonian Lanczos Process March 2007 – p.14

  26. Hamiltonian Lanczos Process H 2 skew Hamiltonian. H Hamiltonian ⇒ March 2007 – p.14

  27. Hamiltonian Lanczos Process H 2 skew Hamiltonian. H Hamiltonian ⇒ Apply skew-Hamiltonian Lanczos process to H 2 . March 2007 – p.14

  28. Hamiltonian Lanczos Process H 2 skew Hamiltonian. H Hamiltonian ⇒ Apply skew-Hamiltonian Lanczos process to H 2 . H 2 u j − u j a j d j − u j − 1 b j − 1 d j = u j +1 b j d j H 2 v j − v j d j a j − v j − 1 d j − 1 b j − 1 = v j +1 d j +1 b j March 2007 – p.14

  29. Hamiltonian Lanczos Process H 2 skew Hamiltonian. H Hamiltonian ⇒ Apply skew-Hamiltonian Lanczos process to H 2 . H 2 u j − u j a j d j − u j − 1 b j − 1 d j = u j +1 b j d j H 2 v j − v j d j a j − v j − 1 d j − 1 b j − 1 = v j +1 d j +1 b j Let v j = Hu j d j . ˜ March 2007 – p.14

  30. Hamiltonian Lanczos Process H 2 skew Hamiltonian. H Hamiltonian ⇒ Apply skew-Hamiltonian Lanczos process to H 2 . H 2 u j − u j a j d j − u j − 1 b j − 1 d j = u j +1 b j d j H 2 v j − v j d j a j − v j − 1 d j − 1 b j − 1 = v j +1 d j +1 b j Let v j = Hu j d j . ˜ Multiply first equation by H and by d j . March 2007 – p.14

  31. Hamiltonian Lanczos Process H 2 skew Hamiltonian. H Hamiltonian ⇒ Apply skew-Hamiltonian Lanczos process to H 2 . H 2 u j − u j a j d j − u j − 1 b j − 1 d j = u j +1 b j d j H 2 v j − v j d j a j − v j − 1 d j − 1 b j − 1 = v j +1 d j +1 b j Let v j = Hu j d j . ˜ Multiply first equation by H and by d j . v j +1 d j +1 b j = H 2 ˜ ˜ v j − ˜ v j d j a j − ˜ v j − 1 d j − 1 b j − 1 March 2007 – p.14

  32. Hamiltonian Lanczos Process H 2 u j − u j a j d j − u j − 1 b j − 1 d j = u j +1 b j d j H 2 v j − v j d j a j − v j − 1 d j − 1 b j − 1 = v j +1 d j +1 b j March 2007 – p.15

  33. Hamiltonian Lanczos Process H 2 u j − u j a j d j − u j − 1 b j − 1 d j = u j +1 b j d j H 2 v j − v j d j a j − v j − 1 d j − 1 b j − 1 = v j +1 d j +1 b j Start the process with v 1 = ˜ v 1 = Hu 1 d 1 . March 2007 – p.15

  34. Hamiltonian Lanczos Process H 2 u j − u j a j d j − u j − 1 b j − 1 d j = u j +1 b j d j H 2 v j − v j d j a j − v j − 1 d j − 1 b j − 1 = v j +1 d j +1 b j Start the process with v 1 = ˜ v 1 = Hu 1 d 1 . Then v j = ˜ v j for all j . March 2007 – p.15

  35. Hamiltonian Lanczos Process H 2 u j − u j a j d j − u j − 1 b j − 1 d j = u j +1 b j d j H 2 v j − v j d j a j − v j − 1 d j − 1 b j − 1 = v j +1 d j +1 b j Start the process with v 1 = ˜ v 1 = Hu 1 d 1 . Then v j = ˜ v j for all j . Conclusion: March 2007 – p.15

  36. Hamiltonian Lanczos Process H 2 u j − u j a j d j − u j − 1 b j − 1 d j = u j +1 b j d j H 2 v j − v j d j a j − v j − 1 d j − 1 b j − 1 = v j +1 d j +1 b j Start the process with v 1 = ˜ v 1 = Hu 1 d 1 . Then v j = ˜ v j for all j . Conclusion: The second recurrence is redundant. March 2007 – p.15

  37. Hamiltonian Lanczos Process H 2 u j − u j a j d j − u j − 1 b j − 1 d j = u j +1 b j d j H 2 v j − v j d j a j − v j − 1 d j − 1 b j − 1 = v j +1 d j +1 b j Start the process with v 1 = ˜ v 1 = Hu 1 d 1 . Then v j = ˜ v j for all j . Conclusion: The second recurrence is redundant. Just run the first recurrence March 2007 – p.15

  38. Hamiltonian Lanczos Process H 2 u j − u j a j d j − u j − 1 b j − 1 d j = u j +1 b j d j H 2 v j − v j d j a j − v j − 1 d j − 1 b j − 1 = v j +1 d j +1 b j Start the process with v 1 = ˜ v 1 = Hu 1 d 1 . Then v j = ˜ v j for all j . Conclusion: The second recurrence is redundant. Just run the first recurrence together with the supplementary condition v j d j = Hu j . March 2007 – p.15

  39. Hamiltonian Lanczos Process H 2 u j − u j a j d j − u j − 1 b j − 1 d j = u j +1 b j d j = v j +1 d j +1 Hu j +1 March 2007 – p.16

  40. Hamiltonian Lanczos Process H 2 u j − u j a j d j − u j − 1 b j − 1 d j = u j +1 b j d j = v j +1 d j +1 Hu j +1 March 2007 – p.16

  41. Hamiltonian Lanczos Process = Hv j d j − u j a j d j − u j − 1 b j − 1 d j u j +1 b j d j = v j +1 d j +1 Hu j +1 March 2007 – p.16

  42. Hamiltonian Lanczos Process = Hv j d j − u j a j d j − u j − 1 b j − 1 d j u j +1 b j d j = v j +1 d j +1 Hu j +1 March 2007 – p.16

  43. Hamiltonian Lanczos Process = Hv j − u j a j − u j − 1 b j − 1 u j +1 b j = v j +1 d j +1 Hu j +1 March 2007 – p.16

  44. Hamiltonian Lanczos Process = Hv j − u j a j − u j − 1 b j − 1 u j +1 b j = v j +1 d j +1 Hu j +1 u T Start with v 1 d 1 = Hu 1 1 Jv 1 = 1 . March 2007 – p.16

  45. Hamiltonian Lanczos Process = Hv j − u j a j − u j − 1 b j − 1 u j +1 b j = v j +1 d j +1 Hu j +1 u T Start with v 1 d 1 = Hu 1 1 Jv 1 = 1 . This is easy to arrange. March 2007 – p.16

  46. Recurrences written as matrix products u j +1 b j = Hv j − u j a j − u j − 1 b j − 1 v j +1 d j +1 = Hu j +1 March 2007 – p.17

  47. Recurrences written as matrix products u j +1 b j = Hv j − u j a j − u j − 1 b j − 1 v j +1 d j +1 = Hu j +1 HV j = U j T j + u j +1 b j e T HU j = V j D j j , March 2007 – p.17

  48. Recurrences written as matrix products u j +1 b j = Hv j − u j a j − u j − 1 b j − 1 v j +1 d j +1 = Hu j +1 � � � � � � T j + u j +1 b j e T = H 2 j . U j V j U j V j D j March 2007 – p.17

  49. Recurrences written as matrix products u j +1 b j = Hv j − u j a j − u j − 1 b j − 1 v j +1 d j +1 = Hu j +1 � � � � � � T j + u j +1 b j e T = H 2 j . U j V j U j V j D j Implicit restarts: March 2007 – p.17

  50. Recurrences written as matrix products u j +1 b j = Hv j − u j a j − u j − 1 b j − 1 v j +1 d j +1 = Hu j +1 � � � � � � T j + u j +1 b j e T = H 2 j . U j V j U j V j D j Implicit restarts: Filter with the HR algorithm. March 2007 – p.17

  51. Recurrences written as matrix products u j +1 b j = Hv j − u j a j − u j − 1 b j − 1 v j +1 d j +1 = Hu j +1 � � � � � � T j + u j +1 b j e T = H 2 j . U j V j U j V j D j Implicit restarts: Filter with the HR algorithm. Next up: symplectic Lanczos process. March 2007 – p.17

  52. Symplectic Lanczos Process March 2007 – p.18

  53. Symplectic Lanczos Process S + S − 1 skew Hamiltonian. S symplectic ⇒ March 2007 – p.18

  54. Symplectic Lanczos Process S + S − 1 skew Hamiltonian. S symplectic ⇒ Apply skew-Hamiltonian Lanczos process to S + S − 1 . March 2007 – p.18

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