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Orthogonal similarity reduction of any symmetric matrix into a diagonal-plus-semiseparable one with free choice of the diagonal Ellen Van Camp, Raf Vandebril, Marc Van Barel and Nicola Mastronardi I. Algorithms Orthogonal similarity

  1. × × × 0 ⊠ ⊠ ⊠ × × × 0 ⊠ ⊠ ⊠ × × × 0 ⊠ ⊠ ⊠ 0 ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ 0 ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ × ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ 0 0 0 0 0 ⊠ ⊠

  2. × × × 0 ⊠ ⊠ ⊠ × × × 0 ⊠ ⊠ ⊠ × × × 0 ⊠ ⊠ ⊠ 0 ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ 0 ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ 0 0 0 0 0 ⊠ ⊠

  3. × × × 0 ⊠ ⊠ ⊠ × × × 0 ⊠ ⊠ ⊠ × × × 0 ⊠ ⊠ ⊠ 0 ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ 0 ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ ⊗ ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ 0 0 0 0 0 ⊠ ⊠

  4. × × × 0 ⊠ ⊠ ⊠ × × × 0 ⊠ ⊠ ⊠ × × × 0 ⊠ ⊠ ⊠ 0 ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ 0 ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ 0 ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ × ⊠ ⊠ ⊠ ⊠ ⊠ ⊠

  5. × × × ⊠ ⊠ ⊠ ⊠ × × × ⊠ ⊠ ⊠ ⊠ × × × ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ × × ⊠ ⊠ ⊠ ⊠ ⊠

  6. × × × ⊠ ⊠ ⊠ ⊠ × × × ⊠ ⊠ ⊠ ⊠ × × × ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ × ⊠ ⊠ ⊠ ⊠ ⊠ ⊠

  7. × × × ⊠ ⊠ ⊠ ⊠ × × × ⊠ ⊠ ⊠ ⊠ × × × ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ ⊠

  8. 3. Reduction to diagonal-plus-semiseparable form with free choice of the diagonal Any symmetric matrix can be transformed into a diagonal-plus-semiseparable one where the diagonal can be chosen in advance, by means of orthogonal similarity transformations in order O ( 4 3 n 3 ) . Definition The sum of a symmetric semiseparable matrix and a diagonal matrix is called a diagonal-plus-semiseparable matrix. So choose a diagonal d = [ d 1 , d 2 , . . . , d n ] .

  9. A first algorithm × × × × × × × × × × × × × × × × × × × × × − → D + S × × × × × × × × × × × × × × × × × × × × × × × × × × × ×

  10. Step 1 × × × × × × × 0 0 0 0 0 0 0 × × × × × × × 0 0 0 0 0 0 0 × × × × × × × 0 0 0 0 0 0 0 + × × × × × × × 0 0 0 0 0 0 0 × × × × × × × 0 0 0 0 0 0 0 × × × × × × × 0 0 0 0 0 0 0 × × × × × × × 0 0 0 0 0 0 d 1

  11. × × × × × × ⊗ 0 0 0 0 0 0 0 × × × × × × ⊗ 0 0 0 0 0 0 0 × × × × × × ⊗ 0 0 0 0 0 0 0 + × × × × × × ⊗ 0 0 0 0 0 0 0 × × × × × × ⊗ 0 0 0 0 0 0 0 × × × × × × × 0 0 0 0 0 0 0 ⊗ ⊗ ⊗ ⊗ ⊗ × × 0 0 0 0 0 0 d 1

  12. × × × × × × 0 0 0 0 0 0 0 0 × × × × × × 0 0 0 0 0 0 0 0 × × × × × × 0 0 0 0 0 0 0 0 + × × × × × × 0 0 0 0 0 0 0 0 × × × × × × 0 0 0 0 0 0 0 0 × × × × × × × 0 0 0 0 0 0 0 0 0 0 0 0 × × 0 0 0 0 0 0 d 1

  13. × × × × × × 0 0 0 0 0 0 0 0 × × × × × × 0 0 0 0 0 0 0 0 × × × × × × 0 0 0 0 0 0 0 0 + × × × × × × 0 0 0 0 0 0 0 0 × × × × × × 0 0 0 0 0 0 0 0 × × × × × × ⊗ 0 0 0 0 0 0 0 0 0 0 0 0 ⊗ × 0 0 0 0 0 0 d 1

  14. Problem � � � � � � � � s 2 d 1 0 0 − s c s c csd 1 = c 2 d 1 − s 0 c d 1 s c csd 1 ⇒ ???

  15. Solution � � � � � � 0 − s c s d 1 c − s 0 c d 1 s c � � � � � � 1 0 − s c s c = d 1 − s 0 1 c s c � � 0 d 1 = 0 d 1

  16. × × × × × × 0 0 0 0 0 0 0 0 × × × × × × 0 0 0 0 0 0 0 0 × × × × × × 0 0 0 0 0 0 0 0 + × × × × × × 0 0 0 0 0 0 0 0 × × × × × × 0 0 0 0 0 0 0 0 × × × × × × ⊗ 0 0 0 0 0 0 0 0 0 0 0 0 ⊗ × 0 0 0 0 0 0 d 1

  17. × × × × × × 0 0 0 0 0 0 0 0 × × × × × × 0 0 0 0 0 0 0 0 × × × × × × 0 0 0 0 0 0 0 0 + × × × × × × 0 0 0 0 0 0 0 0 × × × × × × 0 0 0 0 0 0 0 0 × × × × × + ⊗ 0 0 0 0 0 0 d 1 0 0 0 0 0 ⊗ × 0 0 0 0 0 0 d 1

  18. × × × × × 0 0 0 0 0 0 0 ⊠ ⊠ × × × × × 0 0 0 0 0 0 0 ⊠ ⊠ × × × × × 0 0 0 0 0 0 0 ⊠ ⊠ + × × × × × 0 0 0 0 0 0 0 ⊠ ⊠ × × × × × 0 0 0 0 0 0 0 ⊠ ⊠ 0 0 0 0 0 0 d 1 ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ 0 0 0 0 0 0 d 1 ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ ⊠

  19. × × × × × 0 0 0 0 0 0 0 ⊠ ⊠ × × × × × 0 0 0 0 0 0 0 ⊠ ⊠ × × × × × 0 0 0 0 0 0 0 ⊠ ⊠ + × × × × × 0 0 0 0 0 0 0 ⊠ ⊠ × × × × × 0 0 0 0 0 0 0 ⊠ ⊠ 0 0 0 0 0 0 d 1 ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ 0 0 0 0 0 0 d 2 ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ ⊞

  20. × × × × × 0 0 0 0 0 0 0 ⊠ ⊠ × × × × × 0 0 0 0 0 0 0 ⊠ ⊠ × × × × × 0 0 0 0 0 0 0 ⊠ ⊠ + × × × × × 0 0 0 0 0 0 0 ⊠ ⊠ × × × × × 0 0 0 0 0 0 0 ⊠ ⊠ 0 0 0 0 0 0 d 1 ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ 0 0 0 0 0 0 d 2 ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ ⊠

  21. Step 3 × × × × 0 0 0 0 0 0 0 ⊠ ⊠ ⊠ × × × × 0 0 0 0 0 0 0 ⊠ ⊠ ⊠ × × × × 0 0 0 0 0 0 0 ⊠ ⊠ ⊠ + × × × × 0 0 0 0 0 0 0 ⊠ ⊠ ⊠ 0 0 0 0 0 0 d 1 ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ 0 0 0 0 0 0 d 2 ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ 0 0 0 0 0 0 d 3 ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ ⊠

  22. × × × × ⊗ ⊗ ⊗ 0 0 0 0 0 0 0 × × × × 0 0 0 0 0 0 0 ⊠ ⊠ ⊠ × × × × 0 0 0 0 0 0 0 ⊠ ⊠ ⊠ + × × × × 0 0 0 0 0 0 0 ⊠ ⊠ ⊠ ⊗ 0 0 0 0 0 0 ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ d 1 ⊗ 0 0 0 0 0 0 ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ d 2 ⊗ 0 0 0 0 0 0 ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ d 3

  23. × × × × 0 0 0 0 0 0 0 0 0 0 × × × × ⊗ ⊗ ⊗ 0 0 0 0 0 0 0 × × × × 0 0 0 0 0 0 0 ⊠ ⊠ ⊠ + × × × × 0 0 0 0 0 0 0 ⊠ ⊠ ⊠ 0 ⊗ 0 0 0 0 0 0 ⊠ ⊠ ⊠ ⊠ ⊠ d 1 0 ⊗ 0 0 0 0 0 0 ⊠ ⊠ ⊠ ⊠ ⊠ d 2 0 ⊗ 0 0 0 0 0 0 ⊠ ⊠ ⊠ ⊠ ⊠ d 3

  24. × × × × 0 0 0 0 0 0 0 0 0 0 × × × × 0 0 0 0 0 0 0 0 0 0 × × × × ⊗ ⊗ ⊗ 0 0 0 0 0 0 0 + × × × × 0 0 0 0 0 0 0 ⊠ ⊠ ⊠ 0 0 ⊗ 0 0 0 0 0 0 ⊠ ⊠ ⊠ ⊠ d 1 0 0 ⊗ 0 0 0 0 0 0 ⊠ ⊠ ⊠ ⊠ d 2 0 0 ⊗ 0 0 0 0 0 0 ⊠ ⊠ ⊠ ⊠ d 3

  25. × × × × 0 0 0 0 0 0 0 0 0 0 × × × × 0 0 0 0 0 0 0 0 0 0 × × × × 0 0 0 0 0 0 0 0 0 0 + × × × × ⊗ ⊗ ⊗ 0 0 0 0 0 0 0 0 0 0 ⊗ 0 0 0 0 0 0 ⊠ ⊠ ⊠ d 1 0 0 0 ⊗ 0 0 0 0 0 0 ⊠ ⊠ ⊠ d 2 0 0 0 ⊗ 0 0 0 0 0 0 ⊠ ⊠ ⊠ d 3

  26. × × × × 0 0 0 0 0 0 0 0 0 0 × × × × 0 0 0 0 0 0 0 0 0 0 × × × × 0 0 0 0 0 0 0 0 0 0 + × × × + ⊗ ⊗ ⊗ 0 0 0 0 0 0 d 1 0 0 0 ⊗ 0 0 0 0 0 0 ⊠ ⊠ ⊠ d 1 0 0 0 ⊗ 0 0 0 0 0 0 ⊠ ⊠ ⊠ d 2 0 0 0 ⊗ 0 0 0 0 0 0 ⊠ ⊠ ⊠ d 3

  27. × × × 0 0 0 0 0 0 0 0 0 ⊠ ⊠ × × × 0 0 0 0 0 0 0 0 0 ⊠ ⊠ × × × 0 0 0 0 0 0 0 0 0 ⊠ ⊠ + 0 0 0 0 0 0 0 0 ⊠ ⊠ ⊠ ⊠ ⊠ d 1 0 0 0 0 0 0 ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ d 1 0 0 0 0 0 0 0 0 0 0 ⊠ ⊠ ⊠ d 2 0 0 0 0 0 0 0 0 0 0 ⊠ ⊠ ⊠ d 3

  28. × × × 0 0 0 0 0 0 0 0 0 ⊠ ⊠ × × × 0 0 0 0 0 0 0 0 0 ⊠ ⊠ × × × 0 0 0 0 0 0 0 0 0 ⊠ ⊠ + 0 0 0 0 0 0 0 0 ⊠ ⊠ ⊠ ⊠ ⊠ d 1 0 0 0 0 0 0 ⊠ ⊠ ⊠ ⊠ ⊞ ⊠ ⊠ d 2 0 0 0 0 0 0 0 0 0 0 ⊠ ⊠ ⊠ d 2 0 0 0 0 0 0 0 0 0 0 ⊠ ⊠ ⊠ d 3

  29. × × × 0 0 0 0 0 0 0 0 0 ⊠ ⊠ × × × 0 0 0 0 0 0 0 0 0 ⊠ ⊠ × × × 0 0 0 0 0 0 0 0 0 ⊠ ⊠ + 0 0 0 0 0 0 0 0 ⊠ ⊠ ⊠ ⊠ ⊠ d 1 ⊗ ⊗ 0 0 0 0 0 0 ⊠ ⊠ ⊠ ⊠ ⊠ d 2 0 0 0 0 ⊗ 0 0 0 0 0 0 ⊠ ⊠ d 2 0 0 0 0 ⊗ 0 0 0 0 0 0 ⊠ ⊠ d 3

  30. × × × 0 0 0 0 0 0 0 0 ⊠ ⊠ ⊠ × × × 0 0 0 0 0 0 0 0 ⊠ ⊠ ⊠ × × × 0 0 0 0 0 0 0 0 ⊠ ⊠ ⊠ + 0 0 0 0 0 0 0 ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ d 1 0 0 0 0 0 0 0 ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ d 2 0 0 0 0 0 0 ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ d 2 0 0 0 0 0 0 0 0 0 0 0 ⊠ ⊠ d 3

  31. × × × 0 0 0 0 0 0 0 0 ⊠ ⊠ ⊠ × × × 0 0 0 0 0 0 0 0 ⊠ ⊠ ⊠ × × × 0 0 0 0 0 0 0 0 ⊠ ⊠ ⊠ + 0 0 0 0 0 0 0 ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ d 1 0 0 0 0 0 0 0 ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ d 2 0 0 0 0 0 0 ⊠ ⊠ ⊠ ⊠ ⊠ ⊞ ⊠ d 3 0 0 0 0 0 0 0 0 0 0 0 ⊠ ⊠ d 3

  32. × × × 0 0 0 0 0 0 0 0 ⊠ ⊠ ⊠ × × × 0 0 0 0 0 0 0 0 ⊠ ⊠ ⊠ × × × 0 0 0 0 0 0 0 0 ⊠ ⊠ ⊠ + 0 0 0 0 0 0 0 ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ d 1 0 0 0 0 0 0 0 ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ d 2 ⊗ 0 0 0 0 0 0 ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ d 3 0 0 0 0 0 ⊗ 0 0 0 0 0 0 ⊠ d 3

  33. × × × 0 0 0 0 0 0 0 ⊠ ⊠ ⊠ ⊠ × × × 0 0 0 0 0 0 0 ⊠ ⊠ ⊠ ⊠ × × × 0 0 0 0 0 0 0 ⊠ ⊠ ⊠ ⊠ + 0 0 0 0 0 0 ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ d 1 0 0 0 0 0 0 ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ d 2 0 0 0 0 0 0 ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ d 3 0 0 0 0 0 0 ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ d 3

  34. × × × 0 0 0 0 0 0 0 ⊠ ⊠ ⊠ ⊠ × × × 0 0 0 0 0 0 0 ⊠ ⊠ ⊠ ⊠ × × × 0 0 0 0 0 0 0 ⊠ ⊠ ⊠ ⊠ + 0 0 0 0 0 0 ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ d 1 0 0 0 0 0 0 ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ d 2 0 0 0 0 0 0 ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ d 3 0 0 0 0 0 0 ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ ⊞ d 4

  35. × × × 0 0 0 0 0 0 0 ⊠ ⊠ ⊠ ⊠ × × × 0 0 0 0 0 0 0 ⊠ ⊠ ⊠ ⊠ × × × 0 0 0 0 0 0 0 ⊠ ⊠ ⊠ ⊠ + 0 0 0 0 0 0 ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ d 1 0 0 0 0 0 0 ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ d 2 0 0 0 0 0 0 ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ d 3 0 0 0 0 0 0 ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ d 4

  36. A second algorithm Before the last step of the first algorithm: × 0 0 0 0 0 0 0 ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ 0 0 0 0 0 0 d 1 ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ 0 0 0 0 0 0 d 2 ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ + 0 0 0 0 0 0 d 3 ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ 0 0 0 0 0 0 d 4 ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ 0 0 0 0 0 0 d 5 ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ 0 0 0 0 0 0 ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ d 6

  37. When applying the first algorithm starting with D = [ d 2 , d 3 , . . . , d n , ⋆ ] with ⋆ an arbitrary element, instead of [ d 1 , d 2 , . . . , d n ] , we get the following situation before the last step: × 0 0 0 0 0 0 0 ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ 0 0 0 0 0 0 d 2 ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ 0 0 0 0 0 0 d 3 ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ + 0 0 0 0 0 0 d 4 ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ 0 0 0 0 0 0 ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ d 5 0 0 0 0 0 0 ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ d 6 0 0 0 0 0 0 ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ d 7

  38. No last step necessary: 0 0 0 0 0 0 ⊞ ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ d 1 0 0 0 0 0 0 ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ d 2 0 0 0 0 0 0 ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ d 3 + 0 0 0 0 0 0 ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ d 4 0 0 0 0 0 0 ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ d 5 0 0 0 0 0 0 ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ d 6 0 0 0 0 0 0 ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ ⊠ d 7

  39. Any arbitrary symmetric matrix can be transformed into a symmetric diagonal-plus-semiseparable one with free choice of the diagonal by means of an orthogonal similarity transformation Q such that Qe 1 = e 1 .

  40. II. Accuracy −13 10 −14 10 −15 10 Tridiagonal Semiseparable Diagonal−plus−semiseparable −16 10 0 500 1000 1500 2000 2500

  41. III. Computational complexity −7 2.5 x 10 Tridiagonal Semiseparable Diagonal−plus−semiseparable 2 1.5 1 0.5 500 1000 1500 2000 2500

  42. IV. Convergence behavior of reduction algorithm into diagonal-plus-semiseparable form

  43. For the reduction to semiseparable form Eigenvalues are equidistant 1 : 200. 200 180 160 140 120 100 80 60 40 20 0 0 20 40 60 80 100 120 140 160 180 200

  44. Eigenvalues 1 : 100 and 1000 : 1100. 1200 1000 800 600 400 200 0 0 20 40 60 80 100 120 140 160 180 200

  45. For the reduction to diagonal-plus-semiseparable form Some notation A (0) = A A ( m ) m A ( m − 1) Q m Q T =   R T  A m 1 =  ( D + S ) m R 1 Q T = 1: m AQ 1: m where ( D + S ) m is a square diagonal-plus- semiseparable matrix of dimensions ( m + 1 ) × ( m + 1 ) .

  46. Lemma Q 1 : m < e n > = ( A − d m I )( A − d m − 1 I ) . . . ( A − d 1 I ) < e n >, for m = 1 , 2 , . . . and Q 1 : 0 = I.

  47. Proof. . For m = 0 : Q 1 : 0 < e n > = < e n > . . Suppose the theorem is true for m − 1, i.e., Q 1 : m − 1 < e n > = ( A − d m − 1 I ) . . . ( A − d 2 I )( A − d 1 I ) < e n > . m A ( m − 1 ) is of the form: The structure of Q T     × . . . × 0 . . . 0 0     . . . . ... . . . .     . . . .             × × 0 0 0 . . . . . .     Q T +     m     × × × 0 . . . . . . d 1         . . . . ... ...     . . . .     . . . .         × × × × . . . . . . d m Q T = H + m D

  48. Hence, m ( A ( m − 1) ) Q T H + Q T = m D Q T m ( Q T H + Q T 1: m − 1 AQ 1: m − 1 ) = m D ⇒ AQ 1: m − 1 − Q 1: m − 1 D = Q 1: m H Applying the former equality on < e n > and using the induction hypothesis, we derive that: ( AQ 1: m − 1 − Q 1: m − 1 D ) < e n > = Q 1: m H < e n > ( AQ 1: m − 1 − Q 1: m − 1 d m I ) < e n > = Q 1: m < e n > ⇒ ( A − d m I )( A − d m − 1 I ) . . . ( A − d 1 I ) < e n > = Q 1: m < e n >

  49. Lanczos-Ritz convergence behavior a) Lanczos-Ritz values Because AQ 1: m = Q 1: m A ( m ) equals:   R T  A m � ← Q 1: m |− − → � ← Q 1: m |− − → � � 1  . = A Q 1: m Q 1: m ( D + S ) m R 1 Hence, the eigenvalues of ( D + S ) m are the Ritz-values of A with respect to the subspace spanned by the columns of − → Q 1 : m .

  50. b) Connection with the Krylov subspace Some notation K m = < e n , Ae n , A 2 e n , . . . , A m e n >   ˜ hq T H h    0 × × × ˜ . . . H ∈ R ( n − m − 1) × ( n − m − 2)          0 × h ∈ R ( n − m − 1) × 1 Q m + 1 = .     . . .  ...    q ∈ R ( m +1) × 1 . . .    . . .     0 0 0 × ×

  51. We want to prove by induction that: col ( − → � � span Q 1 : m + 1 ) = K m + 1 . We have:   hq T h   × × × . . .     − → [ ← Q 1: m |− − →   0 × = Q 1: m ] Q 1: m +1 .     . .   ... . .   . .     0 0 × ×   × × × . . .   0 ×   ← Q 1: m h [1 , q T ] + − − →   = Q 1: m . . .   ... . .   . .     0 0 × ×

  52. Because: col ( − → � � = K m = < e n , Ae n , . . . , A m e n > • Q 1: m ) span − → • Q 1: m +1 < e n > = ( A − d m +1 I ) . . . ( A − d 1 I ) < e n > ← − ⇒ Q 1: m h ∈ K m +1 \K m We get: col ( − → � � = K m + 1 = < e n , Ae n , . . . , A m + 1 e n > span Q 1 : m + 1 )

  53. Theorem The eigenvalues of ( D + S ) m , the lower diagonal blocks that appear during the reduction algorithm, are the Lanczos-Ritz values of A with respect to the Krylov subspace K m .

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