The Snap-Back Pivoting Method for Symmetric Banded Indefinite Matrices

1. The Snap-Back Pivoting Method for Symmetric Banded Indefinite Matrices Dror Irony and Sivan Toledo Tel-Aviv University

2. The Problem Solve the system Ax=b Where A is banded nn, symmetric with half bandwidth m : ? ? ? ? = ? ? ? ? ?

3. So far … • Choleski/LLT - just positive definite • GEPP - not exploiting symmetry • 2x2 block pivoting (e.g. B-K) - unbounded fill • Reduction to tridiagonal (e.g. aasen) - unbounded fill • Orthogonal reduction to tridiagonal (e.g. SBR) - too expensive: O(n2m)

4. Our Factorization Snap-Back Pivoting: • If possible => simple 11 LLTstep • Otherwise => Eliminate 1 or 2 rows/columns symmetry is ruined – but then snapped back!

5. Benefits • Stable • Exploits symmetry • Keeps band structure => fill is bounded • Running time: O(m2n) • Fast

6. Snap Back Pivoting in Details

7. If A[1,1]is large enough • Simple symmetric Gaussian elimination step: A[1,1]

8. If A[1,1]is large enough • Simple symmetric Gaussian elimination step:

9. If A[1,1]is large enough • Simple symmetric Gaussian elimination step:

10. If A[1,1]is large enough • Simple symmetric Gaussian elimination step:

11. If A[1,1]is large enough • Simple symmetric Gaussian elimination step:

12. If A[1,1]is large enough • Simple symmetric Gaussian elimination step:

13. If A[1,1]is large enough • Simple symmetric Gaussian elimination step:

14. If A[1,1]is large enough • Simple symmetric Gaussian elimination step:

15. If A[1,1]is not large enough Eliminate elements 2 till m1in first row/column (mi = number of structural non-zeros under the diagonal in column i)

16. If A[1,1]is not large enough Eliminate elements 2 till m1in first row/column (mi = number of structural non-zeros under the diagonal in column i) x If x>y swap rows/cols 2 and 3 y

17. If A[1,1]is not large enough Eliminate elements 2 till m1in first row/column (mi = number of structural non-zeros under the diagonal in column i)

18. If A[1,1]is not large enough Eliminate elements 2 till m1in first row/column (mi = number of structural non-zeros under the diagonal in column i)

19. If A[1,1]is not large enough Eliminate elements 2 till m1in first row/column (mi = number of structural non-zeros under the diagonal in column i) x If x>y swap rows/cols 3 and 4 y

20. If A[1,1]is not large enough Eliminate elements 2 till m1in first row/column (mi = number of structural non-zeros under the diagonal in column i)

21. If A[1,1]is not large enough Eliminate elements 2 till m1in first row/column (mi = number of structural non-zeros under the diagonal in column i)

22. If A[1,1]is not large enough Eliminate elements 2 till m1in first row/column (mi = number of structural non-zeros under the diagonal in column i)

23. If A[1,1]is not large enough Eliminate the element m1+1 in the first column by Givens:

24. If A[1,1]is not large enough Eliminate the element m1+1 in the first column by Givens: Cos() Sin() 1 1 -Sin() Cos() 1 1 1 1 1

25. If A[1,1]is not large enough Eliminate the element m1+1 in the first column by Givens:

26. If A[1,1]is not large enough Now, eliminate the first row using its first element:

27. If A[1,1]is not large enough Now, eliminate the first row using its first element:

28. If A[1,1]is not large enough Now, eliminate the first row using its first element:

29. If A[1,1]is not large enough Now, eliminate the first row using its first element:

30. If A[1,1]is not large enough Now, eliminate the first row using its first element:

31. If A[1,1]is not large enough Now, eliminate the first row using its first element:

32. If A[1,1]is not large enough Now, eliminate the first row using its first element:

33. If A[1,1]is not large enough Denote the matrix we got by B: B =

34. If B[m1+1,m1+1] is not the largest element in its row Scale row m1+1: B[m1+1,m1+1] 1 1 1 1/Cos() 1 1 1 1 1

35. If B[m1+1,m1+1] is not the largest element in its row Now we can move forward to the next elimination step…

36. If B[m1+1,m1+1] is the largest element in its row Bring B[m1+1,m1+1] to location [2,2] : B[m1+1,m1+1] B[2,2]

37. If B[m1+1,m1+1] is the largest element in its row Bring B[m1+1,m1+1] to location [2,2] :

38. If B[m1+1,m1+1] is the largest element in its row Bring B[m1+1,m1+1] to location [2,2] :

39. If B[m1+1,m1+1] is the largest element in its row Bring B[m1+1,m1+1] to location [2,2] :

40. If B[m1+1,m1+1] is the largest element in its row Eliminate elements 3 to m1 in second row/column:

41. If B[m1+1,m1+1] is the largest element in its row Eliminate elements 3 to m1 in second row/column: x If x>y swap rows/cols 3 and 4 y

42. If B[m1+1,m1+1] is the largest element in its row Eliminate elements 3 to m1 in second row/column:

43. If B[m1+1,m1+1] is the largest element in its row Eliminate elements 3 to m1 in second row/column:

44. If B[m1+1,m1+1] is the largest element in its row Now eliminate the second row/column:

45. If B[m1+1,m1+1] is the largest element in its row Now eliminate the second row/column:

46. If B[m1+1,m1+1] is the largest element in its row Now eliminate the second row/column:

47. If B[m1+1,m1+1] is the largest element in its row Now eliminate the second row/column:

48. If B[m1+1,m1+1] is the largest element in its row Now eliminate the second row/column:

49. If B[m1+1,m1+1] is the largest element in its row Now eliminate the second row/column:

50. If B[m1+1,m1+1] is the largest element in its row Now eliminate the second row/column: