SLIDE 1
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
Structure-preserving Krylov subspace methods for Hamiltonian and - - PowerPoint PPT Presentation
Structure-preserving Krylov subspace methods for Hamiltonian and - - PowerPoint PPT Presentation
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
SLIDE 2
SLIDE 3
Definitions
matrices in R2n×2n
March 2007 – p.2
SLIDE 4
Definitions
matrices in R2n×2n
J =
- I
−I
- March 2007 – p.2
SLIDE 5
Definitions
matrices in R2n×2n
J =
- I
−I
- S is symplectic if ST JS = J
March 2007 – p.2
SLIDE 6
Definitions
matrices in R2n×2n
J =
- I
−I
- S is symplectic if ST JS = J (Lie group)
March 2007 – p.2
SLIDE 7
Definitions
matrices in R2n×2n
J =
- I
−I
- S is symplectic if ST JS = J (Lie group)
H is Hamiltonian if (JH)T = JH
March 2007 – p.2
SLIDE 8
Definitions
matrices in R2n×2n
J =
- I
−I
- S is symplectic if ST JS = J (Lie group)
H is Hamiltonian if (JH)T = JH (Lie algebra)
March 2007 – p.2
SLIDE 9
Definitions
matrices in R2n×2n
J =
- I
−I
- S is symplectic if ST JS = J (Lie group)
H is Hamiltonian if (JH)T = JH (Lie algebra) B is skew Hamiltonian if (JB)T = −JB
March 2007 – p.2
SLIDE 10
Definitions
matrices in R2n×2n
J =
- I
−I
- S is symplectic if ST JS = J (Lie group)
H is Hamiltonian if (JH)T = JH (Lie algebra) B is skew Hamiltonian if (JB)T = −JB
(Jordan algebra)
March 2007 – p.2
SLIDE 11
Definitions
matrices in R2n×2n
J =
- I
−I
- S is symplectic if ST JS = J (Lie group)
H is Hamiltonian if (JH)T = JH (Lie algebra) B is skew Hamiltonian if (JB)T = −JB
(Jordan algebra) Matrices with these structures arise in various applications.
March 2007 – p.2
SLIDE 12
Definitions
matrices in R2n×2n
J =
- I
−I
- S is symplectic if ST JS = J (Lie group)
H is Hamiltonian if (JH)T = JH (Lie algebra) B is skew Hamiltonian if (JB)T = −JB
(Jordan algebra) Matrices with these structures arise in various applications. Sometimes they are large and sparse.
March 2007 – p.2
SLIDE 13
Objective
March 2007 – p.3
SLIDE 14
Objective
Produce structure-preserving Krylov subspace methods for symplectic, Hamiltonian, and skew-Hamiltonian eigenvalue problems.
March 2007 – p.3
SLIDE 15
Objective
Produce structure-preserving Krylov subspace methods for symplectic, Hamiltonian, and skew-Hamiltonian eigenvalue
- problems. Done!
March 2007 – p.3
SLIDE 16
Objective
Produce structure-preserving Krylov subspace methods for symplectic, Hamiltonian, and skew-Hamiltonian eigenvalue
- problems. Done!
Freund / Mehrmann (unpublished)
March 2007 – p.3
SLIDE 17
Objective
Produce structure-preserving Krylov subspace methods for symplectic, Hamiltonian, and skew-Hamiltonian eigenvalue
- problems. Done!
Freund / Mehrmann (unpublished) Benner / Fassbender (1997,1998)
March 2007 – p.3
SLIDE 18
Objective
Produce structure-preserving Krylov subspace methods for symplectic, Hamiltonian, and skew-Hamiltonian eigenvalue
- problems. Done!
Freund / Mehrmann (unpublished) Benner / Fassbender (1997,1998) Benner / Fassbender / W (1988,1999)
March 2007 – p.3
SLIDE 19
Objective
Produce structure-preserving Krylov subspace methods for symplectic, Hamiltonian, and skew-Hamiltonian eigenvalue
- problems. Done!
Freund / Mehrmann (unpublished) Benner / Fassbender (1997,1998) Benner / Fassbender / W (1988,1999) Mehrmann / W (2001)
March 2007 – p.3
SLIDE 20
Objective
Produce structure-preserving Krylov subspace methods for symplectic, Hamiltonian, and skew-Hamiltonian eigenvalue
- problems. Done!
Freund / Mehrmann (unpublished) Benner / Fassbender (1997,1998) Benner / Fassbender / W (1988,1999) Mehrmann / W (2001) W (2003)
March 2007 – p.3
SLIDE 21
Today’s Objective
March 2007 – p.4
SLIDE 22
Today’s Objective
Show how easy it is!
March 2007 – p.4
SLIDE 23
Today’s Objective
Show how easy it is! unsymmetric Lanczos process
March 2007 – p.4
SLIDE 24
Today’s Objective
Show how easy it is! unsymmetric Lanczos process skew-Hamiltonian structure preserved automatically
March 2007 – p.4
SLIDE 25
Today’s Objective
Show how easy it is! unsymmetric Lanczos process skew-Hamiltonian structure preserved automatically
H2 = B
March 2007 – p.4
SLIDE 26
Today’s Objective
Show how easy it is! unsymmetric Lanczos process skew-Hamiltonian structure preserved automatically
H2 = B S + S−1 = B
March 2007 – p.4
SLIDE 27
Today’s Objective
Show how easy it is! unsymmetric Lanczos process skew-Hamiltonian structure preserved automatically
H2 = B S + S−1 = B
David S. Watkins, The Matrix Eigenvalue Problem: GR and Krylov Subspace Methods, SIAM, to appear.
March 2007 – p.4
SLIDE 28
Unsymmetric Lanczos Process (1950)
March 2007 – p.5
SLIDE 29
Unsymmetric Lanczos Process (1950)
uj+1βj = Auj − ujαj − uj−1γj−1 wj+1γj = ATwj − wjαj − wj−1βj−1
March 2007 – p.5
SLIDE 30
Unsymmetric Lanczos Process (1950)
uj+1βj = Auj − ujαj − uj−1γj−1 wj+1γj = ATwj − wjαj − wj−1βj−1
Start with u1, w1 = 1.
March 2007 – p.5
SLIDE 31
Unsymmetric Lanczos Process (1950)
uj+1βj = Auj − ujαj − uj−1γj−1 wj+1γj = ATwj − wjαj − wj−1βj−1
Start with u1, w1 = 1. sequences are biorthornomal: uj, wk = δjk
March 2007 – p.5
SLIDE 32
Unsymmetric Lanczos Process (1950)
uj+1βj = Auj − ujαj − uj−1γj−1 wj+1γj = ATwj − wjαj − wj−1βj−1
Start with u1, w1 = 1. sequences are biorthornomal: uj, wk = δjk
- mitting simple formulas for the coefficients
March 2007 – p.5
SLIDE 33
Unsymmetric Lanczos Process (1950)
uj+1βj = Auj − ujαj − uj−1γj−1 wj+1γj = ATwj − wjαj − wj−1βj−1
Start with u1, w1 = 1. sequences are biorthornomal: uj, wk = δjk
- mitting simple formulas for the coefficients
|γj | = |βj |
March 2007 – p.5
SLIDE 34
Collect the coefficients
March 2007 – p.6
SLIDE 35
Collect the coefficients
- Tj =
α1 γ1 β1 α2 γ2 β2 α3
... ... ...
γj−1 βj−1 αj
March 2007 – p.6
SLIDE 36
Collect the coefficients
- Tj =
α1 γ1 β1 α2 γ2 β2 α3
... ... ...
γj−1 βj−1 αj
Eigenvalues are estimates of eigenvalues of A.
March 2007 – p.6
SLIDE 37
Collect the coefficients
- Tj =
α1 γ1 β1 α2 γ2 β2 α3
... ... ...
γj−1 βj−1 αj
Eigenvalues are estimates of eigenvalues of A.
- Tj is pseudosymmetric.
(|γj | = |βj |)
March 2007 – p.6
SLIDE 38
Collect the coefficients
- Tj =
α1 γ1 β1 α2 γ2 β2 α3
... ... ...
γj−1 βj−1 αj
Eigenvalues are estimates of eigenvalues of A.
- Tj is pseudosymmetric.
(|γj | = |βj |)
- Tj = TjDj,
where Tj is symmetric and Dj is a signature matrix.
March 2007 – p.6
SLIDE 39
Tj = a1 b1 b1 a2 b2 b2 a3
... ... ...
bj−1 bj−1 aj Dj = d1 d2 d3
...
dj
March 2007 – p.7
SLIDE 40
Lanczos recurrences recast:
March 2007 – p.8
SLIDE 41
Lanczos recurrences recast:
uj+1bjdj = Auj − ujajdj − uj−1bj−1dj wj+1dj+1bj = ATwj − wjdjaj − wj−1dj−1bj−1,
March 2007 – p.8
SLIDE 42
Lanczos recurrences recast:
uj+1bjdj = Auj − ujajdj − uj−1bj−1dj wj+1dj+1bj = ATwj − wjdjaj − wj−1dj−1bj−1,
Recurrences rewritten as matrix equations:
AUj = UjTjDj + uj+1bjdjeT
j
ATWj = WjDjTj + wj+1dj+1bjeT
j .
March 2007 – p.8
SLIDE 43
Lanczos recurrences recast:
uj+1bjdj = Auj − ujajdj − uj−1bj−1dj wj+1dj+1bj = ATwj − wjdjaj − wj−1dj−1bj−1,
Recurrences rewritten as matrix equations:
AUj = UjTjDj + uj+1bjdjeT
j
ATWj = WjDjTj + wj+1dj+1bjeT
j .
Implicit restarts:
March 2007 – p.8
SLIDE 44
Lanczos recurrences recast:
uj+1bjdj = Auj − ujajdj − uj−1bj−1dj wj+1dj+1bj = ATwj − wjdjaj − wj−1dj−1bj−1,
Recurrences rewritten as matrix equations:
AUj = UjTjDj + uj+1bjdjeT
j
ATWj = WjDjTj + wj+1dj+1bjeT
j .
Implicit restarts: Filter using HR algorithm
March 2007 – p.8
SLIDE 45
Lanczos recurrences recast:
uj+1bjdj = Auj − ujajdj − uj−1bj−1dj wj+1dj+1bj = ATwj − wjdjaj − wj−1dj−1bj−1,
Recurrences rewritten as matrix equations:
AUj = UjTjDj + uj+1bjdjeT
j
ATWj = WjDjTj + wj+1dj+1bjeT
j .
Implicit restarts: Filter using HR algorithm Grimme / Sorensen / Van Dooren (1996)
March 2007 – p.8
SLIDE 46
Preservation of Structure
March 2007 – p.9
SLIDE 47
Preservation of Structure
- A = S−1AS
March 2007 – p.9
SLIDE 48
Preservation of Structure
- A = S−1AS
S symplectic ⇒
structure preserved
March 2007 – p.9
SLIDE 49
Preservation of Structure
- A = S−1AS
S symplectic ⇒
structure preserved Lanczos process is a partial similarity transformation.
March 2007 – p.9
SLIDE 50
Preservation of Structure
- A = S−1AS
S symplectic ⇒
structure preserved Lanczos process is a partial similarity transformation. Vectors produced are columns of transforming matrix.
March 2007 – p.9
SLIDE 51
Preservation of Structure
- A = S−1AS
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
SLIDE 52
Preservation of Structure
- A = S−1AS
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
SLIDE 53
Isotropy
March 2007 – p.10
SLIDE 54
Isotropy
Def: U = R(U) is isotropic if xT Jy = 0 for all x, y ∈ U, i.e.
March 2007 – p.10
SLIDE 55
Isotropy
Def: U = R(U) is isotropic if xT Jy = 0 for all x, y ∈ U, i.e.
UTJU = 0
March 2007 – p.10
SLIDE 56
Isotropy
Def: U = R(U) is isotropic if xT Jy = 0 for all x, y ∈ U, i.e.
UTJU = 0
Symplectic matrix S = [ U V ] satisfies ST JS = J, i.e.
March 2007 – p.10
SLIDE 57
Isotropy
Def: U = R(U) is isotropic if xT Jy = 0 for all x, y ∈ U, i.e.
UTJU = 0
Symplectic matrix S = [ U V ] satisfies ST JS = J, i.e.
UTJU = 0, V T JV = 0, UT JV = I.
March 2007 – p.10
SLIDE 58
Isotropy
Def: U = R(U) is isotropic if xT Jy = 0 for all x, y ∈ U, i.e.
UTJU = 0
Symplectic matrix S = [ U V ] satisfies ST JS = J, i.e.
UTJU = 0, V T JV = 0, UT JV = I.
In particular, R(U), R(V ) are isotropic.
March 2007 – p.10
SLIDE 59
Theorem: If B is skew Hamiltonian, then every Krylov subspace
κj(B, x) = span
- x, Bx, B2x, . . . , Bj−1x
- is isotropic.
March 2007 – p.11
SLIDE 60
Theorem: If B is skew Hamiltonian, then every Krylov subspace
κj(B, x) = span
- x, Bx, B2x, . . . , Bj−1x
- is isotropic.
Proof: Mehrmann / W (2001)
March 2007 – p.11
SLIDE 61
Theorem: If B is skew Hamiltonian, then every Krylov subspace
κj(B, x) = span
- x, Bx, B2x, . . . , Bj−1x
- is isotropic.
Proof: Mehrmann / W (2001) Corollary: Every Krylov subspace method automatically preserves skew-Hamiltonian structure.
March 2007 – p.11
SLIDE 62
Skew-Hamiltonian Lanczos Process
March 2007 – p.12
SLIDE 63
Skew-Hamiltonian Lanczos Process
uj+1bjdj = Buj − ujajdj − uj−1bj−1dj wj+1dj+1bj = BT wj − wjdjaj − wj−1dj−1bj−1
March 2007 – p.12
SLIDE 64
Skew-Hamiltonian Lanczos Process
uj+1bjdj = Buj − ujajdj − uj−1bj−1dj wj+1dj+1bj = BT wj − wjdjaj − wj−1dj−1bj−1 UT
j JUj = 0,
W T
j JWj = 0,
UT
j Wj = I
March 2007 – p.12
SLIDE 65
Skew-Hamiltonian Lanczos Process
uj+1bjdj = Buj − ujajdj − uj−1bj−1dj wj+1dj+1bj = BT wj − wjdjaj − wj−1dj−1bj−1 UT
j JUj = 0,
W T
j JWj = 0,
UT
j Wj = I
Let vk = −Jwk (So Wj = JVj)
March 2007 – p.12
SLIDE 66
Skew-Hamiltonian Lanczos Process
uj+1bjdj = Buj − ujajdj − uj−1bj−1dj wj+1dj+1bj = BT wj − wjdjaj − wj−1dj−1bj−1 UT
j JUj = 0,
W T
j JWj = 0,
UT
j Wj = I
Let vk = −Jwk (So Wj = JVj)
(JB)T = −JB ⇒ −JBT = −BJ
March 2007 – p.12
SLIDE 67
Skew-Hamiltonian Lanczos Process
uj+1bjdj = Buj − ujajdj − uj−1bj−1dj vj+1dj+1bj = Bvj − vjdjaj − vj−1dj−1bj−1
March 2007 – p.13
SLIDE 68
Skew-Hamiltonian Lanczos Process
uj+1bjdj = Buj − ujajdj − uj−1bj−1dj vj+1dj+1bj = Bvj − vjdjaj − vj−1dj−1bj−1
Start with uT
1 Jv1 = 1.
March 2007 – p.13
SLIDE 69
Skew-Hamiltonian Lanczos Process
uj+1bjdj = Buj − ujajdj − uj−1bj−1dj vj+1dj+1bj = Bvj − vjdjaj − vj−1dj−1bj−1
Start with uT
1 Jv1 = 1.
UT
j JUj = 0,
V T
j JVj = 0,
UT
j JVj = I
March 2007 – p.13
SLIDE 70
Skew-Hamiltonian Lanczos Process
uj+1bjdj = Buj − ujajdj − uj−1bj−1dj vj+1dj+1bj = Bvj − vjdjaj − vj−1dj−1bj−1
Start with uT
1 Jv1 = 1.
UT
j JUj = 0,
V T
j JVj = 0,
UT
j JVj = I
These are the columns of a symplectic matrix.
March 2007 – p.13
SLIDE 71
Hamiltonian Lanczos Process
March 2007 – p.14
SLIDE 72
Hamiltonian Lanczos Process
H Hamiltonian ⇒ H2 skew Hamiltonian.
March 2007 – p.14
SLIDE 73
Hamiltonian Lanczos Process
H Hamiltonian ⇒ H2 skew Hamiltonian.
Apply skew-Hamiltonian Lanczos process to H2.
March 2007 – p.14
SLIDE 74
Hamiltonian Lanczos Process
H Hamiltonian ⇒ H2 skew Hamiltonian.
Apply skew-Hamiltonian Lanczos process to H2.
uj+1bjdj = H2uj − ujajdj − uj−1bj−1dj vj+1dj+1bj = H2vj − vjdjaj − vj−1dj−1bj−1
March 2007 – p.14
SLIDE 75
Hamiltonian Lanczos Process
H Hamiltonian ⇒ H2 skew Hamiltonian.
Apply skew-Hamiltonian Lanczos process to H2.
uj+1bjdj = H2uj − ujajdj − uj−1bj−1dj vj+1dj+1bj = H2vj − vjdjaj − vj−1dj−1bj−1
Let
˜ vj = Hujdj.
March 2007 – p.14
SLIDE 76
Hamiltonian Lanczos Process
H Hamiltonian ⇒ H2 skew Hamiltonian.
Apply skew-Hamiltonian Lanczos process to H2.
uj+1bjdj = H2uj − ujajdj − uj−1bj−1dj vj+1dj+1bj = H2vj − vjdjaj − vj−1dj−1bj−1
Let
˜ vj = Hujdj.
Multiply first equation by H and by dj.
March 2007 – p.14
SLIDE 77
Hamiltonian Lanczos Process
H Hamiltonian ⇒ H2 skew Hamiltonian.
Apply skew-Hamiltonian Lanczos process to H2.
uj+1bjdj = H2uj − ujajdj − uj−1bj−1dj vj+1dj+1bj = H2vj − vjdjaj − vj−1dj−1bj−1
Let
˜ vj = Hujdj.
Multiply first equation by H and by dj.
˜ vj+1dj+1bj = H2˜ vj − ˜ vjdjaj − ˜ vj−1dj−1bj−1
March 2007 – p.14
SLIDE 78
Hamiltonian Lanczos Process
uj+1bjdj = H2uj − ujajdj − uj−1bj−1dj vj+1dj+1bj = H2vj − vjdjaj − vj−1dj−1bj−1
March 2007 – p.15
SLIDE 79
Hamiltonian Lanczos Process
uj+1bjdj = H2uj − ujajdj − uj−1bj−1dj vj+1dj+1bj = H2vj − vjdjaj − vj−1dj−1bj−1
Start the process with v1 = ˜
v1 = Hu1d1.
March 2007 – p.15
SLIDE 80
Hamiltonian Lanczos Process
uj+1bjdj = H2uj − ujajdj − uj−1bj−1dj vj+1dj+1bj = H2vj − vjdjaj − vj−1dj−1bj−1
Start the process with v1 = ˜
v1 = Hu1d1.
Then vj = ˜
vj for all j.
March 2007 – p.15
SLIDE 81
Hamiltonian Lanczos Process
uj+1bjdj = H2uj − ujajdj − uj−1bj−1dj vj+1dj+1bj = H2vj − vjdjaj − vj−1dj−1bj−1
Start the process with v1 = ˜
v1 = Hu1d1.
Then vj = ˜
vj for all j.
Conclusion:
March 2007 – p.15
SLIDE 82
Hamiltonian Lanczos Process
uj+1bjdj = H2uj − ujajdj − uj−1bj−1dj vj+1dj+1bj = H2vj − vjdjaj − vj−1dj−1bj−1
Start the process with v1 = ˜
v1 = Hu1d1.
Then vj = ˜
vj for all j.
Conclusion: The second recurrence is redundant.
March 2007 – p.15
SLIDE 83
Hamiltonian Lanczos Process
uj+1bjdj = H2uj − ujajdj − uj−1bj−1dj vj+1dj+1bj = H2vj − vjdjaj − vj−1dj−1bj−1
Start the process with v1 = ˜
v1 = Hu1d1.
Then vj = ˜
vj for all j.
Conclusion: The second recurrence is redundant. Just run the first recurrence
March 2007 – p.15
SLIDE 84
Hamiltonian Lanczos Process
uj+1bjdj = H2uj − ujajdj − uj−1bj−1dj vj+1dj+1bj = H2vj − vjdjaj − vj−1dj−1bj−1
Start the process with v1 = ˜
v1 = Hu1d1.
Then vj = ˜
vj for all j.
Conclusion: The second recurrence is redundant. Just run the first recurrence together with the supplementary condition
vjdj = Huj.
March 2007 – p.15
SLIDE 85
Hamiltonian Lanczos Process
uj+1bjdj = H2uj − ujajdj − uj−1bj−1dj vj+1dj+1 = Huj+1
March 2007 – p.16
SLIDE 86
Hamiltonian Lanczos Process
uj+1bjdj = H2uj − ujajdj − uj−1bj−1dj vj+1dj+1 = Huj+1
March 2007 – p.16
SLIDE 87
Hamiltonian Lanczos Process
uj+1bjdj = Hvjdj − ujajdj − uj−1bj−1dj vj+1dj+1 = Huj+1
March 2007 – p.16
SLIDE 88
Hamiltonian Lanczos Process
uj+1bjdj = Hvjdj − ujajdj − uj−1bj−1dj vj+1dj+1 = Huj+1
March 2007 – p.16
SLIDE 89
Hamiltonian Lanczos Process
uj+1bj = Hvj − ujaj − uj−1bj−1 vj+1dj+1 = Huj+1
March 2007 – p.16
SLIDE 90
Hamiltonian Lanczos Process
uj+1bj = Hvj − ujaj − uj−1bj−1 vj+1dj+1 = Huj+1
Start with
v1d1 = Hu1 uT
1 Jv1 = 1.
March 2007 – p.16
SLIDE 91
Hamiltonian Lanczos Process
uj+1bj = Hvj − ujaj − uj−1bj−1 vj+1dj+1 = Huj+1
Start with
v1d1 = Hu1 uT
1 Jv1 = 1.
This is easy to arrange.
March 2007 – p.16
SLIDE 92
Recurrences written as matrix products
uj+1bj = Hvj − ujaj − uj−1bj−1 vj+1dj+1 = Huj+1
March 2007 – p.17
SLIDE 93
Recurrences written as matrix products
uj+1bj = Hvj − ujaj − uj−1bj−1 vj+1dj+1 = Huj+1 HVj = UjTj + uj+1bjeT
j ,
HUj = VjDj
March 2007 – p.17
SLIDE 94
Recurrences written as matrix products
uj+1bj = Hvj − ujaj − uj−1bj−1 vj+1dj+1 = Huj+1 H
- Uj
Vj
- =
- Uj
Vj Tj Dj
- + uj+1bjeT
2j.
March 2007 – p.17
SLIDE 95
Recurrences written as matrix products
uj+1bj = Hvj − ujaj − uj−1bj−1 vj+1dj+1 = Huj+1 H
- Uj
Vj
- =
- Uj
Vj Tj Dj
- + uj+1bjeT
2j.
Implicit restarts:
March 2007 – p.17
SLIDE 96
Recurrences written as matrix products
uj+1bj = Hvj − ujaj − uj−1bj−1 vj+1dj+1 = Huj+1 H
- Uj
Vj
- =
- Uj
Vj Tj Dj
- + uj+1bjeT
2j.
Implicit restarts: Filter with the HR algorithm.
March 2007 – p.17
SLIDE 97
Recurrences written as matrix products
uj+1bj = Hvj − ujaj − uj−1bj−1 vj+1dj+1 = Huj+1 H
- Uj
Vj
- =
- Uj
Vj Tj Dj
- + uj+1bjeT
2j.
Implicit restarts: Filter with the HR algorithm. Next up: symplectic Lanczos process.
March 2007 – p.17
SLIDE 98
Symplectic Lanczos Process
March 2007 – p.18
SLIDE 99
Symplectic Lanczos Process
S symplectic ⇒ S + S−1 skew Hamiltonian.
March 2007 – p.18
SLIDE 100
Symplectic Lanczos Process
S symplectic ⇒ S + S−1 skew Hamiltonian.
Apply skew-Hamiltonian Lanczos process to S + S−1.
March 2007 – p.18
SLIDE 101
Symplectic Lanczos Process
S symplectic ⇒ S + S−1 skew Hamiltonian.
Apply skew-Hamiltonian Lanczos process to S + S−1.
uj+1bjdj = (S + S−1)uj − ujajdj − uj−1bj−1dj vj+1dj+1bj = (S + S−1)vj − vjdjaj − vj−1dj−1bj−1
March 2007 – p.18
SLIDE 102
Symplectic Lanczos Process
S symplectic ⇒ S + S−1 skew Hamiltonian.
Apply skew-Hamiltonian Lanczos process to S + S−1.
uj+1bjdj = (S + S−1)uj − ujajdj − uj−1bj−1dj vj+1dj+1bj = (S + S−1)vj − vjdjaj − vj−1dj−1bj−1
Let
˜ vj = S−1ujdj.
March 2007 – p.18
SLIDE 103
Symplectic Lanczos Process
S symplectic ⇒ S + S−1 skew Hamiltonian.
Apply skew-Hamiltonian Lanczos process to S + S−1.
uj+1bjdj = (S + S−1)uj − ujajdj − uj−1bj−1dj vj+1dj+1bj = (S + S−1)vj − vjdjaj − vj−1dj−1bj−1
Let
˜ vj = S−1ujdj.
Multiply first equation by S−1 and by dj.
March 2007 – p.18
SLIDE 104
Symplectic Lanczos Process
S symplectic ⇒ S + S−1 skew Hamiltonian.
Apply skew-Hamiltonian Lanczos process to S + S−1.
uj+1bjdj = (S + S−1)uj − ujajdj − uj−1bj−1dj vj+1dj+1bj = (S + S−1)vj − vjdjaj − vj−1dj−1bj−1
Let
˜ vj = S−1ujdj.
Multiply first equation by S−1 and by dj.
˜ vj+1dj+1bj = (S + S−1)˜ vj − ˜ vjdjaj − ˜ vj−1dj−1bj−1
March 2007 – p.18
SLIDE 105
Symplectic Lanczos Process
uj+1bjdj = (S + S−1)uj − ujajdj − uj−1bj−1dj vj+1dj+1bj = (S + S−1)vj − vjdjaj − vj−1dj−1bj−1
March 2007 – p.19
SLIDE 106
Symplectic Lanczos Process
uj+1bjdj = (S + S−1)uj − ujajdj − uj−1bj−1dj vj+1dj+1bj = (S + S−1)vj − vjdjaj − vj−1dj−1bj−1
Start the process with v1 = ˜
v1 = S−1u1d1.
March 2007 – p.19
SLIDE 107
Symplectic Lanczos Process
uj+1bjdj = (S + S−1)uj − ujajdj − uj−1bj−1dj vj+1dj+1bj = (S + S−1)vj − vjdjaj − vj−1dj−1bj−1
Start the process with v1 = ˜
v1 = S−1u1d1.
Then vj = ˜
vj for all j.
March 2007 – p.19
SLIDE 108
Symplectic Lanczos Process
uj+1bjdj = (S + S−1)uj − ujajdj − uj−1bj−1dj vj+1dj+1bj = (S + S−1)vj − vjdjaj − vj−1dj−1bj−1
Start the process with v1 = ˜
v1 = S−1u1d1.
Then vj = ˜
vj for all j.
Conclusion:
March 2007 – p.19
SLIDE 109
Symplectic Lanczos Process
uj+1bjdj = (S + S−1)uj − ujajdj − uj−1bj−1dj vj+1dj+1bj = (S + S−1)vj − vjdjaj − vj−1dj−1bj−1
Start the process with v1 = ˜
v1 = S−1u1d1.
Then vj = ˜
vj for all j.
Conclusion: The second recurrence is redundant.
March 2007 – p.19
SLIDE 110
Symplectic Lanczos Process
uj+1bjdj = (S + S−1)uj − ujajdj − uj−1bj−1dj vj+1dj+1bj = (S + S−1)vj − vjdjaj − vj−1dj−1bj−1
Start the process with v1 = ˜
v1 = S−1u1d1.
Then vj = ˜
vj for all j.
Conclusion: The second recurrence is redundant. Just run the first recurrence
March 2007 – p.19
SLIDE 111
Symplectic Lanczos Process
uj+1bjdj = (S + S−1)uj − ujajdj − uj−1bj−1dj vj+1dj+1bj = (S + S−1)vj − vjdjaj − vj−1dj−1bj−1
Start the process with v1 = ˜
v1 = S−1u1d1.
Then vj = ˜
vj for all j.
Conclusion: The second recurrence is redundant. Just run the first recurrence together with the supplementary condition
vjdj = S−1uj.
March 2007 – p.19
SLIDE 112
Symplectic Lanczos Process
uj+1bjdj = (S + S−1)uj − ujajdj − uj−1bj−1dj vj+1dj+1 = S−1uj+1
March 2007 – p.20
SLIDE 113
Symplectic Lanczos Process
uj+1bjdj = (S + S−1)uj − ujajdj − uj−1bj−1dj vj+1dj+1 = S−1uj+1
March 2007 – p.20
SLIDE 114
Symplectic Lanczos Process
uj+1bjdj = Suj + vjdj − ujajdj − uj−1bj−1dj vj+1dj+1 = S−1uj+1
March 2007 – p.20
SLIDE 115
Symplectic Lanczos Process
uj+1bjdj = Suj + vjdj − ujajdj − uj−1bj−1dj vj+1dj+1 = S−1uj+1
March 2007 – p.20
SLIDE 116
Symplectic Lanczos Process
uj+1bj = Sujdj + vj − ujaj − uj−1bj−1 vj+1dj+1 = S−1uj+1
March 2007 – p.20
SLIDE 117
Symplectic Lanczos Process
uj+1bj = Sujdj + vj − ujaj − uj−1bj−1 vj+1dj+1 = S−1uj+1
Start with
v1d1 = S−1u1 uT
1 Jv1 = 1.
March 2007 – p.20
SLIDE 118
Symplectic Lanczos Process
uj+1bj = Sujdj + vj − ujaj − uj−1bj−1 vj+1dj+1 = S−1uj+1
Start with
v1d1 = S−1u1 uT
1 Jv1 = 1.
This is easy to arrange.
March 2007 – p.20
SLIDE 119
Recurrences written as matrix products
uj+1bj = Sujdj + vj − ujaj − uj−1bj−1 vj+1dj+1 = S−1uj+1
March 2007 – p.21
SLIDE 120
Recurrences written as matrix products
uj+1bj = Sujdj + vj − ujaj − uj−1bj−1 vj+1dj+1 = S−1uj+1 SUj = Uj(TjDj) + uj+1bj+1dj+1eT
j − VjDj,
SVj = UjDj
March 2007 – p.21
SLIDE 121
Recurrences written as matrix products
uj+1bj = Sujdj + vj − ujaj − uj−1bj−1 vj+1dj+1 = S−1uj+1 S
- Uj
Vj
- =
- Uj
Vj TjDj Dj −Dj
- + uj+1bj+1dj+1eT
j
March 2007 – p.21
SLIDE 122
Recurrences written as matrix products
uj+1bj = Sujdj + vj − ujaj − uj−1bj−1 vj+1dj+1 = S−1uj+1 S
- Uj
Vj
- =
- Uj
Vj TjDj Dj −Dj
- + uj+1bj+1dj+1eT
j
Implicit restarts:
March 2007 – p.21
SLIDE 123
Recurrences written as matrix products
uj+1bj = Sujdj + vj − ujaj − uj−1bj−1 vj+1dj+1 = S−1uj+1 S
- Uj
Vj
- =
- Uj
Vj TjDj Dj −Dj
- + uj+1bj+1dj+1eT
j
Implicit restarts: Filter with the HR algorithm.
March 2007 – p.21
SLIDE 124
Conclusions
March 2007 – p.22
SLIDE 125
Conclusions
Since Krylov subspaces associated with skew-Hamiltonian matrices are isotropic, . . .
March 2007 – p.22
SLIDE 126
Conclusions
Since Krylov subspaces associated with skew-Hamiltonian matrices are isotropic, . . . . . . the unsymmetric Lanczos process automatically preserves skew-Hamiltonian structure.
March 2007 – p.22
SLIDE 127
Conclusions
Since Krylov subspaces associated with skew-Hamiltonian matrices are isotropic, . . . . . . the unsymmetric Lanczos process automatically preserves skew-Hamiltonian structure. From the skew-Hamiltonian Lanczos process we easily derive . . .
March 2007 – p.22
SLIDE 128
Conclusions
Since Krylov subspaces associated with skew-Hamiltonian matrices are isotropic, . . . . . . the unsymmetric Lanczos process automatically preserves skew-Hamiltonian structure. From the skew-Hamiltonian Lanczos process we easily derive . . . a Hamiltonian Lanczos process
March 2007 – p.22
SLIDE 129
Conclusions
Since Krylov subspaces associated with skew-Hamiltonian matrices are isotropic, . . . . . . the unsymmetric Lanczos process automatically preserves skew-Hamiltonian structure. From the skew-Hamiltonian Lanczos process we easily derive . . . a Hamiltonian Lanczos process using H2.
March 2007 – p.22
SLIDE 130
Conclusions
Since Krylov subspaces associated with skew-Hamiltonian matrices are isotropic, . . . . . . the unsymmetric Lanczos process automatically preserves skew-Hamiltonian structure. From the skew-Hamiltonian Lanczos process we easily derive . . . a Hamiltonian Lanczos process using H2. a symplectic Lanczos process
March 2007 – p.22
SLIDE 131
Conclusions
Since Krylov subspaces associated with skew-Hamiltonian matrices are isotropic, . . . . . . the unsymmetric Lanczos process automatically preserves skew-Hamiltonian structure. From the skew-Hamiltonian Lanczos process we easily derive . . . a Hamiltonian Lanczos process using H2. a symplectic Lanczos process using S + S−1.
March 2007 – p.22
SLIDE 132
Conclusions
Since Krylov subspaces associated with skew-Hamiltonian matrices are isotropic, . . . . . . the unsymmetric Lanczos process automatically preserves skew-Hamiltonian structure. From the skew-Hamiltonian Lanczos process we easily derive . . . a Hamiltonian Lanczos process using H2. a symplectic Lanczos process using S + S−1. These algorithms are not new,
March 2007 – p.22
SLIDE 133
Conclusions
Since Krylov subspaces associated with skew-Hamiltonian matrices are isotropic, . . . . . . the unsymmetric Lanczos process automatically preserves skew-Hamiltonian structure. From the skew-Hamiltonian Lanczos process we easily derive . . . a Hamiltonian Lanczos process using H2. a symplectic Lanczos process using S + S−1. These algorithms are not new, but the derivations are.
March 2007 – p.22
SLIDE 134
Conclusions
Since Krylov subspaces associated with skew-Hamiltonian matrices are isotropic, . . . . . . the unsymmetric Lanczos process automatically preserves skew-Hamiltonian structure. From the skew-Hamiltonian Lanczos process we easily derive . . . a Hamiltonian Lanczos process using H2. a symplectic Lanczos process using S + S−1. These algorithms are not new, but the derivations are.
Thank you for your attention.
March 2007 – p.22