structure-preserving krylov subspace methods for ...structure-preserving krylov subspace methods for...
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Structure-preserving Krylovsubspace methods for Hamiltonianand symplectic eigenvalue problems
David S. Watkins
Department of Mathematics
Washington State University
March 2007 – p.1
Definitions
matrices in R2n×2n
J =
[0 I
−I 0
]
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
Definitionsmatrices in R
2n×2n
J =
[0 I
−I 0
]
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
Definitionsmatrices in R
2n×2n
J =
[0 I
−I 0
]
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
Definitionsmatrices in R
2n×2n
J =
[0 I
−I 0
]
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
Definitionsmatrices in R
2n×2n
J =
[0 I
−I 0
]
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
Definitionsmatrices in R
2n×2n
J =
[0 I
−I 0
]
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
Definitionsmatrices in R
2n×2n
J =
[0 I
−I 0
]
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
Definitionsmatrices in R
2n×2n
J =
[0 I
−I 0
]
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
Definitionsmatrices in R
2n×2n
J =
[0 I
−I 0
]
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
Definitionsmatrices in R
2n×2n
J =
[0 I
−I 0
]
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
Definitionsmatrices in R
2n×2n
J =
[0 I
−I 0
]
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
Objective
Produce structure-preserving Krylov subspace methods forsymplectic, Hamiltonian, and skew-Hamiltonian eigenvalueproblems. Done!
Freund / Mehrmann (unpublished)
Benner / Fassbender (1997,1998)
Benner / Fassbender / W (1988,1999)
Mehrmann / W (2001)
W (2003)
March 2007 – p.3
Objective
Produce structure-preserving Krylov subspace methods forsymplectic, Hamiltonian, and skew-Hamiltonian eigenvalueproblems.
Done!
Freund / Mehrmann (unpublished)
Benner / Fassbender (1997,1998)
Benner / Fassbender / W (1988,1999)
Mehrmann / W (2001)
W (2003)
March 2007 – p.3
Objective
Produce structure-preserving Krylov subspace methods forsymplectic, Hamiltonian, and skew-Hamiltonian eigenvalueproblems. Done!
Freund / Mehrmann (unpublished)
Benner / Fassbender (1997,1998)
Benner / Fassbender / W (1988,1999)
Mehrmann / W (2001)
W (2003)
March 2007 – p.3
Objective
Produce structure-preserving Krylov subspace methods forsymplectic, Hamiltonian, and skew-Hamiltonian eigenvalueproblems. Done!
Freund / Mehrmann (unpublished)
Benner / Fassbender (1997,1998)
Benner / Fassbender / W (1988,1999)
Mehrmann / W (2001)
W (2003)
March 2007 – p.3
Objective
Produce structure-preserving Krylov subspace methods forsymplectic, Hamiltonian, and skew-Hamiltonian eigenvalueproblems. Done!
Freund / Mehrmann (unpublished)
Benner / Fassbender (1997,1998)
Benner / Fassbender / W (1988,1999)
Mehrmann / W (2001)
W (2003)
March 2007 – p.3
Objective
Produce structure-preserving Krylov subspace methods forsymplectic, Hamiltonian, and skew-Hamiltonian eigenvalueproblems. Done!
Freund / Mehrmann (unpublished)
Benner / Fassbender (1997,1998)
Benner / Fassbender / W (1988,1999)
Mehrmann / W (2001)
W (2003)
March 2007 – p.3
Objective
Produce structure-preserving Krylov subspace methods forsymplectic, Hamiltonian, and skew-Hamiltonian eigenvalueproblems. Done!
Freund / Mehrmann (unpublished)
Benner / Fassbender (1997,1998)
Benner / Fassbender / W (1988,1999)
Mehrmann / W (2001)
W (2003)
March 2007 – p.3
Objective
Produce structure-preserving Krylov subspace methods forsymplectic, Hamiltonian, and skew-Hamiltonian eigenvalueproblems. Done!
Freund / Mehrmann (unpublished)
Benner / Fassbender (1997,1998)
Benner / Fassbender / W (1988,1999)
Mehrmann / W (2001)
W (2003)
March 2007 – p.3
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
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
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
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
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
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
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
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
omitting simple formulas for the coefficients
|γj | = |βj |
March 2007 – p.5
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
omitting simple formulas for the coefficients
|γj | = |βj |
March 2007 – p.5
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
omitting simple formulas for the coefficients
|γj | = |βj |
March 2007 – p.5
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
omitting simple formulas for the coefficients
|γj | = |βj |
March 2007 – p.5
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
omitting simple formulas for the coefficients
|γj | = |βj |
March 2007 – p.5
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
omitting simple formulas for the coefficients
|γj | = |βj |
March 2007 – p.5
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
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
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
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
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
Tj =
a1 b1
b1 a2 b2
b2 a3
. . .. . . . . . bj−1
bj−1 aj
Dj =
d1
d2
d3
. . .
dj
March 2007 – p.7
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+1bjdjeTj
ATWj = WjDjTj + wj+1dj+1bjeTj .
Implicit restarts: Filter using HR algorithm
Grimme / Sorensen / Van Dooren (1996)
March 2007 – p.8
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+1bjdjeTj
ATWj = WjDjTj + wj+1dj+1bjeTj .
Implicit restarts: Filter using HR algorithm
Grimme / Sorensen / Van Dooren (1996)
March 2007 – p.8
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+1bjdjeTj
ATWj = WjDjTj + wj+1dj+1bjeTj .
Implicit restarts: Filter using HR algorithm
Grimme / Sorensen / Van Dooren (1996)
March 2007 – p.8
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+1bjdjeTj
ATWj = WjDjTj + wj+1dj+1bjeTj .
Implicit restarts:
Filter using HR algorithm
Grimme / Sorensen / Van Dooren (1996)
March 2007 – p.8
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+1bjdjeTj
ATWj = WjDjTj + wj+1dj+1bjeTj .
Implicit restarts: Filter using HR algorithm
Grimme / Sorensen / Van Dooren (1996)
March 2007 – p.8
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+1bjdjeTj
ATWj = WjDjTj + wj+1dj+1bjeTj .
Implicit restarts: Filter using HR algorithm
Grimme / Sorensen / Van Dooren (1996)
March 2007 – p.8
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 ofa symplectic matrix.
Isotropy!
March 2007 – p.9
Preservation of StructureA = 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 ofa symplectic matrix.
Isotropy!
March 2007 – p.9
Preservation of StructureA = 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 ofa symplectic matrix.
Isotropy!
March 2007 – p.9
Preservation of StructureA = 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 ofa symplectic matrix.
Isotropy!
March 2007 – p.9
Preservation of StructureA = 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 ofa symplectic matrix.
Isotropy!
March 2007 – p.9
Preservation of StructureA = 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 ofa symplectic matrix.
Isotropy!
March 2007 – p.9
Preservation of StructureA = 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 ofa symplectic matrix.
Isotropy!
March 2007 – p.9
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
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
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
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
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
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
Theorem: If B is skew Hamiltonian, then every Krylovsubspace
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
Theorem: If B is skew Hamiltonian, then every Krylovsubspace
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
Theorem: If B is skew Hamiltonian, then every Krylovsubspace
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
Skew-Hamiltonian Lanczos Process
uj+1bjdj = Buj − ujajdj − uj−1bj−1dj
wj+1dj+1bj = BT wj − wjdjaj − wj−1dj−1bj−1
UTj JUj = 0, W T
j JWj = 0, UTj Wj = I
Let vk = −Jwk (So Wj = JVj)
(JB)T = −JB ⇒ −JBT = −BJ
March 2007 – p.12
Skew-Hamiltonian Lanczos Processuj+1bjdj = Buj − ujajdj − uj−1bj−1dj
wj+1dj+1bj = BT wj − wjdjaj − wj−1dj−1bj−1
UTj JUj = 0, W T
j JWj = 0, UTj Wj = I
Let vk = −Jwk (So Wj = JVj)
(JB)T = −JB ⇒ −JBT = −BJ
March 2007 – p.12
Skew-Hamiltonian Lanczos Processuj+1bjdj = Buj − ujajdj − uj−1bj−1dj
wj+1dj+1bj = BT wj − wjdjaj − wj−1dj−1bj−1
UTj JUj = 0, W T
j JWj = 0, UTj Wj = I
Let vk = −Jwk (So Wj = JVj)
(JB)T = −JB ⇒ −JBT = −BJ
March 2007 – p.12
Skew-Hamiltonian Lanczos Processuj+1bjdj = Buj − ujajdj − uj−1bj−1dj
wj+1dj+1bj = BT wj − wjdjaj − wj−1dj−1bj−1
UTj JUj = 0, W T
j JWj = 0, UTj Wj = I
Let vk = −Jwk (So Wj = JVj)
(JB)T = −JB ⇒ −JBT = −BJ
March 2007 – p.12
Skew-Hamiltonian Lanczos Processuj+1bjdj = Buj − ujajdj − uj−1bj−1dj
wj+1dj+1bj = BT wj − wjdjaj − wj−1dj−1bj−1
UTj JUj = 0, W T
j JWj = 0, UTj Wj = I
Let vk = −Jwk (So Wj = JVj)
(JB)T = −JB ⇒ −JBT = −BJ
March 2007 – p.12
Skew-Hamiltonian Lanczos Processuj+1bjdj = Buj − ujajdj − uj−1bj−1dj
vj+1dj+1bj = Bvj − vjdjaj − vj−1dj−1bj−1
Start with uT1 Jv1 = 1.
UTj JUj = 0, V T
j JVj = 0, UTj JVj = I
These are the columns of a symplectic matrix.
March 2007 – p.13
Skew-Hamiltonian Lanczos Processuj+1bjdj = Buj − ujajdj − uj−1bj−1dj
vj+1dj+1bj = Bvj − vjdjaj − vj−1dj−1bj−1
Start with uT1 Jv1 = 1.
UTj JUj = 0, V T
j JVj = 0, UTj JVj = I
These are the columns of a symplectic matrix.
March 2007 – p.13
Skew-Hamiltonian Lanczos Processuj+1bjdj = Buj − ujajdj − uj−1bj−1dj
vj+1dj+1bj = Bvj − vjdjaj − vj−1dj−1bj−1
Start with uT1 Jv1 = 1.
UTj JUj = 0, V T
j JVj = 0, UTj JVj = I
These are the columns of a symplectic matrix.
March 2007 – p.13
Skew-Hamiltonian Lanczos Processuj+1bjdj = Buj − ujajdj − uj−1bj−1dj
vj+1dj+1bj = Bvj − vjdjaj − vj−1dj−1bj−1
Start with uT1 Jv1 = 1.
UTj JUj = 0, V T
j JVj = 0, UTj JVj = I
These are the columns of a symplectic matrix.
March 2007 – p.13
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 = H2vj − vjdjaj − vj−1dj−1bj−1
March 2007 – p.14
Hamiltonian Lanczos ProcessH 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 = H2vj − vjdjaj − vj−1dj−1bj−1
March 2007 – p.14
Hamiltonian Lanczos ProcessH 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 = H2vj − vjdjaj − vj−1dj−1bj−1
March 2007 – p.14
Hamiltonian Lanczos ProcessH 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 = H2vj − vjdjaj − vj−1dj−1bj−1
March 2007 – p.14
Hamiltonian Lanczos ProcessH 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 = H2vj − vjdjaj − vj−1dj−1bj−1
March 2007 – p.14
Hamiltonian Lanczos ProcessH 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 = H2vj − vjdjaj − vj−1dj−1bj−1
March 2007 – p.14
Hamiltonian Lanczos ProcessH 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 = H2vj − vjdjaj − vj−1dj−1bj−1
March 2007 – p.14
Hamiltonian Lanczos Processuj+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 thesupplementary condition
vjdj = Huj .
March 2007 – p.15
Hamiltonian Lanczos Processuj+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 thesupplementary condition
vjdj = Huj .
March 2007 – p.15
Hamiltonian Lanczos Processuj+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 thesupplementary condition
vjdj = Huj .
March 2007 – p.15
Hamiltonian Lanczos Processuj+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 thesupplementary condition
vjdj = Huj .
March 2007 – p.15
Hamiltonian Lanczos Processuj+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 thesupplementary condition
vjdj = Huj .
March 2007 – p.15
Hamiltonian Lanczos Processuj+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 thesupplementary condition
vjdj = Huj .
March 2007 – p.15
Hamiltonian Lanczos Processuj+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 thesupplementary condition
vjdj = Huj .
March 2007 – p.15
Hamiltonian Lanczos Processuj+1bjdj = H2uj − ujajdj − uj−1bj−1dj
vj+1dj+1 = Huj+1
uj+1bj = Hvj − ujaj − uj−1bj−1
vj+1dj+1 = Huj+1
Start with v1d1 = Hu1 uT1 Jv1 = 1.
This is easy to arrange.
March 2007 – p.16
Hamiltonian Lanczos Processuj+1bjdj = H2uj − ujajdj − uj−1bj−1dj
vj+1dj+1 = Huj+1
uj+1bj = Hvj − ujaj − uj−1bj−1
vj+1dj+1 = Huj+1
Start with v1d1 = Hu1 uT1 Jv1 = 1.
This is easy to arrange.
March 2007 – p.16
Hamiltonian Lanczos Processuj+1bjdj = Hvjdj − ujajdj − uj−1bj−1dj
vj+1dj+1 = Huj+1
uj+1bj = Hvj − ujaj − uj−1bj−1
vj+1dj+1 = Huj+1
Start with v1d1 = Hu1 uT1 Jv1 = 1.
This is easy to arrange.
March 2007 – p.16
Hamiltonian Lanczos Processuj+1bjdj = Hvjdj − ujajdj − uj−1bj−1dj
vj+1dj+1 = Huj+1
uj+1bj = Hvj − ujaj − uj−1bj−1
vj+1dj+1 = Huj+1
Start with v1d1 = Hu1 uT1 Jv1 = 1.
This is easy to arrange.
March 2007 – p.16
Hamiltonian Lanczos Processuj+1bj = Hvj − ujaj − uj−1bj−1
vj+1dj+1 = Huj+1
Start with v1d1 = Hu1 uT1 Jv1 = 1.
This is easy to arrange.
March 2007 – p.16
Hamiltonian Lanczos Processuj+1bj = Hvj − ujaj − uj−1bj−1
vj+1dj+1 = Huj+1
Start with v1d1 = Hu1 uT1 Jv1 = 1.
This is easy to arrange.
March 2007 – p.16
Hamiltonian Lanczos Processuj+1bj = Hvj − ujaj − uj−1bj−1
vj+1dj+1 = Huj+1
Start with v1d1 = Hu1 uT1 Jv1 = 1.
This is easy to arrange.
March 2007 – p.16
Recurrences written as matrixproducts
uj+1bj = Hvj − ujaj − uj−1bj−1
vj+1dj+1 = Huj+1
H[
Uj Vj
]=
[Uj Vj
] [Tj
Dj
]+ uj+1bje
T2j .
Implicit restarts: Filter with the HR algorithm.
Next up: symplectic Lanczos process.
March 2007 – p.17
Recurrences written as matrixproducts
uj+1bj = Hvj − ujaj − uj−1bj−1
vj+1dj+1 = Huj+1
HVj = UjTj + uj+1bjeTj , HUj = VjDj
H[
Uj Vj
]=
[Uj Vj
] [Tj
Dj
]+ uj+1bje
T2j .
Implicit restarts: Filter with the HR algorithm.
Next up: symplectic Lanczos process.
March 2007 – p.17
Recurrences written as matrixproducts
uj+1bj = Hvj − ujaj − uj−1bj−1
vj+1dj+1 = Huj+1
H[
Uj Vj
]=
[Uj Vj
] [Tj
Dj
]+ uj+1bje
T2j .
Implicit restarts: Filter with the HR algorithm.
Next up: symplectic Lanczos process.
March 2007 – p.17
Recurrences written as matrixproducts
uj+1bj = Hvj − ujaj − uj−1bj−1
vj+1dj+1 = Huj+1
H[
Uj Vj
]=
[Uj Vj
] [Tj
Dj
]+ uj+1bje
T2j .
Implicit restarts:
Filter with the HR algorithm.
Next up: symplectic Lanczos process.
March 2007 – p.17
Recurrences written as matrixproducts
uj+1bj = Hvj − ujaj − uj−1bj−1
vj+1dj+1 = Huj+1
H[
Uj Vj
]=
[Uj Vj
] [Tj
Dj
]+ uj+1bje
T2j .
Implicit restarts: Filter with the HR algorithm.
Next up: symplectic Lanczos process.
March 2007 – p.17
Recurrences written as matrixproducts
uj+1bj = Hvj − ujaj − uj−1bj−1
vj+1dj+1 = Huj+1
H[
Uj Vj
]=
[Uj Vj
] [Tj
Dj
]+ uj+1bje
T2j .
Implicit restarts: Filter with the HR algorithm.
Next up: symplectic Lanczos process.
March 2007 – p.17
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
Symplectic Lanczos ProcessS 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
Symplectic Lanczos ProcessS 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
Symplectic Lanczos ProcessS 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
Symplectic Lanczos ProcessS 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
Symplectic Lanczos ProcessS 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
Symplectic Lanczos ProcessS 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
Symplectic Lanczos Processuj+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 thesupplementary condition
vjdj = S−1uj .
March 2007 – p.19
Symplectic Lanczos Processuj+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 thesupplementary condition
vjdj = S−1uj .
March 2007 – p.19
Symplectic Lanczos Processuj+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 thesupplementary condition
vjdj = S−1uj .
March 2007 – p.19
Symplectic Lanczos Processuj+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 thesupplementary condition
vjdj = S−1uj .
March 2007 – p.19
Symplectic Lanczos Processuj+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 thesupplementary condition
vjdj = S−1uj .
March 2007 – p.19
Symplectic Lanczos Processuj+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 thesupplementary condition
vjdj = S−1uj .
March 2007 – p.19
Symplectic Lanczos Processuj+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 thesupplementary condition
vjdj = S−1uj .
March 2007 – p.19
Symplectic Lanczos Processuj+1bjdj = (S + S−1)uj − ujajdj − uj−1bj−1dj
vj+1dj+1 = S−1uj+1
uj+1bj = Sujdj + vj − ujaj − uj−1bj−1
vj+1dj+1 = S−1uj+1
Start with v1d1 = S−1u1 uT1 Jv1 = 1.
This is easy to arrange.
March 2007 – p.20
Symplectic Lanczos Processuj+1bjdj = (S + S−1)uj − ujajdj − uj−1bj−1dj
vj+1dj+1 = S−1uj+1
uj+1bj = Sujdj + vj − ujaj − uj−1bj−1
vj+1dj+1 = S−1uj+1
Start with v1d1 = S−1u1 uT1 Jv1 = 1.
This is easy to arrange.
March 2007 – p.20
Symplectic Lanczos Processuj+1bjdj = Suj + vjdj − ujajdj − uj−1bj−1dj
vj+1dj+1 = S−1uj+1
uj+1bj = Sujdj + vj − ujaj − uj−1bj−1
vj+1dj+1 = S−1uj+1
Start with v1d1 = S−1u1 uT1 Jv1 = 1.
This is easy to arrange.
March 2007 – p.20
Symplectic Lanczos Processuj+1bjdj = Suj + vjdj − ujajdj − uj−1bj−1dj
vj+1dj+1 = S−1uj+1
uj+1bj = Sujdj + vj − ujaj − uj−1bj−1
vj+1dj+1 = S−1uj+1
Start with v1d1 = S−1u1 uT1 Jv1 = 1.
This is easy to arrange.
March 2007 – p.20
Symplectic Lanczos Processuj+1bj = Sujdj + vj − ujaj − uj−1bj−1
vj+1dj+1 = S−1uj+1
Start with v1d1 = S−1u1 uT1 Jv1 = 1.
This is easy to arrange.
March 2007 – p.20
Symplectic Lanczos Processuj+1bj = Sujdj + vj − ujaj − uj−1bj−1
vj+1dj+1 = S−1uj+1
Start with v1d1 = S−1u1 uT1 Jv1 = 1.
This is easy to arrange.
March 2007 – p.20
Symplectic Lanczos Processuj+1bj = Sujdj + vj − ujaj − uj−1bj−1
vj+1dj+1 = S−1uj+1
Start with v1d1 = S−1u1 uT1 Jv1 = 1.
This is easy to arrange.
March 2007 – p.20
Recurrences written as matrixproducts
uj+1bj = Sujdj + vj − ujaj − uj−1bj−1
vj+1dj+1 = S−1uj+1
S[
Uj Vj
]=
[Uj Vj
] [TjDj Dj
−Dj 0
]+ uj+1bj+1dj+1e
Tj
Implicit restarts: Filter with the HR algorithm.
That’s it!
March 2007 – p.21
Recurrences written as matrixproducts
uj+1bj = Sujdj + vj − ujaj − uj−1bj−1
vj+1dj+1 = S−1uj+1
SUj = Uj(TjDj) + uj+1bj+1dj+1eTj − VjDj , SVj = UjDj
S[
Uj Vj
]=
[Uj Vj
] [TjDj Dj
−Dj 0
]+ uj+1bj+1dj+1e
Tj
Implicit restarts: Filter with the HR algorithm.
That’s it!
March 2007 – p.21
Recurrences written as matrixproducts
uj+1bj = Sujdj + vj − ujaj − uj−1bj−1
vj+1dj+1 = S−1uj+1
S[
Uj Vj
]=
[Uj Vj
] [TjDj Dj
−Dj 0
]+ uj+1bj+1dj+1e
Tj
Implicit restarts: Filter with the HR algorithm.
That’s it!
March 2007 – p.21
Recurrences written as matrixproducts
uj+1bj = Sujdj + vj − ujaj − uj−1bj−1
vj+1dj+1 = S−1uj+1
S[
Uj Vj
]=
[Uj Vj
] [TjDj Dj
−Dj 0
]+ uj+1bj+1dj+1e
Tj
Implicit restarts:
Filter with the HR algorithm.
That’s it!
March 2007 – p.21
Recurrences written as matrixproducts
uj+1bj = Sujdj + vj − ujaj − uj−1bj−1
vj+1dj+1 = S−1uj+1
S[
Uj Vj
]=
[Uj Vj
] [TjDj Dj
−Dj 0
]+ uj+1bj+1dj+1e
Tj
Implicit restarts: Filter with the HR algorithm.
That’s it!
March 2007 – p.21
Conclusions
Since Krylov subspaces associated withskew-Hamiltonian matrices are isotropic, . . .
. . . the unsymmetric Lanczos process automaticallypreserves skew-Hamiltonian structure.
From the skew-Hamiltonian Lanczos process we easilyderive . . .
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
ConclusionsSince Krylov subspaces associated withskew-Hamiltonian matrices are isotropic, . . .
. . . the unsymmetric Lanczos process automaticallypreserves skew-Hamiltonian structure.
From the skew-Hamiltonian Lanczos process we easilyderive . . .
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
ConclusionsSince Krylov subspaces associated withskew-Hamiltonian matrices are isotropic, . . .
. . . the unsymmetric Lanczos process automaticallypreserves skew-Hamiltonian structure.
From the skew-Hamiltonian Lanczos process we easilyderive . . .
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
ConclusionsSince Krylov subspaces associated withskew-Hamiltonian matrices are isotropic, . . .
. . . the unsymmetric Lanczos process automaticallypreserves skew-Hamiltonian structure.
From the skew-Hamiltonian Lanczos process we easilyderive . . .
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
ConclusionsSince Krylov subspaces associated withskew-Hamiltonian matrices are isotropic, . . .
. . . the unsymmetric Lanczos process automaticallypreserves skew-Hamiltonian structure.
From the skew-Hamiltonian Lanczos process we easilyderive . . .
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
ConclusionsSince Krylov subspaces associated withskew-Hamiltonian matrices are isotropic, . . .
. . . the unsymmetric Lanczos process automaticallypreserves skew-Hamiltonian structure.
From the skew-Hamiltonian Lanczos process we easilyderive . . .
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
ConclusionsSince Krylov subspaces associated withskew-Hamiltonian matrices are isotropic, . . .
. . . the unsymmetric Lanczos process automaticallypreserves skew-Hamiltonian structure.
From the skew-Hamiltonian Lanczos process we easilyderive . . .
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
ConclusionsSince Krylov subspaces associated withskew-Hamiltonian matrices are isotropic, . . .
. . . the unsymmetric Lanczos process automaticallypreserves skew-Hamiltonian structure.
From the skew-Hamiltonian Lanczos process we easilyderive . . .
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
ConclusionsSince Krylov subspaces associated withskew-Hamiltonian matrices are isotropic, . . .
. . . the unsymmetric Lanczos process automaticallypreserves skew-Hamiltonian structure.
From the skew-Hamiltonian Lanczos process we easilyderive . . .
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
ConclusionsSince Krylov subspaces associated withskew-Hamiltonian matrices are isotropic, . . .
. . . the unsymmetric Lanczos process automaticallypreserves skew-Hamiltonian structure.
From the skew-Hamiltonian Lanczos process we easilyderive . . .
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
ConclusionsSince Krylov subspaces associated withskew-Hamiltonian matrices are isotropic, . . .
. . . the unsymmetric Lanczos process automaticallypreserves skew-Hamiltonian structure.
From the skew-Hamiltonian Lanczos process we easilyderive . . .
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