structure-preserving krylov subspace methods for ...structure-preserving krylov subspace methods for...

Structure-preserving Krylovsubspace methods for Hamiltonianand symplectic eigenvalue problems

David S. Watkins

watkins@math.wsu.edu

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