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The implicit QZ algorithm for thepalindromic eigenvalue problem
David S. [email protected]
Department of Mathematics
Washington State University
Luminy, October 2007 – p. 1
Context
Luminy, October 2007 – p. 2
ContextLQG problem in discrete time (optimal control)
Luminy, October 2007 – p. 2
ContextLQG problem in discrete time (optimal control)
⇒ symplectic eigenvalue problem
Luminy, October 2007 – p. 2
ContextLQG problem in discrete time (optimal control)
⇒ symplectic eigenvalue problem
eigenvalue symmetry:λ, λ−1
Luminy, October 2007 – p. 2
ContextLQG problem in discrete time (optimal control)
⇒ symplectic eigenvalue problem
eigenvalue symmetry:λ, λ−1
−2 −1.5 −1 −0.5 0 0.5 1 1.5 2−2
−1.5
−1
−0.5
0
0.5
1
1.5
2
Luminy, October 2007 – p. 2
ContextLQG problem in discrete time (optimal control)
⇒ symplectic eigenvalue problem
eigenvalue symmetry:λ, λ−1
−2 −1.5 −1 −0.5 0 0.5 1 1.5 2−2
−1.5
−1
−0.5
0
0.5
1
1.5
2
want stable invariant subspaceLuminy, October 2007 – p. 2
Palindromic Eigenvalue Problem
Luminy, October 2007 – p. 3
Palindromic Eigenvalue ProblemMackey / Mackey / Mehl / Mehrmann
Luminy, October 2007 – p. 3
Palindromic Eigenvalue ProblemMackey / Mackey / Mehl / Mehrmann
(C − λCT )x = 0
Luminy, October 2007 – p. 3
Palindromic Eigenvalue ProblemMackey / Mackey / Mehl / Mehrmann
(C − λCT )x = 0
generalized eigenvalue problemA − λB
Luminy, October 2007 – p. 3
Palindromic Eigenvalue ProblemMackey / Mackey / Mehl / Mehrmann
(C − λCT )x = 0
generalized eigenvalue problemA − λB
B = AT
Luminy, October 2007 – p. 3
Palindromic Eigenvalue ProblemMackey / Mackey / Mehl / Mehrmann
(C − λCT )x = 0
generalized eigenvalue problemA − λB
B = AT
(C − λCT )x = 0 ⇔ xT (CT− λC) = 0
Luminy, October 2007 – p. 3
Palindromic Eigenvalue ProblemMackey / Mackey / Mehl / Mehrmann
(C − λCT )x = 0
generalized eigenvalue problemA − λB
B = AT
(C − λCT )x = 0 ⇔ xT (CT− λC) = 0
eigenvalue symmetry:λ, λ−1
Luminy, October 2007 – p. 3
Palindromic Eigenvalue ProblemMackey / Mackey / Mehl / Mehrmann
(C − λCT )x = 0
generalized eigenvalue problemA − λB
B = AT
(C − λCT )x = 0 ⇔ xT (CT− λC) = 0
eigenvalue symmetry:λ, λ−1
Formulate control problems as palindromiceigenvalue problems.
Luminy, October 2007 – p. 3
QR-like algorithms
Luminy, October 2007 – p. 4
QR-like algorithmsfor palindromic problems
Luminy, October 2007 – p. 4
QR-like algorithmsfor palindromic problems
C. Schröder (Berlin)
Luminy, October 2007 – p. 4
QR-like algorithmsfor palindromic problems
C. Schröder (Berlin)
explicit QR-like algorithm
Luminy, October 2007 – p. 4
QR-like algorithmsfor palindromic problems
C. Schröder (Berlin)
explicit QR-like algorithm (It works!)
Luminy, October 2007 – p. 4
QR-like algorithmsfor palindromic problems
C. Schröder (Berlin)
explicit QR-like algorithm (It works!)
implicit QR-like algorithm(bulge chase)
Luminy, October 2007 – p. 4
QR-like algorithmsfor palindromic problems
C. Schröder (Berlin)
explicit QR-like algorithm (It works!)
implicit QR-like algorithm(bulge chase) (It works!)
Luminy, October 2007 – p. 4
QR-like algorithmsfor palindromic problems
C. Schröder (Berlin)
explicit QR-like algorithm (It works!)
implicit QR-like algorithm(bulge chase) (It works!)
Are they equivalent?
Luminy, October 2007 – p. 4
The Bulge Chase
Luminy, October 2007 – p. 5
The Bulge Chasecompact form needed
Luminy, October 2007 – p. 5
The Bulge Chasecompact form needed
C =
0 ∗
0 ∗ ∗
0 ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗
Luminy, October 2007 – p. 5
CT =
∗ ∗
∗ ∗ ∗
∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗
0 ∗ ∗ ∗ ∗ ∗
0 ∗ ∗ ∗ ∗ ∗ ∗
0 ∗ ∗ ∗ ∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗
Luminy, October 2007 – p. 6
CT =
∗ ∗
∗ ∗ ∗
∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗
0 ∗ ∗ ∗ ∗ ∗
0 ∗ ∗ ∗ ∗ ∗ ∗
0 ∗ ∗ ∗ ∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗
not always attainable!
Luminy, October 2007 – p. 6
The Bulge Chase
Luminy, October 2007 – p. 7
The Bulge Chasesingle-shift case
Luminy, October 2007 – p. 7
The Bulge Chasesingle-shift case (for simplicity)
Luminy, October 2007 – p. 7
The Bulge Chasesingle-shift case (for simplicity)
Pick a shiftµ.
Luminy, October 2007 – p. 7
The Bulge Chasesingle-shift case (for simplicity)
Pick a shiftµ.
x = (C − µCT )e1 =
0...0
α
β
Luminy, October 2007 – p. 7
The Bulge Chasesingle-shift case (for simplicity)
Pick a shiftµ.
x = (C − µCT )e1 =
0...0
α
β
Build a Givens rotation (for example) thatannihilatesα.
Luminy, October 2007 – p. 7
The Bulge Chase
∗
∗ ∗
∗ ∗ ∗
∗ ∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗
Luminy, October 2007 – p. 8
The Bulge Chase
∗
∗ ∗
∗ ∗ ∗
∗ ∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗
Luminy, October 2007 – p. 9
The Bulge Chase
∗
∗ ∗
∗ ∗ ∗
∗ ∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗
Luminy, October 2007 – p. 10
The Bulge Chase
∗ ∗
∗ ∗
∗ ∗ ∗
∗ ∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗
Luminy, October 2007 – p. 11
The Bulge Chase
∗ ∗
∗ ∗
∗ ∗ ∗
∗ ∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗
Luminy, October 2007 – p. 12
The Bulge Chase
∗
∗ ∗
∗ ∗ ∗
∗ ∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗
Luminy, October 2007 – p. 13
The Bulge Chase
∗
∗ ∗
∗ ∗ ∗
∗ ∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗
Luminy, October 2007 – p. 14
The Bulge Chase
∗
∗ ∗
∗ ∗ ∗
∗ ∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗
Luminy, October 2007 – p. 15
The Bulge Chase
∗
∗ ∗
∗ ∗ ∗
∗ ∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗
Luminy, October 2007 – p. 16
The Bulge Chase
∗
∗ ∗
∗ ∗ ∗
∗ ∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗
Luminy, October 2007 – p. 17
The Bulge Chase
∗
∗ ∗ ∗
∗ ∗ ∗
∗ ∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗
Luminy, October 2007 – p. 18
The Bulge Chase
∗
∗ ∗
∗ ∗ ∗
∗ ∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗
Luminy, October 2007 – p. 19
The Bulge Chase
∗
∗ ∗
∗ ∗ ∗
∗ ∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗
But what happens when we get to the middle?Luminy, October 2007 – p. 20
Bulge Chase in the Pencil
C − λCT =
∗ ∗
∗ ∗ ∗
∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗
Luminy, October 2007 – p. 21
Bulge Chase in the Pencil
C − λCT =
∗ ∗ ∗
∗ ∗ ∗
∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗
Luminy, October 2007 – p. 22
Bulge Chase in the Pencil
C − λCT =
∗ ∗
∗ ∗ ∗ ∗
∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗
Luminy, October 2007 – p. 23
Bulge pencils on a collision course!
Luminy, October 2007 – p. 24
Bulge Chase in the Pencil
C − λCT =
∗ ∗
∗ ∗ ∗ ∗
∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗
Luminy, October 2007 – p. 25
Half a step further:
C − λCT =
∗ ∗
∗ ∗ ∗
∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗
Luminy, October 2007 – p. 26
Swap the shifts
C − λCT =
∗ ∗
∗ ∗ ∗
∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗
Luminy, October 2007 – p. 27
Swap the shifts
C − λCT =
∗ ∗
∗ ∗ ∗
∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗
Luminy, October 2007 – p. 28
Reverse the process
C − λCT =
∗ ∗
∗ ∗ ∗ ∗
∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗
Luminy, October 2007 – p. 29
Reverse the process
C − λCT =
∗ ∗ ∗
∗ ∗ ∗
∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗
Luminy, October 2007 – p. 30
Reverse the process
C − λCT =
∗ ∗
∗ ∗ ∗
∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗
Luminy, October 2007 – p. 31
Bulge Chase is complete
C − λCT =
∗ ∗
∗ ∗ ∗
∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗
Luminy, October 2007 – p. 32
This works very well.
Luminy, October 2007 – p. 33
This works very well.
Question:
Luminy, October 2007 – p. 33
This works very well.
Question: How do we explain it?
Luminy, October 2007 – p. 33
This works very well.
Question: How do we explain it?
Answer:
Luminy, October 2007 – p. 33
This works very well.
Question: How do we explain it?
Answer: I don’t have time . . .
Luminy, October 2007 – p. 33
This works very well.
Question: How do we explain it?
Answer: I don’t have time . . .
. . . but I’ll try.
Luminy, October 2007 – p. 33
This works very well.
Question: How do we explain it?
Answer: I don’t have time . . .
. . . but I’ll try.
Compare with standard QZ.
Luminy, October 2007 – p. 33
QZ algorithm,
Luminy, October 2007 – p. 34
QZ algorithm, implicit version
Luminy, October 2007 – p. 34
QZ algorithm, implicit versionA − λB (Hessenberg, triangular)
Luminy, October 2007 – p. 34
QZ algorithm, implicit versionA − λB (Hessenberg, triangular)
pick shiftsµ1, . . .µm
Luminy, October 2007 – p. 34
QZ algorithm, implicit versionA − λB (Hessenberg, triangular)
pick shiftsµ1, . . .µm
p(z) = (z − µ1) · · · (z − µm)
Luminy, October 2007 – p. 34
QZ algorithm, implicit versionA − λB (Hessenberg, triangular)
pick shiftsµ1, . . .µm
p(z) = (z − µ1) · · · (z − µm)
x = p(AB−1)e1
Luminy, October 2007 – p. 34
QZ algorithm, implicit versionA − λB (Hessenberg, triangular)
pick shiftsµ1, . . .µm
p(z) = (z − µ1) · · · (z − µm)
x = p(AB−1)e1
Make a bulge,
Luminy, October 2007 – p. 34
QZ algorithm, implicit versionA − λB (Hessenberg, triangular)
pick shiftsµ1, . . .µm
p(z) = (z − µ1) · · · (z − µm)
x = p(AB−1)e1
Make a bulge, then chase it.
Luminy, October 2007 – p. 34
QZ algorithm, implicit versionA − λB (Hessenberg, triangular)
pick shiftsµ1, . . .µm
p(z) = (z − µ1) · · · (z − µm)
x = p(AB−1)e1
Make a bulge, then chase it.
GetA − λB
Luminy, October 2007 – p. 34
QZ algorithm,
Luminy, October 2007 – p. 35
QZ algorithm, explicit version
Luminy, October 2007 – p. 35
QZ algorithm, explicit versionp(AB−1) = QR
Luminy, October 2007 – p. 35
QZ algorithm, explicit versionp(AB−1) = QR p(B−1A) = ZR
Luminy, October 2007 – p. 35
QZ algorithm, explicit versionp(AB−1) = QR p(B−1A) = ZR
A = Q∗AZ, B = Q∗BZ
Luminy, October 2007 – p. 35
QZ algorithm, explicit versionp(AB−1) = QR p(B−1A) = ZR
A = Q∗AZ, B = Q∗BZ
Explicit QZ step is complete.
Luminy, October 2007 – p. 35
QZ algorithm, explicit versionp(AB−1) = QR p(B−1A) = ZR
A = Q∗AZ, B = Q∗BZ
Explicit QZ step is complete.
AB−1 = Q∗(AB−1)Q
Luminy, October 2007 – p. 35
QZ algorithm, explicit versionp(AB−1) = QR p(B−1A) = ZR
A = Q∗AZ, B = Q∗BZ
Explicit QZ step is complete.
AB−1 = Q∗(AB−1)Q (QR iteration onAB−1)
Luminy, October 2007 – p. 35
QZ algorithm, explicit versionp(AB−1) = QR p(B−1A) = ZR
A = Q∗AZ, B = Q∗BZ
Explicit QZ step is complete.
AB−1 = Q∗(AB−1)Q (QR iteration onAB−1)
B−1A = Z∗(B−1A)Z
Luminy, October 2007 – p. 35
QZ algorithm, explicit versionp(AB−1) = QR p(B−1A) = ZR
A = Q∗AZ, B = Q∗BZ
Explicit QZ step is complete.
AB−1 = Q∗(AB−1)Q (QR iteration onAB−1)
B−1A = Z∗(B−1A)Z (QR iteration onB−1A)
Luminy, October 2007 – p. 35
Explicit = Implicit?
Luminy, October 2007 – p. 36
Explicit = Implicit?AB−1 andB−1A are upper Hessenberg.
Luminy, October 2007 – p. 36
Explicit = Implicit?AB−1 andB−1A are upper Hessenberg.
Utilize the Hessenberg form.
Luminy, October 2007 – p. 36
Explicit = Implicit?AB−1 andB−1A are upper Hessenberg.
Utilize the Hessenberg form.
implicit-Q theorem, or . . .
Luminy, October 2007 – p. 36
Explicit = Implicit?AB−1 andB−1A are upper Hessenberg.
Utilize the Hessenberg form.
implicit-Q theorem, or . . .
work directly with the Krylov subspaces.
Luminy, October 2007 – p. 36
Explicit = Implicit?AB−1 andB−1A are upper Hessenberg.
Utilize the Hessenberg form.
implicit-Q theorem, or . . .
work directly with the Krylov subspaces.
time permitting . . .
Luminy, October 2007 – p. 36
Back to the palindromic case:
Luminy, October 2007 – p. 37
Back to the palindromic case:Bulge chase givesC = G−1CG−T
Luminy, October 2007 – p. 37
Back to the palindromic case:Bulge chase givesC = G−1CG−T
CC−T = G−1(CC−T )G
Luminy, October 2007 – p. 37
Back to the palindromic case:Bulge chase givesC = G−1CG−T
CC−T = G−1(CC−T )G (similarity)
Luminy, October 2007 – p. 37
Back to the palindromic case:Bulge chase givesC = G−1CG−T
CC−T = G−1(CC−T )G (similarity)
C−T C = GT (C−TC)G−T
Luminy, October 2007 – p. 37
Back to the palindromic case:Bulge chase givesC = G−1CG−T
CC−T = G−1(CC−T )G (similarity)
C−T C = GT (C−TC)G−T (similarity)
Luminy, October 2007 – p. 37
Back to the palindromic case:Bulge chase givesC = G−1CG−T
CC−T = G−1(CC−T )G (similarity)
C−T C = GT (C−TC)G−T (similarity)
Needp(CC−T ) = GR
Luminy, October 2007 – p. 37
Back to the palindromic case:Bulge chase givesC = G−1CG−T
CC−T = G−1(CC−T )G (similarity)
C−T C = GT (C−TC)G−T (similarity)
Needp(CC−T ) = GR andp(C−TC) = G−TR
Luminy, October 2007 – p. 37
Back to the palindromic case:Bulge chase givesC = G−1CG−T
CC−T = G−1(CC−T )G (similarity)
C−T C = GT (C−TC)G−T (similarity)
Needp(CC−T ) = GR andp(C−TC) = G−TR
or something like that.
Luminy, October 2007 – p. 37
What we actually get:
Luminy, October 2007 – p. 38
What we actually get:r(CC−T ) = GL
Luminy, October 2007 – p. 38
What we actually get:r(CC−T ) = GL
wherer(z) =m
∏
k=1
z − µk
µkz − 1
Luminy, October 2007 – p. 38
What we actually get:r(CC−T ) = GL
wherer(z) =m
∏
k=1
z − µk
µkz − 1
L is lower triangular.
Luminy, October 2007 – p. 38
What we actually get:r(CC−T ) = GL
wherer(z) =m
∏
k=1
z − µk
µkz − 1
L is lower triangular.
r(C−TC) = G−TR
Luminy, October 2007 – p. 38
What we actually get:r(CC−T ) = GL
wherer(z) =m
∏
k=1
z − µk
µkz − 1
L is lower triangular.
r(C−TC) = G−TR
r(CC−T )−T = r(C−TC), R = L−T
Luminy, October 2007 – p. 38
Our Approach:
Luminy, October 2007 – p. 39
Our Approach:wantr(CC−T ) = GL
Luminy, October 2007 – p. 39
Our Approach:wantr(CC−T ) = GL
DefineL = G−1r(CC−T )
Luminy, October 2007 – p. 39
Our Approach:wantr(CC−T ) = GL
DefineL = G−1r(CC−T )
Then show thatL is lower triangular.
Luminy, October 2007 – p. 39
Our Approach:wantr(CC−T ) = GL
DefineL = G−1r(CC−T )
Then show thatL is lower triangular.
This is not entirely straightforward.
Luminy, October 2007 – p. 39
Our Approach:wantr(CC−T ) = GL
DefineL = G−1r(CC−T )
Then show thatL is lower triangular.
This is not entirely straightforward.
Utilize the Hessenberg form.
Luminy, October 2007 – p. 39
Recall that
C =
0 ∗
0 ∗ ∗
0 ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗
Luminy, October 2007 – p. 40
This implies
Luminy, October 2007 – p. 41
This implies
CC−T =
∗ ∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗
Luminy, October 2007 – p. 41
This implies
CC−T =
∗ ∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗
partly lower Hessenberg (n/2 − 1 columns)
Luminy, October 2007 – p. 41
and
Luminy, October 2007 – p. 42
and
C−TC =
∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗ ∗
Luminy, October 2007 – p. 42
and
C−TC =
∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗ ∗
∗ ∗ ∗ ∗ ∗ ∗
partly upper Hessenberg (n/2 − 2 columns)
Luminy, October 2007 – p. 42
Use both.
Luminy, October 2007 – p. 43
Use both.
L =
[
L11 L12
L21 L22
]
Luminy, October 2007 – p. 43
Use both.
L =
[
L11 L12
L21 L22
]
UsingCC−T ,
Luminy, October 2007 – p. 43
Use both.
L =
[
L11 L12
L21 L22
]
UsingCC−T , deduceL12 = 0 . . .
Luminy, October 2007 – p. 43
Use both.
L =
[
L11 L12
L21 L22
]
UsingCC−T , deduceL12 = 0 . . .
andL22 is lower triangular.
Luminy, October 2007 – p. 43
L =
[
L11
L21 L22
]
Luminy, October 2007 – p. 44
L =
[
L11
L21 L22
]
just needL11 lower triangular
Luminy, October 2007 – p. 44
L =
[
L11
L21 L22
]
just needL11 lower triangular
R = L−T =
[
R11 R12
R22
]
Luminy, October 2007 – p. 44
L =
[
L11
L21 L22
]
just needL11 lower triangular
R = L−T =
[
R11 R12
R22
]
UsingC−TC,
Luminy, October 2007 – p. 44
L =
[
L11
L21 L22
]
just needL11 lower triangular
R = L−T =
[
R11 R12
R22
]
UsingC−TC, deduce thatR11 is upper triangular.
Luminy, October 2007 – p. 44
L =
[
L11
L21 L22
]
just needL11 lower triangular
R = L−T =
[
R11 R12
R22
]
UsingC−TC, deduce thatR11 is upper triangular.
HenceL11 is lower triangular.
Luminy, October 2007 – p. 44
L =
[
L11
L21 L22
]
just needL11 lower triangular
R = L−T =
[
R11 R12
R22
]
UsingC−TC, deduce thatR11 is upper triangular.
HenceL11 is lower triangular.
done!
Luminy, October 2007 – p. 44
Conclusion:
Luminy, October 2007 – p. 45
Conclusion:
Schröder’s palindromic bulge-chasing algorithm
Luminy, October 2007 – p. 45
Conclusion:
Schröder’s palindromic bulge-chasing algorithmeffectsa combinationQL andQR iteration
Luminy, October 2007 – p. 45
Conclusion:
Schröder’s palindromic bulge-chasing algorithmeffectsa combinationQL andQR iteration driven by a rationalfunction
r(z) =m
∏
k=1
z − µk
µkz − 1
Luminy, October 2007 – p. 45
Conclusion:
Schröder’s palindromic bulge-chasing algorithmeffectsa combinationQL andQR iteration driven by a rationalfunction
r(z) =m
∏
k=1
z − µk
µkz − 1
with shiftsµ1, . . . ,µm in the numerator and shiftsµ−1
1,
. . . ,µ−1
min the denominator.
Luminy, October 2007 – p. 45
Conclusion:
Schröder’s palindromic bulge-chasing algorithmeffectsa combinationQL andQR iteration driven by a rationalfunction
r(z) =m
∏
k=1
z − µk
µkz − 1
with shiftsµ1, . . . ,µm in the numerator and shiftsµ−1
1,
. . . ,µ−1
min the denominator.
That’s all, folks!Luminy, October 2007 – p. 45
