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The implicit QZ algorithm for the palindromic eigenvalue problem David S. Watkins watkins@math.wsu.edu Department of Mathematics Washington State University Luminy, October 2007 p. 1 Context Luminy, October 2007 p. 2 Context LQG

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SLIDE 1

The implicit QZ algorithm for the palindromic eigenvalue problem

David S. Watkins

watkins@math.wsu.edu

Department of Mathematics Washington State University

Luminy, October 2007 – p. 1

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SLIDE 2

Context

Luminy, October 2007 – p. 2

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SLIDE 3

Context

LQG problem in discrete time (optimal control)

Luminy, October 2007 – p. 2

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SLIDE 4

Context

LQG problem in discrete time (optimal control) ⇒ symplectic eigenvalue problem

Luminy, October 2007 – p. 2

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SLIDE 5

Context

LQG problem in discrete time (optimal control) ⇒ symplectic eigenvalue problem eigenvalue symmetry: λ, λ−1

Luminy, October 2007 – p. 2

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SLIDE 6

Context

LQG problem in discrete time (optimal control) ⇒ symplectic eigenvalue problem eigenvalue symmetry: λ, λ−1

−2 −1.5 −1 −0.5 0.5 1 1.5 2 −2 −1.5 −1 −0.5 0.5 1 1.5 2

Luminy, October 2007 – p. 2

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SLIDE 7

Context

LQG problem in discrete time (optimal control) ⇒ symplectic eigenvalue problem eigenvalue symmetry: λ, λ−1

−2 −1.5 −1 −0.5 0.5 1 1.5 2 −2 −1.5 −1 −0.5 0.5 1 1.5 2

want stable invariant subspace

Luminy, October 2007 – p. 2

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SLIDE 8

Palindromic Eigenvalue Problem

Luminy, October 2007 – p. 3

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SLIDE 9

Palindromic Eigenvalue Problem

Mackey / Mackey / Mehl / Mehrmann

Luminy, October 2007 – p. 3

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SLIDE 10

Palindromic Eigenvalue Problem

Mackey / Mackey / Mehl / Mehrmann (C − λCT)x = 0

Luminy, October 2007 – p. 3

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SLIDE 11

Palindromic Eigenvalue Problem

Mackey / Mackey / Mehl / Mehrmann (C − λCT)x = 0 generalized eigenvalue problem A − λB

Luminy, October 2007 – p. 3

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SLIDE 12

Palindromic Eigenvalue Problem

Mackey / Mackey / Mehl / Mehrmann (C − λCT)x = 0 generalized eigenvalue problem A − λB B = AT

Luminy, October 2007 – p. 3

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SLIDE 13

Palindromic Eigenvalue Problem

Mackey / Mackey / Mehl / Mehrmann (C − λCT)x = 0 generalized eigenvalue problem A − λB B = AT (C − λCT)x = 0 ⇔ xT(CT − λC) = 0

Luminy, October 2007 – p. 3

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SLIDE 14

Palindromic Eigenvalue Problem

Mackey / Mackey / Mehl / Mehrmann (C − λCT)x = 0 generalized eigenvalue problem A − λB B = AT (C − λCT)x = 0 ⇔ xT(CT − λC) = 0 eigenvalue symmetry: λ, λ−1

Luminy, October 2007 – p. 3

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SLIDE 15

Palindromic Eigenvalue Problem

Mackey / Mackey / Mehl / Mehrmann (C − λCT)x = 0 generalized eigenvalue problem A − λB B = AT (C − λCT)x = 0 ⇔ xT(CT − λC) = 0 eigenvalue symmetry: λ, λ−1 Formulate control problems as palindromic eigenvalue problems.

Luminy, October 2007 – p. 3

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SLIDE 16

QR-like algorithms

Luminy, October 2007 – p. 4

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SLIDE 17

QR-like algorithms for palindromic problems

Luminy, October 2007 – p. 4

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SLIDE 18

QR-like algorithms for palindromic problems

  • C. Schröder (Berlin)

Luminy, October 2007 – p. 4

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SLIDE 19

QR-like algorithms for palindromic problems

  • C. Schröder (Berlin)

explicit QR-like algorithm

Luminy, October 2007 – p. 4

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SLIDE 20

QR-like algorithms for palindromic problems

  • C. Schröder (Berlin)

explicit QR-like algorithm (It works!)

Luminy, October 2007 – p. 4

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SLIDE 21

QR-like algorithms for palindromic problems

  • C. Schröder (Berlin)

explicit QR-like algorithm (It works!) implicit QR-like algorithm (bulge chase)

Luminy, October 2007 – p. 4

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SLIDE 22

QR-like algorithms for palindromic problems

  • C. Schröder (Berlin)

explicit QR-like algorithm (It works!) implicit QR-like algorithm (bulge chase) (It works!)

Luminy, October 2007 – p. 4

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SLIDE 23

QR-like algorithms for 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

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SLIDE 24

The Bulge Chase

Luminy, October 2007 – p. 5

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SLIDE 25

The Bulge Chase

compact form needed

Luminy, October 2007 – p. 5

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SLIDE 26

The Bulge Chase

compact form needed C =              0 ∗ 0 ∗ ∗ 0 ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗             

Luminy, October 2007 – p. 5

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SLIDE 27

CT =              ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ 0 ∗ ∗ ∗ ∗ ∗ 0 ∗ ∗ ∗ ∗ ∗ ∗ 0 ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗             

Luminy, October 2007 – p. 6

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SLIDE 28

CT =              ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ 0 ∗ ∗ ∗ ∗ ∗ 0 ∗ ∗ ∗ ∗ ∗ ∗ 0 ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗              not always attainable!

Luminy, October 2007 – p. 6

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The Bulge Chase

Luminy, October 2007 – p. 7

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The Bulge Chase

single-shift case

Luminy, October 2007 – p. 7

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SLIDE 31

The Bulge Chase

single-shift case (for simplicity)

Luminy, October 2007 – p. 7

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SLIDE 32

The Bulge Chase

single-shift case (for simplicity) Pick a shift µ.

Luminy, October 2007 – p. 7

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SLIDE 33

The Bulge Chase

single-shift case (for simplicity) Pick a shift µ. x = (C − µCT)e1 =        . . . α β       

Luminy, October 2007 – p. 7

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SLIDE 34

The Bulge Chase

single-shift case (for simplicity) Pick a shift µ. x = (C − µCT)e1 =        . . . α β        Build a Givens rotation (for example) that annihilates α.

Luminy, October 2007 – p. 7

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The Bulge Chase

             ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗             

Luminy, October 2007 – p. 8

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The Bulge Chase

             ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗             

Luminy, October 2007 – p. 9

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SLIDE 37

The Bulge Chase

             ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗             

Luminy, October 2007 – p. 10

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The Bulge Chase

             ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗             

Luminy, October 2007 – p. 11

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SLIDE 39

The Bulge Chase

             ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗             

Luminy, October 2007 – p. 12

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The Bulge Chase

             ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗             

Luminy, October 2007 – p. 13

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The Bulge Chase

             ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗             

Luminy, October 2007 – p. 14

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The Bulge Chase

             ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗             

Luminy, October 2007 – p. 15

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SLIDE 43

The Bulge Chase

             ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗             

Luminy, October 2007 – p. 16

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SLIDE 44

The Bulge Chase

             ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗             

Luminy, October 2007 – p. 17

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SLIDE 45

The Bulge Chase

             ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗             

Luminy, October 2007 – p. 18

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SLIDE 46

The Bulge Chase

             ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗             

Luminy, October 2007 – p. 19

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SLIDE 47

The Bulge Chase

             ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗              But what happens when we get to the middle?

Luminy, October 2007 – p. 20

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SLIDE 48

Bulge Chase in the Pencil

C − λCT =              ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗             

Luminy, October 2007 – p. 21

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SLIDE 49

Bulge Chase in the Pencil

C − λCT =              ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗             

Luminy, October 2007 – p. 22

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Bulge Chase in the Pencil

C − λCT =              ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗             

Luminy, October 2007 – p. 23

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SLIDE 51

Bulge pencils on a collision course!

Luminy, October 2007 – p. 24

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Bulge Chase in the Pencil

C − λCT =              ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗             

Luminy, October 2007 – p. 25

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SLIDE 53

Half a step further:

C − λCT =              ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗             

Luminy, October 2007 – p. 26

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SLIDE 54

Swap the shifts

C − λCT =              ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗             

Luminy, October 2007 – p. 27

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SLIDE 55

Swap the shifts

C − λCT =              ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗             

Luminy, October 2007 – p. 28

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SLIDE 56

Reverse the process

C − λCT =              ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗             

Luminy, October 2007 – p. 29

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SLIDE 57

Reverse the process

C − λCT =              ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗             

Luminy, October 2007 – p. 30

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SLIDE 58

Reverse the process

C − λCT =              ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗             

Luminy, October 2007 – p. 31

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SLIDE 59

Bulge Chase is complete

C − λCT =              ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗             

Luminy, October 2007 – p. 32

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SLIDE 60

This works very well.

Luminy, October 2007 – p. 33

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SLIDE 61

This works very well. Question:

Luminy, October 2007 – p. 33

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SLIDE 62

This works very well. Question: How do we explain it?

Luminy, October 2007 – p. 33

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SLIDE 63

This works very well. Question: How do we explain it? Answer:

Luminy, October 2007 – p. 33

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SLIDE 64

This works very well. Question: How do we explain it? Answer: I don’t have time ...

Luminy, October 2007 – p. 33

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SLIDE 65

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

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SLIDE 66

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

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SLIDE 67

QZ algorithm,

Luminy, October 2007 – p. 34

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SLIDE 68

QZ algorithm, implicit version

Luminy, October 2007 – p. 34

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SLIDE 69

QZ algorithm, implicit version

A − λB (Hessenberg, triangular)

Luminy, October 2007 – p. 34

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SLIDE 70

QZ algorithm, implicit version

A − λB (Hessenberg, triangular) pick shifts µ1, ...µm

Luminy, October 2007 – p. 34

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SLIDE 71

QZ algorithm, implicit version

A − λB (Hessenberg, triangular) pick shifts µ1, ...µm p(z) = (z − µ1) · · · (z − µm)

Luminy, October 2007 – p. 34

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SLIDE 72

QZ algorithm, implicit version

A − λB (Hessenberg, triangular) pick shifts µ1, ...µm p(z) = (z − µ1) · · · (z − µm) x = p(AB−1)e1

Luminy, October 2007 – p. 34

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SLIDE 73

QZ algorithm, implicit version

A − λB (Hessenberg, triangular) pick shifts µ1, ...µm p(z) = (z − µ1) · · · (z − µm) x = p(AB−1)e1 Make a bulge,

Luminy, October 2007 – p. 34

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SLIDE 74

QZ algorithm, implicit version

A − λB (Hessenberg, triangular) pick shifts µ1, ...µm p(z) = (z − µ1) · · · (z − µm) x = p(AB−1)e1 Make a bulge, then chase it.

Luminy, October 2007 – p. 34

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SLIDE 75

QZ algorithm, implicit version

A − λB (Hessenberg, triangular) pick shifts µ1, ...µm p(z) = (z − µ1) · · · (z − µm) x = p(AB−1)e1 Make a bulge, then chase it. Get ˆ A − λ ˆ B

Luminy, October 2007 – p. 34

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SLIDE 76

QZ algorithm,

Luminy, October 2007 – p. 35

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SLIDE 77

QZ algorithm, explicit version

Luminy, October 2007 – p. 35

slide-78
SLIDE 78

QZ algorithm, explicit version

p(AB−1) = QR

Luminy, October 2007 – p. 35

slide-79
SLIDE 79

QZ algorithm, explicit version

p(AB−1) = QR p(B−1A) = Z ˜ R

Luminy, October 2007 – p. 35

slide-80
SLIDE 80

QZ algorithm, explicit version

p(AB−1) = QR p(B−1A) = Z ˜ R ˆ A = Q∗AZ, ˆ B = Q∗BZ

Luminy, October 2007 – p. 35

slide-81
SLIDE 81

QZ algorithm, explicit version

p(AB−1) = QR p(B−1A) = Z ˜ R ˆ A = Q∗AZ, ˆ B = Q∗BZ Explicit QZ step is complete.

Luminy, October 2007 – p. 35

slide-82
SLIDE 82

QZ algorithm, explicit version

p(AB−1) = QR p(B−1A) = Z ˜ R ˆ A = Q∗AZ, ˆ B = Q∗BZ Explicit QZ step is complete. ˆ A ˆ B−1 = Q∗(AB−1)Q

Luminy, October 2007 – p. 35

slide-83
SLIDE 83

QZ algorithm, explicit version

p(AB−1) = QR p(B−1A) = Z ˜ R ˆ A = Q∗AZ, ˆ B = Q∗BZ Explicit QZ step is complete. ˆ A ˆ B−1 = Q∗(AB−1)Q (QR iteration on AB−1)

Luminy, October 2007 – p. 35

slide-84
SLIDE 84

QZ algorithm, explicit version

p(AB−1) = QR p(B−1A) = Z ˜ R ˆ A = Q∗AZ, ˆ B = Q∗BZ Explicit QZ step is complete. ˆ A ˆ B−1 = Q∗(AB−1)Q (QR iteration on AB−1) ˆ B−1 ˆ A = Z∗(B−1A)Z

Luminy, October 2007 – p. 35

slide-85
SLIDE 85

QZ algorithm, explicit version

p(AB−1) = QR p(B−1A) = Z ˜ R ˆ A = Q∗AZ, ˆ B = Q∗BZ Explicit QZ step is complete. ˆ A ˆ B−1 = Q∗(AB−1)Q (QR iteration on AB−1) ˆ B−1 ˆ A = Z∗(B−1A)Z (QR iteration on B−1A)

Luminy, October 2007 – p. 35

slide-86
SLIDE 86

Explicit = Implicit?

Luminy, October 2007 – p. 36

slide-87
SLIDE 87

Explicit = Implicit?

AB−1 and B−1A are upper Hessenberg.

Luminy, October 2007 – p. 36

slide-88
SLIDE 88

Explicit = Implicit?

AB−1 and B−1A are upper Hessenberg. Utilize the Hessenberg form.

Luminy, October 2007 – p. 36

slide-89
SLIDE 89

Explicit = Implicit?

AB−1 and B−1A are upper Hessenberg. Utilize the Hessenberg form. implicit-Q theorem, or ...

Luminy, October 2007 – p. 36

slide-90
SLIDE 90

Explicit = Implicit?

AB−1 and B−1A are upper Hessenberg. Utilize the Hessenberg form. implicit-Q theorem, or ... work directly with the Krylov subspaces.

Luminy, October 2007 – p. 36

slide-91
SLIDE 91

Explicit = Implicit?

AB−1 and B−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

slide-92
SLIDE 92

Back to the palindromic case:

Luminy, October 2007 – p. 37

slide-93
SLIDE 93

Back to the palindromic case:

Bulge chase gives ˆ C = G−1CG−T

Luminy, October 2007 – p. 37

slide-94
SLIDE 94

Back to the palindromic case:

Bulge chase gives ˆ C = G−1CG−T ˆ C ˆ C−T = G−1(CC−T)G

Luminy, October 2007 – p. 37

slide-95
SLIDE 95

Back to the palindromic case:

Bulge chase gives ˆ C = G−1CG−T ˆ C ˆ C−T = G−1(CC−T)G (similarity)

Luminy, October 2007 – p. 37

slide-96
SLIDE 96

Back to the palindromic case:

Bulge chase gives ˆ C = G−1CG−T ˆ C ˆ C−T = G−1(CC−T)G (similarity) ˆ C−T ˆ C = GT(C−TC)G−T

Luminy, October 2007 – p. 37

slide-97
SLIDE 97

Back to the palindromic case:

Bulge chase gives ˆ C = G−1CG−T ˆ C ˆ C−T = G−1(CC−T)G (similarity) ˆ C−T ˆ C = GT(C−TC)G−T (similarity)

Luminy, October 2007 – p. 37

slide-98
SLIDE 98

Back to the palindromic case:

Bulge chase gives ˆ C = G−1CG−T ˆ C ˆ C−T = G−1(CC−T)G (similarity) ˆ C−T ˆ C = GT(C−TC)G−T (similarity) Need p(CC−T) = G ˜ R

Luminy, October 2007 – p. 37

slide-99
SLIDE 99

Back to the palindromic case:

Bulge chase gives ˆ C = G−1CG−T ˆ C ˆ C−T = G−1(CC−T)G (similarity) ˆ C−T ˆ C = GT(C−TC)G−T (similarity) Need p(CC−T) = G ˜ R and p(C−TC) = G−TR

Luminy, October 2007 – p. 37

slide-100
SLIDE 100

Back to the palindromic case:

Bulge chase gives ˆ C = G−1CG−T ˆ C ˆ C−T = G−1(CC−T)G (similarity) ˆ C−T ˆ C = GT(C−TC)G−T (similarity) Need p(CC−T) = G ˜ R and p(C−TC) = G−TR

  • r something like that.

Luminy, October 2007 – p. 37

slide-101
SLIDE 101

What we actually get:

Luminy, October 2007 – p. 38

slide-102
SLIDE 102

What we actually get:

r(CC−T) = GL

Luminy, October 2007 – p. 38

slide-103
SLIDE 103

What we actually get:

r(CC−T) = GL where r(z) =

m

  • k=1

z − µk µkz − 1

Luminy, October 2007 – p. 38

slide-104
SLIDE 104

What we actually get:

r(CC−T) = GL where r(z) =

m

  • k=1

z − µk µkz − 1 L is lower triangular.

Luminy, October 2007 – p. 38

slide-105
SLIDE 105

What we actually get:

r(CC−T) = GL where r(z) =

m

  • k=1

z − µk µkz − 1 L is lower triangular. r(C−TC) = G−TR

Luminy, October 2007 – p. 38

slide-106
SLIDE 106

What we actually get:

r(CC−T) = GL where r(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

slide-107
SLIDE 107

Our Approach:

Luminy, October 2007 – p. 39

slide-108
SLIDE 108

Our Approach:

want r(CC−T) = GL

Luminy, October 2007 – p. 39

slide-109
SLIDE 109

Our Approach:

want r(CC−T) = GL Define L = G−1r(CC−T)

Luminy, October 2007 – p. 39

slide-110
SLIDE 110

Our Approach:

want r(CC−T) = GL Define L = G−1r(CC−T) Then show that L is lower triangular.

Luminy, October 2007 – p. 39

slide-111
SLIDE 111

Our Approach:

want r(CC−T) = GL Define L = G−1r(CC−T) Then show that L is lower triangular. This is not entirely straightforward.

Luminy, October 2007 – p. 39

slide-112
SLIDE 112

Our Approach:

want r(CC−T) = GL Define L = G−1r(CC−T) Then show that L is lower triangular. This is not entirely straightforward. Utilize the Hessenberg form.

Luminy, October 2007 – p. 39

slide-113
SLIDE 113

Recall that

C =              0 ∗ 0 ∗ ∗ 0 ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗             

Luminy, October 2007 – p. 40

slide-114
SLIDE 114

This implies

Luminy, October 2007 – p. 41

slide-115
SLIDE 115

This implies

CC−T =              ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗             

Luminy, October 2007 – p. 41

slide-116
SLIDE 116

This implies

CC−T =              ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗              partly lower Hessenberg (n/2 − 1 columns)

Luminy, October 2007 – p. 41

slide-117
SLIDE 117

and

Luminy, October 2007 – p. 42

slide-118
SLIDE 118

and

C−TC =              ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗             

Luminy, October 2007 – p. 42

slide-119
SLIDE 119

and

C−TC =              ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗              partly upper Hessenberg (n/2 − 2 columns)

Luminy, October 2007 – p. 42

slide-120
SLIDE 120

Use both.

Luminy, October 2007 – p. 43

slide-121
SLIDE 121

Use both. L = L11 L12 L21 L22

  • Luminy, October 2007 – p. 43
slide-122
SLIDE 122

Use both. L = L11 L12 L21 L22

  • Using CC−T,

Luminy, October 2007 – p. 43

slide-123
SLIDE 123

Use both. L = L11 L12 L21 L22

  • Using CC−T, deduce L12 = 0 ...

Luminy, October 2007 – p. 43

slide-124
SLIDE 124

Use both. L = L11 L12 L21 L22

  • Using CC−T, deduce L12 = 0 ...

and L22 is lower triangular.

Luminy, October 2007 – p. 43

slide-125
SLIDE 125

L = L11 L21 L22

  • Luminy, October 2007 – p. 44
slide-126
SLIDE 126

L = L11 L21 L22

  • just need L11 lower triangular

Luminy, October 2007 – p. 44

slide-127
SLIDE 127

L = L11 L21 L22

  • just need L11 lower triangular

R = L−T = R11 R12 R22

  • Luminy, October 2007 – p. 44
slide-128
SLIDE 128

L = L11 L21 L22

  • just need L11 lower triangular

R = L−T = R11 R12 R22

  • Using C−TC,

Luminy, October 2007 – p. 44

slide-129
SLIDE 129

L = L11 L21 L22

  • just need L11 lower triangular

R = L−T = R11 R12 R22

  • Using C−TC, deduce that R11 is upper triangular.

Luminy, October 2007 – p. 44

slide-130
SLIDE 130

L = L11 L21 L22

  • just need L11 lower triangular

R = L−T = R11 R12 R22

  • Using C−TC, deduce that R11 is upper triangular.

Hence L11 is lower triangular.

Luminy, October 2007 – p. 44

slide-131
SLIDE 131

L = L11 L21 L22

  • just need L11 lower triangular

R = L−T = R11 R12 R22

  • Using C−TC, deduce that R11 is upper triangular.

Hence L11 is lower triangular. done!

Luminy, October 2007 – p. 44

slide-132
SLIDE 132

Conclusion:

Luminy, October 2007 – p. 45

slide-133
SLIDE 133

Conclusion:

Schröder’s palindromic bulge-chasing algorithm

Luminy, October 2007 – p. 45

slide-134
SLIDE 134

Conclusion:

Schröder’s palindromic bulge-chasing algorithm effects a combination QL and QR iteration

Luminy, October 2007 – p. 45

slide-135
SLIDE 135

Conclusion:

Schröder’s palindromic bulge-chasing algorithm effects a combination QL and QR iteration driven by a rational function r(z) =

m

  • k=1

z − µk µkz − 1

Luminy, October 2007 – p. 45

slide-136
SLIDE 136

Conclusion:

Schröder’s palindromic bulge-chasing algorithm effects a combination QL and QR iteration driven by a rational function r(z) =

m

  • k=1

z − µk µkz − 1 with shifts µ1, ..., µm in the numerator and shifts µ−1

1 ,

..., µ−1

m in the denominator.

Luminy, October 2007 – p. 45

slide-137
SLIDE 137

Conclusion:

Schröder’s palindromic bulge-chasing algorithm effects a combination QL and QR iteration driven by a rational function r(z) =

m

  • k=1

z − µk µkz − 1 with shifts µ1, ..., µm in the numerator and shifts µ−1

1 ,

..., µ−1

m in the denominator.

That’s all, folks!

Luminy, October 2007 – p. 45