The implicit QZ algorithm for the palindromic eigenvalue problem - PowerPoint PPT Presentation

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 − λC T = ∗ ∗ ∗ ∗ ∗     ∗ ∗ ∗ ∗ ∗ ∗     ∗ ∗ ∗ ∗ ∗ ∗ ∗     ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗   ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ Luminy, October 2007 – p. 21

Bulge Chase in the Pencil   ∗ ∗ ∗ ∗ ∗ ∗     ∗ ∗ ∗ ∗       C − λC T = ∗ ∗ ∗ ∗ ∗     ∗ ∗ ∗ ∗ ∗ ∗     ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗     ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗   ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ Luminy, October 2007 – p. 22

Bulge Chase in the Pencil   ∗ ∗ ∗ ∗ ∗ ∗     ∗ ∗ ∗ ∗       C − λC T = ∗ ∗ ∗ ∗ ∗     ∗ ∗ ∗ ∗ ∗ ∗ ∗     ∗ ∗ ∗ ∗ ∗ ∗ ∗     ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗   ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ Luminy, October 2007 – p. 23

Bulge pencils on a collision course! Luminy, October 2007 – p. 24

Bulge Chase in the Pencil   ∗ ∗ ∗ ∗ ∗ ∗     ∗ ∗ ∗ ∗       C − λC T = ∗ ∗ ∗ ∗ ∗     ∗ ∗ ∗ ∗ ∗ ∗ ∗     ∗ ∗ ∗ ∗ ∗ ∗ ∗     ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗   ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ Luminy, October 2007 – p. 25

Half a step further:   ∗ ∗ ∗ ∗ ∗     ∗ ∗ ∗ ∗       C − λC T = ∗ ∗ ∗ ∗ ∗     ∗ ∗ ∗ ∗ ∗ ∗     ∗ ∗ ∗ ∗ ∗ ∗ ∗     ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗   ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ Luminy, October 2007 – p. 26

Swap the shifts   ∗ ∗ ∗ ∗ ∗     ∗ ∗ ∗ ∗       C − λC T = ∗ ∗ ∗ ∗ ∗     ∗ ∗ ∗ ∗ ∗ ∗     ∗ ∗ ∗ ∗ ∗ ∗ ∗     ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗   ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ Luminy, October 2007 – p. 27

Swap the shifts   ∗ ∗ ∗ ∗ ∗     ∗ ∗ ∗ ∗       C − λC T = ∗ ∗ ∗ ∗ ∗     ∗ ∗ ∗ ∗ ∗ ∗     ∗ ∗ ∗ ∗ ∗ ∗ ∗     ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗   ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ Luminy, October 2007 – p. 28

Reverse the process   ∗ ∗ ∗ ∗ ∗ ∗     ∗ ∗ ∗ ∗       C − λC T = ∗ ∗ ∗ ∗ ∗     ∗ ∗ ∗ ∗ ∗ ∗ ∗     ∗ ∗ ∗ ∗ ∗ ∗ ∗     ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗   ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ Luminy, October 2007 – p. 29

Reverse the process   ∗ ∗ ∗ ∗ ∗ ∗     ∗ ∗ ∗ ∗       C − λC T = ∗ ∗ ∗ ∗ ∗     ∗ ∗ ∗ ∗ ∗ ∗     ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗     ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗   ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ Luminy, October 2007 – p. 30

Reverse the process   ∗ ∗ ∗ ∗ ∗     ∗ ∗ ∗ ∗       C − λC T = ∗ ∗ ∗ ∗ ∗     ∗ ∗ ∗ ∗ ∗ ∗     ∗ ∗ ∗ ∗ ∗ ∗ ∗     ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗   ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ Luminy, October 2007 – p. 31

Bulge Chase is complete   ∗ ∗ ∗ ∗ ∗     ∗ ∗ ∗ ∗       C − λC T = ∗ ∗ ∗ ∗ ∗     ∗ ∗ ∗ ∗ ∗ ∗     ∗ ∗ ∗ ∗ ∗ ∗ ∗     ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗   ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ 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 version A − λB (Hessenberg, triangular) Luminy, October 2007 – p. 34

QZ algorithm, implicit version A − λB (Hessenberg, triangular) pick shifts µ 1 , ... µ m Luminy, October 2007 – p. 34

QZ algorithm, implicit version A − λB (Hessenberg, triangular) pick shifts µ 1 , ... µ m p ( z ) = ( z − µ 1 ) · · · ( z − µ m ) Luminy, October 2007 – p. 34

QZ algorithm, implicit version A − λB (Hessenberg, triangular) pick shifts µ 1 , ... µ m p ( z ) = ( z − µ 1 ) · · · ( z − µ m ) x = p ( AB − 1 ) e 1 Luminy, October 2007 – p. 34

QZ algorithm, implicit version A − λB (Hessenberg, triangular) pick shifts µ 1 , ... µ m p ( z ) = ( z − µ 1 ) · · · ( z − µ m ) x = p ( AB − 1 ) e 1 Make a bulge, Luminy, October 2007 – p. 34

QZ algorithm, implicit version A − λB (Hessenberg, triangular) pick shifts µ 1 , ... µ m p ( z ) = ( z − µ 1 ) · · · ( z − µ m ) x = p ( AB − 1 ) e 1 Make a bulge, then chase it. Luminy, October 2007 – p. 34

QZ algorithm, implicit version A − λB (Hessenberg, triangular) pick shifts µ 1 , ... µ m p ( z ) = ( z − µ 1 ) · · · ( z − µ m ) x = p ( AB − 1 ) e 1 Make a bulge, then chase it. Get ˆ A − λ ˆ B Luminy, October 2007 – p. 34

QZ algorithm, Luminy, October 2007 – p. 35

QZ algorithm, explicit version Luminy, October 2007 – p. 35

QZ algorithm, explicit version p ( AB − 1 ) = QR Luminy, October 2007 – p. 35

QZ algorithm, explicit version p ( B − 1 A ) = Z ˜ p ( AB − 1 ) = QR R Luminy, October 2007 – p. 35

QZ algorithm, explicit version p ( B − 1 A ) = Z ˜ p ( AB − 1 ) = QR R ˆ ˆ A = Q ∗ AZ, B = Q ∗ BZ Luminy, October 2007 – p. 35

QZ algorithm, explicit version p ( B − 1 A ) = Z ˜ p ( AB − 1 ) = QR R ˆ ˆ A = Q ∗ AZ, B = Q ∗ BZ Explicit QZ step is complete. Luminy, October 2007 – p. 35

QZ algorithm, explicit version p ( B − 1 A ) = Z ˜ p ( AB − 1 ) = QR R ˆ ˆ A = Q ∗ AZ, B = Q ∗ BZ Explicit QZ step is complete. B − 1 = Q ∗ ( AB − 1 ) Q A ˆ ˆ Luminy, October 2007 – p. 35

QZ algorithm, explicit version p ( B − 1 A ) = Z ˜ p ( AB − 1 ) = QR R ˆ ˆ A = Q ∗ AZ, B = Q ∗ BZ Explicit QZ step is complete. B − 1 = Q ∗ ( AB − 1 ) Q ( QR iteration on AB − 1 ) A ˆ ˆ Luminy, October 2007 – p. 35

QZ algorithm, explicit version p ( B − 1 A ) = Z ˜ p ( AB − 1 ) = QR R ˆ ˆ A = Q ∗ AZ, B = Q ∗ BZ Explicit QZ step is complete. B − 1 = Q ∗ ( AB − 1 ) Q ( QR iteration on AB − 1 ) A ˆ ˆ B − 1 ˆ ˆ A = Z ∗ ( B − 1 A ) Z Luminy, October 2007 – p. 35

QZ algorithm, explicit version p ( B − 1 A ) = Z ˜ p ( AB − 1 ) = QR R ˆ ˆ A = Q ∗ AZ, B = Q ∗ BZ Explicit QZ step is complete. B − 1 = Q ∗ ( AB − 1 ) Q ( QR iteration on AB − 1 ) A ˆ ˆ B − 1 ˆ ˆ A = Z ∗ ( B − 1 A ) Z ( QR iteration on B − 1 A ) Luminy, October 2007 – p. 35

Explicit = Implicit? Luminy, October 2007 – p. 36

Explicit = Implicit? AB − 1 and B − 1 A are upper Hessenberg. Luminy, October 2007 – p. 36

Explicit = Implicit? AB − 1 and B − 1 A are upper Hessenberg. Utilize the Hessenberg form. Luminy, October 2007 – p. 36

Explicit = Implicit? AB − 1 and B − 1 A are upper Hessenberg. Utilize the Hessenberg form. implicit- Q theorem, or ... Luminy, October 2007 – p. 36

Explicit = Implicit? AB − 1 and B − 1 A 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 and B − 1 A 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 gives ˆ C = G − 1 CG − T Luminy, October 2007 – p. 37

Back to the palindromic case: Bulge chase gives ˆ C = G − 1 CG − T C − T = G − 1 ( CC − T ) G C ˆ ˆ Luminy, October 2007 – p. 37

Back to the palindromic case: Bulge chase gives ˆ C = G − 1 CG − T C − T = G − 1 ( CC − T ) G C ˆ ˆ (similarity) Luminy, October 2007 – p. 37

Back to the palindromic case: Bulge chase gives ˆ C = G − 1 CG − T C − T = G − 1 ( CC − T ) G C ˆ ˆ (similarity) C − T ˆ ˆ C = G T ( C − T C ) G − T Luminy, October 2007 – p. 37

Back to the palindromic case: Bulge chase gives ˆ C = G − 1 CG − T C − T = G − 1 ( CC − T ) G C ˆ ˆ (similarity) C − T ˆ ˆ C = G T ( C − T C ) G − T (similarity) Luminy, October 2007 – p. 37

Back to the palindromic case: Bulge chase gives ˆ C = G − 1 CG − T C − T = G − 1 ( CC − T ) G C ˆ ˆ (similarity) C − T ˆ ˆ C = G T ( C − T C ) G − T (similarity) Need p ( CC − T ) = G ˜ R Luminy, October 2007 – p. 37

Back to the palindromic case: Bulge chase gives ˆ C = G − 1 CG − T C − T = G − 1 ( CC − T ) G C ˆ ˆ (similarity) C − T ˆ ˆ C = G T ( C − T C ) G − T (similarity) Need p ( CC − T ) = G ˜ R and p ( C − T C ) = G − T R Luminy, October 2007 – p. 37

Back to the palindromic case: Bulge chase gives ˆ C = G − 1 CG − T C − T = G − 1 ( CC − T ) G C ˆ ˆ (similarity) C − T ˆ ˆ C = G T ( C − T C ) G − T (similarity) Need p ( CC − T ) = G ˜ R and p ( C − T C ) = G − T R or something like that. Luminy, October 2007 – p. 37