Example: Introduction to Krylov Subspace Methods DEF: Krylov sequence 10 -1 2 0 -1 11 -1 3 2 -1 10...

Example:

nnn RbRA ,

Introduction to Krylov Subspace Methods

DEF:

bAx

,,, 2bAAbb Krylov sequence

Krylov sequence

10 -1 2 0 -1 11 -1 3 2 -1 10 -1 0 3 -1 8

A1 11 118 1239 127171 12 141 1651 194461 10 100 989 95461 10 106 1171 13332

Ab bA2 bA3 bA4b

Example:

Example:


DEF:

} ,,, {),( 1bAAbbspanbA mm

Krylov subspace

10 -1 2 0 -1 11 -1 3 2 -1 10 -1 0 3 -1 8

A

Krylov subspace

),(

106

100

141

118

,

10

10

12

11

,

1

1

1

1

3 spanbA

DEF:

bAAbbbAK mm

1,,,),(

Krylov matrix

106

100

141

118

10

10

12

11

1

1

1

1

),(3 bA

Example:


Remark: ))((3 AbAAbA

DEF:

bAAbbbAK mm

1,,,),(

Krylov matrix

106

100

141

118

10

10

12

11

1

1

1

1

),(3 bA

Conjugate Gradient Method

nnn RbRA ,


We want to solve the following linear system

bAx

definite) positive (symmetric SPD A

0

0

x

AxxT

end

210for

111

111

1

1

00

00

kkkk

kT

kkT

kk

kkkk

kkkk

kTkk

Tkk

pβrp

r/rrrβ

Apαr r

pαxx

Ap/prrα

,..,, k

rp

Axbr

0 K=1 K=2 K=3 K=4x1

x2

x3

X4

)(kr

1

1

2

1

*x


Example: Solve:

15

11

25

6

4

3

2

1

x

x

x

x 10 -1 2 0 -1 11 -1 3 2 -1 10 -1 0 3 -1 8


end

210for

111

111

1

1

00

00

kkkk

kT

kkT

kk

kkkk

kkkk

kTkk

Tkk

pβrp

r/rrrβ

Apαr r

pαxx

Ap/prrα

,..,, k

rp

Axbr

0 0.4716 0.9964 1.0015 1.00000 1.9651 1.9766 1.9833 2.00000 -0.8646 -0.9098 -1.0099 -1.00000 1.1791 1.0976 1.0197 1.0000

31.7 5.1503 1.0433 0.1929 0.0000

constants

vectors



end

210for

111

111

1

1

00

00

kkkk

kTkk

Tkk

kkkk

kkkk

kTkk

Tkk

pβrp

r/rrrβ

Apαr r

pαxx

Ap/prrα

,..,, k

rp

Axbr

,,, 210 rrr

,,, 210 ppp

,,, 210 xxx

,,, 210

,,, 210

nnn RbRA ,


We want to solve the following linear system

bAx

Define: bxAxxxf TT 2

1)( quadratic function

Example:

21

12A

definite) positive (symmetric SPD A

1

1b 21

2221

2121 ),( xxxxxxxxf

0

0

x

AxxT


Example:

21

12A

1

1b 21

2221

2121 ),( xxxxxxxxf

2

1

x

fx

f

f

12

12

12

21

xx

xx

12

21

2

2

1

1

xx

xx

Axb

Remark: bxAxxxf TT 2

1)( rf

0 when minimum has )( rxf

bAxxf when minimum has )(Why not max?


Remark: bxAxxxf TT 2

1)( rf

0 when minimum has )( rxf

bAxxf when minimum has )(

IDEA: bxAxxxf TT 2

1)(Search for the minimum

minimum is )(

s.t Find

xf

x

bAx

x

s.t Find

Problem (1) Problem (1)


Example:

21

12A

1

1b

212221

2121 ),( xxxxxxxxf

minimum


Method: )0(given x

“search direction”

“step length”

,,, 210 ppp

,,, 210

Method:

*x

0x

1x

0p

0

2x1p

1kkkk pxx 1

find toHow direction given p

ppick weHow


Method: kxgiven

kkk pxx 1

kp direction given

minimized is )( that so find xf

111 )()( kT

kk xd

dxfxf

d

d

11 kTk x

d

dr

kTk pr 1 01 k

Tk pr

01 kTk pr 0)( 1 k

Tk pAxb 0)( k

Tkk pApAxb

0)( kT

kk pApr k

Tk

kTk

App

pr


Method: kxgiven

kkk pxx 1

kp direction given


kTk

kTk

App

pr


end

210for

111

111

1

1

00

00

kkkk

kT

kkT

kk

kkkk

kkkk

kTkk

Tkk

pβrp

r/rrrβ

Apαr r

pαxx

Ap/prrα

,..,, k

rp

Axbr

ppick weHow

INNERPRODUCT

Inner Product

DEF: nnn RRR :,We say that

Is an inner product if

0 iff 0 , and 0 0 ,

,,,,,,

,, ,

xxxxxx

RRzyxzyzxzyx

Ryxyxyxn

n

Example: xyyx T ,

Example:

, 2RIn12212211

2

1

2

1 )(2, yxyxyxyxy

y

x

x

Inner Product


Is an inner product if

0 iff 0 , and 0 0 ,

,,,,,,

,, ,

xxxxxx

RRzyxzyzxzyx

Ryxyxyxn

n

Example:

, nRIn Hxyyx TH ,

where H is SPD

We define the normHH

xxx ,

Inner Product


Is symmetric bilinear form if

RRzyxzyzxzyx

Ryxyxyxn

n

,,,,,,

,, ,

Example:

, nRIn Hxyyx TH ,

where H is Symmetric

Inner Product

DEF: )0( 0, if orthogonal are and uvvuvu T

Example:

DEF: )0( 0, if orthogonal are and HuvvuHvu TH

where H is SPD

conjugateorthogonal HH

2

1u

4

5v

21

12H

ConjugateGradient


Method: kxgiven

kkk pxx 1

kp direction given


kTk

kTk

App

pr


end

210for

111

111

1

1

00

00

kkkk

kT

kkT

kk

kkkk

kkkk

kTkk

Tkk

pβrp

r/rrrβ

Apαr r

pαxx

Ap/prrα

,..,, k

rp

Axbr

ppick weHow


Method: 0given x

direction) (good 00 rp

) ofgradient theis ( frrf

kTkAkk Apppp 11,0

kxgiven

kkkk prp 1 Conjugate-A are ,1 kk pp

kT

kkk Appr )( k

Tk

kT

kk App

Apr


Method: kxgiven

kkk pxx 1

kp direction given


kTk

kTk

App

pr


end

210for

111

111

1

1

00

00

kkkk

kT

kkT

kk

kkkk

kkkk

kTkk

Tkk

pβrp

r/rrrβ

Apαr r

pαxx

Ap/prrα

,..,, k

rp

Axbr

kTk

kT

kk App

Apr


Lemma:[Elman,Silvester,Wathen Book]

),(},,{},,{)(

,0,)(

,0,,

satisfy methodCG by the generated vectors the,such that any For

)0()1()0()1()0(

)()(

)()()()(

rAKppspanrrspaniii

kjpApii

kjprpr(i)

x*xk

kkk

jk

jkjk

(k)


kk

6.0000 5.1362 0.0427 -0.0108 25.0000 0.1121 0.0562 0.1185 -11.0000 -0.4424 -0.8384 0.0698 15.0000 -0.7974 -0.6530 -0.1395

kp

6.0000 4.9781 -0.1681 -0.0123 0.0000 25.0000 -0.5464 0.0516 0.1166 -0.0000 -11.0000 -0.1526 -0.8202 0.0985 0.0000 15.0000 -1.1925 -0.6203 -0.1172 -0.0000

kr

0.0000 0.4716 0.9964 1.0015 1.0000 0.0000 1.9651 1.9766 1.9833 2.0000 0.0000 -0.8646 -0.9098 -1.0099 -1.0000 0.0000 1.1791 1.0976 1.0197 1.0000

kx

0.0786 0.1022 0.1193 0.1411 0.0713

0.0263 0.0410 0.0342

2,1,0k

Example: Introduction to Krylov Subspace Methods DEF: Krylov sequence 10 -1 2 0 -1 11 -1 3 2 -1 10...

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Transcript of Example: Introduction to Krylov Subspace Methods DEF: Krylov sequence 10 -1 2 0 -1 11 -1 3 2 -1 10...