lanczos method ct. - dipartimento di informatica · lanczos method lanczos method ct. the...

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Lanczos Method Lanczos method ct. The coefficients are obtained from β k 1 = (v k 1 ) T Av k =(Av k 1 ) T v k = (γ k 1 v k + α k 1 v k 1 + β k 2 v k 2 ) T v k = γ k 1 , i.e. Av k = β k v k +1 + α k v k + β k 1 v k 1 . (1) Thus, α k =(v k ) T Av k , TUHH Heinrich Voss Krylov subspace methods Eigenvalue problems 2012 8 / 95

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Page 1: Lanczos method ct. - Dipartimento di Informatica · Lanczos Method Lanczos method ct. The coefficients are obtained from

Lanczos Method

Lanczos method ct.

The coefficients are obtained from

βk−1 = (vk−1)T Avk = (Avk−1)T vk

= (γk−1vk + αk−1vk−1 + βk−2vk−2)T vk = γk−1,

i.e.

Avk = βk vk+1 + αk vk + βk−1vk−1. (1)

Thus,

αk = (vk )T Avk ,

TUHH Heinrich Voss Krylov subspace methods Eigenvalue problems 2012 8 / 95

Page 2: Lanczos method ct. - Dipartimento di Informatica · Lanczos Method Lanczos method ct. The coefficients are obtained from

Lanczos Method

Lanczos method ct.

The coefficients are obtained from

βk−1 = (vk−1)T Avk = (Avk−1)T vk

= (γk−1vk + αk−1vk−1 + βk−2vk−2)T vk = γk−1,

i.e.

Avk = βk vk+1 + αk vk + βk−1vk−1. (1)

Thus,

αk = (vk )T Avk ,

and the condition ‖vk+1‖2 = 1 yields

βk = 1/‖Avk − αk vk − βk−1vk−1‖2. (2)

If the denominator in (2) vanishes, then Avk ∈ Kk (v ,A), and therefore,

Kk (v ,A) is a k -dimensional invariant subspace of A.

TUHH Heinrich Voss Krylov subspace methods Eigenvalue problems 2012 8 / 95

Page 3: Lanczos method ct. - Dipartimento di Informatica · Lanczos Method Lanczos method ct. The coefficients are obtained from

Lanczos Method

Lanczos method ct.

1: v0 = 0; k = 1

2: β0 = ‖r0‖3: while r k−1 6= 0 do

4: vk = r k−1/βk−1

5: r k = Avk

6: r k = r k − βk−1vk−1

7: αk = (vk )T r k

8: r k = r k − αk vk

9: βk = ‖r k‖10: end while

TUHH Heinrich Voss Krylov subspace methods Eigenvalue problems 2012 9 / 95

Page 4: Lanczos method ct. - Dipartimento di Informatica · Lanczos Method Lanczos method ct. The coefficients are obtained from

Lanczos Method

Lanczos method ct.

1: v0 = 0; k = 1

2: β0 = ‖r0‖3: while r k−1 6= 0 do

4: vk = r k−1/βk−1

5: r k = Avk

6: r k = r k − βk−1vk−1

7: αk = (vk )T r k

8: r k = r k − αk vk

9: βk = ‖r k‖10: end while

Then with Tk = tridiag{βj−1, αj , βj}

AVk = Vk Tk + r k (ek )T , r k = Avk − αk vk − βk−1vk−1.

and r k = ‖r k‖2vk+1 = βk vk+1

TUHH Heinrich Voss Krylov subspace methods Eigenvalue problems 2012 9 / 95

Page 5: Lanczos method ct. - Dipartimento di Informatica · Lanczos Method Lanczos method ct. The coefficients are obtained from

Lanczos Method

Lucky termination

The Lanczos method may terminate with βj = 0 for some j .

TUHH Heinrich Voss Krylov subspace methods Eigenvalue problems 2012 10 / 95

Page 6: Lanczos method ct. - Dipartimento di Informatica · Lanczos Method Lanczos method ct. The coefficients are obtained from

Lanczos Method

Lucky termination

The Lanczos method may terminate with βj = 0 for some j .

Then v j+1 = 0, and therefore

Av j = αjvj + βj−1v j−1 ∈ Kj(A, v

1).

TUHH Heinrich Voss Krylov subspace methods Eigenvalue problems 2012 10 / 95

Page 7: Lanczos method ct. - Dipartimento di Informatica · Lanczos Method Lanczos method ct. The coefficients are obtained from

Lanczos Method

Lucky termination

The Lanczos method may terminate with βj = 0 for some j .

Then v j+1 = 0, and therefore

Av j = αjvj + βj−1v j−1 ∈ Kj(A, v

1).

For i < j it holds by construction

Av i ∈ Kj(A, v1).

Hence, Kj(A, v1) is an invariant subspace of A, and therefore every

eigenvalue θ(j)i of Tj is an eigenvalue of A, and the corresponding Ritz vectors

are eigenvectors of A.

TUHH Heinrich Voss Krylov subspace methods Eigenvalue problems 2012 10 / 95

Page 8: Lanczos method ct. - Dipartimento di Informatica · Lanczos Method Lanczos method ct. The coefficients are obtained from

Lanczos Method

Error bound

If θ(m)i are the Ritz values (eigenvalues of Tm), s

(m)i the corresponding

eigenvectors, and x(m)i = Vms

(m)i the Ritz vectors, it holds

(A − θ(m)i I)x

(m)i = (A − θ

(m)i I)Vms

(m)i = AVms

(m)i − Vm(θ

(m)i s

(m)i )

= VmTms(m)i + βmvm+1(em)T s

(m)i − VmTms

(m)i = βmvm+1(em)T s

(m)i ,

which implies

‖(A − θ(m)i I)x

(m)i ‖2 = βm|s

(m)m,i |.

TUHH Heinrich Voss Krylov subspace methods Eigenvalue problems 2012 11 / 95

Page 9: Lanczos method ct. - Dipartimento di Informatica · Lanczos Method Lanczos method ct. The coefficients are obtained from

Lanczos Method

Error bound

If θ(m)i are the Ritz values (eigenvalues of Tm), s

(m)i the corresponding

eigenvectors, and x(m)i = Vms

(m)i the Ritz vectors, it holds

(A − θ(m)i I)x

(m)i = (A − θ

(m)i I)Vms

(m)i = AVms

(m)i − Vm(θ

(m)i s

(m)i )

= VmTms(m)i + βmvm+1(em)T s

(m)i − VmTms

(m)i = βmvm+1(em)T s

(m)i ,

which implies

‖(A − θ(m)i I)x

(m)i ‖2 = βm|s

(m)m,i |.

Then by the Krylov & Bogoliubov Theorem there exists an eigenvalue λ of A

such that

|λ− θ(m)i | ≤ βm|s

(m)m,i |.

TUHH Heinrich Voss Krylov subspace methods Eigenvalue problems 2012 11 / 95

Page 10: Lanczos method ct. - Dipartimento di Informatica · Lanczos Method Lanczos method ct. The coefficients are obtained from

Lanczos Method

Error bound

If θ(m)i are the Ritz values (eigenvalues of Tm), s

(m)i the corresponding

eigenvectors, and x(m)i = Vms

(m)i the Ritz vectors, it holds

(A − θ(m)i I)x

(m)i = (A − θ

(m)i I)Vms

(m)i = AVms

(m)i − Vm(θ

(m)i s

(m)i )

= VmTms(m)i + βmvm+1(em)T s

(m)i − VmTms

(m)i = βmvm+1(em)T s

(m)i ,

which implies

‖(A − θ(m)i I)x

(m)i ‖2 = βm|s

(m)m,i |.

Then by the Krylov & Bogoliubov Theorem there exists an eigenvalue λ of A

such that

|λ− θ(m)i | ≤ βm|s

(m)m,i |.

Notice, that this error bound can be computet without determining the Ritz

vector x(m)i = Vms

(m)i .

TUHH Heinrich Voss Krylov subspace methods Eigenvalue problems 2012 11 / 95

Page 11: Lanczos method ct. - Dipartimento di Informatica · Lanczos Method Lanczos method ct. The coefficients are obtained from

Lanczos Method

Convergence

The Lanczos method is a generalization of the power method where

approximations to eigenvectors are obtained from the last iterate only.

TUHH Heinrich Voss Krylov subspace methods Eigenvalue problems 2012 12 / 95

Page 12: Lanczos method ct. - Dipartimento di Informatica · Lanczos Method Lanczos method ct. The coefficients are obtained from

Lanczos Method

Convergence

The Lanczos method is a generalization of the power method where

approximations to eigenvectors are obtained from the last iterate only.

As for the power method we therefore can expect fast convergence to the

eigenvalue which is maximal in modulus and to the corresponding

eigenvector.

TUHH Heinrich Voss Krylov subspace methods Eigenvalue problems 2012 12 / 95

Page 13: Lanczos method ct. - Dipartimento di Informatica · Lanczos Method Lanczos method ct. The coefficients are obtained from

Lanczos Method

Convergence

The Lanczos method is a generalization of the power method where

approximations to eigenvectors are obtained from the last iterate only.

As for the power method we therefore can expect fast convergence to the

eigenvalue which is maximal in modulus and to the corresponding

eigenvector.

One can influence the convergence of the power method by shifts, either to

separate the wanted eigenvalue more from the remaining spectrum or to

enforce convergence to a different eigenvalue.

TUHH Heinrich Voss Krylov subspace methods Eigenvalue problems 2012 12 / 95

Page 14: Lanczos method ct. - Dipartimento di Informatica · Lanczos Method Lanczos method ct. The coefficients are obtained from

Lanczos Method

Convergence

The Lanczos method is a generalization of the power method where

approximations to eigenvectors are obtained from the last iterate only.

As for the power method we therefore can expect fast convergence to the

eigenvalue which is maximal in modulus and to the corresponding

eigenvector.

One can influence the convergence of the power method by shifts, either to

separate the wanted eigenvalue more from the remaining spectrum or to

enforce convergence to a different eigenvalue.

The Lanczos method is independent of shifts since

Km(v1,A) = Km(v

1,A + αI) for all α ∈ R.

Hence, we can expect convergence of the Lanczos method to extreme

eigenvalues first.

TUHH Heinrich Voss Krylov subspace methods Eigenvalue problems 2012 12 / 95

Page 15: Lanczos method ct. - Dipartimento di Informatica · Lanczos Method Lanczos method ct. The coefficients are obtained from

Lanczos Method

Monotonicity

Assume that the eigenvalues of A and the Ritz values with respect to Km are

ordered by magnitude:

λ1 ≤ λ2 ≤ · · · ≤ λn, θ(m)1 ≤ θ

(m)2 ≤ · · · ≤ θ

(m)m ,

TUHH Heinrich Voss Krylov subspace methods Eigenvalue problems 2012 13 / 95

Page 16: Lanczos method ct. - Dipartimento di Informatica · Lanczos Method Lanczos method ct. The coefficients are obtained from

Lanczos Method

Monotonicity

Assume that the eigenvalues of A and the Ritz values with respect to Km are

ordered by magnitude:

λ1 ≤ λ2 ≤ · · · ≤ λn, θ(m)1 ≤ θ

(m)2 ≤ · · · ≤ θ

(m)m ,

Then the minmax principle yields (S denotes a subspace of Cm and S a

subspace of Cn)

θ(m)j = min

dim S=jmaxy∈S

yHTmy

yHy= min

dim S=jmaxy∈S

yHV Hm AVmy

yV Hm Vmy

= mindim S=j, S⊂Km

maxx∈S

xHAx

xHx≥ min

dim S=j, S⊂Km+1

maxx∈S

xHAx

xHx

= θ(m+1)j ≥ min

dim S=j

maxx∈S

xHAx

xHx= λj .

TUHH Heinrich Voss Krylov subspace methods Eigenvalue problems 2012 13 / 95

Page 17: Lanczos method ct. - Dipartimento di Informatica · Lanczos Method Lanczos method ct. The coefficients are obtained from

Lanczos Method

Monotonicity

Assume that the eigenvalues of A and the Ritz values with respect to Km are

ordered by magnitude:

λ1 ≤ λ2 ≤ · · · ≤ λn, θ(m)1 ≤ θ

(m)2 ≤ · · · ≤ θ

(m)m ,

Then the minmax principle yields (S denotes a subspace of Cm and S a

subspace of Cn)

θ(m)j = min

dim S=jmaxy∈S

yHTmy

yHy= min

dim S=jmaxy∈S

yHV Hm AVmy

yV Hm Vmy

= mindim S=j, S⊂Km

maxx∈S

xHAx

xHx≥ min

dim S=j, S⊂Km+1

maxx∈S

xHAx

xHx

= θ(m+1)j ≥ min

dim S=j

maxx∈S

xHAx

xHx= λj .

Each (finite) sequence {θ(m)j }m=j,j+1,j+2 therefore is monotonically decreasing

and bounded below by λj .

Likewise the sequence of the j-largest eigenvalues of Tm are monotonically

increasing and bounded above by the j-largest eigenvalue of A.

TUHH Heinrich Voss Krylov subspace methods Eigenvalue problems 2012 13 / 95

Page 18: Lanczos method ct. - Dipartimento di Informatica · Lanczos Method Lanczos method ct. The coefficients are obtained from

Lanczos Method

Example

TUHH Heinrich Voss Krylov subspace methods Eigenvalue problems 2012 14 / 95

Page 19: Lanczos method ct. - Dipartimento di Informatica · Lanczos Method Lanczos method ct. The coefficients are obtained from

Lanczos Method

Convergence

Before getting results about the speed of convergence of the Lanczos method

we first prove a bound for the angle between an eigenvector of A and a Krylov

space Km(v1,A). Denote by ui a system of orthonormal eigenvectors

corresponding to the eigenvalues λi .

TUHH Heinrich Voss Krylov subspace methods Eigenvalue problems 2012 15 / 95

Page 20: Lanczos method ct. - Dipartimento di Informatica · Lanczos Method Lanczos method ct. The coefficients are obtained from

Lanczos Method

Convergence

Before getting results about the speed of convergence of the Lanczos method

we first prove a bound for the angle between an eigenvector of A and a Krylov

space Km(v1,A). Denote by ui a system of orthonormal eigenvectors

corresponding to the eigenvalues λi .

LEMMA 10.1

Let Pi be the orthogonal projector onto the eigenspace corresponding to λi . If

Piv1 6= 0 then

tan δ(ui ,Km) = minp∈Πm−1, p(λi )=1

‖p(A)y i‖2 tan δ(ui , v1),

where

y i =

{

(I−Pi )v1

‖(I−Pi )v1‖2if (I − Pi)v

1 6= 0

0 otherwise

TUHH Heinrich Voss Krylov subspace methods Eigenvalue problems 2012 15 / 95

Page 21: Lanczos method ct. - Dipartimento di Informatica · Lanczos Method Lanczos method ct. The coefficients are obtained from

Lanczos Method

Proof

The Krylov space Km(v1,A) consists of all vectors which can be written as

x = q(A)v1 where q ∈ Πm−1 is any polynomial of degree m − 1.

TUHH Heinrich Voss Krylov subspace methods Eigenvalue problems 2012 16 / 95

Page 22: Lanczos method ct. - Dipartimento di Informatica · Lanczos Method Lanczos method ct. The coefficients are obtained from

Lanczos Method

Proof

The Krylov space Km(v1,A) consists of all vectors which can be written as

x = q(A)v1 where q ∈ Πm−1 is any polynomial of degree m − 1.

With the orthogonal decomposition

x = q(A)v1 = q(A)Piv1 + q(A)(I − Pi)v

1

it holds for the angle δ(x , ui) between x and ui

tan δ(x , ui) =‖q(A)(I − Pi)v

1‖2

‖q(A)Piv1‖2

=‖q(A)y i‖2

|q(λi)|

‖(I − Pi)v1‖2

‖Piv1‖2

,

TUHH Heinrich Voss Krylov subspace methods Eigenvalue problems 2012 16 / 95

Page 23: Lanczos method ct. - Dipartimento di Informatica · Lanczos Method Lanczos method ct. The coefficients are obtained from

Lanczos Method

Proof

The Krylov space Km(v1,A) consists of all vectors which can be written as

x = q(A)v1 where q ∈ Πm−1 is any polynomial of degree m − 1.

With the orthogonal decomposition

x = q(A)v1 = q(A)Piv1 + q(A)(I − Pi)v

1

it holds for the angle δ(x , ui) between x and ui

tan δ(x , ui) =‖q(A)(I − Pi)v

1‖2

‖q(A)Piv1‖2

=‖q(A)y i‖2

|q(λi)|

‖(I − Pi)v1‖2

‖Piv1‖2

,

and the scaling p(λ) := q(λ)/q(λi) yields

tan δ(x , ui) = ‖p(A)y i‖2 tan δ(v1, ui)

from which we get the statement by minimizing over all x ∈ K(v1,A).�

TUHH Heinrich Voss Krylov subspace methods Eigenvalue problems 2012 16 / 95

Page 24: Lanczos method ct. - Dipartimento di Informatica · Lanczos Method Lanczos method ct. The coefficients are obtained from

Lanczos Method

Convergence ct.

Inserting any polynomial of degree m − 1 which satisfies p(λi) = 1 one

obtains an upper bound for tan δ(ui ,Km(v1,A)) from the last lemma.

TUHH Heinrich Voss Krylov subspace methods Eigenvalue problems 2012 17 / 95

Page 25: Lanczos method ct. - Dipartimento di Informatica · Lanczos Method Lanczos method ct. The coefficients are obtained from

Lanczos Method

Convergence ct.

Inserting any polynomial of degree m − 1 which satisfies p(λi) = 1 one

obtains an upper bound for tan δ(ui ,Km(v1,A)) from the last lemma.

Definition

The Chebyshev polynomials are defined as

ck (t) := cos(k · arccos t), |t | ≤ 1, k ∈ N ∪ {0}.

TUHH Heinrich Voss Krylov subspace methods Eigenvalue problems 2012 17 / 95

Page 26: Lanczos method ct. - Dipartimento di Informatica · Lanczos Method Lanczos method ct. The coefficients are obtained from

Lanczos Method

Convergence ct.

Inserting any polynomial of degree m − 1 which satisfies p(λi) = 1 one

obtains an upper bound for tan δ(ui ,Km(v1,A)) from the last lemma.

Definition

The Chebyshev polynomials are defined as

ck (t) := cos(k · arccos t), |t | ≤ 1, k ∈ N ∪ {0}.

In the following lemma we collect some properties of the Chebyshev

polynomials that are needed in the sequel.

TUHH Heinrich Voss Krylov subspace methods Eigenvalue problems 2012 17 / 95

Page 27: Lanczos method ct. - Dipartimento di Informatica · Lanczos Method Lanczos method ct. The coefficients are obtained from

Lanczos Method

Lemma 3.1

(i) The functions ck (t) satisfy the recurrence formula

c0(t) ≡ 1, c1(t) = t , ck+1(t) = 2tck (t)− ck−1(t), k ≥ 1.

In particular, ck is a polynomial of degree k .

TUHH Heinrich Voss Krylov subspace methods Eigenvalue problems 2012 18 / 95

Page 28: Lanczos method ct. - Dipartimento di Informatica · Lanczos Method Lanczos method ct. The coefficients are obtained from

Lanczos Method

Lemma 3.1

(i) The functions ck (t) satisfy the recurrence formula

c0(t) ≡ 1, c1(t) = t , ck+1(t) = 2tck (t)− ck−1(t), k ≥ 1.

In particular, ck is a polynomial of degree k .

(ii) For |t | ≥ 1 the Chebyshev polynomials have the following representation

ck (t) = cosh(k · Arcosh t), t ∈ R, |t | ≥ 1.

TUHH Heinrich Voss Krylov subspace methods Eigenvalue problems 2012 18 / 95

Page 29: Lanczos method ct. - Dipartimento di Informatica · Lanczos Method Lanczos method ct. The coefficients are obtained from

Lanczos Method

Lemma 3.1

(i) The functions ck (t) satisfy the recurrence formula

c0(t) ≡ 1, c1(t) = t , ck+1(t) = 2tck (t)− ck−1(t), k ≥ 1.

In particular, ck is a polynomial of degree k .

(ii) For |t | ≥ 1 the Chebyshev polynomials have the following representation

ck (t) = cosh(k · Arcosh t), t ∈ R, |t | ≥ 1.

Proof: (i) follows from the properties of the cosine function, in particular from

cos((k + 1)t) + cos((k − 1)t) = 2 cos(kt) · cos t .

TUHH Heinrich Voss Krylov subspace methods Eigenvalue problems 2012 18 / 95

Page 30: Lanczos method ct. - Dipartimento di Informatica · Lanczos Method Lanczos method ct. The coefficients are obtained from

Lanczos Method

Lemma 3.1

(i) The functions ck (t) satisfy the recurrence formula

c0(t) ≡ 1, c1(t) = t , ck+1(t) = 2tck (t)− ck−1(t), k ≥ 1.

In particular, ck is a polynomial of degree k .

(ii) For |t | ≥ 1 the Chebyshev polynomials have the following representation

ck (t) = cosh(k · Arcosh t), t ∈ R, |t | ≥ 1.

Proof: (i) follows from the properties of the cosine function, in particular from

cos((k + 1)t) + cos((k − 1)t) = 2 cos(kt) · cos t .

(ii) follows from the fact that the functions cosh(k · Arcosh t) satisfy the same

recurrence formula. �

TUHH Heinrich Voss Krylov subspace methods Eigenvalue problems 2012 18 / 95

Page 31: Lanczos method ct. - Dipartimento di Informatica · Lanczos Method Lanczos method ct. The coefficients are obtained from

Lanczos Method

Theorem 10.2

Let α, β, γ ∈ R with α < β and γ 6∈ (α, β). Then the minimization problem

minp∈Πm, p(γ)=1

maxt∈[α,β]

|p(t)|

has a unique solution, and is solved by the scaled Chebyshev polynomial

cm(t) :=

{

cm(1 + 2 t−ββ−α

)/cm(1 + 2 γ−ββ−α

) für γ > β

cm(1 + 2 α−tβ−α

)/cm(1 + 2α−γβ−α

) für γ < α

TUHH Heinrich Voss Krylov subspace methods Eigenvalue problems 2012 19 / 95

Page 32: Lanczos method ct. - Dipartimento di Informatica · Lanczos Method Lanczos method ct. The coefficients are obtained from

Lanczos Method

Proof

We restrict ourselves to the case γ > β.

TUHH Heinrich Voss Krylov subspace methods Eigenvalue problems 2012 20 / 95

Page 33: Lanczos method ct. - Dipartimento di Informatica · Lanczos Method Lanczos method ct. The coefficients are obtained from

Lanczos Method

Proof

We restrict ourselves to the case γ > β.

From t := 1 + 2(γ − β)/(β − α) 6∈ [−1, 1] it follows that cm is defined, and

obviously, cm is a polynomial of degree m, and cm(γ) = 1 holds.

TUHH Heinrich Voss Krylov subspace methods Eigenvalue problems 2012 20 / 95

Page 34: Lanczos method ct. - Dipartimento di Informatica · Lanczos Method Lanczos method ct. The coefficients are obtained from

Lanczos Method

Proof

We restrict ourselves to the case γ > β.

From t := 1 + 2(γ − β)/(β − α) 6∈ [−1, 1] it follows that cm is defined, and

obviously, cm is a polynomial of degree m, and cm(γ) = 1 holds.

It remains to show, that cm is the unique solution of the minimum problem.

TUHH Heinrich Voss Krylov subspace methods Eigenvalue problems 2012 20 / 95

Page 35: Lanczos method ct. - Dipartimento di Informatica · Lanczos Method Lanczos method ct. The coefficients are obtained from

Lanczos Method

Proof

We restrict ourselves to the case γ > β.

From t := 1 + 2(γ − β)/(β − α) 6∈ [−1, 1] it follows that cm is defined, and

obviously, cm is a polynomial of degree m, and cm(γ) = 1 holds.

It remains to show, that cm is the unique solution of the minimum problem.

Assume that qm is a polynomial of degree m such that qm(γ) = 1 and

maxα≤t≤β

|qm(t)| ≤1

cm(1 + 2(γ − β)/(β − α))=: cm.

TUHH Heinrich Voss Krylov subspace methods Eigenvalue problems 2012 20 / 95

Page 36: Lanczos method ct. - Dipartimento di Informatica · Lanczos Method Lanczos method ct. The coefficients are obtained from

Lanczos Method

Proof

We restrict ourselves to the case γ > β.

From t := 1 + 2(γ − β)/(β − α) 6∈ [−1, 1] it follows that cm is defined, and

obviously, cm is a polynomial of degree m, and cm(γ) = 1 holds.

It remains to show, that cm is the unique solution of the minimum problem.

Assume that qm is a polynomial of degree m such that qm(γ) = 1 and

maxα≤t≤β

|qm(t)| ≤1

cm(1 + 2(γ − β)/(β − α))=: cm.

Obviously, the Chebyshev polynomial cm(τ) attains the values +1 and −1

alternately at the arguments τj := cos(jπ/k), j = 0, . . . , k . Hence,

cm(tj) =(−1)j

cmfor tj :=

2τj − α− β

β − α, j = 0, . . . ,m.

TUHH Heinrich Voss Krylov subspace methods Eigenvalue problems 2012 20 / 95

Page 37: Lanczos method ct. - Dipartimento di Informatica · Lanczos Method Lanczos method ct. The coefficients are obtained from

Lanczos Method

Proof ct.

Let rm := cm − qm. Then from

|qm(tj)| ≤1

cm= |cm(tj)|, j = 0, . . . ,m,

one obtains that

r(tj) ≥ 0, if j is even, r(tj) ≤ 0, if j is odd.

TUHH Heinrich Voss Krylov subspace methods Eigenvalue problems 2012 21 / 95

Page 38: Lanczos method ct. - Dipartimento di Informatica · Lanczos Method Lanczos method ct. The coefficients are obtained from

Lanczos Method

Proof ct.

Let rm := cm − qm. Then from

|qm(tj)| ≤1

cm= |cm(tj)|, j = 0, . . . ,m,

one obtains that

r(tj) ≥ 0, if j is even, r(tj) ≤ 0, if j is odd.

From the continuity of rm we get the existence of a root tj in every interval

[tj , tj−1], j = 1, . . . , k . Moreover, rm(γ) = cm(γ)− qm(γ) = 0.

TUHH Heinrich Voss Krylov subspace methods Eigenvalue problems 2012 21 / 95

Page 39: Lanczos method ct. - Dipartimento di Informatica · Lanczos Method Lanczos method ct. The coefficients are obtained from

Lanczos Method

Proof ct.

Let rm := cm − qm. Then from

|qm(tj)| ≤1

cm= |cm(tj)|, j = 0, . . . ,m,

one obtains that

r(tj) ≥ 0, if j is even, r(tj) ≤ 0, if j is odd.

From the continuity of rm we get the existence of a root tj in every interval

[tj , tj−1], j = 1, . . . , k . Moreover, rm(γ) = cm(γ)− qm(γ) = 0.

Therefore, the polynomial rm of degree m has at least m + 1 roots, and it

follows that rm(t) ≡ 0, i.e. qm = cm. �

TUHH Heinrich Voss Krylov subspace methods Eigenvalue problems 2012 21 / 95

Page 40: Lanczos method ct. - Dipartimento di Informatica · Lanczos Method Lanczos method ct. The coefficients are obtained from

Lanczos Method

THEOREM 10.3

The angle δ(ui ,Km(v1,A)) between the exact eigenvector ui and the m-th

Krylov space satisfies the inequality

tan δ(ui ,Km) ≤κi

cm−i(1 + 2ρi)tan δ(ui , v1) (1)

where

κ1 = 1, κi =

i−1∏

j=1

λn − λj

λi − λj

für i > 1 (2)

and

ρi =λi+1 − λi

λn − λi+1

. (3)

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Lanczos Method

THEOREM 10.3

The angle δ(ui ,Km(v1,A)) between the exact eigenvector ui and the m-th

Krylov space satisfies the inequality

tan δ(ui ,Km) ≤κi

cm−i(1 + 2ρi)tan δ(ui , v1) (1)

where

κ1 = 1, κi =

i−1∏

j=1

λn − λj

λi − λj

für i > 1 (2)

and

ρi =λi+1 − λi

λn − λi+1

. (3)

In particular for i = 1 one gets the estimate

tan δ(u1,Km(v1,A)) ≤

1

cm−1(1 + 2ρ1)tan δ(v1, u1) where ρ1 =

λ2 − λ1

λn − λ2

.

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Lanczos Method

THEOREM 10.3

The angle δ(ui ,Km(v1,A)) between the exact eigenvector ui and the m-th

Krylov space satisfies the inequality

tan δ(ui ,Km) ≤κi

cm−i(1 + 2ρi)tan δ(ui , v1) (1)

where

κ1 = 1, κi =

i−1∏

j=1

λn − λj

λi − λj

für i > 1 (2)

and

ρi =λi+1 − λi

λn − λi+1

. (3)

In particular for i = 1 one gets the estimate

tan δ(u1,Km(v1,A)) ≤

1

cm−1(1 + 2ρ1)tan δ(v1, u1) where ρ1 =

λ2 − λ1

λn − λ2

.

Crucial for the convergence is the distance of the two smallest eigenvalues

relative to the width of the entire spectrum.

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Lanczos Method

Proof

We consider the case i = 1 first.

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Lanczos Method

Proof

We consider the case i = 1 first.

Expanding the vector y i in the basis {ui} of eigenvectors yields

y i =

n∑

j=1

αjuj , where

n∑

j=1

|αj |2 = 1

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Lanczos Method

Proof

We consider the case i = 1 first.

Expanding the vector y i in the basis {ui} of eigenvectors yields

y i =

n∑

j=1

αjuj , where

n∑

j=1

|αj |2 = 1

from which we get

‖p(A)y1‖2 =

n∑

j=2

|p(λj)αj |2 ≤ max

j=2,...,n|p(λj)|

2 ≤ maxλ∈[λ2,λn]

|p(λ)|2,

and the statement follows from Theorem 10.2.

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Lanczos Method

Proof ct.

For i > 1 we consider in Lemma 10.1 polynomials of the form

p(λ) :=(λ− λ1) · · · · · (λ− λi−1)

(λi − λ1) · · · · · (λi − λi−1)q(λ)

with q ∈ Πm−i and q(λi) = 1.

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Lanczos Method

Proof ct.

For i > 1 we consider in Lemma 10.1 polynomials of the form

p(λ) :=(λ− λ1) · · · · · (λ− λi−1)

(λi − λ1) · · · · · (λi − λi−1)q(λ)

with q ∈ Πm−i and q(λi) = 1.

Then one gets as before

‖p(A)y i‖2 ≤ maxλ∈[λi+1,λn]

i−1∏

j=1

λ− λj

λi − λj

q(λ)

≤i−1∏

j=1

λn − λj

λi − λj

maxλ∈[λi+1,λn]

|q(λ)|.

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Lanczos Method

Proof ct.

For i > 1 we consider in Lemma 10.1 polynomials of the form

p(λ) :=(λ− λ1) · · · · · (λ− λi−1)

(λi − λ1) · · · · · (λi − λi−1)q(λ)

with q ∈ Πm−i and q(λi) = 1.

Then one gets as before

‖p(A)y i‖2 ≤ maxλ∈[λi+1,λn]

i−1∏

j=1

λ− λj

λi − λj

q(λ)

≤i−1∏

j=1

λn − λj

λi − λj

maxλ∈[λi+1,λn]

|q(λ)|.

The result follows by minimizing this expression over all polynomials q

satisfying the constraint q(λi) = 1.�

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Lanczos Method

THEOREM 10.4 (Kaniel & Paige; 1. eigenvalue)

Let A ∈ Cn×n be Hermitean with eigenvalues λ1 ≤ λ2 ≤ · · · ≤ λn and

corresponding orthonormal eigenvectors u1, . . . , un.

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Lanczos Method

THEOREM 10.4 (Kaniel & Paige; 1. eigenvalue)

Let A ∈ Cn×n be Hermitean with eigenvalues λ1 ≤ λ2 ≤ · · · ≤ λn and

corresponding orthonormal eigenvectors u1, . . . , un.

If θ(m)1 ≤ · · · ≤ θ

(m)m denote the eigenvalues of the matrix Tm obtained after m

steps of Lanczos’ method, then

0 ≤ θ(m)1 − λ1 ≤ (λn − λ1)

(

tan δ(u1, v1)

cm−1(1 + 2ρ1)

)2

,

where ρ1 = (λ2 − λ1)/(λn − λ2).

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Lanczos Method

THEOREM 10.4 (Kaniel & Paige; 1. eigenvalue)

Let A ∈ Cn×n be Hermitean with eigenvalues λ1 ≤ λ2 ≤ · · · ≤ λn and

corresponding orthonormal eigenvectors u1, . . . , un.

If θ(m)1 ≤ · · · ≤ θ

(m)m denote the eigenvalues of the matrix Tm obtained after m

steps of Lanczos’ method, then

0 ≤ θ(m)1 − λ1 ≤ (λn − λ1)

(

tan δ(u1, v1)

cm−1(1 + 2ρ1)

)2

,

where ρ1 = (λ2 − λ1)/(λn − λ2).

Crucial for the speed of convergence is the growth of cj−1(1 + 2ρ1), i.e. the

separation of the first two eigenvalues relative to the width of the entire

spectrum of A.

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Lanczos Method

Proof

The left inequality follows from Rayleigh’s principle.

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Lanczos Method

Proof

The left inequality follows from Rayleigh’s principle.

We have

θ(m)1 = min

x∈Km(v1,A),x 6=0

xHAx

xHx,

and, since each x ∈ Km(v1,A) can be represented as x = q(A)v1 for some

q ∈ Πm−1, it follows

θ(m)1 − λ1 = min

x∈Km(v1,A),x 6=0

xH(A − λ1I)x

xHx

= minq∈Πm−1,q 6=0

(v1)Hq(A)H(A − λ1I)q(A)v1

(v1)Hq(A)2v1.

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Lanczos Method

Proof ct.

With v1 =∑n

j=1 αjuj it holds

θ(m)1 − λ1 = min

q∈Πm−1,q 6=0

∑nj=2(λj − λ1)|αjq(λj)|

2

∑nj=1 |αjq(λj)|2

≤ (λn − λ1) minq∈Πm−1,q 6=0

∑nj=2 |αjq(λj)|

2

|α1q(λ1)|2

≤ (λn − λ1) minq∈Πm−1,q 6=0

maxj=2,...,n

|q(λj)|2

|q(λ1)|2·

∑nj=2 |αj |

2

|α1|2

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Lanczos Method

Proof ct.

With v1 =∑n

j=1 αjuj it holds

θ(m)1 − λ1 = min

q∈Πm−1,q 6=0

∑nj=2(λj − λ1)|αjq(λj)|

2

∑nj=1 |αjq(λj)|2

≤ (λn − λ1) minq∈Πm−1,q 6=0

∑nj=2 |αjq(λj)|

2

|α1q(λ1)|2

≤ (λn − λ1) minq∈Πm−1,q 6=0

maxj=2,...,n

|q(λj)|2

|q(λ1)|2·

∑nj=2 |αj |

2

|α1|2

Defining p(λ) = q(λ)/q(λ1), and observing that the set of all p:s when q

passes through the set Πm−1 is the set of all polynomials of degree not

exceeding m − 1 and satisfying the constraint p(λ1) = 1 we get

θ(m)1 − λ1 ≤ (λn − λ1) tan2 δ(u1, v1) min

p∈Πm−1,p(λ1)=1max

λ∈[λ2,λn]|p(λ)|2

and the statement follows from Theorem 10.2.�

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Lanczos Method

Example

Matrix A has eigenvalue λ1 = 1 and 99 eigenvalues uniformly distributed in

[α, β].it. [α, β]=[20,100] [α, β]=[2,100]

error bound error bound

2 3.3890e+001 3.4832e+003 1.9123e+001 1.1385e+005

3 1.9708e+001 1.0090e+002 1.0884e+001 7.5862e+004

4 6.6363e+000 1.5083e+001 5.8039e+000 5.5241e+004

5 1.5352e+000 2.2440e+000 4.1850e+000 3.8224e+004

10 7.9258e-005 1.6293e-004 1.8948e+000 4.2934e+003

15 4.3730e-009 1.1827e-008 1.0945e+000 4.1605e+002

20 2.9843e-013 8.5861e-013 1.6148e-001 3.9725e+001

25 1.4843e-002 3.7876e+000

30 1.6509e-003 3.6109e-001

35 2.0658e-004 3.4423e-002

40 5.8481e-006 3.2816e-003

45 9.5770e-007 3.1285e-004

50 3.0155e-009 2.9824e-005

55 8.0487e-012 2.8432e-006

60 3.1530e-014 2.7105e-007

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Lanczos Method

THEOREM 10.5 (Kaniel & Paige; higher eigenvalues)

Under the conditions of Theorem 10.4 it holds

0 ≤ θ(m)j − λj ≤ (λn − λ1)

(

κ(m)j tan δ(v1, ui)

cm−j(1 + 2ρj)

)2

with

ρj = (λj+1 − λj)/(λn − λj+1),

and

κ(m)1 ≡ 1, κ

(m)j =

j−1∏

i=1

λn − θ(m)i

λj − θ(m)i

.

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Lanczos Method

THEOREM 10.5 (Kaniel & Paige; higher eigenvalues)

Under the conditions of Theorem 10.4 it holds

0 ≤ θ(m)j − λj ≤ (λn − λ1)

(

κ(m)j tan δ(v1, ui)

cm−j(1 + 2ρj)

)2

with

ρj = (λj+1 − λj)/(λn − λj+1),

and

κ(m)1 ≡ 1, κ

(m)j =

j−1∏

i=1

λn − θ(m)i

λj − θ(m)i

.

The general case j > 1 can be proved by using the maxmin characterization

of θ(m)j of Courant and Fischer.

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Lanczos Method

THEOREM 10.5 (Kaniel & Paige; higher eigenvalues)

Under the conditions of Theorem 10.4 it holds

0 ≤ θ(m)j − λj ≤ (λn − λ1)

(

κ(m)j tan δ(v1, ui)

cm−j(1 + 2ρj)

)2

with

ρj = (λj+1 − λj)/(λn − λj+1),

and

κ(m)1 ≡ 1, κ

(m)j =

j−1∏

i=1

λn − θ(m)i

λj − θ(m)i

.

The general case j > 1 can be proved by using the maxmin characterization

of θ(m)j of Courant and Fischer.

Analogous results hold for the largest eigenvalues and Ritz values.

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Lanczos Method

Orthogonality of basis vectors

In exact arithmetic the Lanczos method generates an orthonormal basis of the

Krylov space Km(v1,A). In the algorithm only the orthogonality with respect to

two basis vectors v j and v j−1 obtained in the last two steps is enforced, with

respect to the previous v i :s it follows from the symmetry of A.

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Lanczos Method

Orthogonality of basis vectors

In exact arithmetic the Lanczos method generates an orthonormal basis of the

Krylov space Km(v1,A). In the algorithm only the orthogonality with respect to

two basis vectors v j and v j−1 obtained in the last two steps is enforced, with

respect to the previous v i :s it follows from the symmetry of A.

It can be shown that in floating point arithmetic the orthogonality is destroyed

when a sequence θ(j)i , j = 1, 2, . . . , of Ritz values has converged to an

eigenvalue λ of A, i.e. if the residual βjs(j)j,i has become small.

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Lanczos Method

Orthogonality of basis vectors

In exact arithmetic the Lanczos method generates an orthonormal basis of the

Krylov space Km(v1,A). In the algorithm only the orthogonality with respect to

two basis vectors v j and v j−1 obtained in the last two steps is enforced, with

respect to the previous v i :s it follows from the symmetry of A.

It can be shown that in floating point arithmetic the orthogonality is destroyed

when a sequence θ(j)i , j = 1, 2, . . . , of Ritz values has converged to an

eigenvalue λ of A, i.e. if the residual βjs(j)j,i has become small.

Thereafter all v j :s obtain a component in the direction of the eigenspace of

the converged eigenvalue, and a duplicate copy of that eigenvalue will show

up in the spectrum of the tridiagonal matrix Tm.

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Lanczos Method

Orthogonality of basis vectors

In exact arithmetic the Lanczos method generates an orthonormal basis of the

Krylov space Km(v1,A). In the algorithm only the orthogonality with respect to

two basis vectors v j and v j−1 obtained in the last two steps is enforced, with

respect to the previous v i :s it follows from the symmetry of A.

It can be shown that in floating point arithmetic the orthogonality is destroyed

when a sequence θ(j)i , j = 1, 2, . . . , of Ritz values has converged to an

eigenvalue λ of A, i.e. if the residual βjs(j)j,i has become small.

Thereafter all v j :s obtain a component in the direction of the eigenspace of

the converged eigenvalue, and a duplicate copy of that eigenvalue will show

up in the spectrum of the tridiagonal matrix Tm.

This effect was first observed and studied by Paige (1971). A detailed

discussion is contained in the monograph of Parlett (1998).

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Lanczos Method

Orthogonality of basis vectors ct.

Note that these multiple Ritz values have nothing to do with possible multiple

eigenvalues of a given matrix, they occur simply as a result of a converged

eigenvalue.

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Lanczos Method

Orthogonality of basis vectors ct.

Note that these multiple Ritz values have nothing to do with possible multiple

eigenvalues of a given matrix, they occur simply as a result of a converged

eigenvalue.

For multiple eigenvalues the Lanczos (and Arnoldi) method in exact arithmetic

can only detect one eigenvector, namely the projection of the initial vector v1

to the corresponding eigenspace. Further eigenvectors can only be obtained

restarting with a different initial vector or by a block Lanczos (Arnoldi) method.

TUHH Heinrich Voss Krylov subspace methods Eigenvalue problems 2012 31 / 95

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Lanczos Method

Orthogonality of basis vectors ct.

Note that these multiple Ritz values have nothing to do with possible multiple

eigenvalues of a given matrix, they occur simply as a result of a converged

eigenvalue.

For multiple eigenvalues the Lanczos (and Arnoldi) method in exact arithmetic

can only detect one eigenvector, namely the projection of the initial vector v1

to the corresponding eigenspace. Further eigenvectors can only be obtained

restarting with a different initial vector or by a block Lanczos (Arnoldi) method.

A simple trick to detect duplicate copies of an eigenvalue advocated by

Cullum & Willoughby (1986) is the following.

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Lanczos Method

Orthogonality of basis vectors ct.

Note that these multiple Ritz values have nothing to do with possible multiple

eigenvalues of a given matrix, they occur simply as a result of a converged

eigenvalue.

For multiple eigenvalues the Lanczos (and Arnoldi) method in exact arithmetic

can only detect one eigenvector, namely the projection of the initial vector v1

to the corresponding eigenspace. Further eigenvectors can only be obtained

restarting with a different initial vector or by a block Lanczos (Arnoldi) method.

A simple trick to detect duplicate copies of an eigenvalue advocated by

Cullum & Willoughby (1986) is the following.

Compute the eigenvalues of the reduced matrix Tm ∈ Rm−1×m−1 obtained

from Tm by detracting the first row and column. Those eigenvalues that differ

less than a small multiple times machine precision from the eigenvalues of Tm

are the unwanted eigenvalues, i.e. the ones due to loss of orthogonality.

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Lanczos Method

Example

Convergence of the Lanczos process for A = diag (rand(100,1))

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