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Lecture Notes on Solving Large Scale Eigenvalue Problems.pdf
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172 CHAPTER 9. ARNOLDI AND LANCZOS ALGORITHMS

√

Reorthogonalization takes place in the j-th Lanczos step if maxk(ωj+1,k) > macheps. qj+1 is orthogonalized against all vectors qk with ωj+1,k > macheps3/4. In the following iteration step also qj+2 is orthogonalized against these vectors. In Table 9.2 the base10 logarithms of the values |wi,j |/macheps obtained with this procedure are listed where |wi,j | = |qi qj |, 1 ≤ j ≤ i ≤ 50 and macheps ≈ 2.2 · 10−16. In Table 9.3 the base-10 logarithms of the estimates |ωi,j |/macheps are given. The estimates are too high by (only) an order of magnitude. However, the procedure succeeds in that the resulting {qk} are semi-orthogonal.

9.7Block Lanczos

As we have seen, the Lanczos algorithm produces a sequence {qi} of orthonormal vectors. These Lanczos vectors build an orthonormal basis for the Krylov subspace Kj (x) = span{q1, . . . , qj } Rn. The restriction of A to Kj (x) is an unreduced tridiagonal matrix. However the Lanczos algorithm cannot detect the multiplicity of the eigenvalues it computes. This limitation prompted the development of the block version of the Lanczos process (Block Lanczos algorithm), which is capable of determining multiplicities of eigenvalues up to the block size.

The idea is not to start with a single vector q1 Rn but with a set of mutually orthogonal vectors which we take as the columns of the matrix Q1 Rn×p with the block size p > 1.

Associated with Q1 is the ‘big’ Krylov subspace

(9.36)

Kjp(Q1) = span{Q1, AQ1, . . . , Aj−1Q1}.

(We suppose, for simplicity, that Aj−1Q1 has rank p. Otherwise we would have to consider variable block sizes.)

The approach is similar to the scalar case with p = 1: Let Q1, . . . , Qj Rn×p be pairwise orthogonal block matrices (Qi Qk = O for i 6= k) with orthonormal columns (Qi Qi = Ip for all i ≤ j). Then, in the j-th iteration step, we obtain the matrix AQj and orthogonalize it against matrices Qi, i ≤ j. The columns of the matrices are obtained by means of the QR factorization or with the Gram–Schmidt orthonormalization process. We obtained the following:

Algorithm 9.5 Block Lanczos algorithm

1: Choose Q1 Fn×p such that Q1Q1 = Ip. Set j := 0 and Fn×p V := 0.

ˆ

This algorithm generates a block tridiagonal matrix Tj with the diagonal blocks Ai,

i ≤ j, the lower diagonal blocks Bi, i < j, and the Krylov basis [Q1, . . . , Qj ] of Kjp(Q1).

2:for j ≥ 0 do

3:if j > 0 then

4:V =: Qj+1Bj ; /* QR decomposition */

5:V := −Qj Bj ;

6:end if

7:j := j + 1;

8:Aj := Qj V ;

9:V := V − Qj Aj ;

10:Test for convergence (Ritz pairs, evaluation of error)

11:end for

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LANCZOS BLOCK .7.9

173

0

2 0

Table 9.2: MATLAB demo on the loss of orthogonality among Lanczos vectors: Lanczos with partial reorthogonalization. round(log10(abs(I-Q50Q50)/eps))

CHAPTER 9. ARNOLDI AND LANCZOS ALGORITHMS

174

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0

2

0

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

1

2

2

2

2

2

2

2

2

1

2

1

2

1

1

0

2

0

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

1

2

2

2

2

2

2

2

2

2

2

2

1

2

1

1

0

2

0

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2

2

2

2

2

2

2

2

2

2

2

2

2

1

2

1

1

0

2

0

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2

2

2

2

2

2

2

2

2

2

2

2

2

2

1

2

1

1

0

2

0

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

1

2

1

1

0

2

0

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2

2

2

3

2

3

2

3

2

2

2

2

2

2

2

2

2

2

1

2

0

2

0

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2

2

3

3

3

3

3

3

3

2

3

2

2

2

2

2

2

2

2

1

2

0

2

0

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2

3

3

3

3

3

3

3

3

3

3

3

2

3

2

2

2

2

2

2

1

2

0

2

0

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

3

3

3

3

3

3

3

3

3

3

3

3

3

3

3

2

2

2

2

2

2

1

2

0

2

0

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

3

3

3

3

3

3

3

3

3

3

3

3

3

3

3

3

2

3

2

2

2

2

1

2

0

2

0

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

3

3

4

3

4

3

4

3

4

3

3

3

3

3

3

3

3

3

3

2

2

2

2

1

2

0

2

0

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

3

4

4

4

4

4

4

4

4

4

4

4

3

3

3

3

3

3

3

3

2

2

2

2

1

2

0

2

0

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

4

4

4

4

4

4

4

4

4

4

4

4

4

4

4

3

3

3

3

3

3

2

2

2

2

1

2

0

2

0

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

4

4

4

4

4

4

4

4

4

4

4

4

4

4

4

4

3

4

3

3

3

3

2

3

2

2

1

2

0

2

0

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

4

4

5

5

5

5

5

5

5

4

5

4

4

4

4

4

4

4

4

3

3

3

3

3

3

2

2

1

2

0

2

0

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

4

5

5

5

5

5

5

5

5

5

5

5

4

5

4

4

4

4

4

4

3

3

3

3

3

3

2

2

2

2

0

2

0

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

5

5

5

5

5

5

5

5

5

5

5

5

5

5

5

4

5

4

4

4

4

4

4

3

3

3

3

2

2

2

2

0

2

0

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

5

6

5

6

6

6

6

6

5

6

5

5

5

5

5

5

5

5

4

4

4

4

4

4

3

3

3

3

2

2

2

2

0

2

0

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

6

6

6

6

6

6

6

6

6

6

6

6

6

5

5

5

5

5

5

5

5

4

4

4

4

3

3

3

3

2

2

2

2

0

2

0

 

 

 

 

 

 

 

 

 

 

 

 

 

 

6

6

6

6

6

6

6

6

6

6

6

6

6

6

6

6

5

5

5

5

5

5

4

4

4

4

4

4

3

3

2

3

2

2

0

2

0

 

 

 

 

 

 

 

 

 

 

 

 

 

6

6

7

7

7

7

7

7

7

7

7

6

6

6

6

6

6

6

6

5

5

5

5

5

5

4

4

4

4

3

3

3

3

2

2

0

2

0

 

 

 

 

 

 

 

 

 

 

 

 

7

7

7

7

7

7

7

7

7

7

7

7

7

7

6

7

6

6

6

6

6

6

5

5

5

5

4

4

4

4

3

3

3

3

2

2

0

2

0

 

 

 

 

 

 

 

 

 

 

 

7

7

8

7

8

8

8

7

8

7

7

7

7

7

7

7

7

7

7

6

6

6

6

5

5

5

5

5

5

4

4

3

3

3

3

2

2

0

2

0

 

 

 

 

 

 

 

 

 

 

8

8

8

8

8

8

8

8

8

8

8

8

8

8

7

7

7

7

7

7

7

7

6

6

6

6

5

5

5

5

4

4

4

4

3

3

2

2

0

2

0

 

 

 

 

 

 

 

 

 

7

7

8

8

8

8

8

7

8

7

7

7

7

7

7

7

7

7

7

6

6

6

6

5

6

5

5

5

5

4

4

3

3

4

4

3

3

2

2

0

2

0

 

 

 

 

 

 

 

 

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

3

3

4

4

4

3

3

2

2

0

2

0

 

 

 

 

 

 

 

2

3

3

3

3

3

3

3

3

3

3

3

3

3

3

3

3

3

3

3

3

3

3

3

3

2

2

2

2

2

4

4

4

4

4

4

4

3

3

2

2

0

2

0

 

 

 

 

 

 

3

3

3

3

3

3

3

3

3

3

3

3

3

3

3

3

3

3

3

3

3

3

3

3

3

3

3

3

3

4

4

4

4

4

4

4

4

4

3

3

2

2

0

2

0

 

 

 

 

 

4

4

4

4

4

4

4

4

4

4

4

4

4

4

4

4

4

4

4

4

4

4

4

4

4

4

4

3

4

4

5

5

5

5

5

5

5

4

4

3

4

2

3

0

2

0

 

 

 

 

4

4

5

5

5

5

5

5

5

5

5

5

4

4

4

4

4

4

4

4

4

4

4

4

4

4

4

5

4

5

5

5

5

5

5

5

5

5

5

5

4

4

3

3

0

2

0

 

 

 

5

5

5

5

5

5

5

5

5

5

5

5

5

5

5

5

5

5

5

5

5

5

5

5

5

5

5

4

6

5

6

6

6

6

6

6

6

5

5

5

5

4

4

3

3

0

2

0

 

 

6

6

6

6

6

6

6

6

6

6

6

6

6

6

6

6

6

6

6

6

6

6

6

6

6

6

5

6

6

6

6

7

6

7

6

7

6

6

6

6

5

5

4

4

3

3

0

2

0

 

7

7

7

7

7

7

7

7

7

7

7

7

7

7

7

7

7

7

7

7

7

7

7

7

7

6

7

6

7

7

7

7

8

7

7

7

7

7

7

6

6

6

6

5

5

3

3

0

2

0

Table

9.3:

 

MATLAB

demo

 

on

the

loss

 

of

orthogonality

among

Lanczos

vectors:

 

Lanczos

with

partial

reorthogonalization.

 

 

 

round(log10(abs(I-W50)/eps))

9.8. EXTERNAL SELECTIVE REORTHOGONALIZATION

175

ˆ

, . . . , Qj ] be the Krylov basis generated by Algorithm 9.5. Then, in

Let Qj := [Q1, Q2

 

 

 

 

 

 

 

 

ˆ

 

this basis, the projection of A is the block tridiagonal matrix Tj

 

 

 

B1

A2 ...

 

 

 

 

 

 

 

A1

B1

 

 

 

 

 

 

Qˆj AQˆj

= Tˆj =

 

... ...

B

 

, Ai, Bi Rp×p.

 

 

 

 

 

 

j−1

 

 

 

 

 

 

B

j−1

A

j

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

ˆ

If matrices Bi are chosen to be upper triangular, then Tj is a band matrix with bandwidth 2p + 1!

Similarly as in scalar case, in the j-th iteration step we obtain the equation

ˆ − ˆ ˆ ˆ

AQj Qj Tj = Qj+1Bj Ej + Fj ,

O

.

.

Ej = . ,

O

Ip

ˆ

where Fj accounts for the e ect of roundo error. Let (ϑi, yi) be a Ritz pair of A in Kjp(Q1). Then

ˆ

ˆ

yi = Qj si,

Tj si = ϑisi.

As before, we can consider the residual norm to study the accuracy of the Ritz pair (ϑi, yi) of A

 

Ayi

 

ϑiyi

 

=

 

AQˆj si

 

ϑiQˆj si

 

Qj+1Bj Ej si

 

=

Bj

 

..

.

k

 

−

 

k

 

k

 

−

 

k ≈ k

 

k

 

 

 

s

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

sj(p−1)+1,i

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

jp+1,i

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

We have to compute the bottom p components of the eigenvectors si in order to test for convergence.

Similarly as in the scalar case, the mutual orthogonality of the Lanczos vectors (i.e.,

ˆ

the columns of Qj ) is lost, as soon as convergence sets in. The remedies described earlier are available: full reorthogonalization or selective orthogonalization.

9.8External selective reorthogonalization

If many eigenvalues are to be computed with the Lanczos algorithm, it is usually advisable to execute shift-and-invert Lanczos with varying shifts [4].

In each new start of a Lanczos procedure, one has to prevent the algorithm from finding already computed eigenpairs. We have encountered this problem when we tried to compute multiple eigenpairs by simple vector iteration. Here, the remedy is the same as there. In the second and further runs of the Lanczos algorithm, the starting vectors are made orthogonal to the already computed eigenvectors. We know that in theory all Lanczos vectors will be orthogonal to the previously computed eigenvectors. However, because the previous eigenvectors have been computed only approximately the initial vectors are not orthogonal to the true eigenvectors. Because of this and because of floating point errors loss of orthogonality is observed. The loss of orthogonality can be monitored similarly as with partial reorthogonalization. For details see [4].

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