¡ ¢

§ ¬ ¥w

¢ ¢ ¢¡ ¢

¡

DA 7 ¡¤4 3¤2

65

C

§

The action of on gives plus a rank-

¢ ¢ ¢

one matrix.

u¥

As was noted earlier, the algorithm may break down in case the norm of vanishes at

u¦

a certain step . In this case, the vector cannot be computed and the algorithm stops.

"U

W

¨ ¡¡µ£ „ ¢

§|5¥ yq¢| ¢ £ ¥§

„ 5| j C¦£¥

5§

tC

9 "–© ¡"

§

1B

Still to be determined are the conditions under which this situation occurs.

h¤ ¡ v A£

£¦ A

' ' ©' )$ )

§

¬

u `u

Arnoldi™s algorithm breaks down at step (i.e., in line Y

£

"U

W

5 of Algorithm 6.1), if and only if the minimal polynomial of is of degree . Moreover, ¦

W

§

u

in this case the subspace is invariant under .

£ A™¤ ¢¡6

£

u¥

If the degree of the minimal polynomial is , then must be equal to zero.

u¦ u

Indeed, otherwise can be de¬ned and as a result would be of dimension . w

’U

W "U

W y

Then Proposition 6.2 would imply that , which is a contradiction. To prove w

y

¬

u¥

the converse, assume that . Then the degree of the minimal polynomial of is ¦

Y

W

such that . Moreover, it is impossible that . Otherwise, by the ¬rst part of this

proof, the vector would be zero and the algorithm would have stopped at the earlier

¥

step number . The rest of the result follows from Proposition 6.1.

u

A corollary of the proposition is that a projection method onto the subspace will

be exact when a breakdown occurs at step . This result follows from Proposition 5.6 seen

in Chapter 5. It is for this reason that such breakdowns are often called lucky breakdowns.

§¨6§

F ©DC776P5 P¡& I B & 8#DCA ) 8 6

3

)B

E 'B A 8 A E I 5

¢

In the previous description of the Arnoldi process, exact arithmetic was assumed, mainly

for simplicity. In practice, much can be gained by using the Modi¬ed Gram-Schmidt or the

Householder algorithm instead of the standard Gram-Schmidt algorithm. With the Modi-

¬ed Gram-Schmidt alternative the algorithm takes the following form:

˜ • R—˜

˜Q¤ v ¢

¡¦ ¤ ¤ ¥E

¦

¥ P P¥ D P¥ B 76 8 R¦Q

S¥

D

Q

5Q V2

@8

R¤A

3£

' 7$

%# ) 2

1

1. Choose a vector of norm 1

¦

W #V¡™(

}

¬

2. For Do:

! ( 3 ( ˜

u¥ y (¬

§ u ¦§

3. Compute

T

¬v

4. For Do:

E

Y(V(

XR

u ¥ y vR £ ¬

u

5. ¦(

u 3 u ¥ ¬ ² R ¦ vR £

u

6. ¥

7. EndDo

£u ¥ ¬ ¬

u u£ u $u

8. . If Stop

Y

£

u u ¤ u ¥ W’U u ¦ ’U