Page 141 - Numerical methods for chemical engineering

Computing extremal eigenvalues 127

[0]
coefficients {c 1 , c 2 ,..., c N } to be nonzero. From v , we obtain a sequence of new vectors
[1]
[2]
v , v ,...bythe rule
Av [k]
[k+1]
v = (3.130)
Av
[k]
Writing these vectors as linear combinations of eigenvectors, we have

Av [0] = A c 1 w [1] +· · · + c N w [N] = c 1 λ 1 w [1] +· · · + c N λ N w [N] (3.131)
so that
Av [0] c 1 λ 1 w [1] + c 2 λ 2 w [2] + ··· + c N λ N w [N]
v [1] = = (3.132)
Av Av
[0] [0]
Next,
[1] [N] 2 [1] 2 [N]
A c 1 λ 1 w +· · · + c N λ N w c 1 λ w +· · · + c N λ w
Av [1] = = 1 N (3.133)
Av Av
[0] [0]
so that
2
2
2
c 1 λ w [1] + c 2 λ w [2] +· · · + c N λ w [N]
v [2] = 1 2 N (3.134)
Av × Av
[1] [0]
or in general,
k
k
k
c 1 λ w [1] + c 2 λ w [2] +· · · + c N λ w [N]
1
2
v [k] = N (3.135)
k
k
k
c 1 λ w [1] + c 2 λ w [2] +· · · + c N λ w
[N]
1 2 N
Let us assume that the eigenvalues are ordered by decreasing modulus, and that λ 1 is both
distinct and has a larger modulus than λ 2 :
|λ 1 | > |λ 2 |≥|λ 3 |≥· · ·≥|λ N−1 |≥|λ N | (3.136)
Note that this is a stricter statement than saying that λ 1 is distinct, as if λ 2 = λ 1 ,λ 1 is
distinct, but |λ 2 |=|λ 1 |. If (3.136) holds, as k →∞,
k

λ λ ≥ λ ≥· · · ≥ λ ≥ λ (3.137)
k k k k
1 2 3 N−1 N
and for any finite {c 1 , c 2 ,..., c N }, we eventually have
k

c 1 λ k c 2 λ k ≥ c 3 λ k ≥· · · ≥ c N−1 λ N−1 ≥ c N λ k (3.138)

3
1
N
2
Therefore, as k →∞,
k
c 1 λ w [1]
v [k] ≈ 1 (3.139)
k
c 1 λ w
1
[1]
[k]
In this limit, as Av [k] ≈ λ 1 v ,
λ k ≈ v [k] · Av [k] w [1] ≈ v [k+1] (3.140)
We have assumed above that c 1 = 0; however, this is not really necessary. In the presence of
round-off error, which mixes in some w [1] component, this algorithm will find λ 1 and w [1]
even if initially c 1 were zero.