Unformatted text preview: vector associated with λ1 . But rather than successively powering
A, the sequence An x0 /m(An x0 ) is more eﬃciently generated by starting with
x0 ∈ R (A − λ1 I) and setting
yn = Axn , νn = m(yn ), xn+1 = yn
E for n = 0, 1, 2, . . . . (7.3.17) Not only does xn → x, but as a bonus we get νn → λ1 because for all n,
Axn+1 = A2 xn /νn , so if νn → ν as n → ∞, the limit on the left-hand side
is Ax = λ1 x, while the limit on the right-hand side is A2 x/ν = λ2 x/ν. Since
these two limits must agree, λ1 x = (λ2 /ν )x, and this implies ν = λ1 .
H Summary. The sequence (νn , xn ) deﬁned by (7.3.17) converges to an eigenpair
(λ1 , x) for A provided that G1 x0 = 0 or, equivalently, x0 ∈ R (A − λ1 I).
Advantages. Each iteration requires only one matrix–vector product, and
this can be exploited to reduce the computational eﬀort when A is large
and sparse—assuming that a dominant eigenpair is the only one of interest.
Disadvantages. Only a dominant eigenpair is determined—something else
must be done if others are desired. Furthermore, it’s clear from (7.3.16) that
the rate at which (7.3.17) converges depends on how fast (λ2 /λ1 )n → 0, so
convergence is slow when |λ1 | is close to |λ2 |. IG
R Example 7.3.8 Y
C Inverse Power Method. Given a real approximation α ∈ σ (A) to any real
λ ∈ σ (A), this algorithm (also called the inverse iteration ) determines an
eigenpair (λ, x) for a diagonalizable matrix A ∈ m×m by applying the power
method to B = (A − αI)−1 . Recall from Exercise 7.1.9 that
x is an eigenvector for A ⇐⇒ x is an eigenvector for B,
λ ∈ σ (A) ⇐⇒ (λ − α)−1 ∈ σ (B). (7.3.18) If |λ − α| < |λi − α| for all other λi ∈ σ (A), then (λ − α)−1 is the dominant
eigenvalue of B because |λ − α|−1 > |λi − α|−1 . Therefore, applying the power
75 The relation between the power method and inverse iteration is clear to us now, but it originally
took 15 y...
View Full Document