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Unformatted text preview: of the Neumann series. Y
C Neumann Series For A ∈ C n×n , the following statements are equivalent.
• The Neumann series I + A + A2 + · · · converges. (7.10.8) • ρ(A) < 1. (7.10.9) • lim Ak = 0. (7.10.10) k→∞ In which case, (I − A)−1 exists and ∞
k=0 Ak = (I − A)−1 . (7.10.11) Proof. We know from (7.10.5) that (7.10.9) and (7.10.10) are equivalent, and it
was argued on p. 126 that (7.10.10) implies (7.10.8), so the theorem can be estab∞
lished by proving that (7.10.8) implies (7.10.9). If
k=0 A converges, it follows
that k=0 J∗ must converge for each Jordan block J∗ in the Jordan form for A.
This together with (7.10.7) implies that
k=0 J∗ ii =
k=0 λ converges for Copyright c 2000 SIAM Buy online from SIAM
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7.10 Diﬀerence Equations, Limits, and Summability
http://www.amazon.com/exec/obidos/ASIN/0898714540 619 each λ ∈ σ (A) , and this scalar geometric series converges if and only if |λ| < 1.
Thus the convergence of
implies ρ(A) < 1. When it converges,
because (I − A)(I + A + A2 + · · · + Ak−1 ) = I − Ak →
k=0 A = (I − A)
I as k → ∞.
The following examples illustrate the utility of the previous results for establishing some useful (and elegant) statements concerning spectral radius. It is illegal to print, duplicate, or distribute this material
Please report violations to [email protected] Example 7.10.1
Spectral Radius as a Limit. It was shown in Example 7.1.4 (p. 497) that
if A ∈ C n×n , then ρ (A) ≤ A for every matrix norm. But this was just the
precursor to the following elegant relationship between spectral radius and norm. D
E Problem: Prove that for every matrix norm,
Ak ρ(A) = lim k→∞ 1/k . T
H k Solution: First note that ρ (A) = ρ Ak ≤ Ak
Next, observe that ρ A/(ρ (A) + ) < 1 for every
ρ (A) + lim k→∞ =⇒ ρ (A) ≤ Ak
> 0, so, by (7.10.5), lim Consequently...
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