Linear Algebra, Theory And Applications, 2012a

14.7. EXERCISES 363
Proof: From Lemma 14.6.4 there exists a unique fixed point for T N denoted here as x.
Therefore, T N x = x. NowdoingT to both sides,
T N T x = T x.
By uniqueness, T x = x because the above equation shows T x is a fixed point of T N and
there is only one fixed point of T N . In fact, there is only one fixed point of T because a
fixed point of T is automatically a fixed point of T N .
It remains to show T k x 1 → x, the unique fixed point of T N . If this does not happen,
there exists ε>0 and a subsequence, still denoted by T k such that
∣
∣T k x 1 − x ∣ ∣ ≥ ε
Now k = j k N + r k where r k ∈ {0, ··· ,N − 1} and j k is a positive integer such that
lim k→∞ j k = ∞. Then there exists a single r ∈ {0, ··· ,N − 1} such that for infinitely
many k, r k = r. Taking a further subsequence, still denoted by T k it follows
∣
∣T jkN+r x 1 − x ∣ ≥ ε (14.25)
However,
T jkN+r x 1 = T r T jkN x 1 → T r x = x
and this contradicts (14.25).
Theorem 14.6.6 Suppose ρ ( B −1 C ) < 1. Then the iterates in (14.18) converge to the
unique solution of (14.17).
Proof: Consider the iterates in (14.18). Let T x = B −1 Cx + b. Then
∣
∣T k x − T k y∣ = ∣ ( B −1 C ) k (
x − B −1 C ) ∣ ∣∣ k ∣∣ ∣∣ ∣∣ (
y ≤ B −1 C ) ∣ ∣
k ∣∣ ∣ |x − y| .
Here ||·|| refers to any of the operator norms. It doesn’t matter which one you pick because
they are all equivalent. I am writing the proof to indicate the operator norm taken with
respect to the usual norm on E. Since ρ ( B −1 C ) < 1, it follows from Gelfand’s theorem,
Theorem 14.3.3 on Page 349, there exists N such that if k ≥ N, then for some r 1/k < 1,
∣∣ ( B −1 C ) ∣ ∣
k ∣∣
1/k
∣