Linear Algebra, Theory And Applications, 2012a

346 NORMS FOR FINITE DIMENSIONAL VECTOR SPACES
little bit? This is clearly an interesting question because you often do not know A and b
exactly. If a small change in these quantities results in a large change in the solution, x,
then it seems clear this would be undesirable. In what follows ||·|| when applied to a linear
transformation will always refer to the operator norm.
Lemma 14.2.1 Let A, B ∈L(X, X) where X is a normed vector space as above. Then for
||·|| denoting the operator norm,
||AB|| ≤ ||A|| ||B|| .
Proof: This follows from the definition. Letting ||x|| ≤ 1, it follows from Theorem
14.0.10
||ABx||≤||A|| ||Bx|| ≤ ||A|| ||B|| ||x|| ≤ ||A|| ||B||
and so
||AB|| ≡ sup ||ABx|| ≤ ||A|| ||B|| .
||x||≤1
Lemma 14.2.2 Let A, B ∈L(X, X) ,A −1 ∈L(X, X) , and suppose ||B|| < 1/ ∣ ∣ ∣ A
−1 ∣ .
Then (A + B) −1 exists and
∣
∣∣(A + B) −1∣ ∣
∣ ∣∣ ∣∣ ≤ A −1∣ ∣ ∣ 1
∣1 −||A −1 B|| ∣ .
The above formula makes sense because ∣ ∣ ∣ ∣A −1 B ∣ ∣ ∣ ∣ < 1.
Proof: By Lemma 14.2.1,
∣ ∣ ∣A −1 B ∣ ∣ ≤ ∣ ∣ ∣A −1∣ ∣ ∣ ||B|| < ∣ ∣ ∣A −1∣ ∣ ∣ 1
||A −1 || =1
Suppose (A + B) x = 0. Then 0 = A ( I + A −1 B ) x and so since A is one to one,
(
I + A −1 B ) x =0. Therefore,
0 = ∣ ∣ ( I + A −1 B ) x ∣ ∣ ≥||x|| − ∣ ∣ ∣A −1 Bx ∣ ∣ ≥ ||x|| − ∣ ∣ ∣A −1 B ∣ ∣ ||x|| = ( 1 − ∣ ∣ ∣A −1 B ∣ ∣ ) ||x|| > 0
a contradiction. This also shows ( I + A −1 B ) is one to one. Therefore, both (A + B) −1 and
(
I + A −1 B ) −1
are in L (X, X). Hence
(A + B) −1 = ( A ( I + A −1 B )) −1
=
(
I + A −1 B ) −1
A
−1
Now if
for ||y|| ≤ 1, then
and so
and so
x = ( I + A −1 B ) −1
y
(
I + A −1 B ) x = y
||x|| ( 1 − ∣ ∣ A −1 B ∣ ∣ ) ≤ ∣ ∣ x + A −1 Bx ∣ ∣ ≤||y|| =1
||x|| = ∣∣ ( I + A −1 B ) ∣∣ −1 ∣∣ ∣∣ 1
y ≤
1 −||A −1 B||