334 CHAPTER 13. NORMS

∥∥∥(A+B)−1∥∥∥ ≤

∥∥A−1∥∥ ∣∣∣∣ 1

1− ∥A−1B∥

∣∣∣∣ . (13.7)

The above formula makes sense because∣∣∣∣A−1B

∣∣∣∣ < 1.

Proof: By Lemma 13.0.10,∥∥A−1B∥∥ ≤

∥∥A−1∥∥ ∥B∥ <

∥∥A−1∥∥ 1

∥A−1∥= 1 (13.8)

Then from the triangle inequality,∥∥(I +A−1B)x∥∥ ≥ ∥x∥ −

∥∥A−1Bx∥∥

≥ ∥x∥ −∥∥A−1B

∥∥ ∥x∥ =(1−

∥∥A−1B∥∥) ∥x∥

It follows that I + A−1B is one to one because from 13.8, 1 −∣∣∣∣A−1B

∣∣∣∣ > 0. Thus if(I +A−1B

)x = 0, then x = 0. Thus I +A−1B is also onto, taking a basis to a basis. Then

a generic y ∈ X is of the form y =(I +A−1B

)x and the above shows that∥∥∥(I +A−1B

)−1y∥∥∥ ≤

(1−

∣∣∣∣A−1B∣∣∣∣)−1 ∥y∥

which verifies 13.6. Thus (A+B) = A(I +A−1B

)is one to one and this with Lemma

13.0.10 implies 13.7. ■

Proposition 13.2.3 Suppose A is invertible, b ̸= 0, Ax = b, and (A+B)x1 = b1 where||B|| < 1/

∣∣∣∣A−1∣∣∣∣. Then

∥x1 − x∥∥x∥

≤∥∥A−1

∥∥ ∥A∥1− ∥A−1B∥

(∥b1 − b∥

∥b∥+

∥B∥∥A∥

)Proof: This follows from the above lemma.

∥x1 − x∥∥x∥

=

∥∥∥(I +A−1B)−1

A−1b1 −A−1b∥∥∥

∥A−1b∥

≤ 1

1− ∥A−1B∥

∥∥A−1b1 −(I +A−1B

)A−1b

∥∥∥A−1b∥

≤ 1

1− ∥A−1B∥

∥∥A−1 (b1 − b)∥∥+ ∥∥A−1BA−1b

∥∥∥A−1b∥

≤∥∥A−1

∥∥1− ∥A−1B∥

(∥b1 − b∥∥A−1b∥

+ ∥B∥)

because A−1b/∥∥A−1b

∥∥ is a unit vector. Now multiply and divide by ∥A∥ . Then

≤∥∥A−1

∥∥ ∥A∥1− ∥A−1B∥

(∥b1 − b∥

∥A∥ ∥A−1b∥+

∥B∥∥A∥

)≤

∥∥A−1∥∥ ∥A∥

1− ∥A−1B∥

(∥b1 − b∥

∥b∥+

∥B∥∥A∥

). ■

This shows that the number,∥∥A−1

∥∥ ∥A∥ , controls how sensitive the relative change inthe solution of Ax = b is to small changes in A and b. This number is called the condition