344 CHAPTER 14. NUMERICAL SOLUTIONS OF LINEAR SYSTEMS

7. If you are considering a system of the form Ax = b and A−1 does not exist, will eitherthe Gauss Seidel or Jacobi methods work? Explain. What does this indicate aboutusing either of these methods for finding eigenvectors for a given eigenvalue?

8. Verify that∥x∥

∞≡max{|xi| , i = 1, · · · ,n : x = (x1, · · · ,xn)}

is a norm. Next verify that

∥x∥1 ≡n

∑i=1|xi| , x = (x1, · · · ,xn)

is also a norm on Fn.

9. Let A be an n× n matrix. Denote by ∥A∥2 the operator norm taken with respect tothe usual norm on Fn. Show that

∥A∥2 = σ1

where σ1 is the largest singular value. Next explain why∥∥A−1

∥∥2 = 1/σn where σn is

the smallest singular value of A. Explain why the condition number reduces to σ1/σn

if the operator norm is defined in terms of the usual norm, |x|=(

∑nj=1

∣∣x j∣∣2)1/2

.

10. Let p,q > 1 and 1/p+1/q = 1. Consider the following picture.

b

a

x

t

x = t p−1

t = xq−1

Using elementary calculus, verify that for a,b > 0,

ap

p+

bq

q≥ ab.

11. ↑For p > 1, the p norm on Fn is defined by

∥x∥p ≡

(n

∑k=1|xk|p

)1/p

In fact, this is a norm and this will be shown in this and the next problem. Using theabove problem in the context stated there where p,q > 1 and 1/p+1/q = 1, verifyHolder’s inequality

n

∑k=1|xk| |yk| ≤ ∥x∥p ∥y∥q

Hint: You ought to consider the following.

n

∑k=1

|xk|∥x∥p

|yk|∥y∥q

Now use the result of the above problem.

34410.11.CHAPTER 14. NUMERICAL SOLUTIONS OF LINEAR SYSTEMSIf you are considering a system of the form Ax = b and A~! does not exist, will eitherthe Gauss Seidel or Jacobi methods work? Explain. What does this indicate aboutusing either of these methods for finding eigenvectors for a given eigenvalue?. Verify that\|x||,, = max {|x;|,i=1,---,2:x = (41,--- ,x)}is anorm. Next verify thatnIIx|| =P) bx, X= (X10 ,Xn)i=lis also anorm on F”.Let A be an n x n matrix. Denote by ||A||, the operator norm taken with respect tothe usual norm on F”. Show that|All, = 01where 0; is the largest singular value. Next explain why |A7! Il, = 1/0, where 0, isthe smallest singular value of A. Explain why the condition number reduces to 0/0,1/2. . . 2if the operator norm is defined in terms of the usual norm, |x| = ( ia] |x (| ) .Let p,q > 1 and 1/p+1/q = 1. Consider the following picture.b [x=}t=x!taUsing elementary calculus, verify that for a,b > 0,ap pf—+— >ab.Pp qor p > |, the p norm on 1s defineF 1, th F” is defined byn \/pIIx||, = e nl)k=1In fact, this is a norm and this will be shown in this and the next problem. Using theabove problem in the context stated there where p,q > 1 and 1/p+1/q = 1, verifyHolder’s inequalitynY lxel lye! < UIxll, llyllgk=1Hint: You ought to consider the following.y |xx| lyek=1 IIx\|,, IIy\lqnNow use the result of the above problem.