31CHALLENGE 5.20.(a) SinceUΣV∗x=b,wehavex=VΣ−1U∗b.If we letc=U∗b, thenα1=c1/σ1,andα2=c2/σ2.(b) Here is one way to look at it. This system is very ill-conditioned. The conditionnumber is the ratio of the largest singular value to the smallest, so this must belarge. In other words,σ2is quite small compared toσ1.For the perturbed problems,Ax(i)=b−E(i)x(i),so it is as if we solve the linear system with a slightly perturbed right-hand side.So, lettingf(i)=U∗E(i)x(i), the computed solution isx(i)=α(i)1v1+α(i)2v2,withα(i)1=(c1+f(i)1)/σ1,α(i)2c2+f(i)2)/σ2.From the ±gure, we know thatf(i)must be small, soα(i)1≈α1. But becauseσ2isclose to zero,α(i)2can be quite di²erent fromα2,sothe solutions lie almost on astraight line in the directionv2.CHALLENGE 5.21.Some possibilities:•Any right eigenvectoruofAcorresponding to a zero eigenvalue satis±es
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This note was uploaded on 01/21/2012 for the course MAP 3302 taught by Professor Dr.robin during the Fall '11 term at University of Florida.