Introduction to Linear algebra-Strang-Solutions-Manual_ver13

A d 01 10 has j2 j d j1 j and no convergence 14 the

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: art’s Shake a Stick activity has long sticks representing the graphs of two linear equations in the x -y plane. The matrix is nearly singular and Section 9:2 shows how to compute its condition number c D kAkkA 1 k D max =min  80; 000: " # " # kA 1 k  20000 1 1:0001 1 1:0001 1 AD kAk  2 A D 10000 1 1:0000 1 1 c  40000: Problem Set 9.2, page 478 p k D 2, c D 4; kAk D 3, kA 1 k D 1,p D 3; kpk D 2 C 2 D c A max for positive definite A, kA 1 k D 1=min , c D .2 C 2/=.2 2/ D 5:83. p p 2 kAk D 2; c D 1I kAk D 2; c D infinite (singular matrix); AT A D 2I , kAk D 2, c D 1. 1 k Ak D 2 , k A 1 3 For the first inequality replace x by B x in kAx k  kAkkxk; the second inequality is just kB x k  kB kkx k. Then kAB k D max.kAB x k=kx k/  kAkkB k. 4 1 D kI k D kAA 1 k  kAkkA 1k D c.A/. 5 If ƒmax D ƒmin D 1 then all ƒi D 1 and A D SIS k Ak D k A 1 k D 1 are orthogonal matrices. 1 D I . The only matrices with 6 All orthogonal matrices have norm 1, so kAk  kQkkRk D kRk and in reverse kRk  kQ 1 kkAk D kAk, then kAk D kRk. Inequality is usual in kAk < kLkkU k when AT A ¤ AAT . Use norm on a random A. 7 The triangle inequality gives kAx C B x k  kAx k C kB x k. Divide by kx k and take the maximum over all nonzero vectors to find kA C B k  kAk C kB k. 8 If Ax D x then kAx k=kxk D jj for that particular vector x . When we maximize the ratio over all vectors we get kAk  jj.     01 00 01 9 ACB D C D has .A/ D 0 and .B/ D 0 but .A C B/ D 1. 00 10 10   10 The triangle inequality kA C B k  kAk C kB k fails for .A/. AB D also has 00 .AB/ D 1; thus .A/ D max j.A/j = spectral radius is not a norm. Solutions to Exercises 92 1 is kA 1 kk.A 1 / 1 k which is kA 1 kkAk D c.A/. (b) Since AT A and AAT have the same nonzero eigenvalues, A and AT have the same norm. 10 (a) The condition number of A 11 Use the quadratic formula for max =min , which is c D max =min since this A D AT is positive definite:  p c.A/ D 1:00005 C .1:00005/2  :0001 = 1:00005 p   40; 000: 12 det.2A/ is not 2 det AI det.A C B/ is not always less than det A C det B ; taking j det Aj does not help. The only reasonable property is det AB D .det A/.det B/. The condition number should not change when A is multiplied by 10. Ay D .10 7 ; 0/ is much smaller than b Az D .:0013; :0016/. But z is much closer to the solution than y .   659 563 14 det A D 10 6 so A 1 D 103 : kAk > 1, kA 1 k > 106 , then c > 106 . 913 780 p 15 x D .1; 1; 1; 1; 1/ has kx k D 5; kx k1 D 5; kx k1 D 1. x D .:1; :7; :3; :4; :5/ has kx k D 1; kx k1 D 2 (sum) kx k1 D :7 (largest). 13 The residual b 2 2 2 16 x1 C  Cxn is not smaller than max.xi / and not larger than .jx1 jC  Cjxn j/2 D kx k2 . 1 p 2 2 x1 C    C xn  n max.xi2 / so kx k  nkx k1 . Choose yi D sign xi D ˙1 to get p p kx k1 D x  y  kx kky k D nkx k. x D .1; : : : ; 1/ has kx k1 D n kx k. 17 For the `1 norm, the largest component of x plus the largest component of y is not less than kx C y k1 D largest component of x C y . For the `1 norm, each component has...
View Full Document

Ask a homework question - tutors are online