Introduction to Linear algebra-Strang-Solutions-Manual_ver13

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

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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: &quot; # &quot; # 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 &lt; 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 jj for that particular vector x . When we maximize the ratio over all vectors we get kAk  jj.     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 &gt; 1, kA 1 k &gt; 106 , then c &gt; 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...
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