Introduction to Linear algebra-Strang-Solutions-Manual_ver13

# P t ap i d 0 then each new guess x k c1 is the

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Unformatted text preview: jxi C yi j  jxi j C jyi j. Sum on i D 1 to n: kx C y k1  kx k1 C ky k1 . 18 jx1 j C 2jx2 j is a norm but min.jx1 j; jx2 j/ is not a norm. kx k C kx k1 is a norm; kAx k is a norm provided A is invertible (otherwise a nonzero vector has norm zero; for rectangular A we require independent columns to avoid kAx k D 0). 19 x T y D x1 y1 C x2 y2 C     .max jyi j/.jx1 j C jx2 j C    / D jjx jj1 jjy jj1 . 20 With j D 2 2 cos.j=nC1/, the largest eigenvalue is n  2 C 2 D 4. The smallest  2 is 1 D 2 2 cos.=nC1/  nC1 , using 2 cos   2  2 . So the condition number is c D max =min  .4= 2 / n2 , growing with n.     11 1 q n 21 A D has A D with q D 1 C 1:1 C    C .1:1/n 1 D 0 1:1 0 .1:1/n   n n n n 0 10 .1:1 1/=.1:1 1/  1:1 =:1. So the growing part of A is 1:1 with 01 p jjAn jj  101 times 1:1n for larger n. Problem Set 9.3, page 489 1 The iteration x k C1 D .I A/x k C b has S D I and T D I A and S 1 T DI A. Solutions to Exercises 93 2 If Ax D x then .I A/x D .1 /x . Real eigenvalues of B D I A have j1 j < 1 provided  is between 0 and 2.   1 1 3 This matrix A has I A D which has jj D 2. The iteration diverges. 1 1 4 Always kAB k  kAkkB k. Choose A D B to ﬁnd kB 2 k  kB k2 . Then choose A D 5 6 7 8 9 10 11 12 B 2 to ﬁnd kB 3 k  kB 2 kkB k  kB k3 . Continue (or use induction) to ﬁnd kB k k  kB kk . Since kB k  max j.B/j it is no surprise that kB k < 1 gives convergence. Ax D 0 gives .S T /x D 0. Then S x D T x and S 1 T x D x . Then  D 1 means that the errors do not approach zero. We can’t expect convergence when A is singular and Ax D b is unsolvable!   101 1 with jjmax D 1 . Small problem, fast convergence. Jacobi has S T D 3 3 10 " # 1 03 1 Gauss-Seidel has S 1 T D with jjmax D 9 which is .jjmax for Jacobi)2 . 01 9   1   a 0 b 0 b =a 1 Jacobi has S T D D with jj D jbc=ad j1=2 . d c 0 c =d 0   1   a0 0 b 0 b =a Gauss-Seidel has S 1 T D D with jj D jbc=ad j. cd 0 0 0 b c=ad So Gauss-Seidel is twice as fast to converge .or to explode if jbc j > jad j/. p 3/  1:07. Set the trace 2 2! C 1 ! 2 equal to .! 1/ C .! 1/ to ﬁnd !opt D 4.2 4 The eigenvalues ! 1 are about .07, a big improvement. Gauss-Seidel will converge for the 1; 2; 1 matrix. jjmax D cos2 .=n C 1/ is given on page 485, with the improvement from successive over relaxation. If the iteration gives all xinew D xiold then the quantity in parentheses is zero, which means Ax D b. For Jacobi change x new on the right side to x old . A lot of energy went into SOR in the 1950’s! Now incomplete LU is simpler and preferred. 13 uk =k D c1 x 1 C c2 x 2 .2 =1 /k C    C cn x n .n =1 /k ! c1 x 1 if all ratios j=1 j < i 1  1. The largest ratio controls the rate of convergence (when k is large). A D 01 10 has j2 j D j1 j and no convergence. 14 The eigenvectors of A and also A 1 are x 1 D .:75; :25/ and x 2 D .1; 1/. The inverse power method converges to a multiple of x 2 , since j1=2 j > j1=1 j. j = 2 sin nj1 nC1 C j  cos nC1 . Then nC1 1/ C1/ sin .jnC1 sin .jnC1 .  The last two terms combine into 2 sin...
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## This note was uploaded on 09/25/2012 for the course PHY 103 taught by Professor Minki during the Spring '12 term at Korea Advanced Institute of Science and Technology.

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