Introduction to Linear algebra-Strang-Solutions-Manual_ver13

Introduction to Linear algebra-Strang-Solutions-Manual_ver13

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: ; #" # 1 11=6 1:833 0 1:80 6 A 1 D 13=12 D 1:083 compares with A D 1:10 . kbk < :04 but 1 47=60 0:783 3:6 0:78 kx k > 6: The difference .1; 1; 1/ .0; 6; 3:6/ is in a direction x that has Ax near zero. The largest kx k D kA 1 bk is kA 1 k D 1=min since AT D A; largest error 10 16 =min . Each row of U has at most w entries. Then w multiplications to substitute components of x (already known from below) and divide by the pivot. Total for n rows < wn. The triangular L 1 , U 1 , R 1 need 1 n2 multiplications. Q needs n2 to multiply the 2 right side by Q 1 D QT . So QRx D b takes 1.5 times longer than LU x D b. Solutions to Exercises 90 D I : Back substitution needs 1 j 2 multiplications on column j , using the j 2 by j upper left block. Then 1 .12 C 22 C    C n2 /  1 . 1 n3 / D total to find U 1 . 2 23           10 22 2 2 01 10 8 ! ! D U with P D and L D ; 22 10 0 1 10 :5 1 " # " # " # " # 220 2 20 2 20 220 11!0 2 0 ! 0 2 0 D U with A! 1 0 1 ! 0 020 0 20 0 11 001 " # " # 010 1 00 1 0. P D 0 0 1 and L D 0 100 :5 :5 1 2 3 1100 61 1 1 07 9 AD4 has cofactors C13 D C31 D C24 D C42 D 1 and C14 D C41 D 0 1 1 15 0011 1. A 1 is a full matrix! 1 7 UU x computedk for " D 10 3 , 10 6 , 10 , 10 , 10 are of order 10 , 10 , 10 , 10 4 , 10 3 .      p p 131 1 10 14 D p1 . 11 (a) cos  D 1= 10, sin  D 3= 10, R D p1 10 10 313 5 08 (b) A haseigenvalues 4 and 2. Put one of the unit eigenvectors in row 1 of Q: either      1 1 2 4 1 3 1 QDp and QAQ 1 D or Q D p1 and QAQ 1 D 21 10 3 1 0 4 1   4 4 . 0 2 10 With 16-digit floating point arithmetic the errors kx 9 12 15 16 11 7 12 When A is multiplied by a plane rotation Qij , this changes the 2n (not n2 ) entries in rows i and j . Then multiplying on the right by .Qij / entries in columns i and j . 1 D .Qij /T changes the 2n 13 Qij A uses 4n multiplications (2 for each entry in rows i and j /. By factoring out cos  , 2 the entries 1 and ˙ tan  need only 2n multiplications, which leads to 3 n3 for QR. 14 The .2; 1/ entry of Q21 A is tan  D 2. Then the 2; 1; p 1 . 3 sin  C 2 cos  /. This is zero if sin  D 2 cos  or p p 5 right triangle has sin  D 2= 5 and cos  D 1= 5. Every 3 by 3 rotation with det Q D C1 is the product of 3 plane rotations. 15 This problem shows how elimination is more expensive (the nonzero multipliers are counted by nnz(L) and nnz(LL)) when we spoil the tridiagonal K by a random permutation. If on the other hand we start with a poorly ordered matrix K , an improved ordering is found by the code symamd discussed in this section. 16 The “red-black ordering” puts rows and columns 1 to 10 in the odd-even order 1; 3; 5; 7, 9; 2; 4; 6; 8; 10. When K is the 1; 2; 1 tridiagonal matrix, odd points are connected Solutions to Exercises 91 only to even points (and 2 stays on the diagonal, connecting every point to itself): 2 2 61 KD4 2 1 61 6 DD6 0 4 1 2  1 1 3  2I 1 7 5 and PKP T D D T  12 3 1 to 2 7 3 to 2; 4 7 1 7 5 to 4; 6 5 7 to 6; 8 1 1 1 1 9 to 8; 10 D 2I  with 17 Jeff Stu...
View Full Document

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.

Ask a homework question - tutors are online