Introduction to Linear algebra-Strang-Solutions-Manual_ver13

# Introduction to Linear algebra-Strang-Solutions-Manual_ver13

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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 haseigenvalues 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...
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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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