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#"
#
1
11=6
1:833
0
1:80
6
A 1 D 13=12 D 1:083 compares with A
D 1:10 . kbk < :04 but
1
47=60
0:783
3:6
0:78 kx k > 6:
The difference .1; 1; 1/ .0; 6; 3:6/ is in a direction x that has Ax 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 16digit 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 “redblack ordering” puts rows and columns 1 to 10 in the oddeven 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.
 Spring '12
 Minki
 Mass

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