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1
140
41
0 4 4D4 1
4
0 1 1 D LDLT .
011
004
011
4
001
2
32
32
3
aaaa
1
a
a
a
a
a ¤ 0 All of the
b a b a b a7
b ¤ a multipliers
6a b b b 7 61 1
76
. Need
4 a b c c 5 D 4 1 1 1 54
c b c b5
c ¤ b are `ij D 1
abcd
1111
dc
d ¤ c for this A
2
32
32
3
arrr
1
a
r
r
r
a¤0
b r s r s r7
b¤r
6a b s s 7 61 1
76
. Need
4a b c t 5 D 41 1 1
54
c s t s5
c¤s
abcd
1111
dt
d ¤t
5
2
24
2
2
10
.
gives x D
xD
. Then
gives c D
cD
3
3
01
3
11
41
2
24
2
24
D c.
xD
. Forward to
xD
Ax D b is LU x D
3
01
11
8 17
"#
#
"#
"
"#
#
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"
3
4
111
4
4
100
1 1 0 c D 5 gives c D 1 . Then 0 1 1 x D 1 gives x D 0 .
1
1
001
1
6
111
Those are the forward elimination and back substitution steps for
#
"#
#"
"
4
111
1
1 1 xD 5 .
Ax D 1 1
6
1
111 (b) I goes to L 1 (c) LU goes to U . Elimination multiply by L 1 !
18 (a) Multiply LDU D L1 D1 U1 by inverses to get L1 1 LD D D1 U1 U 1 . The left
side is lower triangular, the right side is upper triangular ) both sides are diagonal.
(b) L; U; L1 ; U1 have diagonal 1’s so D D D1 . Then L1 1 L and U1 U 1 are both I .
#
"
#
#
"
"
#"
a
1
110
a
a
0
b
b
1 1 D LI U I a a C b
D (same L)
19 1 1
0
b
bCc
c
1
011
(same U ). A tridiagonal matrix A has bidiagonal factors L and U .
20 A tridiagonal T has 2 nonzeros in the pivot row and only one nonzero below the pivot
(one operation to ﬁnd ` and then one for the new pivot!). T D bidiagonal L times
bidiagonal U .
17 (a) L goes to I Solutions to Exercises 23 21 For the ﬁrst matrix A; L keeps the 3 lower zeros at the start of rows. But U may not 22 23
24
25
26 have the upper zero where A24 D 0. For the second matrix B; L keeps the bottom left
zero at the start of row 4. U keeps the upper right zero at the start of column 4. One
zero in A and two zeros in B are ﬁlled in.
"
#
"
#
"
#
531
420
200
Eliminating upwards, 3 3 1 ! 2 2 0 ! 2 2 0 D L. We reach
111
111
111
a lower triangular L, and the multipliers are in an upper triangular U . A D UL with
"
#
111
UD 0 1 1 .
001
The 2 by 2 upper submatrix A2 has the ﬁrst two pivots 5; 9. Reason: Elimination on A
starts in the upper left corner with elimination on A2 .
The upper left blocks all factor at the same time as A: Ak is Lk Uk .
The i; j entry of L 1 is j= i for i j . And Li i 1 is .1 i /= i below the diagonal
.K 1 /ij D j.n i C 1/=.n C 1/ for i j (and symmetric): .n C 1/K 1 looks good. Problem Set 2.7, page 115
1
1 AD
9
1
AD
c
0
19
1
0
T
1
has A D
;A D
; .A 1 /T D .AT /
3
03
3 1=3
10
c
c
has AT D A and A 1 D 2
D .A 1 /T .
0
1
cc 1
1
3
D
;
0 1=3 2 .AB/T is not AT B T except when AB D BA. Transpose that to ﬁnd: B T AT D AT B T . 1T
/ D .B 1 A 1 /T D .A 1 /T .B 1 /T . This is also .AT / 1 .B T / 1 .
(b) If U is upper triangular, so is U 1 : then .U 1 /T is lower triangular.
01
AD
has A2 D 0. The diagonal of AT A has dot products of columns of A with
00
themselves. If AT A D 0, zero dot products ) zero columns ) A D zero matrix.
"0#
123...
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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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