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9 12 0
:6
:8
0
T
:6 , v3 D 0 .
A A D 12 16 0 has D 25; 0; 0 and v1 D :8 , v2 D
0 00
0
0
1
Here A D Œ 3 4 0 has rank 1 and AAT D Œ 25 and 1 D 5 is the only singular value
in † D Œ 5 0 0 . 1 AT A D
2 3
4
5 6
7
8 9
10 Solutions to Exercises 11
12
13
14
15 16
17
18
19 20
21 81 "#" #
"
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:2
:12
:36 :48 0
A D Œ 1 Œ 5 0 0 V T and ACDV 0 D :16 ; AC AD :48 :64 0 I AACDŒ 1
0
0
0
00
The zero matrix has no pivots or singular values. Then † D same 2 by 3 zero matrix
and the pseudoinverse is the 3 by 2 zero matrix.
If det A D 0 then rank.A/ < n; thus rank.AC / < n and det AC D 0.
A must be symmetric and positive deﬁnite, if † D ƒ and U D V D eigenvector matrix
Q is orthogonal.
(a) AT A is singular (b) This x C in the row space does give AT Ax C D AT b (c) If
.1; 1/ in the nullspace of A is added to x C , we get another solution to AT Ab D AT b.
x
But this b is longer than x C because the added part is orthogonal to x C in the row
x
space.
x C in the row space of A is perpendicular to b x C in the nullspace of AT A D
x
nullspace of A. The right triangle has c 2 D a2 C b 2 .
AAC p D p, AAC e D 0, AC Ax r D x r , AC Ax n D 0.
:36 :48
AC D V †C U T is 1 Œ :6 :8 D Œ :12 :16 and AC A D Œ 1 and AAC D
D
5
:48 :64
projection.
L is determined by `21 . Each eigenvector in S is determined by one number. The
counts are 1 C 3 for LU , 1 C 2 C 1 for LDU , 1 C 3 for QR, 1 C 2 C 1 for U †V T ,
2 C 2 C 0 for SƒS 1 .
LDLT and QƒQT are determined by 1 C 2 C 0 numbers because A is symmetric.
P
Column times row multiplication gives A D U †V T D
i ui vT and also AC D
i
P1
CT
T
C
V† U D
i vi ui . Multiplying A A and using orthogonality of each ui to all
P
C
C
other
1vi vT . Similarly AAC D
i
P ujT leaves theTprojection matrix A A: A A D
1ui ui from V V D I . 22 Keep only the r by r corner †r of † (the rest is all zero). Then A D U †V T has the
Tb
T
b
required form A D U M1 †r M2 V T with an invertible M D M1 †r M2 in the middle.
0Au
Av
u
The singular values of A are
23
D
D
.
v
v
eigenvalues of this block matrix.
AT 0
AT u Problem Set 8.1, page 418
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c1 C c2
c2
0
c2
c2 C c3
c3
1 Det AT C0 A0 D
is by direct calculation. Set c4 D 0 to
0
0
c3
c3 C c4
ﬁnd det AT C1 A1 D c1 c2 c3 .
1
3"
"
#2 1
#
c1
100
111
5011D
2 .AT C1 A1 / 1 D 1 1 0 4
c2 1
1
1
111
001
c
21
33
1
1
c1
c1
c1
5.
4c 1 c 1 C c 1
c1 1 C c2 1
1
1
2
1
1
1
1
1
1
c1
c1 C c2
c1 C c2 C c3 Solutions to Exercises 82 3 The rows of the freefree matrix in equation (9) add to Œ 0 0 0 so the right side needs f1 C f2 C f3 D 0. f D . 1; 0; 1/ gives c2 u1 c2 u2 D 1; c3 u2 c3 u3 D 1; 0 D 0.
Then uparticular D . c2 1 c3 1 ; c3 1 ; 0/. Add any multiple of unullspace D .1; 1; 1/.
Z
Z
d
du
du 1
4
c.x/
dx D
c.x/
D 0 (bdry cond) so we need f .x/ dx D 0.
dx
dx
dx 0
Zx
Z1
dy
5
D f .x/ gives y.x/ D C
f .t /dt . Then y.1/ D 0 gives C D
f .t /dt
dx
0
0
Z1
and y.x/ D
f .t /dt . If the load is f .x/ D 1 then the displacement is y.x/ D 1 x .
x as columns of AT times c ’s times rows of A1 . The ﬁrst...
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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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