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.1; 1; 1; 1/T x D 0, a basis for S ? is .1; 1; 1; 1/
(c) Split .1; 1; 1; 1/ D b1 C b2
11
3
by projection on S ? and S : b2 D . 1 ; 1 ; 1 ; 1 / and b1 D . 2 ; 2 ; 1 ; 2 /.
222
2
2
This questionshows 2 2 formulas for QR breakdown R22 D 0 when A is sinby
;
12
1 53
11
21
1
11
1
gular.
Dp
p
. Singular
Dp
11
2
11
1
51
501
21
1 22
p
. The GramSchmidt process breaks down when ad b c D 0.
200
T
.q T C /q 2 D BT c B because q 2 D B and the extra q 1 in C is orthogonal to q 2 .
2
kB k
BB
When a and b are not orthogonal, the projections onto these lines do not add to the projection onto the plane of a and b. We must use the orthogonal A and B (or orthonormal
q 1 and q 2 ) to be allowed to add 1D projections. 28 There are mn multiplications in (11) and 1 m2 n multiplications in each part of (12).
2
29 q 1 D 1
.2; 2;
3 1
1/, q 2 D 1 .2; 1; 2/, q 3 D 3 .1; 2 ; 2/.
3 31 32
33
34
35 1 D W T . See
Section 7.2 for more about wavelets : a useful orthonormal basis with many zeros.
1
(a) c D 2 normalizes all the orthogonal columns to have unit length
(b) The pro1
jection .aT b=aT a/a of b D .1; 1; 1; 1/ onto the ﬁrst column is p1 D 2 . 1; 1; 1; 1/.
(Check e D 0.) To project onto the plane, add p2 D 1 .1; 1; 1; 1/ to get .0; 0; 1; 1/.
2
"
#
1
0
0
1
0
0
1 across plane y C z D 0.
Q1 D
reﬂects across x axis, Q2 D 0
0
1
0
1
0
Orthogonal and lower triangular ) ˙1 on the main diagonal and zeros elsewhere.
(a) Qu D .I
2uuT /u D u 2uuT u. This is u, provided that uT u equals 1
(b) Qv D .I 2uuT /v D u 2uuT v D u, provided that uT v D 0.
Starting from A D .1; 1; 0; 0/, the orthogonal (not orthonormal) vectors B D
.1; 1; 2 ; 0/ and C D .1; 1; 1; 3/ and D D .1; 1; 1; 1/ are in the directions of q 2 ; q 3 ; q 4 .
The 4 by 4 and 5 by 5 matrices with integer orthogonal columns (not orthogonal3
rows,
2
2
3
1
1
11
1
1 17
61
since not orthonormal Q!) are 4 A B C D 5 D 4
and
0
2
1 15
0
0
31
2
3
1
1
1
11
61
1
1
1 17
6
7
2
1
1 17
60
40
0
3
1 15
0
0
0
41 30 The columns of the wavelet matrix W are orthonormal. Then W Solutions to Exercises 51 36 ŒQ; R D q r .A/ produces from A (m by n of rank n) a “fullsize” square Q D Œ Q1 Q2
R
. The columns of Q1 are the orthonormal basis from GramSchmidt of the
0
column space of A. The m n columns of Q2 are an orthonormal basis for the left
nullspace of A. Together the columns of Q D Œ Q1 Q2 are an orthonormal basis
for Rm .
and 37 This question describes the next q nC1 in GramSchmidt using the matrix Q with the columns q 1 ; : : : ; q n (instead of using those q ’s separately). Start from a, subtract its
projection p D QT a onto the earlier q ’s, divide by the length of e D a QT a to get
q nC1 D e =ke k. Problem Set 5.1, page 251
1 det.2A/ D 8I det. A/ D . 1/4 det A D
2 det. 1 A/
2
1 D
/D det.A . 1 /3
2
1. det A D 1
8 1
I
2 det.A2 / D 1 I det.A
4
3 1 / D 2 D det.AT / 1 2 and det. A/ D . 1/ det A D 1; det.A / D 1; 3 (a) False: det.I C I / is not 1 C 1 (b) True: The product rule extends...
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 Spring '12
 Minki
 Mass

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