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Unformatted text preview: Math 235 Assignment 6 Solutions 1. Find a and b to obtain the best ﬁtting equation of the form y = a + bt for the given
data. t 1 2 3 4 5
y 9 6 5 31
9 11
6 12
Solution: We take ﬂ 2: 5 andX = 1 3 .Then
3 1 4
1 1 5
9
~1 6
a_T1T,,__515 11111 __1~_55~1524*10.5
[61‘(XX) X “[1555] 1123451§“50 ~15 5 53”~1.9
1 Hence, the best ﬁtting line is y = 10.5 ~ 1.916. 2. Verify that the following system A2? =2 5 is inconsistent, then determine for the vector :3
that minimizes [[1455 ~ £131 ‘1‘ 21'2 Z 5
2961 ~ 3332 2: 6
£191 "‘ 12112 I: “*4
1 2 5 1 2 5
Solution: Row reducing gives 2 ~13 6 N 0 ~7 ~41 Hence, the system is
1 ~12 «4 0 0 ~1
inconsistent.
1 2 ..
We have A = 2 ~23 . Hence, the vector :5 that minimizes “As? —— [9H is
1 ~12 f:(ATA)“1A5 “ 6 ~16 "1 1 2 1 g
“ ~16 157 2 ~3 ~12 __4
____1__ 157 16 13 “685 16 6 40 2: [3385/4998] 2 3. Prove that if U and W are subspaces of a vector space V such that U 0 W = {6}, then
U 63 W is a subspace of V. Moreover, if {171, . . . ,ﬂk} is a basis for U and {2171,  ,w’g} is a
basis for W, then {271, . . . £19,131, . . . ,u‘ig} is a basis for U EB W. Solution: We Will prove that {171, . . . ,17k+g} is a linearly independent spanning set for UEEW.
Consider C1111 +    +Ck’17k + 0194,1131 + ' ' ‘ + Ck+gtvg == 0
Then, we have 61771 + ' ' ' + Ck'Uk : “CMﬂBi '* ' ' ' ‘“ Ck+€117£
The vector on the right is in U and the vector on the left is in W. But, the only vector that is
both in U and W is the zero vector. Therefore, each oz 2 O and hence {171, . . . , 17k, 21517 . . . , 117g} is linearly independent.
For any 17 E U EB W we have that 17 =2 {i + ’u')’ by deﬁnition of U G9 W. We can write
11’: 01171 + ' ' ‘ + Ckﬂk and U7: d1131+~~+ dew} and hence 17201771+"'+Ck’17k+d1231+"‘+dg’tvg and so {171,...,17k,161,...,tb’g} also spans UEBW. 4. Let U be a subspace of a ﬁnite dimensional vector space V and suppose that there are
subspaces W1 and W2 such that V = U @ W1 and V = U EB W2.
(a) Prove that dim W1 = dim W2. Solution: By Theorem 3.5.1 (problem 3), we have that dimV 2: dimU + dile and
dimV :2 dimU + dim W2. Thus, dim W1 :2 dimV —— dimU =2 dim W2 (b) Give an example of a vector space V With subspaces U, W1, and W2 such that V =
UEBW1 andV—TUEBWQ, but W1 Solution: In R2, let U = Span { }, W1 :2 Span{ } and W2 2 Span{ Then,
by Theorem 3.5.1, we have V 2 U EB W1 and V = U 89 W2, but clearly W1 % W2. 5. Let A be an m x 71 real matrix. Prove that the nullspace of AT is the orthogonal
complement of the column space of A. 17?
Solution: Let A = [171 17”], so AT :2 _
5:5
Ufa:
HfEQﬂAHJMnQdHJLEmmmATH= ; zﬁhmmfeNmmﬂy
173%
On the other hand, let :3 E Null(AT), then AT” :2 6 so 13}  a? 2: 0 for all 2'. Now, pick any 56 Col(A). Then 5: 01171 +  ~  + cuff”. But then
szkﬁﬁ~~+%%yfzm
hence a? E Col(A)i. MATH 235 Assignment 6 MATLAB Solutions Least Squares (a)
X = [0; 10; 20; 30; 40;
X =
0
10
20
30
40
50
60
70
80
90
y = [4.77; 16.47; 19.76;
y =
4.7700
16.4700
19.7600
22.3500
23.6700
25.0300
27.0100
30.9300
40.9800
53.8600
plOt(XI Y! H”);
(b)
X = [x X.“2 x.“3]
X 7:.
0 O
10 100
20 400
30 900
40 1600
50 2500
60 3600
70 4900
80 6400
90 8100 50; 60; 70; 22.35; 23.67; 0 1000
8000
27000
64000
125000
216000
343000
512000
729000 80; 90] 25.03; 27.01; 30.93; 40.98; 53.86] rref(X'*X)
ans =
l O O
0 1 0
0 0 1
beta = inV(X'*X)*X'*y
beta =
1.6288
—0.0366
0.0003 (d) xnrange = [0: l: 901'; ymfitnline = beta(l,l)*x_range + beta(2,l)*x_range.“2 + beta(3,1)*x_range.“3;
hold on plot(x_range, y_fit_line, 'r—') MATH 235 Mm fa: Assignment 5
£933: Sqaams: Safes v3. Costs costs (méltimg) wk 5313 minis
WWW i235! sqeareg {it {I E} 2:3 33 413 Sit 80 m 833 238
sales (méiﬁtms) ...
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This note was uploaded on 09/30/2011 for the course MATH 235 taught by Professor Celmin during the Spring '08 term at Waterloo.
 Spring '08
 CELMIN

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