MATH 425 Hour Exam I Solution Radford 02/22/08
1. (20 points) First
(a) observe that a+bx+cx2∈S⊥if and only if 0 = <a +bx +cx2, x> =Z2
0
(a+bx +cx2)x dx
=Z2
0
(ax +bx2+cx3)dx = (ax2
2+bx3
3+cx4
4)¯
¯
¯
¯
¯
2
0
= 2a+8
3b+ 4c; 0 = a+4
3b+ 2c. (10)
(b) Solutions to this equation, which is equivalent to the system
a=−4
3b−2c
b=b
c=c
, is
given by
a
b
c
=b
−4/3
1
0
+c
−2
0
1
. A basis for S⊥is {−4
3+x, −2 + x2}. (10)
2. (20 points) Recall that s0is a closest vector in Sto vmeans s∈Sand ||v−s0|| ≤ ||v−s||
for all s∈S.
(a) Let s∈S. Then
||v−s||2=||(v−s0)+(s0−s)||2=||v−s0||2+||s0−s||2≥ ||v−s0||2,(1)
where the second equation follows since s0−s∈S. By (1) we deduce ||v−s||2≥ ||v−s0||2,
hence ||v−s|| ≥ ||v−s0||. (10)
(b) ||v−s|| ≥ ||v−s1|| for all s∈S; in particular for s=s0. Thus by (1), with s=s1,
we deduce ||v−s0||2≥ ||v−s1||2=||v−s0||2+||s0−s1||2≥ ||v−s0||2,from which
||v−s0||2=||v−s0||2+||s0−s1||2,or equivalently ||s0−s1||2= 0, follows. Therefore
||s0−s1|| = 0 and consequently s0=s1. (10)
3. (20 points) Let {q1,q2}be the orthonormal basis, and let v=
2
1
1
1
.
(a) By inspection <v,q1>=−2
13 and <v,q2>=7
13. Thus the closest vector is
<v,q1>q1+<v,q2>q2=1
169
−6−28
24 + 0
−8 + 21
0 + 84
=1
169
−34
24
13
84
.
(10)
1
