Homework # 13 (Written) Solutions
Math 152, Fall 2014
Instructor: Dr. Doreen De Leon
p. 352-3 (Section 6.3): 14, 20, 24
14. Find the best approximation to z by vectors of the form c1 v1 + c2 v2 , where
 
 
 
2
2
5
 4
 0
−2

 
 
z=
 0 , v1 = −1 , v2 =  4 .
−1
−3
2
Solution: First, note that v1 · v2 = (2, 0, −1, −3) · (5, −2, 4, 2) = 0, so v1 and v2 are
orthogonal vectors. Therefore, we know that
z=
z · v2
z · v1
v1 +
v2 ,
v1 · v1
v2 · v2
i.e.,
z · v1
v1 · v1
(2, 4, 0, −1) · (2, 0, −1, −3)
=
(2, 0, −1, −3) · (2, 0, −1, −3)
2(2) + 4(0) + 0(−1) + (−1)(−3)
=
22 + 02 + (−1)2 + (−3)2
7
1
=
= .
14
2
z · v2
c2 =
v2 · v2
(2, 4, 0, −1) · (5, −2, 4, 2)
=
(5, −2, 4, 2) · (5, −2, 4, 2)
2(5) + 4(−2) + 0(4) + (−1)(2)
=
52 + (−2)2 + 42 + 22
= 0.
c1 =
So, the best approximation to z is c1 v1 + c2 v2 , which equals
   
1
2



1
 0 =  01  .
2 −1 − 2 
−3
− 32
1
 
0
20. Let u1 and u2 be as in Exercise 19, and let u4 = 1. It can be shown that u4 is not in
0
the subspace spanned by u1 and u2 . Use this fact to construct a nonzero vector v ∈ R3
that is orthogonal to u1 and u2 .
Solution: Let W = Span{u1 , u2 }. Since u4 ∈
/ W and u1 and u2 are orthogonal, we may
apply the Orthogonal Decomposition Theorem to write
u4 = û4 + z,
where z ∈ W ⊥ (and so is orthogonal to u1 and u2 ).
u4 · u2
u4 · u1
u1 +
u2
u1 · u1
u2 · u2
(0, 1, 0) · (1, 1, −2)
(0, 1, 0) · (5, −1, 2)
=
(1, 1, −2) +
(5, −1, 2)
(1, 1, −2) · (1, 1, −2)
(5, −1, 2) · (5, −1, 2)
0(1) + 1(1) + 0(−2)
0(5) + 1(−1) + 0(2)
=
(1, 1, −2) +
(5, −1, 2)
2
2
2
1 + 1 + (−2)
52 + (−1)2 + 22
1
−1
= (1, 1, −2) +
(5, −1, 2)
6
30
1 1 1
1
1 1
=
, ,−
+ − , ,−
6 6 3
6 30 15
1 2
= 0, , −
.
5 5
1 2
z = u4 − û4 = (0, 1, 0) − 0, , −
5 5
4 2
= 0, ,
.
5 5
û4 = projW u4 =
So,
 
0
v = 4 .
5
2
5
24. Let W be a subspace of Rn with an orthogonal basis {w1 , . . . , wp }, and let {v1 , . . . , vq }
be an orthogonal basis for W ⊥ .
(a) Explain why {w1 , . . . , wp , v1 , . . . , vq } is an orthogonal set.
Solution: Since vi ∈ W ⊥ for i = 1, 2, . . . , q, we know that for each i, vi is orthogonal to every vector in a basis for W . Specifically, vi is orthogonal to wj
for all i = 1, 2, . . . , q and all j = 1, 2, . . . , p. Since {w1 , . . . , wp } is an orthogonal basis, we know that every pair of vectors is orthogonal and similarly with
{v1 , . . . , vq }. Therefore, every pair of vectors in {w1 , . . . , wp , v1 , . . . , vq } is orthogonal and {w1 , . . . , wp , v1 , . . . , vq } is an orthogonal set.
2
(b) Explain why the set in part (a) spans Rn .
Solution: Let x ∈ Rn . By the Orthogonal Decomposition Theorem, we can write
x = x̂ + z, where x̂ ∈ W and z ∈ W ⊥ . Since {w1 , . . . , wp } is a basis for W , we can
write
x̂ = c1 w1 + · · · + cp wp ,
and since {v1 , . . . , vq } is a basis for W ⊥ , we can write
z = d1 v1 + · · · + dq vq .
Therefore, we can write x as
x = x̂ + z = c1 w1 + · · · + cp wp + d1 v1 + · · · + dq vq ,
a linear combination of the vectors in {w1 , . . . , wp , v1 , . . . , vq }.
(c) Show that dim W + dim W ⊥ = n.
Solution: We know that {w1 , . . . , wp , v1 , . . . , vp } spans Rn . Since it is an orthogonal
set, it is also linearly independent. Therefore, {w1 , . . . , wp , v1 , . . . , vq } is a basis of
Rn , and since dim Rn = n, we have p + q = n. But, p = dim W and q = dim W ⊥ .
Therefore, dim W + dim W ⊥ = n.
3