# Math 222 - Selected Homework Solutions from Chapter 5.5, 5.6. ```Math 222 - Selected Homework Solutions from
Chapter 5.5, 5.6.
Instructor - Al Boggess
Fall 1998
Section 5
10. Prove that the transpose of an orthogonal matrix is orthogonal. If Q
is orthogonal, then Qt Q = I . Therefore Qt is the inverse of Q. To
show that Qt is orthogonal, we must show that
(Qt )t Qt = I
Now (Qt )t = Q. So equation is equivalent to showing
QQt
=I
However, since Qt is the inverse of Q, clearly QQt = I as well.
11. The acute angle between two vectors x and y can be determined by
the Law of Cosines:
cos = jjxxjjjjyyjj
Likewise, let be the acute angle between Qx and Qy. It is determined
by
Qx Qy
cos = jjQx
jj jjQyjj
Since Q is orthogonal, Qx Qy is the same as x y and jjQxjj = jjxjj;
jjQyjj = jjyjj. Thus cos = cos , which implies = .
25. a) Let f = 1 and g = 2x , 1. Then
hf; gi =
Z
1
0
f g dx =
Z
1
0
1
(2x , 1) dx = x2 , xj10 = 0
So, f and g are orthogonal.
b)
Z 1
jj1jj2 = 12 dx = 1
0
So jj1jj = 1. We also have
jjgjj =
2
Z
1
0
g dx =
2
Z
1
(2x , 1)2 dx = (1=6)(2x , 1)3 j10 = 1=3
0
p
So jj2x , 1jj = 1= 3.
p
p
c) Let f1 = f =jjf jj = 1 and f2 = g=jjgjj = 3(1 , 2x). Let h = x.
Then the best least squares approximation of h by an element in S =
the span of f1 and f2 is
H
= hh; f1 if1 + hh; f2 if2
We now compute the inner products:
hh; f1 i =
and
Thus
Z
1
0
0
p
px dx = (2=3)x3=2 j1 = 2=3
hh; f2 i = 3
Z
1
0
px(1 , 2x) dx = ,2p3
15
p
2 3 f = (2=3) , (2=5)(1 , 2x)
H = (2=3) ,
15 2
Section 6
11. If v3 is in the span of v1 ; v2 , then the Gram Schmidt orthogonalization process will fail to produce three orthonormal vectors which span
v1 ; v2 ; v3 . Note that this space is at most two dimensional (since
v1 , v2 and v3 are linearly dependent). However, three orthonormal
vectors must be linearly independent and therefore the span of these
vectors would be three dimensional. The step that fails in the Gram
Schmidt process is that if p is the projection of v3 onto the space
spanned by v1 , v2 , then p = v3 (since v3 is contained in this space).
Therefore v3 , p = 0 and one cannot divide by its length to get the
third orthonormal vector.
2
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