Math 2270 Spring 2004
Homework 16 Solutions
Section 5.5) 8, 10, 16, 20, 23, 24
(8) We define T (v) =< v, w > and take elements v, x ∈ V and a scalar k. Then by properties
of inner products,
T (v + x) =< v + x, w >=< v, w > + < x, w >= T (v) + T (x)
T (kv) =< kv, w >= k < v, w >= kT (v)
so T is linear. If w is the zero element of V , then T (v) =< v, w >=< v, 0 >= 0 for all
v ∈ V . In this case im(T )= {0} and ker(T )= V . If w 6= 0, then im(T )= ℜ and ker(T ) is the
set of all v ∈ V that are orthogonal to w.
(10) First, we find the general form for polynomials in P2 that are orthogonal to f (t) = t
with respect to the given inner product.
1
< a + bt + ct , t >=
2
2
Z
1
−1
2
3
at + bt + ct
1 1 2 1 3 1 4
dt =
at + bt + ct
2 2
3
4
1
−1
1 2
b
=
2 3
The previous inner product is equal to zero iff b = 0. Therefore, polynomials orthogonal
to f (t) = t are of the form g(t) = a + ct2 . We start with the basis {1, t2 } and use the
Gram-Schmidt process to find an orthonormal basis.
||1|| =
1Z 1
dt = 1
2 −1
The first element of our basis is g1 (t) = 1.
1Z 1 2
1
< 1, t >=
t dt =
2 −1
3
s
s s Z 1
2
1
1 2 4 2
4
2
1 1
2
t − =
t4 − t2 +
dt =
=
− +
= √
3
2 −1
3
9
2 5 9 9
45
3 5
√
√ 3 5 2 1
5 2
t −
=
(3t − 1)
g2 (t) =
2
3
2
2
An orthonormal basis is
(
√
5 2
1,
(3t − 1)
2
1
)
Math 2270 Spring 2004
Homework 16 Solutions
(16) We start with the standard basis {1, t} and use the Gram-Schmidt process to find an
orthonormal basis for the given inner product.
< 1, 1 >=
1
Z
dt = 1
0
< 1, t >=
t −
1 2
=
s
Z
1
0
t2 − t +
1
dt =
4
s
1 1 1
− + =
3 2 4
1
Z
t dt =
s0
1
2
1
1
= √
12
2 3
The basis we are looking for is
n
1,
√
3 (2t − 1)
o
Now, we project the function f (t) = t2 onto the linear space defined by the orthonormal
basis we found above.
1
t2 dt =
0
√3
Z 1√
√
√
3
1 1
=
−
3(2t3 − t2 ) dt = 3
< t2 , 3(2t − 1) >=
2 3
6
0
√
√
3
1
1
1
1
projV t2 = +
3(2t − 1) = + t − = t −
3
6
3
2
6
< t , 1 >=
Z
h
i
2
1
(20)
<
"
1
0
#
,w
~ >=
h
1 0
i
"
1 2
2 8
#
w
~=
1 2
"
a
b
#
=0
iff a + 2b
e1 are scalar multiples of the
" = 0# or a = −2b. Therefore vectors perpendicular to ~
−2
vector
.
1
To find an orthonormal basis of ℜ2 with respect to this inner product we start with the basis
("
1
0
# "
,
−2
1
#)
as we already know that these two vectors are orthogonal to one another with respect to
this inner product. Therefore, the only work will be to determine their length and normalize
2
Math 2270 Spring 2004
Homework 16 Solutions
them if necessary.
"
"
−2
1
#2
1
0
#2
=
=
h
h
1 0
i
−2 1
i
An orthonormal basis is
"
1 2
2 8
#"
1
0
"
1 2
2 8
#"
−2
1
("
1
0
# "
,
#
−1
1/2
h
=
#
=
1 2
h
i
0 4
"
i
1
0
#
"
−2
1
=1
#
=4
#)
(23) We start with the standard basis {1, t} and use the Gram-Schmidt procedure to construct an orthonormal basis.
t −
1 2
2 1
< 1, 1 >= (1 + 1) = 1
2
1
1
< 1, t >= (0 + 1) =
2
2
1 −1
1 1
1
−1
=
+ ·
=
2
2
2
2 2
4
We have the basis {1, 2t − 1}
(24) (a)
< f, g + h >=< f, g > + < f, h >= 0 + 8 = 8
(b)
||g + h|| =
q
< g + h, g + h > =
q
< g, g > +2 < g, h > + < h, h > =
√
1 + 6 + 50 =
√
57
For part (c) we must first normalize the elements f and √
g. Luckily, they are √
orthogonal to
one another since < f, g >= 0. From the table ||f || = 4 = 2 and ||g|| = 1 = 1. The
orthonormal basis we are looking for is {f /2, g}. Now, we use the formula for projections
projE h =< f /2, h >
f
f
+ < g, h > g =< f, h > + < g, h > g = 2f + 3g
2
4
3
Math 2270 Spring 2004
Homework 16 Solutions
For part (d), we use our answer from part (c).
=
||h − 2f − 3g|| =
q
q
< h − 2f − 3g, h − 2f − 3g >
< h, h > +2 < h, −2f > +2 < h, −3g > + < −2f, −2f > +2 < −2f, −3g > + < −3g, −3g >
=
q
50 − 4(8) − 6(3) + 4(4) + 12(0) + 9(1) = 5
Therefore, an orthonormal basis is
1
1
f, g, (h − 2f − 3g)
2
5
4