Examples:
1. Let
p
(
x
) =
x
2
+ 2
x
+ 3
,
q
(
x
) = 3
x
+ 2
.
With the inner product
< p, q >
=
a
0
b
0
+
a
1
b
1
+
a
2
b
2
for
p
=
a
0
+
a
1
x
+
a
2
x
2
, q
=
b
0
+
b
1
x
+
b
2
x
2
. Find length of p,q
and distance between p and q.
Solution:

p

=
√
< p, p >
=
√
1 + 4 + 9 =
√
14
,

q

=
√
9 + 4 =
√
13
d
(
p, q
) =

p

q

=
√
< p

q, p

q >
=
p
< x
2

x
+ 1
, x
2

x
+ 1
>
=
√
3
2. Let
f
(
x
) =
x
2
and
g
(
x
) =
x
be continuous functions on [1,2]. Consider the inner
product
< f, g >
=
R
2
1
f
(
x
)
g
(
x
)
dx
. Find length of f,g and distance between f and g.
Solution:

f

=
√
< f, f >
=
q
R
2
1
x
4
dx
=
p
31
/
5
,

g

=
√
< g, g >
=
q
R
2
1
x
2
dx
=
p
7
/
3
d
(
f, g
) =

f

g

=
p
< x
2

x, x
2

x >
=
p
31
/
30
.
Theorem 6.1.1:
if u and v are vectors in a real inner product space V, and if k is a
scalar, then:
(a)

v
 ≥
0
with equality if and only if v=0
(b)

kv

=

k

v

(c) d(u,v)=d(v,u)
(d)
d
(
u, v
)
≥
0
with equality if and only if u=v.
Section 6.2 Angle and orthogonality in Inner Spaces
Recall that for vectors in
R
n
we defined angle between two vectors u and v as
cos (
θ
) =
u
·
v

u

v

hence we said u is orthogonal to v if and only if
u
·
v
= 0
. We will generalize this formula
for inner product spaces. That is for a given inner product
<, >
we have
cos (
θ
) =
< u, v >

u

v

where
0
≤
θ
≤
π
and u is orthogonal to v if and only if
< u, v >
= 0
. This formula is valid because of
Theorem 6.2.1 CauchySchwarz Inequality
If u and v are vectors in a real
inner product space V, then

< u, v >
 ≤ 
u

v

Definition:
Two vectors u and v in an inner product space are called orthogonal if
< u, v >
= 0
.
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 Spring '11
 Linear Algebra, Euclidean space, inner product, Inner product space

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