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and the other properties also guarantee that the triangle inequality is satisfied (see
Theorem C.2.2 below), so that we have a
good
definition of length. This aspect needs
to be emphasized. What makes linear algebra beautiful is that it enables us to solve
problems in an elegant, clean way without having to get involved in messy computa-
tions with coordinates. As long as we use only general (and
essential
) properties like
those in the definition for an inner product, we shall be able to give simple proofs.
Definition.
Let
V
be a vector space. Given two points
p
,
q
Œ
V
, define the
vector from
p
to
q
,
pq
, by
pq
=-.
q
p
If
V
has an inner product, then define the
distance from
p
to
q
, dist (
p
,
q
), by
(
)
=
dist
pq
,
pq
.
There are two very important inequalities.
C.2.2. Theorem.
Let
u
and
v
be vectors in a vector space with an inner
product <,>.
(1) (The Cauchy-Schwarz inequality)
|<
u
,
v
>| £ |
u
| |
v
|
We have equality if and only if
u
and
v
are linearly dependent.
(2) (The triangle inequality)
|
u
+
v
| £ |
u
| + |
v
|
We have equality
only if u
and
v
are linearly dependent.
Proof.
We prove (1) first. Let c be any scalar. Then
2
2
2
0
£<
uvuv u
-
c
,
-
c
>=
-
2
c
<
uv
,
>+
c
v
.
(C.1)
If we consider the right-hand side of (C.1) as a quadratic equation in the variable c,
then we can use the discriminant test from the quadratic formula to conclude that
“<” holds (that is, there are no solutions) if and only if
2
2
2
[
]
-
-<
2
uv
,
>
4
u v
<
0
,
which simplifies to what we want. On the other hand, it is easy to see from (C.1) that
equality holds if and only if
u
= c
v
or
v
=
0
(that is,
u
and
v
are linearly dependent).
An alternate way to prove the Cauchy-Schwarz inequality is simply to set c to
1
2
u
<>
,
v
in (C.1) and simplify the resulting expression.
Part (2) of the theorem follows from the Cauchy-Schwarz inequality because
