Math 110 Spring 2008
Solutions to HW10
Section 6.1
6. Part c.:
h
x,
0
i
=
h
x,
0 + 0
i
= 2
h
x,
0
i ⇒ h
x,
0
i
= 0. It follows that
h
0
, x
i
=
h
x,
0
i
= 0.
Part d.:
By part c., if
x
= 0, then
h
0
,
0
i
= 0.
By Axiom IV of an inner product, if
x
6
= 0, then
h
x, x
i
>
0.
Part e.: If
h
x, y
i
=
h
x, z
i
for all
x
∈
V
, then
h
x, y

z
i
= 0 for all
x
∈
V
. Then in particular, taking
x
=
y

z
, we get
h
y

z, y

z
i
= 0. Thus
y

z
= 0 by part d.; i.e.,
y
=
z
.
11.

x
+
y

2
+

x

y

2
=
h
x
+
y, x
+
y
i
+
h
x

y, x

y
i
= 2
h
x, x
i
+ 2
h
y, y
i
+
h
x, y
i
+
h
y, x
i  h
x, y
i  h
y, x
i
= 2

x

2
+ 2

y

2
15. a. Our proof of CS used the following chain of inequalities, where
c
=
h
x, y
i
/
h
y, y
i
(again, we assume
y
6
= 0):
0
≤ h
x

cy, x

cy
i
=
h
x, x
i  h
x, y
i
/
h
y, y
i
.
We get an equality in the CS if and only if the first inequality is an equality. This in turn is true if
and only if
x

cy
= 0, or
x
=
cy
.
b. Similarly, our proof of the triangle inequality used the following chain of inequalities:
h
x
+
y, x
+
y
i
=
h
x, x
i
+ 2
Rh
x, y
i
+
h
y, y
i
≤ h
x, x
i
+ 2
h
x, y
i
+
h
y, y
i
≤ h
x, x
i
+ 2

x
 · 
y

+
h
y, y
i
= (
h
x, x
i
+
h
y, y
i
)
2
.
We get an equality iff these two inequalities are equalities. The second one is an equality iff
x
=
cy
for
some
c
∈
F
, by part a. The first and second are equalities iff
x
=
cy
and
Rh
x, y
i
=
R