Physics 316, Problem Set 1, due Tuesday October 4 in class
1. Schwarz’s inequality:
A vector space
V
has a norm, with the properties listed in the Glossary (see
materials
in web site). Show that its natural scalar product (
x
,
y
) satisﬁes (
x
,
y
)
2
≤
(
x
,
x
)(
y
,
y
). This is
known as the Schwartz inequality.
Solution:
(
x
,
y
)=
1
2
‡

x
+
y
2


x
2

y
2
·
We observe that
(
x
,
x
x
2
for any
x
. According to the triangle inequality,
x
+
y

≤

x
+
y
x
+
y
2
≤

x
2
+
y
2
+2
x
 
y
Expressing
x
+
y
using the deﬁnition of (
x
,
y
),
2(
x
,
y
)+
x
2
+
x
2
=
x
+
y
2

x
2
+
y
2
x
y
2(
x
,
y
)
≤
2
x
y
dividing by 2 and squaring.
..
(
x
,
y
)
2
≤
(
x
,
x
)(
y
,
y
)
QED
2. Aﬃne transformations and linear transformations:
A certain aﬃne space is also a vector space,
and a line in this vector space is given by
y
(
η
)
≡
b
+
η
x
.
a) Find the analogous expression for a line passing through the origin (
0
) of the vector space that is parallel
to the given line .
b) A transformation
f
:
x
∈
R
2
→
y
∈
R
2
in a twodimensional space
R
2
is deﬁned by
(
y
1
,y
2
•
ab
cd
‚
(
x
1
,x
2
)+(
g
1
,g
2
)
,
where
a...d
,
g
1
and
g
2
are constants. Show that the image of the generic line deﬁned in a) is also a line.
Solution:
a) Let’s call the desired line passing through the origin
z
(
η
). If
z
=
b
0
+
η
x
0
includes the origin, then
b
0
must be
0
. To make
z
be parallel to
y
we can take
x
0
=
x
: there is no value of
η,η
0
such that
y
(
η
z
(
η
0
) thus the two lines don’t intersect. The same is true if
x
0
is a scalar multiple of
x
.
b) Since
x
(
η
) is a line, it can be expressed
x
(
η
b
+
η
u
. Using the component language
x
=(
x
1
2
)
, this means (
x
1
2
)(
η
)=(
b
1
+
ηu
1
,b
2
+
2
). We need to show that (
y
1
2
)(
η
) can be expressed in
the form
B
+
η
U
, or in component language, (
y
1
2
B
1
+
ηU
1
,B
2
+
2
). On the other hand the
formula says
(
y
1
2
•
‚
(
b
1