Problem Set 4  Solutions
Economics 204  August 2006
1. Let Θ be the set of all continuous functions whose domain is the unit interval [0
,
1]
and range is
R
. Let Φ be the subset consisting of all real polynomials (whose domain
is restricted to the unit interval) of degree at most two:
Φ
≡ {
a
+
bx
+
cx
2

a, b, c
∈
R
}
Note that the set Θ is a vector space over the field of real numbers and the subset Φ
is a proper subspace.
(a) Are the vectors
{
x,
(
x
2
−
1)
,
(
x
2
+ 2
x
+ 1)
}
linearly independent over
R
?
Apply the usual test for independence of vectors. Solve for
A, B,
and
C
such that
Ax
+
B
(
x
2
−
1) +
C
(
x
2
+ 2
x
+ 1) = 0
Equating like powers of
x
we obtain the following system in the three unknowns:
B
+
C
= 0
A
+ 2
C
= 0
C
−
B
= 0
⇒
C
=
B
=
−
B
⇒
C
=
B
= 0
⇒
A
= 0.
Thus, the three vectors are linearly independent over
R
.
(b) Find a Hamel basis for the subspace Φ.
Clearly,
{
1
, x, x
2
}
is linearly independent and spans Φ, so it is a Hamel basis
and dim Φ = 3. Also, since the set of vectors in (
a
) is linearly independent and
contains three elements, it is a basis.
(c) What is the dimension of Φ ? What is the dimension of Θ ?
Three and
∞
, respectively. To see that the dimension of Θ is infinite note that
the set of vectors
{
1
, x, x
2
, x
3
, ...
}
form a linearly independent set over
R
. Since
Θ contains an infinite linearly indpendent set of vectors and the number of lin
early independent elements of a vector space cannot exceed the dimension, the
dimension of Θ must be infinite.
2. Recall that a reﬂection across the
x
axis can be achieved with the transformation
(
x, y
)
→
(
x,
−
y
).
Derive a transformation,
T
, which reﬂects a point across the line
y
= 3
x
.
1
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
(a) First, calculate the action of
T
on the points (1
,
3) and (
−
3
,
1).
Since (1
,
3) lies on the line
y
= 3
x
it is unchanged by
T
so we know that
T
(1
,
3) =
(1
,
3).
The slope of the line
y
= 3
x
is 3 and the slope of the vector (
−
3
,
1) is
−
1
/
3. Because the vector is perpendicular with the line, reﬂecting it across the
line takes is to (3
,
−
1), so
T
(
−
3
,
1) = (3
,
−
1).
(b) Next, write the matrix representation of
T
using these two vectors as bases.
Let
V
=
{
v
1
, v
2
}
=
1
3
,
−
3
1
. From (2a) we know that
T
(
v
1
) =
v
1
= 1
·
v
1
+0
·
v
2
and
T
(
v
2
) =
−
v
2
= 0
·
v
1
−
1
·
v
2
. So we write
P
=
Mtx
V
(
T
) =
1
0
0
−
1
.
(c) Find
S
and
S

1
, the matrices that changes coordinates under this basis to stan
dard coordinates and back again.
We can easily find
S
, the matrix that changes coordinates in
V
to coordinates un
der the standard basis,
E
, because we already expressed the basis vectors
v
1
and
v
2
in terms of standard basis coordinates. We have
S
=
Mtx
E,V
(
id
) = (
v
1
v
2
) =
1
−
3
3
1
.
The matrix that changes coordinates from
E
to
V
is simply the
inverse of
S
. So
S

1
=
Mtx
V,E
(
id
) =
1
10
1
3
−
3
1
=
1
/
10
3
/
10
−
3
/
10
1
/
10
.
This is the end of the preview.
Sign up
to
access the rest of the document.
 Summer '08
 Staff
 Economics

Click to edit the document details