Homework assignment 4
pp. 49
Exercise 8.
Let
V
be the space of
2
×
2
matrices over
F
. Find a basis
{
A
1
,A
2
,A
3
,A
4
}
for
V
such that
A
2
j
=
A
j
for each
j
.
Solution
If we start with the canonical basis for
V
, namely
B
=
‰
B
1
=
±
1 0
0 0
¶
,B
2
=
±
0 1
0 0
¶
,B
3
=
±
0 0
1 0
¶
,B
4
=
±
0 0
0 1
¶²
we notice that the ﬁrst and the last elements satisfy the required condition.
Therefore we only need to ﬁnd other two matrices, such that the four matrices
generate
V
. Let
A
=
‰
A
1
=
±
1 0
0 0
¶
,A
2
=
±
1 1
0 0
¶
,A
3
=
±
0 0
1 1
¶
,A
4
=
±
0 0
0 1
¶²
Notice that
A
2
j
=
A
j
and
A
2

A
1
=
B
2
and
A
3

A
1
=
B
3
. Therefore
A
is a basis for
V
Bonus exercise 14.
Let
V
be the set of real numbers. Regard
V
as a
vector space over the ﬁeld of
rational
numbers, with the usual operations. Prove
that this vector space is
not
ﬁnitedimensional.
Solution:
By contradiction. Suppose that
V
is ﬁnite dimensional, this implies
that
V
is a countable set (see the lemma below), but the set of real numbers is
not countable!
Lemma 1.
A ﬁnite dimensional vector space
V
over the rational numbers is a
countable set.
Proof.
By induction on
n
=
the number of elements in a basis for
V
.
Base:
n
= 1
Let
{
α
}
be a basis for
V
.
Let
{
a
1
,a
2
,...,a
n
,...
}
be an enumeration of the rational numbers, then any
element of
V
may be written as
a
i
α
. That is,
V
⊂ {
a
1
α,a
2
α,.
..,a
n
α,.
..
}
(actually
the two sets are equal, but we don’t need that fact). Therefore
V
is countable.
Inductive Step:
Suppose that
V
is countable if its dimension is less than or
equal to
n
. We will prove then that
if the dimension of
V
is
n
+ 1
then it is a
countable set
. Let
{
β
1
,...,β
n
,α
}
be a basis for
V
. By the induction hypothesis
we know that the subspace
W
generated by
{
β
1
,...,β
n
}
is a countable set. Let
1