Linear Algebra  Problem Set 1 A
Notation: For a matrix or linear operator
X
, det
X
denotes the determinant of
X
, Tr(
X
) the trace of
X
,
and rank(
X
) the rank of
X
. We will denote by
X
t
the transpose of
X
, by
X
the conjugate of
X
, and by
X
*
the conjugate transpose of
X
. If
F
is a field and
n
a positive integer, we will denote by
M
n
(
F
) the set of all
n
×
n
matrices with entries in
F
, and denote by
F
n
the vector space of
n
×
1 column vectors with entries in
F
. We will also denote by
I
∈
M
n
(
F
) the identity matrix and by diag(
λ
1
, λ
2
,
· · ·
, λ
n
)
∈
M
n
(
F
) the diagonal
matrix with main diagonal entries
λ
1
, λ
2
,
· · ·
, λ
n
.
1. Let
V
be a vector space over a field
F
.
(a) If
L
is a linearly independent subset of
V
, show that
V
has a basis containing
L
. Conclude that
every nonzero vector space has a basis.
(b) Let
W
be a subspace of
V
. Prove that there exists a subspace
U
of
V
such that
V
=
U
⊕
W
.
Solution:
(a) We will use Zorn’s Lemma. Let
S
be the set of all linearly independent subsets of
V
containing
L
, which is nonempty as
L
∈ S
. Note that
S
is partially ordered by inclusion. For any chain
C ⊆ S
, let
U
=
[
C
∈C
C
which clearly contains
L
. To show that
U
is a linearly independent subset of
V
, let
x
1
, x
2
,
· · ·
, x
n
∈
U
such that
α
1
x
1
+
α
2
x
2
+
· · ·
α
n
x
n
= 0
for some
α
i
∈
F
. Since
C
is a chain, there exists
C
∈ C
such that
x
i
∈
C
for all
i
∈ {
1
,
2
,
· · ·
, n
}
.
Since
C
is linearly independent,
α
i
= 0 for all
i
∈ {
1
,
2
,
· · ·
, n
}
. Hence
U
is a linearly indepen
dent subset of
V
containing
L
which is an upper bound for
C
.
Therefore by Zorn’s Lemma,
S
contains a maximal element
B
, which is a linearly independent
subset of
V
containing
L
. We claim that
B
is a basis for
V
. If span(
B
)
6
=
V
then there exists
an element
x
∈
V
such that
B
0
=
B
∪ {
x
}
is linearly independent, and
B
0
is an element of
S
. This contradicts the maximality of
B
, and so span(
B
) =
V
, meaning that
B
is a linearly
independent set that spans
V
. That is,
B
is a basis for
V
containing
L
.
Lastly, if
V
is any nonzero vector space, choose any
x
∈
V
with
x
6
= 0. Then
{
x
}
is a linearly
independent subset of
V
. By the previous result,
V
has a basis containing
{
x
}
, finishing the
proof.
(b) Let
B
W
be a basis for
W
. Since this is a linearly independent subset of
V
, there exists a basis
B
V
of
V
which contains
B
W
. Let
C
=
B
V
\ B
W
, which is another linearly independent subset
of
V
disjoint from
B
W
.
Let
U
= span (
C
) which is clearly a subspace of
V
not containing
W
. Now if
v
∈
V
, we can
write a unique finite sum
v
=
X
v
∈B
V
α
v
v
where
α
v
∈
F
. And so
v
=
X
w
∈B
W
α
w
w
+
X
v
∈C
α
v
v
which shows that
V
=
U
+
W
.
But because
B
V
is a disjoint union of
B
W
and
C
, and this
expression is unique, we have that
V
=
U
⊕
W
.
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2. Let
V
and
W
be vector spaces over the field
F
, with dim
V
=
n
and dim
W
=
m
. Let
T
:
V
→
W
be a
linear transformation. Prove that there exists bases
{
v
1
, v
2
,
· · ·
, v
n
}
for
V
and
{
w
1
, w
2
,
· · ·
, w
m
}
for
W
over
F
and an integer
p
≤
n
such that
T
(
c
1
v
1
+
c
2
v
2
+
· · ·
+
c
n
v
n
) =
c
1
w
1
+
c
2
w
2
+
· · ·
+
c
p
w
p
for all
c
1
, c
2
,
· · ·
, c
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 Spring '10
 HALL
 Linear Algebra, Determinant, linearly independent subset

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