Direct Sum of Subspaces
Paul Skoufranis
September 29, 2011
The purpose of this document is to demonstrate that vector spaces can have ’nice’ decompositions. In
particular, we will present some proofs relating to spaces of vector spaces. We begin with a definition for
the sum of two sets of vectors.
Definition)
Let
V
be a vector space.
Suppose that
S
1
and
S
2
are nonempty subsets of
V
.
The sum
of
S
1
and
S
2
, denoted
S
1
+
S
2
, is
{
~x
+
~
y

~x
∈
S
1
, ~
y
∈
S
2
}
(that is,
S
1
+
S
2
is the set of vectors of
V
that
can be obtained by adding a vector in
S
1
to a vector in
S
2
).
Example)
Let
S
1
be the
x
axis in
R
3
and let
S
2
be the
y
axis in
R
3
.
Then
S
1
+
S
2
is the
xy
plane in
R
3
. To see this, we notice that if we add a vector in the direction of the
x
axis to a vector in the direction
of the
y
axis, we obtain a vector in the
xy
plane (that is (
x,
0
,
0) + (0
, y,
0) = (
x, y,
0) is in the
xy
plane).
Moreover, every vector in the
xy
plane can be written as the sum of a vector in the direction of the
x
axis
and a vector in the direction of the
y
axis (that is, (
x, y,
0) = (
x,
0
,
0) + (0
, y,
0)).
Example)
Let
S
1
=
{
(
x,
0)

x
∈
R
} ⊆
R
2
and
S
2
=
{
(
y, y
)

y
∈
R
} ⊆
R
2
}
.
Then
S
1
+
S
2
∈
R
2
(as
span
{
(1
,
0)
,
(1
,
1)
}
=
R
2
).
The following is our first result about the sum of two subspaces of a vector space.
Proposition)
Let
V
be a vector space and suppose
W
1
and
W
2
are subspaces of
V
.
Then
W
1
+
W
2
is
a subspace of
V
that contains
W
1
and
W
2
.
Proof
:
Clearly
W
1
+
W
2
⊆
V
.
To show that
W
1
+
W
2
is a subspace of
V
, we need to demonstrate
the three properties of being a subspace.
1. We need to demonstrate that
~
0
∈
W
1
+
W
2
.
We recall that since
W
1
and
W
2
are subspaces of
V
,
~
0
∈
W
1
and
~
0
∈
W
2
. Therefore, since
~
0 =
~
0 +
~
0,
~
0
∈
W
1
, and
~
0
∈
W
2
,
~
0
∈
W
1
+
W
2
by the definition
of
W
1
+
W
2
.
2. We need to demonstrate that
W
1
+
W
2
is closed under addition. Suppose
~x, ~
y
∈
W
1
+
W
2
are arbitrary
vectors. Therefore, by the definition of
W
1
+
W
2
, there exists vectors
~x
1
, ~
y
1
∈
W
1
and
~x
2
, ~
y
2
∈
W
2
such
that
~x
=
~x
1
+
~x
2
and
~
y
=
~
y
1
+
~
y
2
. Therefore
~x
+
~
y
= (
~x
1
+
~x
2
) + (
~
y
1
+
~
y
2
) = (
~x
1
+
~
y
1
) + (
~x
2
+
~
y
2
)
Since
~x
1
, ~
y
1
∈
W
1
and
W
1
is a subspace of
V
,
~x
1
+
~
y
1
∈
W
1
. Since
~x
2
, ~
y
2
∈
W
2
and
W
2
is a subspace
of
V
,
~x
2
+
~
y
2
∈
W
2
. Therefore, we have shown that
~x
+
~
y
is the sum of a vector in
W
1
and a vector
in
W
2
. Hence
~x
+
~
y
∈
W
1
+
W
2
by the definition of
W
1
+
W
2
. Therefore, since
~x, ~
y
∈
W
1
+
W
2
were
arbitrary,
W
1
+
W
2
is closed under addition.