2Vectors

IA Vectors and Matrices

2.7 Vector subspaces
Definition
(Vector subspace)
.
A vector subspace of a vector space
V
is a subset
of
V
that is also a vector space under the same operations. Both
V
and
{0}
are
subspaces of V . All others are proper subspaces.
A useful criterion is that a subset U V is a subspace iff
(i) x, y U (x + y) U.
(ii) x U λx U for all scalars λ.
(iii) 0 U .
This can be more concisely written as
U
is non-empty and for all
x, y U
,
(λx + µy) U”.
Example.
(i)
If
{a, b, c}
is a basis of
R
3
, then
{a + c, b + c}
is a basis of a 2D subspace.
Suppose x, y span{a + c, b + c}. Let
x = α
1
(a + c) + β
1
(b + c);
y = α
2
(a + c) + β
2
(b + c).
Then
λx + µy = (λα
1
+ µα
2
)(a + c) + (λβ
1
+ µβ
2
)(b + c) span{a + c, b + c}.
Thus this is a subspace of R
3
.
Now check that
a + c, b + c
is a basis. We only need to check linear
independence. If
α
(
a + c
) +
β
(
b + c
) =
0
, then
αa
+
βb
+ (
α
+
β
)
c
=
0
.
Since
{a, b, c}
is a basis of
R
3
, therefore
a, b, c
are linearly independent
and
α
=
β
= 0. Therefore
a + c, b + c
is a basis and the subspace has
dimension 2.
(ii)
Given a set of numbers
α
i
, let
U
=
{x R
n
:
P
n
i=1
α
i
x
i
= 0
}
. We show
that this is a vector subspace of
R
n
: Take
x, y U
, then consider
λx
+
µy
.
We have
P
α
i
(
λx
i
+
µy
i
) =
λ
P
α
i
x
i
+
µ
P
α
i
y
i
= 0. Thus
λx
+
µy U
.
The dimension of the subspace is
n
1 as we can freely choose
x
i
for
i = 1, ··· , n 1 and then x
n
is uniquely determined by the previous x
i
’s.
(iii)
Let
W
=
{x R
n
:
P
α
i
x
i
= 1
}
. Then
P
α
i
(
λx
i
+
µy
i
) =
λ
+
µ 6
= 1.
Therefore W is not a vector subspace.