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.