all bases for a vector space have the same cardinality

all bases for a vector space have the same cardinality

In this entry, we want to show the following

property of bases

for a

vector space

:

Theorem 1

.

All bases for a vector space

V

have the same

cardinality

.

Let

B

be a basis for

V

(

B

exists, see this link (

http://planetmath.org/ZornsLemmaAndBasesForVectorSpaces

)). If

B

is

infinite

, then all bases for

V

have the same cardinality as that of

B

(proof (

http://planetmath.org/CardinalitiesOfBasesForModules

)). So all we really need to show is where

V

has a finite basis.

Before proving this important property, we want to prove something that is almost as important:

Lemma 1

.

If

A

and

B

are subsets of a vector space

V

such that

A

is

linearly independent

and

B

spans

V

, then

|

A

|

≤

|

B

|

.

Proof.

If

A

is finite and

B

is infinite, then we are done. Suppose now that

A

is infinite. Since

A

is linearly independent, there is a

superset

C

of

A

that is a basis for

V

. Since

A

is infinite, so is

C

, and therefore all bases for

V

are infinite, and have the same cardinality as that of

C

. Since

B

spans

V

, there is a subset

D

of

B

that is a basis for

V

. As a result, we have

|

A

|

≤

|

C

|

=

|

D

|

≤

|

B

|

.

Now, we suppose that

A

and

B

are both finite. The case where

A

=

∅

is clear. So assume

A

≠

∅

. As

B

spans

V

,

B

≠

∅

. Let

A

=

{

a

1

,

…

,

a

n

}

and

B

=

{

b

1

,

…

,

b

m

}

and assume

m

<

n

. So

a

i

≠

0

for all

i

=

1

,

…

,

n

. Since

B

spans

V

,

a

1

can be expressed as a

linear combination

of elements of

B

. In this expression, at least one of the coefficients (in the field

k

) can not be

0

(or else

a

1

=

0

). Rename the

elements

if possible, so that

b

1

has a non-zero coefficient in the expression of

a

1

. This means that

b

1

can be written as a linear combination of

a

and the remaining

b

’s. Set

B

1

=

{

a

1

,

b

2

,

…

,

b

m

}

.

As every element in

V

is a linear combination of elements of

B

, it is therefore a linear combination of elements of

B

1

. Thus,

B

1

spans

V

. Next, express

a

2

as a linear combination of elements in

B

1

. In this expression, if the only non-zero coefficient is in front of

a

1

, then

a

1

and

a

2

would be linearly dependent, a

contradiction

! Therefore, there must be a non-zero coefficient in front of one of the

b

’s, and after some renaming once more, we have that

b

2

is the one with a non-zero coefficient. Therefore,

b

2

, likewise, can be expressed as a linear combination of

a

1

,

a

2

and the remaining

b

’s. It is easy to see that

B

2

=

{

a

1

,

a

2

,

b

3

,

…

,

b

m

}

spans

V

as well. Continue this process until all of the

b

’s have been replaced, which is possible since

m

<

n

. We have finally arrived at the set

B

m

=

{

a

1

,

…

,

a

m

}

which is a

proper subset

of

A

. In

addition

,

B

m

spans

V

. But this would imply that

A

is linearly dependent, a contradiction.
∎

Now we can

complete

the proof of

theorem

1.

Proof.

Suppose

A

and

B

are bases for

V

. We apply the lemma. Then

|

A

|

≤

|

B

|

since

A

is linearly independent and

B

spans

V

. Similarly,

|

B

|

≤

|

A

|

since

B

is linearly independent and

A

spans

V

. An application of

Schroeder-Bernstein theorem

completes the proof.
∎

Title

all bases for a vector space have the same cardinality

Canonical name

AllBasesForAVectorSpaceHaveTheSameCardinality

Date of creation

2013-03-22 18:06:51

Last modified on

2013-03-22 18:06:51

Owner

CWoo (3771)

Last modified by

CWoo (3771)

Numerical id

6

Author

CWoo (3771)

Entry type

Result

Classification

msc 16D40

Classification

msc 13C05

Classification

msc 15A03

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