vector spaces are isomorphic iff their bases are equipollent

vector spaces are isomorphic iff their bases are equipollent

Theorem 1

.

Vector spaces

V

and

W

are

isomorphic

iff their bases are

equipollent

(have the same cardinality).

Proof.

(

⟹

) Let

ϕ

:

V

→

W

be a

linear isomorphism

. Let

A

and

B

be bases for

V

and

W

respectively. The set

ϕ

⁢

(

A

)

:=

{

ϕ

⁢

(

a

)

∣

a

∈

A

}

is a basis for

W

. If

r

1

⁢

ϕ

⁢

(

a

1

)

+

⋯

+

r

n

⁢

ϕ

⁢

(

a

n

)

=

0

,

with

a

i

∈

A

. Then

ϕ

⁢

(

r

1

⁢

a

1

+

⋯

+

r

n

⁢

a

n

)

=

0

since

ϕ

is linear. Furthermore, since

ϕ

is

one-to-one

, we have

r

1

⁢

a

1

+

⋯

+

r

n

⁢

a

n

=

0

,

hence

r

i

=

0

for

i

=

1

,

…

,

n

, since

A

is

linearly independent

. This shows that

ϕ

⁢

(

A

)

is linearly independent. Next, pick any

w

∈

W

, then there is

v

∈

V

such that

ϕ

⁢

(

v

)

=

w

since

ϕ

is onto. Since

A

spans

V

, we can write

v

=

r

1

⁢

a

1

+

⋯

+

r

n

⁢

a

n

,

so that

w

=

ϕ

⁢

(

v

)

=

r

1

⁢

ϕ

⁢

(

a

1

)

+

⋯

+

r

n

⁢

ϕ

⁢

(

a

n

)

.

This shows that

ϕ

⁢

(

A

)

spans

W

. As a result,

ϕ

⁢

(

A

)

is a basis for

W

.

A

and

ϕ

⁢

(

A

)

are equipollent because

ϕ

is one-to-one. But since

B

is also a basis for

W

,

ϕ

⁢

(

A

)

and

B

are equipollent. Therefore

|

A

|

=

|

ϕ

⁢

(

A

)

|

=

|

B

|

.

(

⟸

)

Conversely

, suppose

A

is a basis for

V

,

B

is a basis for

W

, and

|

A

|

=

|

B

|

. Let

f

be a

bijection

from

A

to

B

. We extend the domain of

f

to all of

A

, and call this

extension

ϕ

, as follows:

ϕ

⁢

(

a

)

=

f

⁢

(

a

)

for any

a

∈

A

. For

v

∈

V

, write

v

=

r

1

⁢

a

1

+

⋯

+

r

n

⁢

a

n

with

a

i

∈

A

, set

ϕ

⁢

(

v

)

=

r

1

⁢

ϕ

⁢

(

a

1

)

+

⋯

+

r

n

⁢

ϕ

⁢

(

a

n

)

.

ϕ

is a well-defined

function

since the expression of

v

as a

linear combination

of elements of

A

is unique. It is a routine verification to check that

ϕ

is indeed a

linear transformation

. To see that

ϕ

is one-to-one, let

ϕ

⁢

(

v

)

=

0

. But this means that

v

=

0

, again by the uniqueness of expression of

0

as a linear combination of elements of

A

. If

w

∈

W

, write it as a linear combination of elements of

B

:

w

=

s

1

⁢

b

1

+

⋯

+

s

m

⁢

b

m

.

Each

b

i

∈

B

is the image of some

a

∈

A

via

f

. For simplicity, let

f

⁢

(

a

i

)

=

b

i

. Then

w

=

s

1

⁢

f

⁢

(

a

1

)

+

⋯

+

s

m

⁢

f

⁢

(

a

m

)

=

s

1

⁢

ϕ

⁢

(

a

1

)

+

⋯

+

s

m

⁢

ϕ

⁢

(

a

m

)

=

ϕ

⁢

(

s

1

⁢

a

1

+

⋯

+

s

m

⁢

a

m

)

,

which shows that

ϕ

is onto. Hence

ϕ

is a linear isomorphism between

V

and

W

.
∎

Title

vector spaces are isomorphic iff their bases are equipollent

Canonical name

VectorSpacesAreIsomorphicIffTheirBasesAreEquipollent

Date of creation

2013-03-22 18:06:55

Last modified on

2013-03-22 18:06:55

Owner

CWoo (3771)

Last modified by

CWoo (3771)

Numerical id

7

Author

CWoo (3771)

Entry type

Result

Classification

msc 13C05

Classification

msc 15A03

Classification

msc 16D40

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