tensor product basis

tensor product basis

The following

theorem

describes a basis of the

tensor product

(

http://planetmath.org/TensorProduct

)
of two

vector spaces

, in terms of given bases of the

spaces. In passing, it also gives a construction of this tensor
product. The exact same method can be used also for free
modules over a commutative ring with unit.

\PMlinkescapephrase

tensor product

Theorem.

Let

U

and

V

be vector spaces over a field

𝒦

with bases

{

𝐞

i

}

i

∈

I

and

{

𝐟

j

}

j

∈

J

respectively. Then

{

𝐞

i

⊗

𝐟

j

}

(

i

,

j

)

∈

I

×

J

(1)

is a basis for the tensor product space

U

⊗

V

.

Proof.

Let

W

=

{

ψ

:

I

×

J

⟶

𝒦

⁢

.

f

-

1

⁢

(

𝒦

∖

{

0

}

)

⁢

is finite

}

⁢

;

this set is obviously a

𝒦

-vector-space under

pointwise addition

and

multiplication

by scalar (see also
this (

http://planetmath.org/FreeVectorSpaceOverASet

) article).
Let

p

:

U

×

V

⟶

W

be
the bilinear map which

satisfies

p

⁢

(

𝐞

i

,

𝐟

j

)

⁢

(

k

,

l

)

=

{

1

if

i

=

k

and

j

=

l

,

0

otherwise

(2)

for all

i

,

k

∈

I

and

j

,

l

∈

J

, i.e.,

p

⁢

(

𝐞

i

,

𝐟

j

)

∈

W

is the

characteristic function

of

{

(

i

,

j

)

}

. The reasons (

2

) uniquely
defines

p

on the whole of

U

×

V

are that

{

𝐞

i

}

i

∈

I

is a basis of

U

,

{

𝐟

i

}

j

∈

J

is a basis of

V

, and

p

is

bilinear

.

Observe that

{

p

⁢

(

𝐞

i

,

𝐟

j

)

}

(

i

,

j

)

∈

I

×

J

is a basis of

W

. Since one may always define a

linear map

by giving its values on the basis elements, this implies that there for every

𝒦

-vector-space

X

and every map

γ

:

U

×

V

⟶

X

exists a unique linear map

γ

^

:

W

⟶

X

such that

γ

^

⁢

(

p

⁢

(

𝐞

i

,

𝐟

j

)

)

=

γ

⁢

(

𝐞

i

,

𝐟

j

)

for all

i

∈

I

and

j

∈

J

.

For

γ

that are bilinear it holds for arbitrary

𝐮

=

∑

i

∈

I

′

u

i

⁢

𝐞

i

∈

U

and

𝐯

=

∑

j

∈

J

′

v

j

⁢

𝐟

j

∈

V

that

γ

⁢

(

𝐮

,

𝐯

)

=

(

γ

^

∘

p

)

⁢

(

𝐮

,

𝐯

)

, since

γ

⁢

(

𝐮

,

𝐯

)

=

γ

⁢

(

∑

i

∈

I

′

u

i

⁢

𝐞

i

,

∑

j

∈

J

′

v

j

⁢

𝐟

j

)

=

∑

i

∈

I

′

∑

j

∈

J

′

u

i

⁢

v

j

⁢

γ

⁢

(

𝐞

i

,

𝐟

j

)

=

=

∑

i

∈

I

′

∑

j

∈

J

′

u

i

⁢

v

j

⁢

γ

^

⁢

(

p

⁢

(

𝐞

i

,

𝐟

j

)

)

=

γ

^

⁢

(

∑

i

∈

I

′

∑

j

∈

J

′

u

i

⁢

v

j

⁢

p

⁢

(

𝐞

i

,

𝐟

j

)

)

=

=

γ

^

⁢

(

p

⁢

(

∑

i

∈

I

′

u

i

⁢

𝐞

i

,

∑

j

∈

J

′

v

j

⁢

𝐟

j

)

)

=

γ

^

⁢

(

p

⁢

(

𝐮

,

𝐯

)

)

⁢

.

As this is the

defining property

of the tensor product

U

⊗

V

however, it follows that

W

is (an incarnation of) this tensor
product, with

𝐮

⊗

𝐯

:=

p

⁢

(

𝐮

,

𝐯

)

.
Hence the claim in the theorem is

equivalent

to the observation
about the basis of

W

.
∎

Title

tensor product basis

Canonical

name

TensorProductBasis

Date of creation

2013-03-22 15:24:48

Last modified on

2013-03-22 15:24:48

Owner

lars_h (9802)

Last modified by

lars_h (9802)

Numerical id

11

Author

lars_h (9802)

Entry type

Theorem

Classification

msc 15A69

Synonym

basis construction of tensor product

Related topic

TensorProduct

Related topic

FreeVectorSpaceOverASet

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