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C.6.1. Theorem.
Let V i , 1 £ i £ k, be vector spaces.
(1) (Existence) A tensor product ( A ,a) of the V i exists. The image of a will actu-
ally span A .
(2) (Uniqueness) Given another tensor product ( B ,b), then there is a unique iso-
morphism m : A Æ B with b=m
a.
Proof. To prove part (1), here is how one can define A and a when k = 2. It should
be obvious how to generalize the construction to handle the case of an arbitrary k.
Let M be the free vector space with basis ( v 1 , v 2 ), where v i ΠV i . Let N be the vector
subspace of M generated by all elements of M of the form
(
) - (
)
(
)
(
) - (
) -
(
)
v
v
,
v
v
,
v
v
,
v
,
v
,
v
v
v
,
v
v
,
v
¢
,
1
1
2
1
2
1
2
1
2
2
1
2
1
2
(
) -
(
)
(
) -
(
)
r
vv
,
r
vv
,
,
v v
,
r
r
vv
,
,
12
12
1 2
12
where v i , v i ¢Œ V i and r Œ R . Define A = M / N and
a : VV A
¥Æ
1
by
(
) = (
) +
a vv
,
vv
,
N
.
12
12
Given a map f : V 1 ¥ V 2 Æ W , define g 0 : M Æ W by
Ê
Á
ˆ
˜ =
Â
Â
(
)
(
)
g
a
vv
,
a
f
vv
,
.
0
vv
,
1
2
vv
,
1
2
12
12
(
vv
,
)δ
V V
(
vv
,
)δ
V V
12
1 2
12
1 2
One can show that g 0 sends N to 0 and hence induces a map g : A Æ W . Clearly,
f = g
a. The map g is unique because the image of a spans A . It follows that the pair
( A ,a) is a tensor product for V 1 and V 2 . See [AusM63].
To prove the uniqueness of the tensor product, let ( B ,b) be another such. See
Figure C.4. Since b is multilinear, the universal factorization property of ( A ,a) implies
that there is a unique linear map m : A Æ B with b=m
a. Similarly, the universal fac-
torization property of ( B ,b) implies that there is a unique linear map m¢ : B Æ A with
a=m¢
b. Therefore, b=m
b and a=m¢
m
a. This implies that m is an isomorphism.
The theorem is proved.
Theorem C.6.1(2) shows that it is the universal factorization property of a tensor
product that is important and not the particular construction that is used. For that
reason one usually talks about “the” tensor product and uses a uniform notation.
a
...
V 1 ¥ V 2 ¥¥ V k
A
m
b
Figure C.4.
The uniqueness of tensor
products.
B
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