This is on the tensor product space. It is a way to consider multilinear forms. There is a space
called the tensor product space which has a norm. Then when you have a multilinear form
with values in some space V , it can be represented as a linear map from this tensor product
space to V . Some people like this notation because the map just mentioned is linear. To do
this right, you need some material which is equivalent to the axiom of choice. This is in an
appendix.
Definition 18.12.1Denote by B
(X × Y,K )
the bilinear forms having values in K a vector space. Define⊗ : X × Y →ℒ
(B (X × Y,K ),K)
x⊗ y (A ) ≡ A (x,y)
where A is a bilinear form. Often K is a field, but it could be any vector space with no change.
Remark 18.12.2⊗ is bilinear. That is
(ax1 + bx2)⊗ y = a(x1 ⊗ y)+ b(x2 ⊗y)
x⊗ (ay + by ) = a(x⊗ y )+ b(x ⊗ y)
1 2 1 2
This follows right away from the definition. Note that x⊗y ∈ℒ
Definition 18.12.3Define X ⊗ Y as the span of the x ⊗ y in ℒ
(B (X × Y,K) ,K )
.
Lemma 18.12.4Let V ≤ X. That is, V≠0 is a subspace of X. Let B_{V }be a Hamel basis for V .Then there exists a Hamel basis for X,B_{X}which includes B_{V }.
Proof:Consider
(Y,BY )
where Y ⊇ V and B_{Y }⊇ B_{V }. Call it ℱ. Let
(Y,BY )
≤
(ˆ )
Y,BYˆ
if and only
if Y ≤Ŷ and B_{Y }⊆ B_{Ŷ}. Then let C denote a maximal chain. Let
( )
ˆX,B ˆX
consist of the union of all Y in
the chain along with the union of all B_{Y } in the chain. Then in fact this is something in ℱ thanks to this
being a chain. Thus
Xˆ
= span
(B ˆ)
X
. Then it must be the case that
Xˆ
= X. If not, there exists
y
∕∈
span
(B )
Xˆ
. But then
(span (B ,y) ,{B } ,{y})
ˆX ˆX
could be added to C and obtain a strictly longer
chain. ■
Next,
x ⊗ (ay+ bz)(A) ≡ A (x,ay + bz) = aA (x,y)+ bA (x,z)
= ax ⊗y (A)+ bx⊗ z (A ) ≡ (ax⊗ y +bx ⊗ z)(A )
and so ⊗ distributes across addition from the left and you can factor out scalars. Similarly, it will work the
same way from the right.
Lemma 18.12.5Let B_{X}be a Hamel basis for X. Then define L : X → Y as follows. For eachx ∈ B_{X}, let Lx ≡ y_{x}. Then for arbitrary z, define Lz as follows. For z = ∑_{x∈BX}c_{x}x where thisdenotes a finite sum consisting of the unique linear combination of basis vectors which equalsx,
∑
Lz ≡ czyx
x∈BX
Then L is linear.
Proof:There is only one finite linear combination equal to z thanks to linear independence. Thus L is
well defined. Why is it linear? Let z = ∑_{x∈BX}c_{z}x,w = ∑_{x∈BX}c_{w}x. Then L
(az + bw)
=
( ∑ ∑ ) ∑ ∑
L a czx+ b cwx ≡ a czyx + b cwyx = aLz + bLw
x∈BX x∈BX x∈BX x∈BX
Thus L is linear. ■
Proposition 18.12.6Let X,Y be vector spaces and let E and F be linearly independent subsetsof X,Y respectively. Then
{e⊗ f : e ∈ E,f ∈ F}
is linearly independent in X ⊗ Y .
Proof:Let κ ∈ K. Say ∑_{i=1}^{n}λ_{i}e_{i}⊗f_{i} = 0 in X ⊗Y . This means it sends everything in B
(X × Y;K )
to 0. Let ψ ∈ F^{∗},ψ
(fk)
= 1 and ψ
(fi)
= 0 for i≠k. Let ϕ ∈ E^{∗} be defined similarly. What
you do is extend
{ei}
to a Hamel basis and then define ϕ to equal 1 at e_{k} and ϕ sends every
other thing in the Hamel basis to 0. Then you look at ϕ
(x)
ψ
(y)
κ ≡ A
(x,y)
. Then you have
0 = ∑_{i=1}^{n}λ_{i}e_{i}⊗f_{i}
(A)
≡ λ_{k}ϕ
(ek)
ψ
(fk)
κ = λ_{k}κ. Since κ is arbitrary it must be that λ_{k} = 0. Thus these
are linearly independent. ■
Proposition 18.12.7Suppose u = ∑_{i=1}^{n}x_{i}⊗ y_{i}is in X ⊗ Y and is a shortest representation.Then
{xi}
,
{yi}
are linearly independent. All such shortest representations have the same length. If∑_{i}x_{i}⊗ y_{i} = 0 and
{yi}
are independent, then x_{i} = 0 for each i. In particular, x ⊗ y = 0 iff x = 0
or y = 0.
Proof:Suppose the first part. If
{y}
i
are not linearly independent, then one is a linear combination of
the others. Say y_{n} = ∑_{j=1}^{n−1}a_{j}y_{j}. Then
∑n n∑−1 n−∑ 1
xi ⊗ yi = xi ⊗ yi + xn ⊗ ajyj
i=1 i=1 j=1
n∑−1 n∑−1
= xi ⊗ yi + aixn ⊗ yi
i=1 i=1
n∑−1
= xi ⊗ yi + aixn ⊗ yi
i=1
n∑−1
= (xi + aixn)⊗ yi
i=1
and so n was not smallest after all. Similarly
{xi}
must be linearly independent. Now suppose
that
∑n ∑m
xi ⊗ yi = uj ⊗vj
i=1 j=1
and both are of minimal length. Why is n = m? We know that
{xi}
,
{yi}
,
{uj}
,
{vj}
are independent.
Let ψ_{k}
(yk)
= 1, ψ sends all other vectors to 0. Then do both sides to A
(x,y)
≡ ϕ
(x)
ψ_{k}
(y)
κ where κ ∈ K
is given, ϕ ∈ X^{∗} arbitrary.
∑n n∑
xi ⊗ yi(A) = ϕ(xi)ψk(yi)κ =
i=1 i=1
∑m
ϕ (xk)κ = ϕ (uj) ψ(vj)κ
j=1
Hence
( m )
ϕ (x − ∑ u ψ(v )) κ = 0
k j=1 j j
and this holds for any ϕ ∈ X^{∗} and for any κ. Hence x_{k}∈span
({uj})
. Thus m ≥ n since this can be done
for each k.
{x1,⋅⋅⋅,xn}
⊆span
(u1,⋅⋅⋅,um )
and the left side is independent and is contained in
the span of the right side. Similarly,
{u1,⋅⋅⋅,um}
⊆span
(x1,⋅⋅⋅,xn)
and so m ≤ n. Thus
m = n.
Next suppose that ∑_{i=1}^{n}x_{i}⊗y_{i} = 0 and the
{yi}
are linearly independent. Then let ψ_{k}
(yk)
= 1 and
ψ_{k} sends the other vectors to 0. Then do the sum to A
(x,y)
= ϕ
(x)
ψ
(y)
κ for κ ∈ K. This yields
ϕ
(xk)
κ = 0 for every ϕ. Hence x_{k} = 0. This is so for each k. Similarly, if ∑_{i=1}^{n}x_{i}⊗ y_{i} = 0 and
{xi}
independent, then each y_{i} = 0. ■
The next theorem is very interesting. It is concerned with bilinear forms having values in V a vector
space ψ : X × Y → V . Roughly, it says there is a unique linear map from X ⊗ Y which delivers the given
bilinear form.
Theorem 18.12.8Suppose ψ is a bilinear map in B
(X × Y;V )
where V is a vector space. Let⊗ : X ×Y → X ⊗Y be given by ⊗
(x,y)
≡ x⊗y. Then there exists a unique linear map
ˆψ
∈ℒ
(X ⊗ Y;V )
such that
ˆ
ψ
∘⊗ = ψ. In other words, the following diagram commutes.
X × Y
ψ
⊗ ↓ ↘
↺ˆψ
X ⊗ Y −−→ V
That is,
ˆψ
(x⊗ y)
= ψ
(x,y)
.
Proof: Let E be a Hamel basis for X and let F be one for Y . Then by definition,
{e⊗ f : e ∈ E,f ∈ F}
spans X ⊗ Y. To see this, suppose you have x ⊗ y. Then x = ∑_{i}a_{i}e_{i},y = ∑_{j}b_{j}e_{j}
and so
∑ ∑ ∑
x⊗ y = aiei ⊗ bjfj = aibjei ⊗ fj
i j i,j
Also, it was shown above in Proposition 18.12.6 that this set is linearly independent. Therefore, it is a
Hamel basis and you can define
ψˆ
as follows:
ˆψ
(e ⊗ f)
≡ ψ
(e,f)
and extend
ˆψ
linearly to
get the desired linear transformation
ˆψ
. It is unique because its value on a Hamel basis is
completely determined by the values of ψ. Thus ψ gives the right thing on E × F. It gives the right
thing on X × Y also. To see this, suppose you have x ⊗ y = ∑_{i,j}a_{i}b_{j}e_{i}⊗ f_{j} as above. Then