# Tensor Product question.

1. Jul 5, 2008

### MathematicalPhysicist

There's a question that asks me to show that there exists a unique linear transformation
from: $$f\otimes g: V_1\otimes W_1\rightarrow V_2\otimes W_2$$
where f and g are linear transformations f:V1->V2, g:W1->W2
that satisfies: $$(f\otimes g)(u\otimes v)=f(u)\otimes g(v)$$

well I think that what I need to show is only that
it's linear by first componenet and second componenet which I did.
well if that's it for linear (including multplication by a scalar), then now I need to show uniqueness, well I guess this depends on f and g is it not?
well it's unique by our choice of f and g, with other functions we would have a different linear transformation.
What do I miss here?

2. Jul 5, 2008

### Hurkyl

Staff Emeritus
What definition of the tensor product of two vector spaces are you using? This sounds like it's the sort of thing that should be a nearly immediate consequence of the definition.

3. Jul 5, 2008

### MathematicalPhysicist

well it does seem like that.

basically what I'm using here that
the tensor product satisfies that each terms v1,v2,v,w,w1,w2 in V,W:
$$v\otimes(w_1+w_2)=v\otimes w_1+v\\otimes w_2$$
$$(v_1+v_2)\otimes w=v_1\otimes w+v_2\otimes w$$
$$(\lambda v)\otimes w=\lambda (v\otimes w)$$
lambda is a scalar.

basically the uniquness part distrubs me, if ofcourse I got it correct for proof of linearity of map.

cheers.

4. Jul 5, 2008

### mrandersdk

a little hint, try to define the function on a basis of V_1 and W_1 by

$$(f\otimes g)(v_i\otimes w_j)=f(v_i)\otimes g(w_j)$$

and then extend by linearity. Then you need to show that

$$(f\otimes g)(u\otimes v)=f(u)\otimes g(v)$$

is satified (which also gives you that it doesn't depend on the basis chosen, which helps with the uniqueness part), and you need to show that given another linear function on V_1 \otimes W_1, satisfying this it is equal to the one defined above.

Last edited: Jul 5, 2008
5. Jul 5, 2008

### MathematicalPhysicist

but wait if it's defined by f and g already, then it's unique is it not?
if we were to replace f by another linear map, say for example f(u)=u and h(u)=2u
then: f x g(u x v)=u x g(v) and h x g(u x v)=2(u x g(v))
which arent necessarily the same.

and for the hint with linearity, basically if v is decomposed of a linear combination of the bases of V1 and W1, f x g(u x v) is a summation over i and j of f x g (u_i x v_j)=f(u_i) x g(v_j) and then you recombine it again, as I said it's just by using the definition.

6. Jul 5, 2008

### MathematicalPhysicist

what I want to say is that another function would be different than the one defined here, if we change at least one of f or g.

7. Jul 5, 2008

### mrandersdk

you need not to show that it is linear in the first component and in the second component, this doesn't actually make sence because it is a function of only one component, you need to show that there exist a function satisfying

$$(f\otimes g)(\lambda u_1\otimes v_1+u_2\otimes v_2)=\lambda (f\otimes g)(u_1\otimes v_1)+(f\otimes g)(u_2\otimes v_2)$$

I give you that the uniqueness part is clear.

8. Jul 5, 2008

### MathematicalPhysicist

Yes I thought that I need to show linearity by addition and by scalar (scalar is easy).

so linearity by addition should go something like this:

$$f\otimes g(u_1 \otimes v_1+u_2 \otimes v_2)=f\otimes g(\sum_{i,j} a_ju_j\otimes b_iv_i+\sum_{i,j} a'_ju_j\otimes b'_iv_i)=f\otimes g(\sum_{i,j}(a_j+a'_j)u_j\otimes (b_i+b'_i)v_j)$$
and from here we use the definition.

Well when I started with it I didn't see the need for resorting to the basis elements.

9. Jul 5, 2008

### mrandersdk

exactly, that was my point, you actually need to define it on basis, and then do as you say in the post.

A little point about the uniqueness part. It seems that if the function satisfy

$$(f\otimes g)(u\otimes v)=f(u)\otimes g(v)$$

then it is only on the elements of the form $$u\otimes v$$, but all elements in V_1 \otimes W_1 need not to be on this form, so it may not be completely obivous. You actually to make it complete need to say, that given a linear function h such that

$$h(u\otimes v)=f(u)\otimes g(v)$$

then this h agree with (f\otimes g) on basis elements of V_1 \otimes W_1, which is all elements on the form v_j \otimes w_i, where v_j and w_i are baes for V_1 and W_1. And when two linear functions agree on a basis they are equal.

Actually it is first at this step, you have shown that, the definition you/I made of (f\otimes g) is not depending on the basis, chosen in the definition. I was a bit confused because I forgot that all elements isn't on the form v \otimes u, maybe you did too ?

10. Jul 5, 2008

### mrandersdk

There was something that, is missing in the last post, so here is a corrected one.

exactly, that was my point, you actually need to define it on basis, and then do as you say in the post.

A little point about the uniqueness part. It seems that if the function satisfy

$$(f\otimes g)(u\otimes v)=f(u)\otimes g(v)$$

then it is unique, but this is only on the elements of the form $$u\otimes v$$, but all elements in V_1 \otimes W_1 need not to be on this form, so it may not be completely obivous. You actually to need to say, that given a linear function h such that

$$h(u\otimes v)=f(u)\otimes g(v)$$

then this h agree with (f\otimes g) on basis elements of V_1 \otimes W_1, which is all elements on the form v_j \otimes w_i, where v_j and w_i are baes for V_1 and W_1. And when two linear functions agree on a basis they are equal.

Actually it is first at this step, you have shown that, the definition you/I made of (f\otimes g) is not depending on the basis, chosen in the definition. I was a bit confused because I forgot that all elements isn't on the form v \otimes u, maybe you did too ?