Tensor Product question.

Gold Member
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?

Staff Emeritus
Gold Member
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.

Gold Member
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.

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.

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Gold Member
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.

Gold Member
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.

mrandersdk
you need not to show that it is linear in the first component and in the second component, this doesn't actually make sense 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.

Gold Member
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.

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 ?

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 ?