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Tensor Product question.

  1. Jul 5, 2008 #1
    There's a question that asks me to show that there exists a unique linear transformation
    from: [tex]f\otimes g: V_1\otimes W_1\rightarrow V_2\otimes W_2[/tex]
    where f and g are linear transformations f:V1->V2, g:W1->W2
    that satisfies: [tex](f\otimes g)(u\otimes v)=f(u)\otimes g(v)[/tex]

    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?

    thanks in advance.
     
  2. jcsd
  3. Jul 5, 2008 #2

    Hurkyl

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    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.
     
  4. Jul 5, 2008 #3
    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:
    [tex]v\otimes(w_1+w_2)=v\otimes w_1+v\\otimes w_2[/tex]
    [tex](v_1+v_2)\otimes w=v_1\otimes w+v_2\otimes w[/tex]
    [tex](\lambda v)\otimes w=\lambda (v\otimes w)[/tex]
    lambda is a scalar.

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

    cheers.
     
  5. Jul 5, 2008 #4
    a little hint, try to define the function on a basis of V_1 and W_1 by

    [tex](f\otimes g)(v_i\otimes w_j)=f(v_i)\otimes g(w_j)[/tex]

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

    [tex](f\otimes g)(u\otimes v)=f(u)\otimes g(v)[/tex]

    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
  6. Jul 5, 2008 #5
    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.
     
  7. Jul 5, 2008 #6
    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.
     
  8. Jul 5, 2008 #7
    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

    [tex](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)[/tex]

    I give you that the uniqueness part is clear.
     
  9. Jul 5, 2008 #8
    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:

    [tex]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)[/tex]
    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.
     
  10. Jul 5, 2008 #9
    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

    [tex](f\otimes g)(u\otimes v)=f(u)\otimes g(v)[/tex]

    then it is only on the elements of the form [tex]u\otimes v[/tex], 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

    [tex]h(u\otimes v)=f(u)\otimes g(v)[/tex]

    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 ?
     
  11. Jul 5, 2008 #10
    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

    [tex](f\otimes g)(u\otimes v)=f(u)\otimes g(v)[/tex]

    then it is unique, but this is only on the elements of the form [tex]u\otimes v[/tex], 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

    [tex]h(u\otimes v)=f(u)\otimes g(v)[/tex]

    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 ?
     
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