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Tensor product

  1. Jan 12, 2009 #1
    If V1 and V2 are both subspaces of a vector space V, then in order for their tensor product to be defined, does the intersection of V1 and V2 have to be 0?
     
    Last edited: Jan 12, 2009
  2. jcsd
  3. Jan 13, 2009 #2

    lurflurf

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    No, infact an important example of tensor product is that between a vector space and itself.
    V^2=VxV

    One possible definition of tensor product is that it maps two vector spaces onto ordered pairs.
    UxV contains ordered pairs (u,v) with u from U and v from V.
    Edited to add. UxV also contains linear combinations of ordered pairs.
     
    Last edited: Jan 13, 2009
  4. Jan 13, 2009 #3
    I see. But I have to say, I'm puzzled about this: according to one definition I have, the operation on pairs (u,v), u belonging to U and v belonging to V, that is used to construct the tensor product has to be bilinear. But that certainly isn't a sufficient condition, is it? One must also have that the operation in question applied to the whole of the cartesian product of a basis of U and a basis of V forms a basis for the tensor product, right? Does the tensor product always exist (i.e. is there always such a bilinear operation and vector space for any ordered pair of vector spaces)?
     
  5. Jan 13, 2009 #4

    lurflurf

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    The tensor product always exists.
    Existence is not a problem because tensor product is more a construction than an operation. That is the operation always endows the product with the required properties.
    lets assume U and V are vector spaces over the same field F.
    we want
    (a1u1+a2v2,b1v1+b2v2)=a1b1(u1,v1)+a1b2(u1,v1)+a2b1(u2,v1)+a2b2(u2,v2)
    so in declaring the tensor product has this property we determin the tensor product uniquely.
    Since the tensor product is bilinear we can obtain a basis from the bases of the spaces used to construct it.
    if
    {u(i)} is a basis for U
    and
    {v(j)} is a basis for V
    {u(i),v(j)} is a basis for UxV

    it is important to notice that
    X is an element of UxV does not mean X is of the form (u,v)
    UxV contains all elements of the form (u,v), but also all linear combinations of such terms.
    By a counting argument if
    dim(U)=m
    dim(V)=n
    dim(UxV)=mn
    dim(elements of the form (u,v))=m+n
    I see I was unclear above I said
    UxV contains ordered pairs (u,v) with u from U and v from V
    which is true but it contains more ie sums of such.
     
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