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Linear transformations as tensor.

  1. Feb 26, 2012 #1
    I was looking at this table here: http://en.wikipedia.org/wiki/Tensor#Examples

    And i didn't understand why a (1,1) tensor is a linear transformation, I was wondering if someone could explain why this is.

    A (1,1) tensor takes a vector and a one-form to a scalar.
    But a linear transformation takes a vector to a vector.

  2. jcsd
  3. Feb 26, 2012 #2


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    A bilinear form can take a vector to a vector or a covector to a covector:

    Auvζv = γu
    Auvωu = κv
  4. Feb 26, 2012 #3
    I thought a bilinear form was the tensor product of 2 one-forms/linear functionals, so it would take two vectors to a scalar.
  5. Feb 26, 2012 #4


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    If your form is T(u,v) where u is a vector and v is a co-vector, consider the function f(u)=T(u,-). f(u) is a function which, when you plug in a co-vector you get a scalar. Those are exactly what vectors are (co-co vectors)
  6. Feb 28, 2012 #5


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    Assume all spaces are finite diml.

    by plugging into one variable at a time, you see that bilinear maps from VxW to U is the same as linear maps from V to linear maps from W to U.

    I.e. Bil(VxW,U) ≈ Lin(V,Lin(W,U)).

    Write V* for linear maps from V to k, where k is the scalar field.
    For finite diml spaces then Lin(Lin(V,k),k) = V** ≈ V.

    essentially by definition of tensors, V*tensorW* ≈ (VtensorW)* ≈ Bil(VxW,k).

    Thus V*tensorW ≈ Bil(VxW*,k) ≈ Lin(V,W**) ≈ Lin(V,W).

    I.e. if ftensorw is an elementary (1,1) tensor in V*tensW,

    then it maps the vector v in V to the vector

    f(v).w in W. A general (1,1) tensor is a sum of such elementary ones and maps v to the corresponding sum of vectors in W.
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