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Tensor product for vectors

  1. Aug 25, 2011 #1
    Hello. I keep on encountering the need to find the Tensor or Kronecker product of two vectors. Based on the definition, If I found the product of two 2D vectors, I would get a 4-dimensional vector. Some authors claim this is the correct interpretation.

    However the dyadic product, which many claim is just the 1st order case of the tensor product, would generate a second-order tensor. In other words, if I multiplied my two 2D vectors, I would get a 2x2 matrix. This is essentially finding the tensor product between the first vector and the transpose of the second vector.

    This troubles me. The first interpretation is consistent with the fundamental definition of the vector product, but the latter version is necessary for things like the rotation tensor definition to be true:

    [tex] R = I\cos\theta + \sin\theta[\mathbf u]_{\times} + (1-\cos\theta)\mathbf{u}\otimes\mathbf{u} [/tex]

    Which version is correct, and how can it be made consistent with the other version? Thanks for any help.
  2. jcsd
  3. Aug 25, 2011 #2
    I don't know much at all about tensors, but I was under the impression that a 2x2 matrix is a vector in a 4-dimensional space [itex]\mathcal{L}(V,V)[/itex] if [itex]V[/itex] is 2-dimensional.
    Last edited: Aug 25, 2011
  4. Aug 25, 2011 #3
    That idea sounds promising, but I'm afraid I don't quite understand what you mean yet. Would you mind elucidating things a little? I'm new to tensors, and so there could be a critical spatial relationship that I am missing out on here.
  5. Aug 25, 2011 #4
    I made a little bit of a mistake but I'll clear it all up.

    [itex]\mathcal{L}(V,W)[/itex] is the set of all linear transformations [itex]T:V\to W[/itex]. If [itex]V[/itex] has dimension [itex]m[/itex] and [itex]W[/itex] has dimension [itex]n[/itex], then [itex]\mathcal{L}(V,W)[/itex] can be shown to be a vector space with dimension [itex]mn[/itex].

    The set [itex]M_{m,n}(\mathbb{R})[/itex] of [itex]m[/itex]-by-[itex]n[/itex] matrices with entries from [itex]\mathbb{R}[/itex] is isomorphic to [itex]\mathcal{L}(V,W)[/itex] and thus, is also a vector space of dimension [itex]mn[/itex]. If [itex]n=m=2[/itex], then the dimension of [itex]M_{2,2}(\mathbb{R})[/itex] is 4, and its elements are 2-by-2 matrices.

    You can also find an isomorphism [itex]T:M_{2,2}(\mathbb{R})\to\mathbb{R}^{4}[/itex], such that [itex]T(\left[ {\begin{array}{cc}
    x & y \\
    z & w \\
    \end{array} } \right])=(x,y,z,w)[/itex]

    I hope that clears some things up.
  6. Aug 25, 2011 #5
    That actually very much does clear things up... I had never learned about the equivalency of, for example, spaces of 2x2 tensors and spaces of 1x4 tensors. Where could I find an example of a relationship that establishes the isomorphism? (ie, is there any sort of tensor A would make your relation
    T(\left[ {\begin{array}{cc}
    x & y \\
    z & w \\
    \end{array} } \right])=(x,y,z,w)

    true, or is it a more abstract sort of transformation?
  7. Aug 25, 2011 #6
    That is the isomorphism, you just need to prove it to be one.
  8. Aug 25, 2011 #7
    ah, I see. Thanks very much!
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