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What does this notation mean? Linear map A = [A^\mu \nu]_\mu \nu

  1. Jan 18, 2012 #1
    It's used in a certain proof that I'm reading. A is a linear map from a vectorspace V onto itself.

    They say they can rewrite the vector space as [itex]\mathcal V = \bigoplus_\mu \mathbb C^{m_\mu} \otimes \mathcal V^\mu[/itex] and I understand this, but they then claim one can (always, as any linear map) rewrite A as [itex]A = [A^{\mu \nu}]_{\mu \nu}[/itex] "where [itex]A^{\mu \nu}[/itex] is a linear map of [itex]\mathbb C^{m_\nu} \otimes \mathcal V^\nu[/itex] to [itex]\mathbb C^{m_\mu} \otimes \mathcal V^\mu[/itex]."

    I don't understand the nature of this decomposition/rewriting. Note that this rewriting has to be possible for any A, it doesn't use any special properties of A (that comes later in the proof).
     
  2. jcsd
  3. Jan 18, 2012 #2

    morphism

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    This seems to be an extremely obtuse way of writing the matrix of A in a given basis (indexed by \mu in the domain and by \nu in the range).
     
  4. Jan 18, 2012 #3
    Writing a matrix in a given basis is possible, but they also fix the codomain for each part of the domain (it seems); I don't see why that is possible.
     
  5. Jan 18, 2012 #4

    morphism

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    Do they explicitly state what they mean by [itex]\mathcal V^\mu[/itex]?
     
  6. Jan 18, 2012 #5
    Simply different vector spaces labelled by mu, it's as general as that.
     
  7. Jan 18, 2012 #6

    morphism

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    Okay, so they are just writing A in terms of a given basis.

    For simplicity, let's suppose [itex]V = V^1 \oplus V^2[/itex]. Then, with respect to this decomposition, we can write
    [tex]A = \begin{pmatrix} A^{11} & A^{12} \\ A^{21} & A^{22}\end{pmatrix}.[/tex]
    Here A^{11} will take V^1 to V^1, and A^{12} will take V^2 to V^1, etc.
     
  8. Jan 18, 2012 #7
    Oh that makes sense, thank you :)
     
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