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Inverse matrix notation question

  1. Apr 17, 2014 #1
    I'm hoping that you can help me settle an argument. For a matrix [itex]\textbf{M}[/itex] with elements [itex]m_{ij}[/itex], is there any sitaution where the notation [itex](M_{ij})^{-1}[/itex] could be correctly interpreted as a matrix with elements [itex]1/m_{ij}[/itex]?

    Personally I interpret [itex](M_{ij})^{-1}[/itex] in the usual sense of an inverse matrix, where it would have the property [itex]\textbf M \textbf M^{-1} = I[/itex], but perhaps there are other interpretations that I don't know about. Thanks!
     
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  3. Apr 17, 2014 #2

    disregardthat

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    The only interpretation I can think of is when you define matrix multiplication component-wise. In this case, the invertible matrices would be the ones with non-zero values, and the matrix you describe would be the inverse. Note that the identity matrix would be the one with 1's as values. I don't think this is commonly used.
     
  4. Apr 17, 2014 #3

    Fredrik

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    I'm not a fan of the notation ##(M_{ij})^{-1}##, mainly because ##M_{ij}## should refer to the component on row i, column j, not the matrix itself. I'm also not a fan of using a lowercase m for the components, because that prevents us from writing the definition of matrix multiplication in what I consider the obviously best way: ##(AB)_{ij}=\sum_k A_{ik}B_{kj}##. I find it very puzzling that some authors go out of their way to avoid this notation, by writing things like "if ##C=AB##, then ##c_{ij}=\sum_k a_{ik}b_{kj}##".

    If M is a diagonal matrix, for example
    \begin{pmatrix}2 & 0 & 0\\ 0 & 3 & 0\\ 0 & 0 & 4\end{pmatrix} then its inverse is simply
    \begin{pmatrix}\frac 1 2 & 0 & 0\\ 0 & \frac 1 3 & 0\\ 0 & 0 & \frac 1 4\end{pmatrix} But even if M is diagonal, and we use horrible notation, we still don't quite have ##(M_{ij})^{-1}=1/m_{ij}## because the off-diagonal elements of ##M^{-1}## aren't 1/0.
     
    Last edited: Apr 17, 2014
  5. Apr 18, 2014 #4
    This is helpful. The matrix in question isn't diagonal, and that's a good point about 1/0. Hopefully I can convince this other person to change their notation!
     
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