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Inverting a tensor?

  1. Aug 22, 2010 #1
    Thank you for reading.

    If I have an object (is it correct to call it a tensor?) whose components are defined by:


    where R and S are invertible matrices. I want to find the "inverse" of X, i.e. to find (X^{-1}) such that,


    Is there a way to find out whether (X^{-1}) exists? A matrix doesn't have an inverse if its determinant is zero - is there a similar rule here? I've tried some trial functions for X^{-1} but nothing works, and I want to know whether there's a better way of tackling this problem.
  2. jcsd
  3. Aug 25, 2010 #2
    With no further constraints the tensor you have proposed is, in general, not invertible.

    Consider for example the case [tex]R=S[/tex]. Then [tex]X[/tex] will be the null tensor. Or even the less trivial case [tex]R=S^{-1}[/tex] will lead also to [tex]X=0[/tex] if [tex]R^2=1[/tex] (idempotent).

    Therefore, in general, it is not invertible.
    Last edited: Aug 25, 2010
  4. Aug 25, 2010 #3
    R is not equal to S, or to the inverse of S. It is clearly true that for special cases where X=0 there is no inverse, but I don't see how this tells us that X is generally non-invertible. Can you explain?
  5. Aug 26, 2010 #4
    If for the problem stated there are cases where X is not invertible (even if there is just one of such cases) then it can not exist a theorem, a result, or an algorithm which allows us to invert the given expression with generelity, i.e. regardless of the values of R and S, (i.e., operating with "letters" in full generality insted of with "numbers"). Hence, the invertibility or not of X will depend on the particular values of R and S, and so we say that "in general" X is not invertible.

    So, the meaning of "in general X is not invertible" is not "X is non-invertible more often than not"; the meaning is actually "the invertibility of X has to be determined on a particular case basis, it can not be determined generally (no matter the values of R and S)"
    Last edited: Aug 26, 2010
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