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LINEAR ALG.: A^2 is invertible, is the inverse of a^2 equal to the square of inv(A)?

  1. Oct 2, 2006 #1
    Given that [itex]A^2[/itex] is invertible, does it neccesarily mean that

    [tex]\left(A^2\right)^{-1}\,=\,\left(A^{-1}\right)^2[/tex]?



    I know that this is true, but I have no idea on where to even start a proof of this!

    Maybe:

    [tex]\left(A^2\right)^{-1}\,=\,\left(A\,A\right)^{-1}[/tex]

    But how would I operate on infinite matrices (i.e. - [itex]a_{i\,j}[/itex])?
     
    Last edited: Oct 3, 2006
  2. jcsd
  3. Oct 3, 2006 #2

    kreil

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    [tex](A^2)^{-1}=(AA)^{-1}=A^{-1}A^{-1}=(A^{-1})^2[/tex]

    This proves that the inverse of the square of the matrix is equal to the square of the inverse of the matrix.
    I'm not sure if that's sufficient, because I'm not exactly sure what the little a is supposed to be.
     
    Last edited: Oct 3, 2006
  4. Oct 3, 2006 #3

    matt grime

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    X is the in inverse of Y if and only if XY=YX=1.
     
  5. Oct 3, 2006 #4

    HallsofIvy

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    Please do NOT mix "A" and "a"! matt grimes' suggestion is exactly what you need. Use the definition of inverse.
     
  6. Oct 3, 2006 #5
    Sorry, a little typo I made!

    It's fixed now.

    How to prove something? Is there am algorithim for proofs? The concept is too abstract for me to reason about it or start a problem!

    Can someone list the different methods for "proving" something? OIr maybe provide a good link to this type of information?
     
  7. Oct 3, 2006 #6

    matt grime

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    that would depend on the 'something'

    [quote[Is there am algorithim for proofs?[/quote]

    no, otherwise it would be a pointless subject.


    There is no such list. But in this case it is trivial. You only need to show something satisfies some rule, and that is trivially easy to do in this case.
     
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