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Thread: Inverse of a Squared Matrix

  1. #1
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    Inverse of a Squared Matrix

    This was a question on one of my prelims for linear algebra that I'm reviewing for the final:

    If A has an inverse, then so does A squared and the inverse of A squared = A inverse, squared:

    (A^(2))-1 = (A^(-1))2

    I started to prove that A squared has an inverse but didn't make much headway. What should I have done?
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  2. #2
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    Use the definition of inverse.

    The inverse of A^2, call this B, has the property B * A^2 = I
    Now let B = (A^-1)^2
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  3. #3
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    \displaystyle (A*A)^{-1}=A^{-1}*A^{-1}=(A^{-1})^2
    Last edited by dwsmith; Dec 11th 2010 at 01:19 PM. Reason: Mistake corrected.
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  4. #4
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    reposted in proper formal below, sorry.
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  5. #5
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    Quote Originally Posted by dwsmith View Post
    \displaystyle (A*A)^{-1}=A^{-1}*A=I
    \displaystyle (A*A)^{-1}= A^{-1}* A^{-1}?
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  6. #6
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    Ah, yes. Sorry.

    \displaystyle (A*A)^{-1}=A^{-1}*A^{-1}=(A^{-1})^2
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  7. #7
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    I corrected the original post too.
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  8. #8
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    \displaystyle (A*A)^{-1}=A^{-1}*A^{-1}=(A^{-1})^2 ....

    \displaystyle (A^{2})^{-1}= (A*A)^{-1}= A^{-1}*A^{-1}=(A^{-1})^2

    Is that all?? Still confused about the first part, a proof that if A is invertible, so is A squared. Also, not seeing how the Identity works here..
    Last edited by verka; Dec 11th 2010 at 01:27 PM. Reason: inversible =/= invertible..
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  9. #9
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    Our goal is to show that both sides are equal. Once we have shown that we are done.
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  10. #10
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    Always, if you already have a guess for the inverse, just multiply the inverse guess by the original matrix, if the result is the identity matrix, then you have shown conclusively it is the inverse. A more low level proof would be something like this:

    We know that A^{-1} exists.
    This means that A^{-1} * A^{-1} = (A^{-1})^2 exists.

    Now we have
    (A^2) * (A^{-1})^2 = A * A * A^{-1} * A^{-1} = A * I * A^{-1} = A * A^{-1} = I

    Thus, (A^{-1})^2 satisfies the definition of the inverse of A^2 or (A^2)^{-1}
    so (A^2)^{-1} = (A^{-1})^2

    (Every thing posted before is correct, but I just thought it also good to provide a proof using basic definitions)
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