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Math Help - Prove that if A is diagonalizable, so is its inverse

  1. #1
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    Prove that if A is diagonalizable, so is its inverse

    Let A be an invertible matrix. Prove that if A is diagonalizable, so is A^{-1}.

    I'm afraid I don't quite know how this could be proven. This is the closest thing I have: proving that if A is diagonalizable, so is A^T.
    P^{-1}AP=D\Rightarrow P^TA^T(P^T)^{-1}=D Since D=D^T=(P^{-1}AP)^T=P^TA^T(P^T)^{-1}
    I doubt that proof is even relevant to the question. Does anyone know how to prove it?
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  2. #2
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    A matrix is diagonizable iff it acts on a basis by scaling each vector by a certain amount. Then its inverse acts on that basis by scaling each vector by the inverse amount. Equivalently, if a_{kk} are the diagonal entries of a diagonal matrix, its inverse is the diagonal matrix with diagonal entries 1/x_{kk}.
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  3. #3
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    [ Different wording from maddas but essentially the same thing: A matrix, A, is diagonalizable if and only if there is a basis for the vector space consisting of eigenvectors of A. That is, for every basis vector v_i, Av_i= \lambda_i v_i (maddas' "scaling" each vector).

    Applying A^{-1} to both sides of that, A^{-1}(Av_i)= A^{-1}(\lambda_iv_i)= \lambda_i A^{-1}v_i so that A^{-1}v_i= \frac{1}{\lambda_i} v_i.

    Of course, in order to have an inverse, A must not have any 0 eigenvalues. Then every eigenvector of A is an eigenvector of A^{-1}.
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  4. #4
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    That's a good one, but not in terms of what we were given. I found another source which put in this way.

    A=Q^{-1}DQ, where D is diagonal.
    Since A is an invertible matrix, Q^{-1}DQ has an inverse as well, which means D is invertible.
    A^{-1}=(Q^{-1}DQ)^{-1}=Q^{-1}D^{-1}(Q^{-1})^{-1}=Q^{-1}D^{-1}Q
    Since A is written in such a form, we can say that it is diagonalizable.

    Is this proof solid enough?
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  5. #5
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    You must observe that D^{-1} is diagonal (which is trivial) but yea, looks like a fine proof.
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