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Math Help - identity proof

  1. #1
    Newbie
    Joined
    Apr 2011
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    Unhappy identity proof

    hey guys,

    i need help in proving this identity. i've been working on it for 3 days now

    some notation:
    A is a matrix;
    A^T is A transpose;
    A^-1 is A inverse;
    v is a vector;
    v^-1 is v inverse;
    v^T is v transpose;

    Assume A is a non singular matrix and v is a vector with comparable dimension. Prove:

    (A + vv^T)^-1 = A^-1 - [ (A^-1v)(v^TA^-1) / (1 + v^TA^-1v) ]


    Need help please. or at least some tips. Thanks!
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  2. #2
    Super Member TheChaz's Avatar
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    Northwest Arkansas
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    Quote Originally Posted by appsmail12 View Post
    hey guys,

    i need help in proving this identity. i've been working on it for 3 days now

    some notation:
    A is a matrix;
    A^T is A transpose;
    A^-1 is A inverse;
    v is a vector;
    v^-1 is v inverse;
    v^T is v transpose;

    Assume A is a non singular matrix and v is a vector with comparable dimension. Prove:

    (A + vv^T)^-1 = A^-1 - [ (A^-1v)(v^TA^-1) / (1 + v^TA^-1v) ]


    Need help please. or at least some tips. Thanks!
    What work do you have from your three days of effort?
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  3. #3
    A Plied Mathematician
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    Multiply out

    \left[A^{-1}-\dfrac{A^{-1}\mathbf{v}\mathbf{v}^{T}A^{-1}}{1+\mathbf{v}^{T}A^{-1}\mathbf{v}}\right]\left(A+\mathbf{v}\mathbf{v}^{T}\right), and

    \left(A+\mathbf{v}\mathbf{v}^{T}\right)\left[A^{-1}-\dfrac{A^{-1}\mathbf{v}\mathbf{v}^{T}A^{-1}}{1+\mathbf{v}^{T}A^{-1}\mathbf{v}}\right],

    and show that you get I both times. Keep in mind that the product \mathbf{v}^{T}A^{-1}\mathbf{v} is a scalar, and thus commutes with any multiplication.

    How does that work for you?
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  4. #4
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    Solved it. Thanks Ackbeet!
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  5. #5
    A Plied Mathematician
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    You're welcome! Have a good one.
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