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Thread: Total derivative of f(A)=A^-1 at I

  1. #1
    Junior Member
    Jul 2008

    Total derivative of f(A)=A^-1 at I

    Work in M(n, R), the space of nxn matrices over the real numbers.


    Find the total derivative at I, the identity matrix. That is, find an expression

    f(I+H)-f(I)= linear in H + small (quadratic and above) in H

    Thoughts: This is pretty easy to do if you assume you can expand (I+H)^-1 binomially, but a later question I have to do tells me to use Taylor's Theorem to prove that formula, so I obviously can't use it in this question.

    So, how on earth do I find (I+H)^-1 -I without expanding the bracket out?

    Thanks for any help. Also, sorry if this is in the wrong place. Couldn't decide if it belonged here or in Analysis.
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  2. #2
    Jan 2010
    I'd do it like so:
    1. For arbitrary H and epsilon sufficiently close to 0,
    $\displaystyle (I + \epsilon H)^{-1}(I+\epsilon H) = I.$

    2. Taking a derivative of both sides with respect to epsilon,
    $\displaystyle \left( \frac{d}{d\epsilon} \left[ (I+\epsilon H)^{-1} \right]\right)(I+\epsilon H) + (I+\epsilon H)^{-1}\frac{d}{d\epsilon}(I+\epsilon H) = 0$.

    3. We can evaluate this equation at epsilon = 0 to get
    $\displaystyle \frac{d}{d\epsilon}\left[ (I+\epsilon H)^{-1}\right] \Big\vert_{\epsilon = 0} = -H,$
    where the LHS is conveniently the directional derivative of f:
    $\displaystyle \nabla f(I) H = -H$.

    4. Since this last equation holds for arbitrary H, we have the desired
    $\displaystyle \nabla f(I) = -I.$

    Of course, there are some technical details left to fill in: for instance in step 1 we assumed that the set of invertible matrices is open, and in step 3, that matrix inversion is smooth (follows from Cramer's Rule).
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