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Banach lemma proof

  1. Sep 30, 2015 #1

    FOIWATER

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    Gold Member

    Hi,

    I found the following relationship in a proof for gradient of log det x

    $$(I+A)^{-1}=I-A$$ When A is a "small" matrix (?? eigenvalues)

    I am not sure how to prove it, any ideas?
     
  2. jcsd
  3. Sep 30, 2015 #2
    My opinion:
    If ##(I+A)^{-1}=I-A,## then ##I=(I+A)(I-A)=I-A^2,## or ##A^2=O.##
    I'm not sure if this helps.
     
  4. Sep 30, 2015 #3

    lavinia

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    $$(I+A)^{-1}=I-A $$ plus an error term which can be neglected if A is small. Look at the Taylor expansion around A = 0.
     
  5. Sep 30, 2015 #4

    FOIWATER

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    Probably should be seeing it, but I'm not
     
  6. Sep 30, 2015 #5
    (I + A)^-1 = I - A + A^2 - A^3 + A^4 - .....
    [verify by multiplying each side by (I+A)]

    Then if A is "small" the terms of order A^2 or above are neglected. Yes, small means small eigenvalues
     
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