Banach lemma proof

1. Sep 30, 2015

FOIWATER

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. Sep 30, 2015

tommyxu3

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.

3. Sep 30, 2015

lavinia

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

4. Sep 30, 2015

FOIWATER

Probably should be seeing it, but I'm not

5. Sep 30, 2015

davidmoore63@y

(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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