Proof of Banach Lemma: Small Matrix Eigenvalues

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The discussion centers on the proof of the relationship $$(I+A)^{-1}=I-A$$ for small matrices A, specifically when considering their eigenvalues. Participants emphasize the importance of the Taylor expansion around A = 0, which allows for the neglect of higher-order terms when A is small. The conclusion drawn is that small matrices refer to those with small eigenvalues, validating the approximation used in the proof.

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FOIWATER
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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?
 
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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.
 
FOIWATER said:
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?
$$(I+A)^{-1}=I-A $$ plus an error term which can be neglected if A is small. Look at the Taylor expansion around A = 0.
 
Probably should be seeing it, but I'm not
 
(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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