Proof of Banach Lemma: Small Matrix Eigenvalues

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    Banach Proof
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Discussion Overview

The discussion revolves around proving the relationship $$(I+A)^{-1}=I-A$$ for a "small" matrix A, particularly in the context of its application to the gradient of the log determinant. Participants explore the implications of this relationship and the conditions under which it holds.

Discussion Character

  • Exploratory
  • Mathematical reasoning

Main Points Raised

  • One participant expresses uncertainty about how to prove the relationship, suggesting that if $$(I+A)^{-1}=I-A$$, then it leads to the conclusion that $$A^2=O$$.
  • Another participant proposes that the relationship can be understood through a Taylor expansion around A = 0, indicating that an error term can be neglected if A is small.
  • A further contribution notes that the series expansion $$(I + A)^{-1} = I - A + A^2 - A^3 + A^4 - ...$$ can be verified by multiplying both sides by $$(I+A)$$, and suggests that neglecting terms of order $$A^2$$ or higher is valid when A is small, specifically referring to small eigenvalues.

Areas of Agreement / Disagreement

Participants appear to agree on the notion that the relationship holds under the condition of A being a "small" matrix, particularly in terms of its eigenvalues. However, there is no consensus on the exact proof or the implications of the relationship.

Contextual Notes

Participants have not fully resolved the mathematical steps required to prove the relationship, and there are varying interpretations of what constitutes a "small" matrix in this context.

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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