Proof of Banach Lemma: Small Matrix Eigenvalues

[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?

 

[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.

 

[lavinia]

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.

 

[FOIWATER]

Probably should be seeing it, but I'm not

 

[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