1.1.16

It's easy to explicitly construct the inverse if you for some reason know the expression $$1 + x^n = (1 + x)(1 - x + x^2 - \dots + (-1)^{n-1}x^{n-1})$$, as shown in Prove that $A+I$ is invertible if $A$ is nilpotent.

= Remarks =

As I didn't recall the above formula I instead, after a while, found that $$(I + A)(I - A) = I - A^2$$, $$(I + A)(I - A)(I + A^2) = (I - A^2)(I + A^2) = (I - A^4)$$, $$(I + A)(I - A)(I + A^2)(I + A^4) = (I - A^4)(I + A^4) = (I - A^8)$$, and so on. Using this we can express the inverse $$(I + A)^{-1}$$ with the less pretty expression $$(I-A)\prod_{i=1}^\infty (I + A^{2^i})$$. --Cicss 22:56, June 3, 2012 (UTC)