If $A^2=O$, then prove that $I+A$ is invertible and find $(I+A)^{-1}$

Question

So I'm stuck on this Linear Algebra question. My first (naive) train of thought here was to go through with an implication that $A^2=O$ implies $A=O$. Then this got quickly debunked having read up on nilpotent matrices.

So I'm back to square one without a clue as to how to proceed. Any hints will be much appreciated on where to begin with this, has been pestering me for at least one day now.

EDIT: Thank you everyone, very happy for all your assistance.

Use https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference to properly format equations. — Arbuja, Apr 20 '17 at 13:10
Note that in general, $AB\ne BA$, but $AI=IA=A$ is true for every matrix $A$, so the terms $AI$ and $IA$ actually cancel out. — Peter, Apr 20 '17 at 13:19
See also: If $A$ is an $n \times n$ matrix such that $A^2=0$, is $A+I_{n}$ invertible? This is related, to some extent, too: What is inverse of $I+A$? — Martin Sleziak, May 16 '24 at 04:01

score 5 · Accepted Answer · answered Apr 20 '17 at 13:23

5

Given that $A^2=0$, the inverse of $I+A$ is apparent from the identity $$(I+A)(I-A)=I-A^2=I.$$

answered Apr 20 '17 at 13:23

Servaes

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Ah yes! That is brilliant. Thank you so much, very grateful for that. Very silly of me to have missed that. – thesmallprint Apr 20 '17 at 13:32

score 0 · Answer 2 · answered Apr 21 '17 at 08:07

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$A^2=O$ suggests that the only eigenvalue of $A$ is $0$ with multiplicity $n$, where $A$ is a square of order $n$. Now, eigenvalues of $A+I$ will be $0+1=1$ (Why?), which tells us that $A+I$ is invertible (Why?)

answered Apr 21 '17 at 08:07

Nitin Uniyal

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If $A^2=O$, then prove that $I+A$ is invertible and find $(I+A)^{-1}$

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