# What is inverse of $I+A$ given that $A^2=2\mathbb{I}$?

I have the next problem:

Let $A$ be a real square matrix such that $A ^ 2 = 2\mathbb{I}$. Prove that $A +\mathbb{I}$ is an invertible matrix and find its inverse.

I tried with the answers given here:What is inverse of $I+A$?

Any hints?

HINT: What is $(A+\Bbb I)(A-\Bbb I)$?

We need to find X where $$(A+I)X = I$$

$$AX + X = I$$ Multiplying on both sides by $A$ $$A^2 X + AX = 2IX + AX = A$$ $$IX + (A+I)X = X + I = A$$

Therefore $$X = A-I$$

This is the brute force solution.

Note that

$(\Bbb I + A)(\Bbb I - A) = \Bbb I - A^2 = \Bbb I - 2 \Bbb I = -\Bbb I, \tag{1}$

or

$(\Bbb I + A)(A - \Bbb I) = \Bbb I. \tag{2}$

(2) shows both that $(\Bbb I + A)^{-1}$ exists and that it is given by

$(\Bbb I + A)^{-1} = A - \Bbb I; \tag{3}$

without further knowledge of $A$, not much more can be said. One can of course find all $B$ such that $B^2 = \Bbb I$ and then take $A - \Bbb I = \sqrt 2 B- \Bbb I$ for any such $B$, of which there are many, but since we can't further specify $A$ or $B$ based on what is given here, this seems like a good place to leave off.

Hope this helps. Cheers,

and as always,

Fiat Lux!!!