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?

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!!!
