TITLE: $A^2=I,\det A>0$ implies $A+I$ is non-singular
QUESTION [8 upvotes]: Question:
If a square matrix $A$ satisfies $A^2=I$ and $\det A>0$, show that $A+I$ is non-singular.
I have tried to suppose a non-zero vector $x$ s.t. $Ax=x$ but fail to make a contradiction.
And I tried to find the inverse matrix of $A+I$ directly, suppose $(A+I)^{-1}=\alpha I +\beta A$, but it still doesn't work.
(Update: According to the two answers, this question itself is incorrect.)

REPLY [10 votes]: $(I+A)(I-A)=0$, so $I+A$ is invertible (for $A$ satisfying $A^2=I$) if and only if $A=I$.
This works for $A$ in any ring with unit [in which $2$ is invertible], not only a ring of square matrices.