Simple linear algebra problem: prove a matrix is invertible

I'm preparing for a test in linear algebra and I've come across a problem I'm having trouble with for some reason:

Given a square matrix A, $A^2=2I$, prove that $A-I$ is invertible.

I know this is pretty simple but I can't seem to play with the equations to get it so that for some $B$, $B(A-I)=I$

It's pretty easy to see that $A^{-1}=\frac{1}{2}A$, but beyond that I haven't been able to get very far.

Can anyone help with this?

Solution 1:

From your original equation subtract $\;I\;$ from both sides, and you'll get at once what you want:

$$A^2=2I\implies A^2-I=2I-I=I\implies (A-I)(A+I)=I\;\;\color{green}\checkmark$$

Solution 2:

No “guess and check” is needed. Set $B=A-I$, so $A=B+I$; then $$ (B+I)^2=2I $$ that becomes $$ B^2+2B+I=2I $$ or $$ I=B^2+2B=B(B+2I) $$ Thus $$ (A-I)^{-1}=B^{-1}=B+2I=A+I $$