TITLE: Simple linear algebra problem: prove a matrix is invertible
QUESTION [18 upvotes]: 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?

REPLY [2 votes]: Suppose that $A-I$ is not invertible. Then $\lambda = 1$ is an eigenvalue of $A$, but then $1$ is an eigenvalue of $A^{2}$. So that $v = A^{2}v = 2v \implies 1 = 2$ a contradiction.