I have this problem:
Let $A$ be any $n \times n$ matrix, defined over the real numbers, such that $A-A^2=I$. Then prove that $A$ does not have any real eigenvalues.
What I did:
$$A-A^2=I$$ $$A-A^2-I=0$$ $$A(A-I)-I=0$$
Now I need to show that $A(A-I) \neq I$. Assume that $A=2I$. (If $A=I$ then $0 \neq I$.) Then $$A(A-I)=I$$ $$2I(2I-I)=I$$ $$2I(I)=I$$ $$2I \neq I$$
This equation doesn't have a solution, but I don't think I proved correctly.
Any help will be appreciated, thanks.