TITLE: Linear Algebra - eigenvalues
QUESTION [8 upvotes]: 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.

REPLY [10 votes]: If $v$ is an eigenvector of $A$ with eigenvalue $\lambda$, then $(A-A^2)v = \lambda v - \lambda^2 v = (\lambda - \lambda^2)v$; hence if $A-A^2 = I$, then you'd be solving $(\lambda - \lambda^2)v = v$ implying that $\lambda-\lambda^2 = 1$, which has no real solutions (but certainly does have complex  solutions).