Prove that if $A - A^2 = I$ then $A$ has no real eigenvalues Given:
$$ A \in M_{n\times n}(\mathbb R) \; , \; A - A^2 = I $$
Then we have to prove that $A$ does not have real eigenvalues.
How do we prove such a thing?

By using index notation, $A-A^2=I$ can be written as $A_{ij}-A_{ik}A_{kj}=\delta_{ij}$. By definition: $A_{ij}n_i=\lambda n_j$. So that, $A_{ij}n_i-A_{ik}A_{kj}n_i=\delta_{ij}n_i$, hence $\lambda n_j -\lambda n_k A_{kj}=n_j$, whence $\lambda n_j -\lambda^2 n_j=n_j$, or $(\lambda^2-\lambda+1)n_j=0$, $n_j\neq 0$ an eigenvector. $\lambda^2-\lambda+1=0$ has not real roots. 
