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?
 A: $A-A^2=I\to A^2-A=-I\to A^2-A+I=0$
Since A satisfies its own characteristic equation (Cayley-Hamilton Theorem), replace A with $\lambda$ and see what you get....
A: Hint: Is there a real number $\lambda$ with $\lambda-\lambda^2=1$?
A: Assume $A-A^2=1$. Let $Q(X):=X^2-X+1$. Then $Q(A) = 0$.
We also denote the characteristic polynomial of $A$ by $\chi_A$.
Furthermore, we denote the minimal polynomial of $A$ by $\mu_A$. This is the smallest polynomial, w.r.t. degree, that satisfies $\mu_A(A)=0$. This polynomial divides all other polynomials $P$ that fulfil $P(A)=0$.
Let $\lambda$ be an arbitrary root of $\chi_A$. Then $\lambda$ is an eigenvalue of $A$ to some eigenvector $v\neq 0$. For any polynomial $P$, $P(\lambda)$ is then an eigenvalue of $P(A)$. In particular, $\mu_A(\lambda)$ is an eigenvalue of $\mu_A(A)$, i.e. $\mu_A(A)\cdot v = \mu_A(\lambda)\cdot v$. However, $\mu_A(A)=0$ by definition of $\mu_A$, thus $\mu_A(\lambda)\cdot v = 0\cdot v = 0$. Since $v \neq 0$, we conclude $\mu_A(\lambda)=0$.
Therefore, any root of $\chi_A$ is also a root of $\mu_A$. Furthermore, since $Q(A)=0$, $\mu_A$ divides $Q$, therefore any root of $\mu_A$ must be a root of $Q$. However, $Q$ has no root in $\mathbb{R}$.
A: $A$ satisfies $x^2-x+1=0$ whose discriminant is negative.
A: Hint: Apply your equation to an eigenvector.
A: 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. 
