If $A$ is a $3\times3$ matrix, then $A-A^2 \neq I$

Question

Suppose a matrix $A \in M_{3\times3}(\mathbb R)$, then $A-A^2 \neq I$.

I know that I should contradict that statement, and use the fact that a $3\times3$ matrix has at least one real eigenvalue. However, I can only think of substituting $A$ for $$\begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix}$$

and solving a system of $9$ equations in $9$ variables.

Any ideas? Thanks ahead.

Answer 1

Since 3 is odd, the characteristic polynomial of $A $ has a real root. Is this possible ?

Answer 2

The minimal polynomial of $A$ would be $\mu(x)=x^2-x+1$. However, $\mu(x)\neq0$ for all $x\in\mathbb R$. So, $A$ has no real eigenvalue.