Let $A \in M_n(\mathbb{R})$ with a minimal polynomial: $m_A = x^2+1$.
Let $f \in \mathbb{R}[x]$ such that $f(A)$ is a non-scalar matrix ( $\forall \lambda \in \mathbb{R} : f(A) \ne \lambda I_n$).
I need to prove that the matrix $f(A)$ does not have eigenvalues in $\mathbb{R}$.
I know that $A$ does not have any Real eigenvalues, because it's minimal polynomial does not have any roots in $\mathbb{R}$.
I'm just not sure how to proceed to prove this.
could anyone explain how to approch this?
Let $f(X)=(X^2+1)q(X)+r(X)$ where $\mathrm{deg}(r)\leq1$. $r(X)$ must be linear otherwise we would have $f(A)=r(A)=\lambda I_n$ for some $\lambda\in \mathbb{R}$. Hence $\exists$ $a(\neq0),b\in\mathbb{R}$ such that $r(X)=aX+b$. Then we have $f(A)=aA+bI_n$. If $\exists$ $\lambda_0\in\mathbb{R}$ and $\nu(\neq0)\in\mathbb{R}^n$ such that $f(A)\nu=\lambda_0\nu$ then we have $\lambda_0\nu=aA\nu+b\nu\implies A\nu=\frac{\lambda_0-b}{a}\nu$. Which means $A$ has a real eigenvalue. A contradiction! Hence $f(A)$ does not have any eigenvalue in $\mathbb{R}$.
Hint: write the euclidean division of $f$ by $m_A$, and look at the rest. The fact that $A$ has no real eigenvalue and the assumption that $f(A)$ is not a scalar matrix should help you conclude.
