# Prove: a matrix $f(A)$ does not have eigenvalues in $\mathbb{R}$

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.