Polynomial with no real roots implies that $\det(P(A))\ge 0$

Question

Let $P \in \mathbb{R}[X]$ be a polynomial of degree $n$, $n\in \mathbb{N}$, which has no real roots. If $A\in \mathcal{M_n(\mathbb{R})}$, then prove that $\det(P(A))\ge 0$.
I don't know how to approach the general case, but for $n=2$ I was able to prove this by using the canonical form of a quadratic. However, I don't think that this can be generalised to any $n\in \mathbb{N}$ and I don't have any other approaches to this question.

Answer 1

Since $P$ is real and has no real roots we see that $n$ is even. Since $A$ is real we see that it has an even number of real eigenvalues and the remainder are conjugate pairs.

If $R$ are the indices of the real eigenvalues of $A$ then we see that $\Pi_{k \in R} P(\lambda_k) >0$ (since there are an even number) and if $\lambda_k$ is not real, we see that $P(\lambda_k) P(\overline{\lambda_k}) = P(\lambda_k) \overline{ P(\lambda_k) } = |P(\lambda_k)|^2$. Hence $\det P(A) \ge 0$.

Answer 2

Since $P(x)=0$ has not any real roots, its roots must occur in conjugate pairs. Hence we may write $P(x)=\prod_{i=1}^m(x-z_i)(x-\overline{z}_i)$ for some complex numbers $z_1,z_2,\ldots,z_m$. It follows that \begin{aligned} \det P(A) &=\det\prod_i(A-z_iI)(A-\overline{z}_iI)\\ &=\det\prod_i(A-z_iI)(\overline{A-z_iI})\\ &=\prod_i\det(A-z_iI)\det(\overline{A-z_iI})\\ &=\prod_i\det(A-z_iI)\,\overline{\det(A-z_iI)}\\ &=\prod_i|\det(A-z_iI)|^2\ge0. \end{aligned}