Is the set of real matrices diagonalizable in $\mathcal M_n(C)$ dense in the set of all companion matrices?

Question

Let $\mathcal E=\{A \in \mathcal M_n(\mathbb R): A \text{ is of companion form} \}$. That is, $A$ has the following form \begin{align*} A=\begin{pmatrix} 0 & \cdots & 0& -a_{0} \\ 1 & \cdots & 0 & -a_{1}\\ \vdots &\ddots & \vdots &\vdots \\ 0 &\cdots & 1 & -a_{n-1} \end{pmatrix}. \end{align*} Let $\mathcal D = \{A \in \mathcal M_n(\mathbb R): A \text{ is diagonalizable in $\mathcal M_n(\mathbb C)$}\}$. We know $\mathcal D$ is dense in $\mathcal M_n(\mathbb R)$. I am wondering whether $\mathcal E \cap \mathcal D$ is dense in $\mathcal E$. The set $\mathcal E$ is closed. It is not clear to me whether this statement is true.

Answer 1

This companion matrix is diagonalisable over $\Bbb C$, iff $f(x)=x^n+a_{n-1}x^{n-1}+\cdots+a_0$ has $n$ distinct complex zeros. So, the question is whether real polynomials with distinct zeros are dense in the set of all real polynomials of a certain degree.

The answer is yes. If $f$ has a repeated zero, then there is a complex number $z_0$ with $f(z_0)=f'(z_0)=0$. For a small nonzero real $\epsilon$, $g(x)=f(x)+\epsilon$ has $g'(z_0)=0$ and $g(z_0)\ne0$. For small enough $\epsilon$. This holds for all common roots of $f$ and $f'$. But if $f(z_1)\ne0$ and $f'(z_1)=0$ choosing $\epsilon $ small enough will ensure $g(z_1)\ne0$. Then $g$ and $g'=f'$ have no common roots. So for any $f$, there are $g$ with no repeated roots close to $f$.