Are matrices diagonalizable dense in a specific parametrization of $M_5(\mathbb R)$?

The problem seems to be difficult.

$(\star)$ is an affine space of dimension $10$.

Let $A=[a_{i,j}]\in\mathcal{E}$ and $spectrum(A)=(\lambda_i)$. If we move a little $A$ through $\mathcal{E}$ to $A'$ with $spectrum(A')=(\lambda'_i)$, then, with probability $1$, the eigenvalues of $A'$ are distinct and when $Re(\lambda_i)<0$, $Re(\lambda'_i)$ too. However, when $Re(\lambda_i)=0$, we cannot conclude about the signum of $Re(\lambda'_i)$.

I think that it is equivalent to choose for $\mathcal{F}$

$\{C=[c_{i,j}]\in\mathcal{E};\text{its eigenvalues are distinct and have negative real part}\}$

Then the elements of $\mathcal{F}$ are stable and satisfy the inequalities $(*)$ of Routh-Hurwitz; cf. chapter "Using matrices" in

https://en.wikipedia.org/wiki/Routh%E2%80%93Hurwitz_stability_criterion

$(*)$ is in the form $P_k((c_{i,j})_{i,j})>0,k=1,\cdots,r$ where the $P_k$'s are polynomial.

In general $A$ is not stable (but almost) and satisfy relations in the form $Q_1=0,\cdots,Q_s=0,Q_{s+1}>0,\cdots,Q_r>0$ where the $(Q_i)$'s are polynomials in the $(a_i,b_j)$'s. Clearly, $A'$ satisfy the conditions $Q_{s+1}>0,\cdots,Q_r>0$.

But what about the signum of $Q_1,\cdots,Q_s$ ? It is reasonable to think that $A'$ satisfies, with probabiity $1$, $Q_1\not= 0,\cdots,Q_s\not= 0$. However, can we choose small variations of our $10$ parameters so that all the $(Q_i)_{i\leq s}$ become $>0$ ?

A bad new; the above polynomial $(Q_i)$ are very complicated.