Is this subset of $M_4(\mathbb R)$ connected? Let us consider an affine structure $\star$ of $M_4(\mathbb R)$ which has following form
\begin{align*}
\begin{pmatrix}
0 & * & 0 & * \\
1 & * & 0 & * \\
0 & * & 0 & * \\
0 & * & 1 & *
\end{pmatrix},
\end{align*}
where $*$ can assume any real number. It is also clear for any monic $4^{th}$ degree real polynomial, we can at least find one realization in $\star$ since the upper left block and lower right block can be considered as in companion form. We further concern this set
\begin{align*}
\mathcal E = \{ A \in \star: \max_i \left( \lambda_i(A) \right) < 0 \}.
\end{align*}
In other words, all elements in $\mathcal E$ has above defined structure and all eigenvalues on the left open half plane. I am trying to determine whether the set $\mathcal E$ is connected.

My first try was to determine for a fixed monic polynomial, whether all realizations in $\star$ is connected. If this is true, for any $A_1, A_2 \in \mathcal E$, we may first connect them to a companion form in $\star$ (by making the $32$ entry to be $1$ and other entries in the second column to be $0$) without changing the eigenvalues, and then the companion forms are connected as a consequence of property of polynomials. But as the question I asked a while ago, this is only true if it has at least one real eigenvalue.
I strongly believe the set is connected. The question is related but I am not sure it is that related. Because the condition there is much more strict. I was asking to connect all realizations in $\star$ yielding the same characteristic equation without changing its eigenvalues. Here we allow this to vary inside $\mathcal E$.
 A: One may prove the set is connected by stability analysis of matrices. In following, I will concern the transpose of $\star$ and this does not change the connectedness.
Note $\star \approx \mathbb R^{2 \times 4}$. Putting $A= \pmatrix{0 & 1 & 0 & 0\\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0}$ and $B = \pmatrix{0 & 0 \\ 1 & 0 \\0 & 0 \\ 0 & 1}$, the set $\mathcal E$ can be identified by a subset $\mathcal F \subset \mathbb R^{2 \times 4}$ with $\mathcal F = \{X \subset \mathbb R^{2 \times 4}: A-BX \text{ is stable}\}$.(see definition of stable matrix.) Now by Lyapunov matrix theory, $X \in \mathcal F$ if and only if there exists some positive definite matrix $P > 0$, such that $(A-BX) P + P(A-BX)^T < 0$ ($< 0$ means negative definite.) A change of variable $Y=XP$ yields
\begin{align}
\label{eq:1}
\tag{$\dagger$}
 AP + PA^T - BY - Y^TB^T < 0.
\end{align}
Observe the solution set $(P, Y)$ of $\dagger$ is a convex set and thus connected. So $\mathcal F$ is a continuous image under the map $(P, Y) \mapsto YP^{-1}$.
This method does not reveal too much of set structure which I think should be very nice. Hope someone can give a more transparent argument.
