This is a cross-post to the [question][1] I asked at MSE.

Let $A \in M_4(\mathbb R)$ and $A = (e_2, x, e_4, y)$ where $e_2, e_4$ are standard basis in $\mathbb R^4$ and $x,y$ are undetermined variables. Let $\phi, \psi: M_4(\mathbb R) \to \mathbb R^4$ be linear maps defined by
\begin{align*}
&\phi: B \mapsto (AB-BA) e_1, \\
&\psi: B \mapsto (AB-BA)e_3.
\end{align*}
Let $S$ be the intersection of kernels of the two linear maps, i.e., $S :=\text{ker}(\phi) \cap \text{ker}{\psi}$. In other words, the elements in $S \cap GL_4(\mathbb R)$ would preserve the structure of first and third columns of $A$ by conjugation, i.e., $(B^{-1}AB) e_1 = e_2, (B^{-1}AB)e_3 = e_4$ for $B \in S \cap GL_4(\mathbb R)$. I would like to determine whether there exists $A$ <strike>with eigenvalues all lying on the left open half plane of $\mathbb C$, i.e., with negative real parts </strike>( we can freely choose $x, y$) such that $S \cap GL_4(\mathbb R)$ has precisely two connected components or precisely one component.

[1]:https://math.stackexchange.com/questions/2858989/when-does-s-cap-gl-4-mathbb-r-have-precisely-two-connected-components-where