# When does $S \cap GL_4(\mathbb R)$ have precisely two connected components where $S \subset M_4(\mathbb R)$ is a linear subspace?

This is a cross-post to the question 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:

1. whether there exists $$A$$ (we can freely choose $$x, y$$) such that $$S \cap GL_4(\mathbb R)$$ has precisely two connected components or precisely one component.
2. If there exists $$A$$, such that $$\{V^{-1} A V: V \in S \cap GL_4(\mathbb R)\}$$ is connected.

Edit: If the intersection only have one component, then the $$2^{\text{nd}}$$ question is immediate. Or if $$A$$ has two components but with a real eigenvalue, then $$2$$ should hold too. However, it is possible question $$2$$ can be solve directly which I could not see.