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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 $A$ with eigenvalues all lying on the left open half plane of $\mathbb C$, i.e., with negative real parts ( 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 1: 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 $2$ can be solve directly which I could not see.

Edit2: I crossed out the restrictions I put on $A$ although I feel this should not matter too much. The second question is newly added which is actually my end question. Before I had a feeling there should be some "special" $A$ such that the intersection would only give $1$ or $2$ components. As mentioned above, it's highly possible we can directly attack the second question.

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.

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 whether there exists $A$ with eigenvalues all lying on the left open half plane of $\mathbb C$, i.e., with negative real parts ( we can freely choose $x, y$) such that $S \cap GL_4(\mathbb R)$ has precisely two connected components or precisely one component.

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 $A$ with eigenvalues all lying on the left open half plane of $\mathbb C$, i.e., with negative real parts ( 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 1: 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 $2$ can be solve directly which I could not see.

Edit2: I crossed out the restrictions I put on $A$ although I feel this should not matter too much. The second question is newly added which is actually my end question. Before I had a feeling there should be some "special" $A$ such that the intersection would only give $1$ or $2$ components. As mentioned above, it's highly possible we can directly attack the second question.

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 whether there exists $A$ with eigenvalues all lying on the left open half plane of $\mathbb C$, i.e., with negative real parts ( we can freely choose $x, y$) such that $S \cap GL_4(\mathbb R)$ has precisely two connected components or precisely one component.

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 whether there exists $A$ with eigenvalues all lying on the left open half plane of $\mathbb C$, i.e., with negative real parts ( we can freely choose $x, y$) such that $S \cap GL_4(\mathbb R)$ has precisely two connected components or precisely one component.