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This question is related to the question I asked but the underlying field is $\mathbb C$ instead of $\mathbb R$.

Let $A \in M_n(\mathbb R)$ be a fixed real matrix. The set $\{S^{-1} A S: S \in GL_n(\mathbb R)\}$ is a continuous image of $GL_n(\mathbb R)$ defined by $\phi: GL_n(\mathbb R) \ni S \mapsto S^{-1}AS$, so I think it will have at most two connected components corresponding to general linear maps with positive determinants and negative determinants. Let $A = (a_1, \dots, a_n)$ where $a_j \in \mathbb R^n$ denotes the columns of $A$. Let $$F = \{S^{-1} A S: S \in GL_n(\mathbb R) \text{ and } (S^{-1}AS)_{\cdot,1} = a_1\}.$$ The condition $(S^{-1} A S )_{\cdot,1} = a_1$ is equivalent to $(AS-SA) e_1 = 0$ where $e_1 = (1, 0, \dots, )$. This is also a linear equation over $S$. So if we define $\psi: M_n(\mathbb R) \to \mathbb R^n$ by $S \mapsto (AS-SA)e_1$. Then Equivalently, $$F = \{R^{-1}AR: R \in \text{ker}(\psi) \cap GL_n(\mathbb R)\},$$ where $\text{ker}(\psi)$ is a linear subspace in $M_n(\mathbb R)$.

My question is: How many connected components does the set $F$ have? Is it at most two connected components?


Edit 1: Let $E_{\pm} = \{S \in GL_n(\mathbb R): (AS-SA)e_1 = 0, \pm \det(S) > 0\}$. Let $\phi: GL_n(\mathbb R) \to M_n(\mathbb R)$ be defined by $S \mapsto S^{-1}AS$. As discussed in the comments, if $n$ is odd, then $\phi(E_+) = \phi(E_{\_})$ which says the image does not differentiate the sign of determinants. As demonstrated by amsmath, $E_+$ can have more than $1$ connected component, but is it possible that all the components are mapped to the same connected component by $\phi$?

Edit 2: This question Connectedness of matrix conjugacy class might be helpful here. It concerns whether $\{S^{-1}AS: S \in GL_n(\mathbb R)\}$ is connected (we know it has at most two connected components).

p.s. The answer below provides nice insights but not addressing this particular question.

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  • $\begingroup$ wait.. determinant is invariant, so $det(S^{-1}AS) = det(A)$ and you don't have two components.. $\endgroup$ – Exodd Jul 19 '18 at 18:20
  • $\begingroup$ @Exodd: Why does this imply we only get one component? I was purely thinking in terms of continuous images and so concluded there would be at most two. $\endgroup$ – user1101010 Jul 19 '18 at 18:27
  • $\begingroup$ for example, if $n$ is odd and $T=-S$ then $T^{-1}AT = S^{-1}AS$, but $det(T)=-det(S)$, so you don't have two components $\endgroup$ – Exodd Jul 19 '18 at 18:29
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The answer to your question is no, not necessarily. To see this, let $A$ be such that $\ker A = \operatorname{span}\{e_1\}$. Then the condition $(SA-AS)e_1 = 0$ reduces to $Se_1\in\operatorname{span}\{e_1\}$, that is, $Se_1 = \lambda e_1$ for some $\lambda\in\mathbb R$. The sets $$ E_\pm := \{S\in GL(n,\mathbb R) : Se_1\in\operatorname{span}\{e_1\},\;\pm\det S > 0\} $$ are disconnected. So, the number of connected components of $E := E_+\cup E_-$ is at least two. We will now show that $E_+$ is not connected. For this, choose some $S\in GL(n,\mathbb R)$ with $Se_1 = -e_1$ and $\det S > 0$ and assume that $t\mapsto S(t)$, $z\in [0,1]$, is a continuous path from $S = S(0)$ to the identity $I = S(1)$ in $E_+$. Then for each $t\in [0,1]$ we have $S(t)e_1 = \lambda(t)e_1$ for some $\lambda(t)\in\mathbb R$, where $\lambda(0) = -1$ and $\lambda(1) = 1$. But as $\lambda(t) = \langle S(t)e_1,e_1\rangle$ is continuous, by the intermediate value theorem there exists some $t_0\in (0,1)$ such that $\lambda(t_0) = 0$, i.e., $S(t_0)e_1 = 0$, which is a contradiction.

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    $\begingroup$ I don't follow. How is the the number of connected components of $E$ related to the number of connected components of $F$? When $n=1$, $E=(-\infty,0)\cup(0,+\infty)$ has exactly two connected components, but $F=\{A\}$, being a singleton, has only one connected component. $\endgroup$ – user1551 Jul 20 '18 at 2:31
  • $\begingroup$ You are right. I forgot about the image that we have to take and was stuck to the pre-image. Sorry. So, for $n$ odd we have $F_+ = F_-$. But what about $n$ even? Does the same hold? So, if $S\in GL(n)$ with $\det S > 0$, does there exist $T\in GL(n)$ with $\det T < 0$ such that $S^{-1}AS = T^{-1}AT$? This is equivalent to asking whether there always exists a matrix $T$ with negative determinant such that $A = T^{-1}AT$ (which I guess is true). $\endgroup$ – amsmath Jul 20 '18 at 11:41
  • $\begingroup$ Ok, the answer is no, as is easily seen for $A = \begin{pmatrix}0 & 1\\-1 & 0\end{pmatrix}$. $\endgroup$ – amsmath Jul 20 '18 at 11:51
  • $\begingroup$ @user9527 Have a look at math.stackexchange.com/questions/2857577/…. I asked a similar question there. $\endgroup$ – amsmath Jul 20 '18 at 14:32
  • $\begingroup$ @amsmath: The problem is still a bit different from the one you asked. Yes, the image does not differentiate the sign of determinant. But in our case, we need to know how many components there are in $E_+$ (or $E_{-}$) and determine the behavior of the image of these components if there is more than $1$. $\endgroup$ – user1101010 Jul 20 '18 at 14:55

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