# Given $A\in\Bbb R^{n\times n}$, is $C_A := \{SAS^{-1} : S\in GL(n,\mathbb R)\}$ connected?

Let us define $\phi : GL(n,\Bbb R)\to C_A$, $\;\phi(S) = SAS^{-1}$ and the sets $$E_\pm := \{S\in GL(n,\Bbb R) : \pm\det S > 0\}.$$ If $n$ is odd, then $\phi(E_+) = \phi(E_-) = C_A$ (because $\phi(-S) = \phi(S)$) and hence $C_A$ is connected. But what about even $n$?

EDIT: I just saw that

Lemma 1. If $\det A < 0$, then still $\phi(E_+) = \phi(E_-)$ and hence $C_A$ is connected.

Proof. Indeed, if $T\in\phi(E_-)$, $T = \phi(S_0)$, $\det S_0 < 0$, then $\det(S_0A) > 0$ and $T = \phi(S_0A)\in\phi(E_+)$. The other inclusion is proved similarly.

So the question reduces to $n$ even and $\det A\ge 0$.

EDIT2: Here is another fact.

Lemma 2. If there is $S_0\in E_-$ that commutes with $A$, then $C_A$ is connected.

Proof. Let $T\in C_A$. Let us show that we can find a path within $C_A$ from $T$ to $A$. Let $T = SAS^{-1}$ with $S\in GL(n,\Bbb R)$. If $S\in E_+$, we find a path from $S$ to $I$ in $E_+$ and hence a path in $C_A$ from $T$ to $A$. If $S\in E_-$, we find a path within $E_-$ from $S$ to $S_0$. Its image under $\phi$ is again a path from $T$ to $A$ within $C_A$.

EDIT 3: For arbitrary $\lambda\in\Bbb R$ we have $C_{A-\lambda I} = C_A - \lambda I$. As this is just a translation in $\Bbb R^{n\times n}$ of $C_A$ by $\lambda I$, it follows that $C_A$ is connected if and only if $C_{A-\lambda I}$ is connected.

Therefore we can conclude the following: Let $J$ be the real Jordan form of $A$. Then $C_A = C_J$. If $A$ has a real eigenvalue $\lambda_0$ which appears in $J$ in a $k\times k$ Jordan block with $k$ odd, then $C_A$ is connected. Indeed, due to the above, we can shift $A$ and $J$ simultaneously and thus assume that $A$ and $J$ are invertible. Let $\tilde J$ be $J$, but with $-\lambda_0$'s instead of $\lambda_0$'s on the diagonal of the $k\times k$ Jordan block. Then $\tilde J$ commutes with $J$ and hence so does $\tilde JJ$. Since $\det(\tilde JJ) < 0$, $C_J = C_A$ is connected by Lemma 2.

We summarize for the critical matrices: In the real Jordan form each Jordan block corresponding to a real eigenvalue has size $k\times k$ with $k$ even.

I conjecture that the following are equivalent:

1. $C_A$ is connected
2. There exists $S\in E_-$ that commutes with $A$.
3. There exists a Jordan block $J$ of $A$ for which $C_J$ is connected.
4. There exists a real odd-sized Jordan block of $A$.

I could only prove (4)$\Rightarrow$(3), (3)$\Rightarrow$(1), and (2)$\Rightarrow$(1) so far.

Remark: This question is related to and motivated by Connectedness of matrix conjugacy classes of a fixed real $A$ but with the first column of $A$ invariant

• In Lemma $1$, does $-I$ always commute with $A$? – user1101010 Jul 20 '18 at 16:36
• @user9527 I don't understand your question. $-IA = -A = A(-I)$. – amsmath Jul 20 '18 at 17:09
• @user9527 $\det -I_2 = \det\begin{pmatrix}-1 & 0\\0 & -1\end{pmatrix} = 1$. Therefore, we consider $n$ even only. – amsmath Jul 20 '18 at 17:12
• @user9527 For example, there exists no $2\times 2$ matrix with negative determinant that commutes with $A = \begin{pmatrix}0 & 1\\-1 & 0\end{pmatrix}$. Is $C_A$ connected? – amsmath Jul 20 '18 at 17:18
• A possibly relavent observation: If $A$ is diagonalizable and $A$ has a real eigenvalue, then $C_A$ should be connected. Let $A = P^T D P$ and $\hat{P}$ be such negating the real eigenvector and unchanged on other columns. Then $\hat{P}^T D \hat{P} = A$. This implies $\phi$ maps two connected components of $GL_n(\mathbb R)$ to the same component. – user1101010 Jul 20 '18 at 19:01

The question boils down to: do there exist (for even $n>0$) matrices in $E_+$ whose centraliser in $GL(n,\Bbb R)$ is entirely contained in $E_+$ (in other words your (1) and (2) are equivalent). To see the direction you did not address in the question, remark that $\phi(E_+)$ is certainly a connected component of$~C_A$; if $C_A$ is connected, then it is all of$~C_A$. But then for any chosen $S\in E_-$ one has $\phi(S)\in\phi(E_+)$, say $\phi(S)=\phi(T)$ with $T\in E_+$ whence $T^{-1}S\in E_-$ centralises $A$.
As for your other characterisations of the situation, it would seem to me that a disconnected conjugacy class can happen ever in the presence of imaginary eigenvalues (try a block triangular $4\times4$ matrix with equal rotation matrices as its $2\times2$ diagonal block).
• First of all, thanks for dealing with my question. But why is $\phi(E_+)$ a connected component of $C_A$? This is exactly the point. So, if 2 does not hold, we have $C_A = \phi(E_+)\cup\phi(E_-)$ (disjoint union). But there might be a point on the $C_A$-boundary of, e.g., $\phi(E_+)$, that belongs to $\phi(E_+)$. Then $C_A$ would be connected. The task is to show that such a point cannot exist. Reformulated: If 2 does not hold, are $\phi(E_+)$ and $\phi(E_-)$ relatively open in $C_A$? – amsmath Jul 22 '18 at 13:54
• I don't think that can happen with orbits, because of the transitive continuous group action. If one one point of $\phi(E_-)$ would belong to the closure of $\phi(E_+)$, then by the group action the same would hold for every point of $\phi(E_-)$, and the latter would be contained in the closure of $\phi(E_+)$. But since conjugation by an element of $E_-$ interchanges the $\phi(E_+)$ and $\phi(E_-)$, the opposite also holds. Since these orbits are varieties, this can only be if $\phi(E_+)=\phi(E_-)$. – Marc van Leeuwen Jul 22 '18 at 19:37
• Ok, thanks. I could actually follow everything up to the last sentence. Very nice interchanging argument! So, we would have $C_A = \overline{\phi(E_+)} = \overline{\phi(E_-)}$ (relative closure). Also, I know that both $\phi(E_\pm)$ are contained in the algebraic variety $\{X : \det X = \det A\}$. But why are they themselves varieties? It would be nice if you could define what you mean by "variety" and maybe give a reference to the theorem which says that if $A$ and $B$ are varieties then $\overline A = \overline B$ implies $A = B$. Thanks in advance. – amsmath Jul 22 '18 at 20:10
• There are some things that nice people like you and me simply don't do, or even contemplate, without it being simple to pinpoint which law would forbid us doing them; it is just not compatible with being nice. Two disjoint non-empty parts of a topological space interchanged by a homeomorphism having identical closures is maybe possible, but not compatible with those parts being nice. The orbits considered here are certainly nice sets, and calling them varieties (locally homeomorphic to some $\Bbb R^k$) is one way to express that; possibly not the most useful one to prove the required property. – Marc van Leeuwen Jul 23 '18 at 11:25